AC Power Tut Qns PDF
description
Transcript of AC Power Tut Qns PDF
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 126
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
1
Exam - Questions on Power circuit P Q and S Ind motor simple equiv cct synch gen simple
equiv circuit
EX 1
A 318 mH inductor in series with a 200 Ω resistor are connected across a 50 Hz 100ang0o
supply
(a) Calculate
(i) the impedance of the circuit
(ii) the current in the circuit
(iii) the voltage drops vr and v L across R and L
(b) Draw a phasor diagram showing e vr and v L (c) Det the active power reactive powe and apparent power
Soln
(i) the impedance of the circuit is
= 22361 ang2653o Ω
(ii) the current is IT = eZT = 100ang0o = 0447 ang-2656
o A
22361ang2653o
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 226
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
2
Active power P = V I cosφ = 40 W
Or P = I2R = 0447
2200 = 40W
Reactive power Q = 20 VARAparant power = S = 447VA
Ex 2 (Bird) M A coil of resistance 5 and inductive reactance 12 is connected across a supply
voltage of 52 ang30o volts Determine the active power in the circuit
SolnI = VZ Z = 200 + j100 = 2236ang2656
o ohms
I = 100ang0o 2236ang2656
o = 0447ang -2656
o
P = |I|2R = 0447
2 200 = 40W
Ex 3 (Bird) A coil of resistance R and inductive L henries is connected in series with a 50 microF
capacitor If the supply voltage is 225V at 50 Hz and the current flowing in the circuit is
15 ang-30oA determine the values of R and L Also determine the voltages across the coil
vr and voltage across the capacitor vL and draw a phasor diagram
V1 = 447ang6344oV
IT =0447ang-2656oV
E =100ang60oV
V1 = 894ang-2656oV
j axis
φ
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 326
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
3
Soln
Circuit impedance Z = VI = 150ang 30o
Cap reactance Xc = 6366
Circuit impedance Z = 1299 + j(XL ndash 6366)
R = j75 XL = 13866
Since XL = 2π fL L = 0441 H
Vcoil 285 ang 1687o V = 27274 + j8271 V
Voltage across cap VC = I jXc = -4775 ndashj827 V
Check Vs = vL + vc = 225ang 0o V
Ex 4 A balanced three phase load of 8 kVA with power factor 085 lagging is connected to athree phase 50Hz supply via a line of impedance 05+j1Ω per phase The voltage at the load is
exactly 400V line to line
bull Draw a per phase equivalent circuit of this system
bull Calculate the line current in complex number form
bull Calculate the impedance of the load
bull Calculate the magnitude of the line to line voltage at the supply
VL = 208ang60oV
IT =15ang-30oA
=o
Vr = 19485ang-30oV
j axis
φ
Vc = 955ang-120oV
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 426
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 526
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
5
EX 5 A balanced three phase load of 10 MW with power factor 08 lagging is connected to a
three phase 50Hz supply via a line of impedance 02+j05Ω per phase The voltage at the load is
exactly 10kV line to line
bull Draw a per phase equivalent circuit of this systembull Calculate the line current in complex number form
bull Calculate the impedance of the load
bull Calculate the magnitude of the line t o line voltage at the supply
bull Calculate the amount of power factor VARS (correcting capacitance) that would
need to be fitted at the supply end to achieve 095 power factor at the supply
Soln (a)
(b) Pf = cos φ = 08 = PS
ZLine = 02 + j05
Vn Zload
IL
Vload
P=7008 (W)
j axis
Q =4659 (kVAr)
S = 8392(VA)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 626
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
6
S = Ppf = 10 106 08 = 12 5 MVA
P = radic3 VL IL cos φ = 10MW
A I L 69721=
IL = 72169 ang -3687o A
(c) Z2 = VL IL = 64 + j48
(d) Van = VL + IL Z1 = 610496 + j 2021 V
= 61083 ang 19o
|Van| = 61083 V
(e) pf orginal 08 new 095
P = 10 MW S = Ppf = 10 106 095 = 10526 MVA
Old Q = gt cos-108 = 3686o Old
Q = 10 106 Tan 3686 = 10 106 075 = 75 MVar
New Q = gt cos-1
095 = 18195o
Q = P tan 18195o = 10 10
6 032868 = 3287 MVar
Diff in MVar = 75 ndash 3287 = 4213 MVar required
Ex 6 (Past Exam q)
A balanced three phase load of 8 kVA with power factor 085 lagging is connected to a three
phase 50Hz supply via a line of impedance 05+j1Ω per phase The voltage at the load is exactly
400V line to line
bull Draw a per phase equivalent circuit of this system
bull Calculate the line current in complex number form
bull Calculate the impedance of the load
bull Calculate the magnitude of the line to line voltage at the supply
bull Calculate real power (P) and reactive power (Q) drawn from the supply
Soln (a)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 726
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
7
(b) Pf = cos φ = 085 = PS
P = Spf = 8 103 085 = 68 kW
A I L 5411=
IL = 981 ndash j607 = 1154 ang -3175o A
(c) Z2 = VL IL = Ωdegang 75319319 = 1695 + j1049
(d) Van = VL + IL Z1 = 2411ang 161o
|Van| = 241 V
(e) pf 085
S = 085 S = Ppf = 10526 MVA
P = S pf = 68 kW
S = P + jQ
Q = P tan 3179o = 4214 kVar
Ex 7
Delta to Star Transformation
ZLine = 05 + j1
Vn 8kVA
IL
Vload
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 826
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
8
C B A
B A
Z Z Z
Z Z
Z ++=1
C B A
C B
Z Z Z
Z Z Z
++=2
C B A
C A
Z Z Z
Z Z Z
++=3
Star ndash Delta Transformation
2
133221
Z
Z Z Z Z Z Z Z A
++=
3
133221
Z
Z Z Z Z Z Z Z B
++=
ZA ZB
ZC
1
2 3
Z1
Z2Z3
Delta
Star
ZA ZB
ZC
1
2 3
Z1
Z2Z3
Delta Star
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 926
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1026
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
10
RT = 2 [ 29(2+9)] = 3612 = 32727 Ω
Ind Machines
Ex 9
Why does the rotor of an induction motor turn slower than the revolving field
Ex 10
Describe briefly the key differences between a squirrel cage motor and wound-rotor motor
in terms of construction and method of operation
See lecture notes
Ex 11
Both the voltage and frequency induced in the rotor of an induction motor decrease as the
rotor speeds up Briefly explain the reason for this and include the relationships between
slip the applied stator angular velocity and rotor angular velocity
Ex 12A 3-phase 400V 6-pole 50Hz star-connected induction motor operates on full load at a speed of
1475 revmin Determine the slip s
Ex 13 a) Calculate the synchronous speed of a 3-phase 12-pole induction motor that is excited by a
50Hz source
b) What is the nominal speed (speed at full load) is the slip at full load is 5
9
9
9
2
2
2
a
b
c
RT
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1126
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
11
a) Synchronous speed rpmns = 120 f p revmin
500 revmin
(b) s = (ωs - ωr) ωs = (ns - nr) ns
nr = ns(1 - s) = 475 revmin
Ex 14
An 400 V 50 Hz six-pole star connected induction machine operates with full load slip
of 005 pu and has the following per-phase equivalent circuit parameters
stator resistance Rs = 012Ω stator leakage reactance Xs = 06 Ω and magnetising
reactance Xm = 2215 Ω The rotor referred resistance Rr` = 015 Ω and the rotor referred
leakage reactance Xr` = 0586 Ω The induction machine initially operates as a motoron full load Using an approximate per-phase equivalent circuit determine
(i) synchronous speed and the speed of the motor at rated slip in revmin and
radsec
(ii) The approximate equivalent circuit per-phase input impedance
(iii) The input current
(iv) The developed torque
(v) The starting torque
Soln
Motoring modeSynchronous speed ωs = 2 π fp = 1046 rads
ωm = ωs(1 ndash s) = 9937 rads = 9489 revmin
Rr ` s = 015005 = 3
Z = Rs + jXs+ Rr ` s +jXr
`)
= 3122 + j 1186
The total Z is Zin = ( jXm|| Z)
R`r s
j06ΩΩΩΩ
is i`r
Vs = 400 radic3 V
j0586ΩΩΩΩ 012ΩΩΩΩ
015005 j221515ΩΩΩΩ
A
B
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1226
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
12
= 7398ang110o 2354ang 8238
o
== 3143ang2762o
The input current
I i = 7318 ang-2762o A
The rotor current
I r = 6886 ang-208o A
6886 A at a pf of 0935 lagging
The developed torque
T = Pm ωm = ||3
r
2`r
s
R I
sω = 408 Nm
Starting Torque
T = ||3
r
2`r
s
R I
sω with s=1
= 204 Nm
Ex 15 A 4 pole induction motor running from a 3 phase 400V line to line 50 Hz
supply has the following parametersStator Winding Resistance (Rs) = 005Ω
Stator Leakage Inductance (Xs) = 01 Ω
Magnetising Inductance (Xm) = 20 Ω Rotor Winding Resistance referred to stator (Rrrsquo) = 01 Ω
Rotor Leakage Inductance referred to stator (Xrrsquo) = 02 Ω
Rotational losses may be assumed to be negligible
(i) Draw a per phase equivalent circuit of this machine
(ii) Calculate the rotor speed at which the machine will have a slip of 2
(iii) Calculate the Power over the gap (Pog) in this machine when it is running at a slip of 2
(iv) Calculate the mechanical power (Pm) and the torque (Tm) delivered by the machine
running at a slip of 2
Soln
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1326
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
13
Synchronous speed ωs = 2 π fp = 1500 revmin
ωm = ωs(1 ndash s) = 15386 rads = 146925 revmin
Rr` s = 01002 = 5
Z = Rs + jXs+ Rr ` s +jXr
`)
= 505 + j 03
= 5059ang34o
The rotor referred current
I r = 4546 ang-34o A
4546 A at a pf of 099 lagging
The air-gap power
Pg = ||3 r2`
r
s R I = 31kW
Prot = 0 ndash no rotational losses
Pm = Pg (1-s) = 3038 kW
T = 3038 103 15386
= 197 Nm
Ex 16 A 6 pole induction motor running from a 3 phase 400V line to line 50 Hz supply has
the following parameters
Stator Winding Resistance (Rs) = 015Ω
Stator Leakage Inductance (Xs) = 03 Ω
Magnetising Inductance (Xm) = 60 Ω
R`r s
j01ΩΩΩΩ
is i`r
Vs = 400 radic3 V
j02ΩΩΩΩ 005ΩΩΩΩ
01002 j20ΩΩΩΩ
A
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1426
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
14
Rotor Winding Resistance referred to stator (Rrrsquo) = 03 Ω
Rotor Leakage Inductance referred to stator (Xrrsquo) = 06 Ω
Rotational losses have previously been measured to be 1000W at 5 slip
bull Draw a per phase equivalent circuit of this machinebull Calculate the rotor speed at which the machine will have a slip of 5
bull Calculate the air-gap Power (Pg) in this machine when it is running at a slip of 5
bull Calculate the efficiency of the induction motor from electrical input to mechanical
output when the motor is running with a slip of 5
Soln(a)
(b) Synchronous speed ωs = 2 π fp = 1000 revmin
ωm = ωs(1 ndash s) = 9948 rads = 950 revmin
(c) Rr` s = 03005 = 6
Z = Rs + jXs+ Rr ` s +jXr
`)
= 615 + j 09
= 622ang832o
(d) The air-gap power
The rotor referred current
I r = 3698 ang-303o A
Pg = ||3
r
2`r
s
R I = 246 kW
(d) I i =inZ
3400
Zin = Xm || Z = j60 || 622ang832o
Zin = 6097 ang1409o
I = VZ = 3772 ang-1409oA
R`r s
j03ΩΩΩΩ
is i`r
Vs = 230V
j06ΩΩΩΩ 015ΩΩΩΩ
03005 j60ΩΩΩΩ
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1526
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
15
Pelec = radic3 VL IL cos φ = 25044 kW
Pm = Pg (1-s) ndash Prot = 2237 kW
η = Pelec Pm = 893s
Ex 17
A large 3-phase 4000V 8-pole 50Hz squirrel-cage induction motor draws a current of
380A and a total active power of 2344kW when operating at full load The speed at full load
is accurately measured and found to be 742rpm The stator is star-connected and the
resistance between two stator terminals is 01Ω The total iron losses are 234kW and the
windage and friction losses are 12kW
Calculate
a) The power factor at full loadb) The active power supplied to the rotor
c) The total i2R losses in the rotor
d) The mechanical power output to the load and torque delivered to the load
e) The motorrsquos net efficiency
a) The active power input to the motor is given as 2344kW The apparent power is
S = radic3 VL IL = 2633kVA
Therefore pf = PS 089 lagging
b) The active power supplied to the rotor is the active power input to the motor minus the ironlosses and copper losses in the stator The stator iron losses are given as 234kW To calculate
the copper losses note that the terminal resistance (01) is twice that of each phase for a wye
connection Therefore the copper losses are
Stator copper losses Pcu = 217kW
Therefore the active power supplied to the rotor is
Pr = Input power ndash stator copper loss ndash iron losses
Pr = 2344 kW ndash 234kW ndash 217kW
= 2298kW
c) The copper losses in the rotor are proportional to the slip (see lecture notes) Prcu sP
Therefore we must calculate the slip Neither the synchronous speed nor number of poles are
explicitly stated in the problem however by inspection we determine that the synchronous speed
must be 750rpm and the number of poles = 8 (For 50Hz supply synchronous speeds occur at
discrete intervals corresponding to pole pairs 2-pole 3000rpm 4-pole 1500rpm 6- pole
1000rpm 8-pole 750rpm etchellip We also know that induction motors operate near synchronous
speed ndash the speed is given as 742 rpm which is close to synchronous speed for an
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1626
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
16
8-pole machine)
Synchronous speed ωs = 2 π fp = 7854 rads
ωm = = 777 radss = 00107
Therefore the rotor copper losses are Pr cu = sPr = 2456 kw
(d)
The mechanical power (Pm) delivered to the load is the active power delivered to the rotor
minus rotor copper losses and windage and friction losses The windage and friction losses are
given as 12kW Therefore
Pm =22989 kW ndash 245kW-12 kW =22624 kW
kNmT L 129
60
2742
42262=
=
π
(e) the motors efficiency 5962344
42262===
kW
kW
P
P
in
out η
Ex 18
A 3-phase 440V 8-pole 50Hz star-connected induction motor has the following equivalent
circuit per-phase parameters
Rs = 01 Ω Rr = 015Ω xs = xr =06 Ω Rc = 100 Ω xm = 10 Ω
(a) Calculate using the equivalent circuit neglecting stator impedance at a slip of 5 the input
stator-current and power factor the rotor current referred to the stator the torque the
mechanical output power and the efficiency Calculate the starting torque Assume the
mechanical loss is 1kW for all speeds
Soln
(a) Synchronous speed ωs = 2 π fp 7854 rads
ωm = ωs(1 - s) = 7618 rads
Stator impedance neglected
Vs = 400 radic3 = 230
Ir` =
60050
10
230
j+ =
= 68 ang-1671 = 651 - j 1954
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1726
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
17
Io =230
100 + j230
10 = 23 ndash j 23
Is = Ir` + Io = 674 - j4294
= 799 A at cos φ = 084
Ir` = 68 A
T = 529 Nm
Pmech = 39365 kW
Pelect = 465 kW
Efficiency = 39365465 = 846
Ts = 39627 Nm
Ex19
Tests conducted at 50 Hz on a three-phase star connected 45 kW 2200V six-pole 50Hz squirrel
cage induction motor are listed below
The average dc resistance per stator phase Rs = 28 Ω (a) Briefly describe the tests required to determine the per-phase equivalent circuit
impedance parameters of an induction machine
(a) Determine the parameters of the per-phase equivalent circuit
(b) Determine the rotational losses of the machine Core losses can be neglected
load
0
No-loadV
R
I1
Ic
I
c X
Im
m
Figure 1 Equivalent Circuit during No-Load Test
torPower Fac0 times= I I C = 2025 A
22
0 cm I I I minus=
= 4019 A
Ω== 627c
c I
V R Ω== 5317
m
m I
V X
No Load Test Locked Rotor Test
Power Factor = 045 Power Factor = 09
Line Voltage = 22kV Line Voltage = 270V
Line Current = 45A Line Current = 25A
Input Power = 16kW Input Power = 9kW
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1826
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
18
For the locked rotor test slip = 1
Ω== 246s
s
LR I
V Z
Locked rotor reactance is
( )2
21
2
21 R R Z X X +minus=+
φ cos21 times=+= Z R R Rtotal = 5616Ω
R2rsquo = Rtot - R1 = 2816Ω
Ω=+ 72221 X X
(a) From the no-load test the no-load power isPNL = 16kW
The no-load rotational loss is
Prot = PNL ndash 3 I s2 Rs
= 14299W
Synch Machines
Ex 20
A 60 MVA 20 kV three-phase 50 Hz synchronous generator has a synchronous reactance
of 8 Ω per phase and negligible resistance The generator delivers rated power at a power
factor of 085 lagging at the rated terminal voltage to an infinite bus(a) Determine the excitation voltage per phase and the power angle δ
(b) With the excitation voltage Eg of the synchronous generator held constant at the
value determined in part (a) the input driving torque is reduced until the generator
delivers 28 MW Determine the armature current and the power factor
(c) If the generator operates at the excitation voltage of Eg obtained in part (a) what is
the steady state maximum power the machine can deliver before losing
jXsΩΩΩΩ jX`r ΩΩΩΩ Rs
R`r s
Is I`r
Vs ph
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1926
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
19
synchronism Determine the armature current corresponding to this maximum
power
(d) With real generated power at 42MW compute the power angle δ the stator
current Ia and power factor when the field excitation current If is adjusted so that
the excitation voltage Eg is decreased to 7919 of the value determined in part
(a)
Soln
(a) Apparent power S = 51 MW ndash j 316Mvar
Rated voltage per phase
V = 1154 ang 0deg kV
Rated current
Ia = S 3V
= 1733 ang -3178deg AExcitation voltage
Eg = 11540 + j9(1733) ang -3178deg)= 237922 ang 3386deg
Power angle = 3386o
(b) When delivering 28 MW
δ = sin-1
)5411)(79223(3
828 = 1578
o
P =
s
g
X
V E 3sinδ
Ia = 13069ang2967o j8
= 16336ang-6033o A
Power factor = cos(6033) = 0495 lagging
(c) The maximum power occurs at δ = 90o
Pmax =s
g
X
V E 3
= 102 MW
Ia = 324885ang25875o A
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2026
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
20
The power factor cos(25875) = 0899 leading
(d) Eg = 13463 V
The power angle δ = sin-1
)5411)(84318(3
842
= 31o
Ia = 86674ang0o A
-Power factor = 0
Ex 21 (Aug 13) A 3 phase synchronous machine with negligible stator winding resistance and a
synchronous reactance of 2 per phase is connected to a 400V 50 Hz 3-phase supply (infinite
bus) and operated as a generator It is currently generating 50kW with power factor 08 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)
(iv) What is the maximum power the machine can generate with this value of excitation
Soln(i)
(ii)P = 50kW
Apparent power S = 625 ang -3687
o
kVA
Rated voltage per phase V = 230ang 0deg VRated current
Ia = S 3V
= 9058ang -3687deg A
LoadGen
Xs =2
Eg Vt =400 radic3 V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2126
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
21
|Ia| = 9058 A
(iii)
Excitation voltage
Eg = 230 ang 0deg + j2(9058ang -3687deg)= 3684 ang 2318degV
|Eg| = 3684 V
Power angle = 2318o
(iv)
P = deg90sin2
23043683
(max occurs when δ = 90o)
= 1271 kW
Ex 22A 3 phase 4 pole synchronous machine with negligible stator winding resistance and a
synchronous reactance of 2 per phase is connected to a 480V 60 Hz 3phase supply
(infinite bus) and operated as a generator It is currently generating 80kW with power
factor 08 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)(iv) What is the maximum power the machine can generate with this value
excitation(v) What is the rotational speed of this generator
Soln (i)
(ii)P = 80kW Apparent power S 100 ang -3687
o kVA
Rated voltage per phase
V 277ang 0deg V
LoadGen
Xs =2
Eg Vt = 480V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2226
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
22
Rated current
Ia = S 3V
= 9058ang -3687deg A
|Ia| = 12028 A
(iii)
Excitation voltage
Eg = 277 ang 0deg + j2(12028ang -3687deg)
= 4632 ang 2455degV|Eg| = 4632 V
Power angle = 2455o
(iv)
P = deg90sin2
27724633
(max occurs when δ = 90o)
= 19238 kW
(v)Synchronous speed ωs = 2 π fp = 1885 rads
= 1800 revmin
Ex 23 (Jan 14)
A 3 phase synchronous machine with negligible stator winding resistance and a synchronous
reactance of 25 per phase is connected to a 20kV 50 Hz 3phase supply (infinite bus) and
operated as a generator It is currently generating 100MW with power factor 085 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)(iv) What minimum value of excitation voltage (E) would be required for this machine to
deliver 200MW into a 20kV bus assuming a 20deg stability margin must be maintained onpower angle (δ)
Soln
(i)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2326
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
23
(ii)
P = 100 MW Apparent power S = 100 085
= 11765 ang -3179o MVA
Rated voltage per phase
V= 11547ang 0deg VRated current
Ia = S 3V
= 3396 103ang -3179deg A
|Ia| = 3396 103A
(iii)
Excitation voltage
Eg = 11547 ang 0deg + j25(3396 103ang -3179deg)
= 17570 ang 2425degV|Eg| = 17570 V
Power angle = 2425o V
(iv)
P = )2090sin(52
3degminusdegt gV E
(max occurs when δ = 90o stability margin 20o)
200 MW = )2090sin(52
547113degminusdegg E
200 MW = 38130209397085613 gg E E =
|Eg| = 15360 V
LoadGen
Xs =25
Eg Vt =11547V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2426
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
24
Ex 24A manufacturing plant fed with a 400V 50Hz three phase runs from 0900 to 1800
Monday to Friday The factoryrsquos load is normally 80kVA with power factor 08 lagging
during these hours Once a month they need to run a special heating cycle which adds a
further 20kW to the factory load for a 2 hour period This heating cycle is timer
controlled and fully automated but the timer is currently set to run on the third Friday
afternoon of the month Outside of normal business hours and at weekends the factory
load is negligible The firm has a contractually agreed Maximum Import Capacity (MIC)
of 120kVA
bull Use the attached table of Low Voltage Maximum Demand charges to calculate the
firmrsquos electricity bill (exclusive of VAT) for the month of February (28 days)
bull Identify three steps that this firm could take to reduce their expenditure on
electricity assuming they stick with the same tariff Calculate the potential
savings
983116983151983159 983126983151983148983156983137983143983141 983117983137983160983145983149983157983149 983108983141983149983137983150983140 983107983144983137983154983143983141983155
983108983137983145983148983161 983123983156983137983150983140983145983150983143 983107983144983137983154983143983141 991404302983140983137983161
983120983123983119 983116983141983158983161
983120983123983119 983116983141983158983161 983152983141983154 983115983126983105 983151983142 983117983113983107 983152983141983154 983140983137983161 9914040036983097983147983126983105983140983137983161
983117983137983160983145983149983157983149 983113983149983152983151983154983156 983107983137983152983137983139983145983156983161 (983117983113983107) 983107983144983137983154983143983141983155
983108983137983145983148983161 983107983144983137983154983143983141 983152983141983154 983147983126983105 983151983142 983117983113983107 9914040072983147983126983105983140983137983161
983108983141983149983137983150983140 983107983144983137983154983143983141983155 983118983151983158983085983110983141983138 983117983137983154983085983119983139983156
983117983151983150983156983144983148983161 983107983144983137983154983143983141 983152983141983154 983147983127 983151983142 983149983141983156983141983154983141983140 983117983108 983145983150 983137
983149983151983150983156983144983148983161 983152983141983154983145983151983140 (983123983157983138983146983141983139983156 983156983151 983137 983149983145983150983145983149983157983149 983139983144983137983154983143983141 983151983142
30983147983127) 991404189830975983147983127983149983151983150983156983144 991404000
983108983137983161983156983145983149983141 983147983127983144 (983125983150983145983156) 983107983144983137983154983143983141 (0898309800 983085 19830979830985983097) 1480983139983147983127983144 1275983139983147983127983144
983118983145983143983144983156983145983149983141 983147983127983144 (983125983150983145983156) 983118983145983143983144983156 983107983144983137983154983143983141983155 (2098309800983085079830985983097) 64983097983139983147983127983144 64983097983139983147983127983144
983127983137983156983156983148983141983155983155 983125983150983145983156 983120983154983145983139983141 (983147983126983105983122983144 983145983150 983141983160983139983141983155983155 983151983142 983151983150983141 983156983144983145983154983140
983151983142 983156983151983156983137983148 983147983127983144) 991404000714 983147983126983105983122983144 991404000714 983147983126983105983122983144
bull Table Low Voltage Maximum Demand Charges (Q6)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2526
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
25
SolnNumber of days in February = 28
Number of working hours = 459 = 180
Normal kW = 80kVA08 = 64kW
Total kWh = 64kW180 hours+20kW2 hours = 11560 kWh
Normal kvar =80kVA sqrt(1-08^2) = 48kvar = sqrt(802-64
2) = 48 kVAr
Total kvarh = 48kvar180 hours = 8640 kvarh
Chargeable kvarh = 8640-(11560)3=4786 kvarh
983116983151983159 983126983151983148983156983137983143983141 983117983137983160983145983149983157983149 983108983141983149983137983150983140
983107983144983137983154983143983141983155 983152983141983154 983122983137983156983141 983121983156983161 983120983141983154983145983151983140 983107983144983137983154983143983141
983108983137983145983148983161 983123983156983137983150983140983145983150983143 983107983144983137983154983143983141 991404302 983140983137983161 991404302000 1 28 991404983096456
983120983123983119 983116983141983158983161
983120983123983119 983116983141983158983161 983152983141983154 983115983126983105 983152983141983154 983140983137983161 9914040036983097 983147983126983105983140983137983161 99140400369830970 120 28 991404123983097983096
983117983137983160983145983149983157983149 983113983149983152983151983154983156 983107983137983152983137983139983145983156983161
(983117983113983107) 983107983144983137983154983143983141983155
983108983137983145983148983161 983107983144983137983154983143983141 983152983141983154 983147983126983105 983151983142 983117983113983107 9914040072 983147983126983105983140983137983161 991404007200 120 28 9914042419830972
983108983141983149983137983150983140 983107983144983137983154983143983141983155 983118983151983158983085983110983141983138
983117983151983150983156983144983148983161 983107983144983137983154983143983141 983152983141983154 983147983127 983151983142
983149983141983156983141983154983141983140 983117983108 983145983150 983137 983149983151983150983156983144983148983161
983152983141983154983145983151983140 (983123983157983138983146983141983139983156 983156983151 983137
983149983145983150983145983149983157983149 983139983144983137983154983143983141 983151983142 30983147983127) 991404189830975
983147983127983149983151983150983156
983144 99140418983097500 84 1 991404159830971983096 983108983137983161983156983145983149983141 983147983127983144 (983125983150983145983156) 983107983144983137983154983143983141
(0898309800 983085 19830979830985983097) 9914040148 983147983127983144 991404014800 11560 1
991404198309510983096
983096
983118983145983143983144983156983145983149983141 983147983127983144 (983125983150983145983156) 983118983145983143983144983156
983107983144983137983154983143983141983155 (2098309800983085079830985983097) 9914040064983097 983147983127983144 99140400649830970
983127983137983156983156983148983141983155983155 983125983150983145983156 983120983154983145983139983141 (983147983126983105983122983144
983145983150 983141983160983139983141983155983155 983151983142 983151983150983141 983156983144983145983154983140 983151983142 983156983151983156983137983148
983147983127983144) 991404000714 983147983126983137983154983144 991404000714 4787 1 991404341983096
983124983151983156983137983148 983106983145983148983148
983142983151983154
983110983141983138983154983157983137983154983161
9914042354983095
0
[16 Marks]
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2626
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
26
Potential Savings
bull Use Power Factor correction to eliminate Watt less unit charge and to reduce MIC
bull Shift Heating load to night time or weekend
bull Negotiate lower MIC
Potential savings from these measures
Eliminate Wattless charge save euro3417 month
Shift heating cycle to night time
Max demand it lowered by 20kW = saving of 20euro1895 = euro3790 month
Saving by moving 20kW to night rate = 20kW2 hours (euro0148-euro00649)= euro332
Negotiating a reduced MIC
If power factor correction is employed and the heating cycle is shifted to night time the
then maximum VA of the plant drops to close to its maximum kW demand of 64kW An
MIC of 80kVA should be sufficient to cover this Potential Saving = 40kVA
(euro0072+euro00369)28 = euro12196 per month
Total potential saving = euro3417+euro3790+euro332+euro12196=euro19735month
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 226
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
2
Active power P = V I cosφ = 40 W
Or P = I2R = 0447
2200 = 40W
Reactive power Q = 20 VARAparant power = S = 447VA
Ex 2 (Bird) M A coil of resistance 5 and inductive reactance 12 is connected across a supply
voltage of 52 ang30o volts Determine the active power in the circuit
SolnI = VZ Z = 200 + j100 = 2236ang2656
o ohms
I = 100ang0o 2236ang2656
o = 0447ang -2656
o
P = |I|2R = 0447
2 200 = 40W
Ex 3 (Bird) A coil of resistance R and inductive L henries is connected in series with a 50 microF
capacitor If the supply voltage is 225V at 50 Hz and the current flowing in the circuit is
15 ang-30oA determine the values of R and L Also determine the voltages across the coil
vr and voltage across the capacitor vL and draw a phasor diagram
V1 = 447ang6344oV
IT =0447ang-2656oV
E =100ang60oV
V1 = 894ang-2656oV
j axis
φ
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 326
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
3
Soln
Circuit impedance Z = VI = 150ang 30o
Cap reactance Xc = 6366
Circuit impedance Z = 1299 + j(XL ndash 6366)
R = j75 XL = 13866
Since XL = 2π fL L = 0441 H
Vcoil 285 ang 1687o V = 27274 + j8271 V
Voltage across cap VC = I jXc = -4775 ndashj827 V
Check Vs = vL + vc = 225ang 0o V
Ex 4 A balanced three phase load of 8 kVA with power factor 085 lagging is connected to athree phase 50Hz supply via a line of impedance 05+j1Ω per phase The voltage at the load is
exactly 400V line to line
bull Draw a per phase equivalent circuit of this system
bull Calculate the line current in complex number form
bull Calculate the impedance of the load
bull Calculate the magnitude of the line to line voltage at the supply
VL = 208ang60oV
IT =15ang-30oA
=o
Vr = 19485ang-30oV
j axis
φ
Vc = 955ang-120oV
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 426
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 526
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
5
EX 5 A balanced three phase load of 10 MW with power factor 08 lagging is connected to a
three phase 50Hz supply via a line of impedance 02+j05Ω per phase The voltage at the load is
exactly 10kV line to line
bull Draw a per phase equivalent circuit of this systembull Calculate the line current in complex number form
bull Calculate the impedance of the load
bull Calculate the magnitude of the line t o line voltage at the supply
bull Calculate the amount of power factor VARS (correcting capacitance) that would
need to be fitted at the supply end to achieve 095 power factor at the supply
Soln (a)
(b) Pf = cos φ = 08 = PS
ZLine = 02 + j05
Vn Zload
IL
Vload
P=7008 (W)
j axis
Q =4659 (kVAr)
S = 8392(VA)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 626
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
6
S = Ppf = 10 106 08 = 12 5 MVA
P = radic3 VL IL cos φ = 10MW
A I L 69721=
IL = 72169 ang -3687o A
(c) Z2 = VL IL = 64 + j48
(d) Van = VL + IL Z1 = 610496 + j 2021 V
= 61083 ang 19o
|Van| = 61083 V
(e) pf orginal 08 new 095
P = 10 MW S = Ppf = 10 106 095 = 10526 MVA
Old Q = gt cos-108 = 3686o Old
Q = 10 106 Tan 3686 = 10 106 075 = 75 MVar
New Q = gt cos-1
095 = 18195o
Q = P tan 18195o = 10 10
6 032868 = 3287 MVar
Diff in MVar = 75 ndash 3287 = 4213 MVar required
Ex 6 (Past Exam q)
A balanced three phase load of 8 kVA with power factor 085 lagging is connected to a three
phase 50Hz supply via a line of impedance 05+j1Ω per phase The voltage at the load is exactly
400V line to line
bull Draw a per phase equivalent circuit of this system
bull Calculate the line current in complex number form
bull Calculate the impedance of the load
bull Calculate the magnitude of the line to line voltage at the supply
bull Calculate real power (P) and reactive power (Q) drawn from the supply
Soln (a)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 726
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
7
(b) Pf = cos φ = 085 = PS
P = Spf = 8 103 085 = 68 kW
A I L 5411=
IL = 981 ndash j607 = 1154 ang -3175o A
(c) Z2 = VL IL = Ωdegang 75319319 = 1695 + j1049
(d) Van = VL + IL Z1 = 2411ang 161o
|Van| = 241 V
(e) pf 085
S = 085 S = Ppf = 10526 MVA
P = S pf = 68 kW
S = P + jQ
Q = P tan 3179o = 4214 kVar
Ex 7
Delta to Star Transformation
ZLine = 05 + j1
Vn 8kVA
IL
Vload
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 826
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
8
C B A
B A
Z Z Z
Z Z
Z ++=1
C B A
C B
Z Z Z
Z Z Z
++=2
C B A
C A
Z Z Z
Z Z Z
++=3
Star ndash Delta Transformation
2
133221
Z
Z Z Z Z Z Z Z A
++=
3
133221
Z
Z Z Z Z Z Z Z B
++=
ZA ZB
ZC
1
2 3
Z1
Z2Z3
Delta
Star
ZA ZB
ZC
1
2 3
Z1
Z2Z3
Delta Star
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 926
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1026
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
10
RT = 2 [ 29(2+9)] = 3612 = 32727 Ω
Ind Machines
Ex 9
Why does the rotor of an induction motor turn slower than the revolving field
Ex 10
Describe briefly the key differences between a squirrel cage motor and wound-rotor motor
in terms of construction and method of operation
See lecture notes
Ex 11
Both the voltage and frequency induced in the rotor of an induction motor decrease as the
rotor speeds up Briefly explain the reason for this and include the relationships between
slip the applied stator angular velocity and rotor angular velocity
Ex 12A 3-phase 400V 6-pole 50Hz star-connected induction motor operates on full load at a speed of
1475 revmin Determine the slip s
Ex 13 a) Calculate the synchronous speed of a 3-phase 12-pole induction motor that is excited by a
50Hz source
b) What is the nominal speed (speed at full load) is the slip at full load is 5
9
9
9
2
2
2
a
b
c
RT
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1126
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
11
a) Synchronous speed rpmns = 120 f p revmin
500 revmin
(b) s = (ωs - ωr) ωs = (ns - nr) ns
nr = ns(1 - s) = 475 revmin
Ex 14
An 400 V 50 Hz six-pole star connected induction machine operates with full load slip
of 005 pu and has the following per-phase equivalent circuit parameters
stator resistance Rs = 012Ω stator leakage reactance Xs = 06 Ω and magnetising
reactance Xm = 2215 Ω The rotor referred resistance Rr` = 015 Ω and the rotor referred
leakage reactance Xr` = 0586 Ω The induction machine initially operates as a motoron full load Using an approximate per-phase equivalent circuit determine
(i) synchronous speed and the speed of the motor at rated slip in revmin and
radsec
(ii) The approximate equivalent circuit per-phase input impedance
(iii) The input current
(iv) The developed torque
(v) The starting torque
Soln
Motoring modeSynchronous speed ωs = 2 π fp = 1046 rads
ωm = ωs(1 ndash s) = 9937 rads = 9489 revmin
Rr ` s = 015005 = 3
Z = Rs + jXs+ Rr ` s +jXr
`)
= 3122 + j 1186
The total Z is Zin = ( jXm|| Z)
R`r s
j06ΩΩΩΩ
is i`r
Vs = 400 radic3 V
j0586ΩΩΩΩ 012ΩΩΩΩ
015005 j221515ΩΩΩΩ
A
B
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1226
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
12
= 7398ang110o 2354ang 8238
o
== 3143ang2762o
The input current
I i = 7318 ang-2762o A
The rotor current
I r = 6886 ang-208o A
6886 A at a pf of 0935 lagging
The developed torque
T = Pm ωm = ||3
r
2`r
s
R I
sω = 408 Nm
Starting Torque
T = ||3
r
2`r
s
R I
sω with s=1
= 204 Nm
Ex 15 A 4 pole induction motor running from a 3 phase 400V line to line 50 Hz
supply has the following parametersStator Winding Resistance (Rs) = 005Ω
Stator Leakage Inductance (Xs) = 01 Ω
Magnetising Inductance (Xm) = 20 Ω Rotor Winding Resistance referred to stator (Rrrsquo) = 01 Ω
Rotor Leakage Inductance referred to stator (Xrrsquo) = 02 Ω
Rotational losses may be assumed to be negligible
(i) Draw a per phase equivalent circuit of this machine
(ii) Calculate the rotor speed at which the machine will have a slip of 2
(iii) Calculate the Power over the gap (Pog) in this machine when it is running at a slip of 2
(iv) Calculate the mechanical power (Pm) and the torque (Tm) delivered by the machine
running at a slip of 2
Soln
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1326
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
13
Synchronous speed ωs = 2 π fp = 1500 revmin
ωm = ωs(1 ndash s) = 15386 rads = 146925 revmin
Rr` s = 01002 = 5
Z = Rs + jXs+ Rr ` s +jXr
`)
= 505 + j 03
= 5059ang34o
The rotor referred current
I r = 4546 ang-34o A
4546 A at a pf of 099 lagging
The air-gap power
Pg = ||3 r2`
r
s R I = 31kW
Prot = 0 ndash no rotational losses
Pm = Pg (1-s) = 3038 kW
T = 3038 103 15386
= 197 Nm
Ex 16 A 6 pole induction motor running from a 3 phase 400V line to line 50 Hz supply has
the following parameters
Stator Winding Resistance (Rs) = 015Ω
Stator Leakage Inductance (Xs) = 03 Ω
Magnetising Inductance (Xm) = 60 Ω
R`r s
j01ΩΩΩΩ
is i`r
Vs = 400 radic3 V
j02ΩΩΩΩ 005ΩΩΩΩ
01002 j20ΩΩΩΩ
A
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1426
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
14
Rotor Winding Resistance referred to stator (Rrrsquo) = 03 Ω
Rotor Leakage Inductance referred to stator (Xrrsquo) = 06 Ω
Rotational losses have previously been measured to be 1000W at 5 slip
bull Draw a per phase equivalent circuit of this machinebull Calculate the rotor speed at which the machine will have a slip of 5
bull Calculate the air-gap Power (Pg) in this machine when it is running at a slip of 5
bull Calculate the efficiency of the induction motor from electrical input to mechanical
output when the motor is running with a slip of 5
Soln(a)
(b) Synchronous speed ωs = 2 π fp = 1000 revmin
ωm = ωs(1 ndash s) = 9948 rads = 950 revmin
(c) Rr` s = 03005 = 6
Z = Rs + jXs+ Rr ` s +jXr
`)
= 615 + j 09
= 622ang832o
(d) The air-gap power
The rotor referred current
I r = 3698 ang-303o A
Pg = ||3
r
2`r
s
R I = 246 kW
(d) I i =inZ
3400
Zin = Xm || Z = j60 || 622ang832o
Zin = 6097 ang1409o
I = VZ = 3772 ang-1409oA
R`r s
j03ΩΩΩΩ
is i`r
Vs = 230V
j06ΩΩΩΩ 015ΩΩΩΩ
03005 j60ΩΩΩΩ
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1526
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
15
Pelec = radic3 VL IL cos φ = 25044 kW
Pm = Pg (1-s) ndash Prot = 2237 kW
η = Pelec Pm = 893s
Ex 17
A large 3-phase 4000V 8-pole 50Hz squirrel-cage induction motor draws a current of
380A and a total active power of 2344kW when operating at full load The speed at full load
is accurately measured and found to be 742rpm The stator is star-connected and the
resistance between two stator terminals is 01Ω The total iron losses are 234kW and the
windage and friction losses are 12kW
Calculate
a) The power factor at full loadb) The active power supplied to the rotor
c) The total i2R losses in the rotor
d) The mechanical power output to the load and torque delivered to the load
e) The motorrsquos net efficiency
a) The active power input to the motor is given as 2344kW The apparent power is
S = radic3 VL IL = 2633kVA
Therefore pf = PS 089 lagging
b) The active power supplied to the rotor is the active power input to the motor minus the ironlosses and copper losses in the stator The stator iron losses are given as 234kW To calculate
the copper losses note that the terminal resistance (01) is twice that of each phase for a wye
connection Therefore the copper losses are
Stator copper losses Pcu = 217kW
Therefore the active power supplied to the rotor is
Pr = Input power ndash stator copper loss ndash iron losses
Pr = 2344 kW ndash 234kW ndash 217kW
= 2298kW
c) The copper losses in the rotor are proportional to the slip (see lecture notes) Prcu sP
Therefore we must calculate the slip Neither the synchronous speed nor number of poles are
explicitly stated in the problem however by inspection we determine that the synchronous speed
must be 750rpm and the number of poles = 8 (For 50Hz supply synchronous speeds occur at
discrete intervals corresponding to pole pairs 2-pole 3000rpm 4-pole 1500rpm 6- pole
1000rpm 8-pole 750rpm etchellip We also know that induction motors operate near synchronous
speed ndash the speed is given as 742 rpm which is close to synchronous speed for an
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1626
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
16
8-pole machine)
Synchronous speed ωs = 2 π fp = 7854 rads
ωm = = 777 radss = 00107
Therefore the rotor copper losses are Pr cu = sPr = 2456 kw
(d)
The mechanical power (Pm) delivered to the load is the active power delivered to the rotor
minus rotor copper losses and windage and friction losses The windage and friction losses are
given as 12kW Therefore
Pm =22989 kW ndash 245kW-12 kW =22624 kW
kNmT L 129
60
2742
42262=
=
π
(e) the motors efficiency 5962344
42262===
kW
kW
P
P
in
out η
Ex 18
A 3-phase 440V 8-pole 50Hz star-connected induction motor has the following equivalent
circuit per-phase parameters
Rs = 01 Ω Rr = 015Ω xs = xr =06 Ω Rc = 100 Ω xm = 10 Ω
(a) Calculate using the equivalent circuit neglecting stator impedance at a slip of 5 the input
stator-current and power factor the rotor current referred to the stator the torque the
mechanical output power and the efficiency Calculate the starting torque Assume the
mechanical loss is 1kW for all speeds
Soln
(a) Synchronous speed ωs = 2 π fp 7854 rads
ωm = ωs(1 - s) = 7618 rads
Stator impedance neglected
Vs = 400 radic3 = 230
Ir` =
60050
10
230
j+ =
= 68 ang-1671 = 651 - j 1954
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1726
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
17
Io =230
100 + j230
10 = 23 ndash j 23
Is = Ir` + Io = 674 - j4294
= 799 A at cos φ = 084
Ir` = 68 A
T = 529 Nm
Pmech = 39365 kW
Pelect = 465 kW
Efficiency = 39365465 = 846
Ts = 39627 Nm
Ex19
Tests conducted at 50 Hz on a three-phase star connected 45 kW 2200V six-pole 50Hz squirrel
cage induction motor are listed below
The average dc resistance per stator phase Rs = 28 Ω (a) Briefly describe the tests required to determine the per-phase equivalent circuit
impedance parameters of an induction machine
(a) Determine the parameters of the per-phase equivalent circuit
(b) Determine the rotational losses of the machine Core losses can be neglected
load
0
No-loadV
R
I1
Ic
I
c X
Im
m
Figure 1 Equivalent Circuit during No-Load Test
torPower Fac0 times= I I C = 2025 A
22
0 cm I I I minus=
= 4019 A
Ω== 627c
c I
V R Ω== 5317
m
m I
V X
No Load Test Locked Rotor Test
Power Factor = 045 Power Factor = 09
Line Voltage = 22kV Line Voltage = 270V
Line Current = 45A Line Current = 25A
Input Power = 16kW Input Power = 9kW
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1826
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
18
For the locked rotor test slip = 1
Ω== 246s
s
LR I
V Z
Locked rotor reactance is
( )2
21
2
21 R R Z X X +minus=+
φ cos21 times=+= Z R R Rtotal = 5616Ω
R2rsquo = Rtot - R1 = 2816Ω
Ω=+ 72221 X X
(a) From the no-load test the no-load power isPNL = 16kW
The no-load rotational loss is
Prot = PNL ndash 3 I s2 Rs
= 14299W
Synch Machines
Ex 20
A 60 MVA 20 kV three-phase 50 Hz synchronous generator has a synchronous reactance
of 8 Ω per phase and negligible resistance The generator delivers rated power at a power
factor of 085 lagging at the rated terminal voltage to an infinite bus(a) Determine the excitation voltage per phase and the power angle δ
(b) With the excitation voltage Eg of the synchronous generator held constant at the
value determined in part (a) the input driving torque is reduced until the generator
delivers 28 MW Determine the armature current and the power factor
(c) If the generator operates at the excitation voltage of Eg obtained in part (a) what is
the steady state maximum power the machine can deliver before losing
jXsΩΩΩΩ jX`r ΩΩΩΩ Rs
R`r s
Is I`r
Vs ph
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1926
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
19
synchronism Determine the armature current corresponding to this maximum
power
(d) With real generated power at 42MW compute the power angle δ the stator
current Ia and power factor when the field excitation current If is adjusted so that
the excitation voltage Eg is decreased to 7919 of the value determined in part
(a)
Soln
(a) Apparent power S = 51 MW ndash j 316Mvar
Rated voltage per phase
V = 1154 ang 0deg kV
Rated current
Ia = S 3V
= 1733 ang -3178deg AExcitation voltage
Eg = 11540 + j9(1733) ang -3178deg)= 237922 ang 3386deg
Power angle = 3386o
(b) When delivering 28 MW
δ = sin-1
)5411)(79223(3
828 = 1578
o
P =
s
g
X
V E 3sinδ
Ia = 13069ang2967o j8
= 16336ang-6033o A
Power factor = cos(6033) = 0495 lagging
(c) The maximum power occurs at δ = 90o
Pmax =s
g
X
V E 3
= 102 MW
Ia = 324885ang25875o A
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2026
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
20
The power factor cos(25875) = 0899 leading
(d) Eg = 13463 V
The power angle δ = sin-1
)5411)(84318(3
842
= 31o
Ia = 86674ang0o A
-Power factor = 0
Ex 21 (Aug 13) A 3 phase synchronous machine with negligible stator winding resistance and a
synchronous reactance of 2 per phase is connected to a 400V 50 Hz 3-phase supply (infinite
bus) and operated as a generator It is currently generating 50kW with power factor 08 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)
(iv) What is the maximum power the machine can generate with this value of excitation
Soln(i)
(ii)P = 50kW
Apparent power S = 625 ang -3687
o
kVA
Rated voltage per phase V = 230ang 0deg VRated current
Ia = S 3V
= 9058ang -3687deg A
LoadGen
Xs =2
Eg Vt =400 radic3 V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2126
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
21
|Ia| = 9058 A
(iii)
Excitation voltage
Eg = 230 ang 0deg + j2(9058ang -3687deg)= 3684 ang 2318degV
|Eg| = 3684 V
Power angle = 2318o
(iv)
P = deg90sin2
23043683
(max occurs when δ = 90o)
= 1271 kW
Ex 22A 3 phase 4 pole synchronous machine with negligible stator winding resistance and a
synchronous reactance of 2 per phase is connected to a 480V 60 Hz 3phase supply
(infinite bus) and operated as a generator It is currently generating 80kW with power
factor 08 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)(iv) What is the maximum power the machine can generate with this value
excitation(v) What is the rotational speed of this generator
Soln (i)
(ii)P = 80kW Apparent power S 100 ang -3687
o kVA
Rated voltage per phase
V 277ang 0deg V
LoadGen
Xs =2
Eg Vt = 480V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2226
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
22
Rated current
Ia = S 3V
= 9058ang -3687deg A
|Ia| = 12028 A
(iii)
Excitation voltage
Eg = 277 ang 0deg + j2(12028ang -3687deg)
= 4632 ang 2455degV|Eg| = 4632 V
Power angle = 2455o
(iv)
P = deg90sin2
27724633
(max occurs when δ = 90o)
= 19238 kW
(v)Synchronous speed ωs = 2 π fp = 1885 rads
= 1800 revmin
Ex 23 (Jan 14)
A 3 phase synchronous machine with negligible stator winding resistance and a synchronous
reactance of 25 per phase is connected to a 20kV 50 Hz 3phase supply (infinite bus) and
operated as a generator It is currently generating 100MW with power factor 085 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)(iv) What minimum value of excitation voltage (E) would be required for this machine to
deliver 200MW into a 20kV bus assuming a 20deg stability margin must be maintained onpower angle (δ)
Soln
(i)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2326
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
23
(ii)
P = 100 MW Apparent power S = 100 085
= 11765 ang -3179o MVA
Rated voltage per phase
V= 11547ang 0deg VRated current
Ia = S 3V
= 3396 103ang -3179deg A
|Ia| = 3396 103A
(iii)
Excitation voltage
Eg = 11547 ang 0deg + j25(3396 103ang -3179deg)
= 17570 ang 2425degV|Eg| = 17570 V
Power angle = 2425o V
(iv)
P = )2090sin(52
3degminusdegt gV E
(max occurs when δ = 90o stability margin 20o)
200 MW = )2090sin(52
547113degminusdegg E
200 MW = 38130209397085613 gg E E =
|Eg| = 15360 V
LoadGen
Xs =25
Eg Vt =11547V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2426
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
24
Ex 24A manufacturing plant fed with a 400V 50Hz three phase runs from 0900 to 1800
Monday to Friday The factoryrsquos load is normally 80kVA with power factor 08 lagging
during these hours Once a month they need to run a special heating cycle which adds a
further 20kW to the factory load for a 2 hour period This heating cycle is timer
controlled and fully automated but the timer is currently set to run on the third Friday
afternoon of the month Outside of normal business hours and at weekends the factory
load is negligible The firm has a contractually agreed Maximum Import Capacity (MIC)
of 120kVA
bull Use the attached table of Low Voltage Maximum Demand charges to calculate the
firmrsquos electricity bill (exclusive of VAT) for the month of February (28 days)
bull Identify three steps that this firm could take to reduce their expenditure on
electricity assuming they stick with the same tariff Calculate the potential
savings
983116983151983159 983126983151983148983156983137983143983141 983117983137983160983145983149983157983149 983108983141983149983137983150983140 983107983144983137983154983143983141983155
983108983137983145983148983161 983123983156983137983150983140983145983150983143 983107983144983137983154983143983141 991404302983140983137983161
983120983123983119 983116983141983158983161
983120983123983119 983116983141983158983161 983152983141983154 983115983126983105 983151983142 983117983113983107 983152983141983154 983140983137983161 9914040036983097983147983126983105983140983137983161
983117983137983160983145983149983157983149 983113983149983152983151983154983156 983107983137983152983137983139983145983156983161 (983117983113983107) 983107983144983137983154983143983141983155
983108983137983145983148983161 983107983144983137983154983143983141 983152983141983154 983147983126983105 983151983142 983117983113983107 9914040072983147983126983105983140983137983161
983108983141983149983137983150983140 983107983144983137983154983143983141983155 983118983151983158983085983110983141983138 983117983137983154983085983119983139983156
983117983151983150983156983144983148983161 983107983144983137983154983143983141 983152983141983154 983147983127 983151983142 983149983141983156983141983154983141983140 983117983108 983145983150 983137
983149983151983150983156983144983148983161 983152983141983154983145983151983140 (983123983157983138983146983141983139983156 983156983151 983137 983149983145983150983145983149983157983149 983139983144983137983154983143983141 983151983142
30983147983127) 991404189830975983147983127983149983151983150983156983144 991404000
983108983137983161983156983145983149983141 983147983127983144 (983125983150983145983156) 983107983144983137983154983143983141 (0898309800 983085 19830979830985983097) 1480983139983147983127983144 1275983139983147983127983144
983118983145983143983144983156983145983149983141 983147983127983144 (983125983150983145983156) 983118983145983143983144983156 983107983144983137983154983143983141983155 (2098309800983085079830985983097) 64983097983139983147983127983144 64983097983139983147983127983144
983127983137983156983156983148983141983155983155 983125983150983145983156 983120983154983145983139983141 (983147983126983105983122983144 983145983150 983141983160983139983141983155983155 983151983142 983151983150983141 983156983144983145983154983140
983151983142 983156983151983156983137983148 983147983127983144) 991404000714 983147983126983105983122983144 991404000714 983147983126983105983122983144
bull Table Low Voltage Maximum Demand Charges (Q6)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2526
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
25
SolnNumber of days in February = 28
Number of working hours = 459 = 180
Normal kW = 80kVA08 = 64kW
Total kWh = 64kW180 hours+20kW2 hours = 11560 kWh
Normal kvar =80kVA sqrt(1-08^2) = 48kvar = sqrt(802-64
2) = 48 kVAr
Total kvarh = 48kvar180 hours = 8640 kvarh
Chargeable kvarh = 8640-(11560)3=4786 kvarh
983116983151983159 983126983151983148983156983137983143983141 983117983137983160983145983149983157983149 983108983141983149983137983150983140
983107983144983137983154983143983141983155 983152983141983154 983122983137983156983141 983121983156983161 983120983141983154983145983151983140 983107983144983137983154983143983141
983108983137983145983148983161 983123983156983137983150983140983145983150983143 983107983144983137983154983143983141 991404302 983140983137983161 991404302000 1 28 991404983096456
983120983123983119 983116983141983158983161
983120983123983119 983116983141983158983161 983152983141983154 983115983126983105 983152983141983154 983140983137983161 9914040036983097 983147983126983105983140983137983161 99140400369830970 120 28 991404123983097983096
983117983137983160983145983149983157983149 983113983149983152983151983154983156 983107983137983152983137983139983145983156983161
(983117983113983107) 983107983144983137983154983143983141983155
983108983137983145983148983161 983107983144983137983154983143983141 983152983141983154 983147983126983105 983151983142 983117983113983107 9914040072 983147983126983105983140983137983161 991404007200 120 28 9914042419830972
983108983141983149983137983150983140 983107983144983137983154983143983141983155 983118983151983158983085983110983141983138
983117983151983150983156983144983148983161 983107983144983137983154983143983141 983152983141983154 983147983127 983151983142
983149983141983156983141983154983141983140 983117983108 983145983150 983137 983149983151983150983156983144983148983161
983152983141983154983145983151983140 (983123983157983138983146983141983139983156 983156983151 983137
983149983145983150983145983149983157983149 983139983144983137983154983143983141 983151983142 30983147983127) 991404189830975
983147983127983149983151983150983156
983144 99140418983097500 84 1 991404159830971983096 983108983137983161983156983145983149983141 983147983127983144 (983125983150983145983156) 983107983144983137983154983143983141
(0898309800 983085 19830979830985983097) 9914040148 983147983127983144 991404014800 11560 1
991404198309510983096
983096
983118983145983143983144983156983145983149983141 983147983127983144 (983125983150983145983156) 983118983145983143983144983156
983107983144983137983154983143983141983155 (2098309800983085079830985983097) 9914040064983097 983147983127983144 99140400649830970
983127983137983156983156983148983141983155983155 983125983150983145983156 983120983154983145983139983141 (983147983126983105983122983144
983145983150 983141983160983139983141983155983155 983151983142 983151983150983141 983156983144983145983154983140 983151983142 983156983151983156983137983148
983147983127983144) 991404000714 983147983126983137983154983144 991404000714 4787 1 991404341983096
983124983151983156983137983148 983106983145983148983148
983142983151983154
983110983141983138983154983157983137983154983161
9914042354983095
0
[16 Marks]
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2626
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
26
Potential Savings
bull Use Power Factor correction to eliminate Watt less unit charge and to reduce MIC
bull Shift Heating load to night time or weekend
bull Negotiate lower MIC
Potential savings from these measures
Eliminate Wattless charge save euro3417 month
Shift heating cycle to night time
Max demand it lowered by 20kW = saving of 20euro1895 = euro3790 month
Saving by moving 20kW to night rate = 20kW2 hours (euro0148-euro00649)= euro332
Negotiating a reduced MIC
If power factor correction is employed and the heating cycle is shifted to night time the
then maximum VA of the plant drops to close to its maximum kW demand of 64kW An
MIC of 80kVA should be sufficient to cover this Potential Saving = 40kVA
(euro0072+euro00369)28 = euro12196 per month
Total potential saving = euro3417+euro3790+euro332+euro12196=euro19735month
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 326
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
3
Soln
Circuit impedance Z = VI = 150ang 30o
Cap reactance Xc = 6366
Circuit impedance Z = 1299 + j(XL ndash 6366)
R = j75 XL = 13866
Since XL = 2π fL L = 0441 H
Vcoil 285 ang 1687o V = 27274 + j8271 V
Voltage across cap VC = I jXc = -4775 ndashj827 V
Check Vs = vL + vc = 225ang 0o V
Ex 4 A balanced three phase load of 8 kVA with power factor 085 lagging is connected to athree phase 50Hz supply via a line of impedance 05+j1Ω per phase The voltage at the load is
exactly 400V line to line
bull Draw a per phase equivalent circuit of this system
bull Calculate the line current in complex number form
bull Calculate the impedance of the load
bull Calculate the magnitude of the line to line voltage at the supply
VL = 208ang60oV
IT =15ang-30oA
=o
Vr = 19485ang-30oV
j axis
φ
Vc = 955ang-120oV
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 426
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 526
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
5
EX 5 A balanced three phase load of 10 MW with power factor 08 lagging is connected to a
three phase 50Hz supply via a line of impedance 02+j05Ω per phase The voltage at the load is
exactly 10kV line to line
bull Draw a per phase equivalent circuit of this systembull Calculate the line current in complex number form
bull Calculate the impedance of the load
bull Calculate the magnitude of the line t o line voltage at the supply
bull Calculate the amount of power factor VARS (correcting capacitance) that would
need to be fitted at the supply end to achieve 095 power factor at the supply
Soln (a)
(b) Pf = cos φ = 08 = PS
ZLine = 02 + j05
Vn Zload
IL
Vload
P=7008 (W)
j axis
Q =4659 (kVAr)
S = 8392(VA)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 626
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
6
S = Ppf = 10 106 08 = 12 5 MVA
P = radic3 VL IL cos φ = 10MW
A I L 69721=
IL = 72169 ang -3687o A
(c) Z2 = VL IL = 64 + j48
(d) Van = VL + IL Z1 = 610496 + j 2021 V
= 61083 ang 19o
|Van| = 61083 V
(e) pf orginal 08 new 095
P = 10 MW S = Ppf = 10 106 095 = 10526 MVA
Old Q = gt cos-108 = 3686o Old
Q = 10 106 Tan 3686 = 10 106 075 = 75 MVar
New Q = gt cos-1
095 = 18195o
Q = P tan 18195o = 10 10
6 032868 = 3287 MVar
Diff in MVar = 75 ndash 3287 = 4213 MVar required
Ex 6 (Past Exam q)
A balanced three phase load of 8 kVA with power factor 085 lagging is connected to a three
phase 50Hz supply via a line of impedance 05+j1Ω per phase The voltage at the load is exactly
400V line to line
bull Draw a per phase equivalent circuit of this system
bull Calculate the line current in complex number form
bull Calculate the impedance of the load
bull Calculate the magnitude of the line to line voltage at the supply
bull Calculate real power (P) and reactive power (Q) drawn from the supply
Soln (a)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 726
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
7
(b) Pf = cos φ = 085 = PS
P = Spf = 8 103 085 = 68 kW
A I L 5411=
IL = 981 ndash j607 = 1154 ang -3175o A
(c) Z2 = VL IL = Ωdegang 75319319 = 1695 + j1049
(d) Van = VL + IL Z1 = 2411ang 161o
|Van| = 241 V
(e) pf 085
S = 085 S = Ppf = 10526 MVA
P = S pf = 68 kW
S = P + jQ
Q = P tan 3179o = 4214 kVar
Ex 7
Delta to Star Transformation
ZLine = 05 + j1
Vn 8kVA
IL
Vload
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 826
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
8
C B A
B A
Z Z Z
Z Z
Z ++=1
C B A
C B
Z Z Z
Z Z Z
++=2
C B A
C A
Z Z Z
Z Z Z
++=3
Star ndash Delta Transformation
2
133221
Z
Z Z Z Z Z Z Z A
++=
3
133221
Z
Z Z Z Z Z Z Z B
++=
ZA ZB
ZC
1
2 3
Z1
Z2Z3
Delta
Star
ZA ZB
ZC
1
2 3
Z1
Z2Z3
Delta Star
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 926
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1026
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
10
RT = 2 [ 29(2+9)] = 3612 = 32727 Ω
Ind Machines
Ex 9
Why does the rotor of an induction motor turn slower than the revolving field
Ex 10
Describe briefly the key differences between a squirrel cage motor and wound-rotor motor
in terms of construction and method of operation
See lecture notes
Ex 11
Both the voltage and frequency induced in the rotor of an induction motor decrease as the
rotor speeds up Briefly explain the reason for this and include the relationships between
slip the applied stator angular velocity and rotor angular velocity
Ex 12A 3-phase 400V 6-pole 50Hz star-connected induction motor operates on full load at a speed of
1475 revmin Determine the slip s
Ex 13 a) Calculate the synchronous speed of a 3-phase 12-pole induction motor that is excited by a
50Hz source
b) What is the nominal speed (speed at full load) is the slip at full load is 5
9
9
9
2
2
2
a
b
c
RT
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1126
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
11
a) Synchronous speed rpmns = 120 f p revmin
500 revmin
(b) s = (ωs - ωr) ωs = (ns - nr) ns
nr = ns(1 - s) = 475 revmin
Ex 14
An 400 V 50 Hz six-pole star connected induction machine operates with full load slip
of 005 pu and has the following per-phase equivalent circuit parameters
stator resistance Rs = 012Ω stator leakage reactance Xs = 06 Ω and magnetising
reactance Xm = 2215 Ω The rotor referred resistance Rr` = 015 Ω and the rotor referred
leakage reactance Xr` = 0586 Ω The induction machine initially operates as a motoron full load Using an approximate per-phase equivalent circuit determine
(i) synchronous speed and the speed of the motor at rated slip in revmin and
radsec
(ii) The approximate equivalent circuit per-phase input impedance
(iii) The input current
(iv) The developed torque
(v) The starting torque
Soln
Motoring modeSynchronous speed ωs = 2 π fp = 1046 rads
ωm = ωs(1 ndash s) = 9937 rads = 9489 revmin
Rr ` s = 015005 = 3
Z = Rs + jXs+ Rr ` s +jXr
`)
= 3122 + j 1186
The total Z is Zin = ( jXm|| Z)
R`r s
j06ΩΩΩΩ
is i`r
Vs = 400 radic3 V
j0586ΩΩΩΩ 012ΩΩΩΩ
015005 j221515ΩΩΩΩ
A
B
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1226
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
12
= 7398ang110o 2354ang 8238
o
== 3143ang2762o
The input current
I i = 7318 ang-2762o A
The rotor current
I r = 6886 ang-208o A
6886 A at a pf of 0935 lagging
The developed torque
T = Pm ωm = ||3
r
2`r
s
R I
sω = 408 Nm
Starting Torque
T = ||3
r
2`r
s
R I
sω with s=1
= 204 Nm
Ex 15 A 4 pole induction motor running from a 3 phase 400V line to line 50 Hz
supply has the following parametersStator Winding Resistance (Rs) = 005Ω
Stator Leakage Inductance (Xs) = 01 Ω
Magnetising Inductance (Xm) = 20 Ω Rotor Winding Resistance referred to stator (Rrrsquo) = 01 Ω
Rotor Leakage Inductance referred to stator (Xrrsquo) = 02 Ω
Rotational losses may be assumed to be negligible
(i) Draw a per phase equivalent circuit of this machine
(ii) Calculate the rotor speed at which the machine will have a slip of 2
(iii) Calculate the Power over the gap (Pog) in this machine when it is running at a slip of 2
(iv) Calculate the mechanical power (Pm) and the torque (Tm) delivered by the machine
running at a slip of 2
Soln
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1326
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
13
Synchronous speed ωs = 2 π fp = 1500 revmin
ωm = ωs(1 ndash s) = 15386 rads = 146925 revmin
Rr` s = 01002 = 5
Z = Rs + jXs+ Rr ` s +jXr
`)
= 505 + j 03
= 5059ang34o
The rotor referred current
I r = 4546 ang-34o A
4546 A at a pf of 099 lagging
The air-gap power
Pg = ||3 r2`
r
s R I = 31kW
Prot = 0 ndash no rotational losses
Pm = Pg (1-s) = 3038 kW
T = 3038 103 15386
= 197 Nm
Ex 16 A 6 pole induction motor running from a 3 phase 400V line to line 50 Hz supply has
the following parameters
Stator Winding Resistance (Rs) = 015Ω
Stator Leakage Inductance (Xs) = 03 Ω
Magnetising Inductance (Xm) = 60 Ω
R`r s
j01ΩΩΩΩ
is i`r
Vs = 400 radic3 V
j02ΩΩΩΩ 005ΩΩΩΩ
01002 j20ΩΩΩΩ
A
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1426
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
14
Rotor Winding Resistance referred to stator (Rrrsquo) = 03 Ω
Rotor Leakage Inductance referred to stator (Xrrsquo) = 06 Ω
Rotational losses have previously been measured to be 1000W at 5 slip
bull Draw a per phase equivalent circuit of this machinebull Calculate the rotor speed at which the machine will have a slip of 5
bull Calculate the air-gap Power (Pg) in this machine when it is running at a slip of 5
bull Calculate the efficiency of the induction motor from electrical input to mechanical
output when the motor is running with a slip of 5
Soln(a)
(b) Synchronous speed ωs = 2 π fp = 1000 revmin
ωm = ωs(1 ndash s) = 9948 rads = 950 revmin
(c) Rr` s = 03005 = 6
Z = Rs + jXs+ Rr ` s +jXr
`)
= 615 + j 09
= 622ang832o
(d) The air-gap power
The rotor referred current
I r = 3698 ang-303o A
Pg = ||3
r
2`r
s
R I = 246 kW
(d) I i =inZ
3400
Zin = Xm || Z = j60 || 622ang832o
Zin = 6097 ang1409o
I = VZ = 3772 ang-1409oA
R`r s
j03ΩΩΩΩ
is i`r
Vs = 230V
j06ΩΩΩΩ 015ΩΩΩΩ
03005 j60ΩΩΩΩ
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1526
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
15
Pelec = radic3 VL IL cos φ = 25044 kW
Pm = Pg (1-s) ndash Prot = 2237 kW
η = Pelec Pm = 893s
Ex 17
A large 3-phase 4000V 8-pole 50Hz squirrel-cage induction motor draws a current of
380A and a total active power of 2344kW when operating at full load The speed at full load
is accurately measured and found to be 742rpm The stator is star-connected and the
resistance between two stator terminals is 01Ω The total iron losses are 234kW and the
windage and friction losses are 12kW
Calculate
a) The power factor at full loadb) The active power supplied to the rotor
c) The total i2R losses in the rotor
d) The mechanical power output to the load and torque delivered to the load
e) The motorrsquos net efficiency
a) The active power input to the motor is given as 2344kW The apparent power is
S = radic3 VL IL = 2633kVA
Therefore pf = PS 089 lagging
b) The active power supplied to the rotor is the active power input to the motor minus the ironlosses and copper losses in the stator The stator iron losses are given as 234kW To calculate
the copper losses note that the terminal resistance (01) is twice that of each phase for a wye
connection Therefore the copper losses are
Stator copper losses Pcu = 217kW
Therefore the active power supplied to the rotor is
Pr = Input power ndash stator copper loss ndash iron losses
Pr = 2344 kW ndash 234kW ndash 217kW
= 2298kW
c) The copper losses in the rotor are proportional to the slip (see lecture notes) Prcu sP
Therefore we must calculate the slip Neither the synchronous speed nor number of poles are
explicitly stated in the problem however by inspection we determine that the synchronous speed
must be 750rpm and the number of poles = 8 (For 50Hz supply synchronous speeds occur at
discrete intervals corresponding to pole pairs 2-pole 3000rpm 4-pole 1500rpm 6- pole
1000rpm 8-pole 750rpm etchellip We also know that induction motors operate near synchronous
speed ndash the speed is given as 742 rpm which is close to synchronous speed for an
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1626
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
16
8-pole machine)
Synchronous speed ωs = 2 π fp = 7854 rads
ωm = = 777 radss = 00107
Therefore the rotor copper losses are Pr cu = sPr = 2456 kw
(d)
The mechanical power (Pm) delivered to the load is the active power delivered to the rotor
minus rotor copper losses and windage and friction losses The windage and friction losses are
given as 12kW Therefore
Pm =22989 kW ndash 245kW-12 kW =22624 kW
kNmT L 129
60
2742
42262=
=
π
(e) the motors efficiency 5962344
42262===
kW
kW
P
P
in
out η
Ex 18
A 3-phase 440V 8-pole 50Hz star-connected induction motor has the following equivalent
circuit per-phase parameters
Rs = 01 Ω Rr = 015Ω xs = xr =06 Ω Rc = 100 Ω xm = 10 Ω
(a) Calculate using the equivalent circuit neglecting stator impedance at a slip of 5 the input
stator-current and power factor the rotor current referred to the stator the torque the
mechanical output power and the efficiency Calculate the starting torque Assume the
mechanical loss is 1kW for all speeds
Soln
(a) Synchronous speed ωs = 2 π fp 7854 rads
ωm = ωs(1 - s) = 7618 rads
Stator impedance neglected
Vs = 400 radic3 = 230
Ir` =
60050
10
230
j+ =
= 68 ang-1671 = 651 - j 1954
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1726
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
17
Io =230
100 + j230
10 = 23 ndash j 23
Is = Ir` + Io = 674 - j4294
= 799 A at cos φ = 084
Ir` = 68 A
T = 529 Nm
Pmech = 39365 kW
Pelect = 465 kW
Efficiency = 39365465 = 846
Ts = 39627 Nm
Ex19
Tests conducted at 50 Hz on a three-phase star connected 45 kW 2200V six-pole 50Hz squirrel
cage induction motor are listed below
The average dc resistance per stator phase Rs = 28 Ω (a) Briefly describe the tests required to determine the per-phase equivalent circuit
impedance parameters of an induction machine
(a) Determine the parameters of the per-phase equivalent circuit
(b) Determine the rotational losses of the machine Core losses can be neglected
load
0
No-loadV
R
I1
Ic
I
c X
Im
m
Figure 1 Equivalent Circuit during No-Load Test
torPower Fac0 times= I I C = 2025 A
22
0 cm I I I minus=
= 4019 A
Ω== 627c
c I
V R Ω== 5317
m
m I
V X
No Load Test Locked Rotor Test
Power Factor = 045 Power Factor = 09
Line Voltage = 22kV Line Voltage = 270V
Line Current = 45A Line Current = 25A
Input Power = 16kW Input Power = 9kW
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1826
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
18
For the locked rotor test slip = 1
Ω== 246s
s
LR I
V Z
Locked rotor reactance is
( )2
21
2
21 R R Z X X +minus=+
φ cos21 times=+= Z R R Rtotal = 5616Ω
R2rsquo = Rtot - R1 = 2816Ω
Ω=+ 72221 X X
(a) From the no-load test the no-load power isPNL = 16kW
The no-load rotational loss is
Prot = PNL ndash 3 I s2 Rs
= 14299W
Synch Machines
Ex 20
A 60 MVA 20 kV three-phase 50 Hz synchronous generator has a synchronous reactance
of 8 Ω per phase and negligible resistance The generator delivers rated power at a power
factor of 085 lagging at the rated terminal voltage to an infinite bus(a) Determine the excitation voltage per phase and the power angle δ
(b) With the excitation voltage Eg of the synchronous generator held constant at the
value determined in part (a) the input driving torque is reduced until the generator
delivers 28 MW Determine the armature current and the power factor
(c) If the generator operates at the excitation voltage of Eg obtained in part (a) what is
the steady state maximum power the machine can deliver before losing
jXsΩΩΩΩ jX`r ΩΩΩΩ Rs
R`r s
Is I`r
Vs ph
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1926
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
19
synchronism Determine the armature current corresponding to this maximum
power
(d) With real generated power at 42MW compute the power angle δ the stator
current Ia and power factor when the field excitation current If is adjusted so that
the excitation voltage Eg is decreased to 7919 of the value determined in part
(a)
Soln
(a) Apparent power S = 51 MW ndash j 316Mvar
Rated voltage per phase
V = 1154 ang 0deg kV
Rated current
Ia = S 3V
= 1733 ang -3178deg AExcitation voltage
Eg = 11540 + j9(1733) ang -3178deg)= 237922 ang 3386deg
Power angle = 3386o
(b) When delivering 28 MW
δ = sin-1
)5411)(79223(3
828 = 1578
o
P =
s
g
X
V E 3sinδ
Ia = 13069ang2967o j8
= 16336ang-6033o A
Power factor = cos(6033) = 0495 lagging
(c) The maximum power occurs at δ = 90o
Pmax =s
g
X
V E 3
= 102 MW
Ia = 324885ang25875o A
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2026
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
20
The power factor cos(25875) = 0899 leading
(d) Eg = 13463 V
The power angle δ = sin-1
)5411)(84318(3
842
= 31o
Ia = 86674ang0o A
-Power factor = 0
Ex 21 (Aug 13) A 3 phase synchronous machine with negligible stator winding resistance and a
synchronous reactance of 2 per phase is connected to a 400V 50 Hz 3-phase supply (infinite
bus) and operated as a generator It is currently generating 50kW with power factor 08 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)
(iv) What is the maximum power the machine can generate with this value of excitation
Soln(i)
(ii)P = 50kW
Apparent power S = 625 ang -3687
o
kVA
Rated voltage per phase V = 230ang 0deg VRated current
Ia = S 3V
= 9058ang -3687deg A
LoadGen
Xs =2
Eg Vt =400 radic3 V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2126
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
21
|Ia| = 9058 A
(iii)
Excitation voltage
Eg = 230 ang 0deg + j2(9058ang -3687deg)= 3684 ang 2318degV
|Eg| = 3684 V
Power angle = 2318o
(iv)
P = deg90sin2
23043683
(max occurs when δ = 90o)
= 1271 kW
Ex 22A 3 phase 4 pole synchronous machine with negligible stator winding resistance and a
synchronous reactance of 2 per phase is connected to a 480V 60 Hz 3phase supply
(infinite bus) and operated as a generator It is currently generating 80kW with power
factor 08 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)(iv) What is the maximum power the machine can generate with this value
excitation(v) What is the rotational speed of this generator
Soln (i)
(ii)P = 80kW Apparent power S 100 ang -3687
o kVA
Rated voltage per phase
V 277ang 0deg V
LoadGen
Xs =2
Eg Vt = 480V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2226
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
22
Rated current
Ia = S 3V
= 9058ang -3687deg A
|Ia| = 12028 A
(iii)
Excitation voltage
Eg = 277 ang 0deg + j2(12028ang -3687deg)
= 4632 ang 2455degV|Eg| = 4632 V
Power angle = 2455o
(iv)
P = deg90sin2
27724633
(max occurs when δ = 90o)
= 19238 kW
(v)Synchronous speed ωs = 2 π fp = 1885 rads
= 1800 revmin
Ex 23 (Jan 14)
A 3 phase synchronous machine with negligible stator winding resistance and a synchronous
reactance of 25 per phase is connected to a 20kV 50 Hz 3phase supply (infinite bus) and
operated as a generator It is currently generating 100MW with power factor 085 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)(iv) What minimum value of excitation voltage (E) would be required for this machine to
deliver 200MW into a 20kV bus assuming a 20deg stability margin must be maintained onpower angle (δ)
Soln
(i)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2326
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
23
(ii)
P = 100 MW Apparent power S = 100 085
= 11765 ang -3179o MVA
Rated voltage per phase
V= 11547ang 0deg VRated current
Ia = S 3V
= 3396 103ang -3179deg A
|Ia| = 3396 103A
(iii)
Excitation voltage
Eg = 11547 ang 0deg + j25(3396 103ang -3179deg)
= 17570 ang 2425degV|Eg| = 17570 V
Power angle = 2425o V
(iv)
P = )2090sin(52
3degminusdegt gV E
(max occurs when δ = 90o stability margin 20o)
200 MW = )2090sin(52
547113degminusdegg E
200 MW = 38130209397085613 gg E E =
|Eg| = 15360 V
LoadGen
Xs =25
Eg Vt =11547V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2426
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
24
Ex 24A manufacturing plant fed with a 400V 50Hz three phase runs from 0900 to 1800
Monday to Friday The factoryrsquos load is normally 80kVA with power factor 08 lagging
during these hours Once a month they need to run a special heating cycle which adds a
further 20kW to the factory load for a 2 hour period This heating cycle is timer
controlled and fully automated but the timer is currently set to run on the third Friday
afternoon of the month Outside of normal business hours and at weekends the factory
load is negligible The firm has a contractually agreed Maximum Import Capacity (MIC)
of 120kVA
bull Use the attached table of Low Voltage Maximum Demand charges to calculate the
firmrsquos electricity bill (exclusive of VAT) for the month of February (28 days)
bull Identify three steps that this firm could take to reduce their expenditure on
electricity assuming they stick with the same tariff Calculate the potential
savings
983116983151983159 983126983151983148983156983137983143983141 983117983137983160983145983149983157983149 983108983141983149983137983150983140 983107983144983137983154983143983141983155
983108983137983145983148983161 983123983156983137983150983140983145983150983143 983107983144983137983154983143983141 991404302983140983137983161
983120983123983119 983116983141983158983161
983120983123983119 983116983141983158983161 983152983141983154 983115983126983105 983151983142 983117983113983107 983152983141983154 983140983137983161 9914040036983097983147983126983105983140983137983161
983117983137983160983145983149983157983149 983113983149983152983151983154983156 983107983137983152983137983139983145983156983161 (983117983113983107) 983107983144983137983154983143983141983155
983108983137983145983148983161 983107983144983137983154983143983141 983152983141983154 983147983126983105 983151983142 983117983113983107 9914040072983147983126983105983140983137983161
983108983141983149983137983150983140 983107983144983137983154983143983141983155 983118983151983158983085983110983141983138 983117983137983154983085983119983139983156
983117983151983150983156983144983148983161 983107983144983137983154983143983141 983152983141983154 983147983127 983151983142 983149983141983156983141983154983141983140 983117983108 983145983150 983137
983149983151983150983156983144983148983161 983152983141983154983145983151983140 (983123983157983138983146983141983139983156 983156983151 983137 983149983145983150983145983149983157983149 983139983144983137983154983143983141 983151983142
30983147983127) 991404189830975983147983127983149983151983150983156983144 991404000
983108983137983161983156983145983149983141 983147983127983144 (983125983150983145983156) 983107983144983137983154983143983141 (0898309800 983085 19830979830985983097) 1480983139983147983127983144 1275983139983147983127983144
983118983145983143983144983156983145983149983141 983147983127983144 (983125983150983145983156) 983118983145983143983144983156 983107983144983137983154983143983141983155 (2098309800983085079830985983097) 64983097983139983147983127983144 64983097983139983147983127983144
983127983137983156983156983148983141983155983155 983125983150983145983156 983120983154983145983139983141 (983147983126983105983122983144 983145983150 983141983160983139983141983155983155 983151983142 983151983150983141 983156983144983145983154983140
983151983142 983156983151983156983137983148 983147983127983144) 991404000714 983147983126983105983122983144 991404000714 983147983126983105983122983144
bull Table Low Voltage Maximum Demand Charges (Q6)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2526
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
25
SolnNumber of days in February = 28
Number of working hours = 459 = 180
Normal kW = 80kVA08 = 64kW
Total kWh = 64kW180 hours+20kW2 hours = 11560 kWh
Normal kvar =80kVA sqrt(1-08^2) = 48kvar = sqrt(802-64
2) = 48 kVAr
Total kvarh = 48kvar180 hours = 8640 kvarh
Chargeable kvarh = 8640-(11560)3=4786 kvarh
983116983151983159 983126983151983148983156983137983143983141 983117983137983160983145983149983157983149 983108983141983149983137983150983140
983107983144983137983154983143983141983155 983152983141983154 983122983137983156983141 983121983156983161 983120983141983154983145983151983140 983107983144983137983154983143983141
983108983137983145983148983161 983123983156983137983150983140983145983150983143 983107983144983137983154983143983141 991404302 983140983137983161 991404302000 1 28 991404983096456
983120983123983119 983116983141983158983161
983120983123983119 983116983141983158983161 983152983141983154 983115983126983105 983152983141983154 983140983137983161 9914040036983097 983147983126983105983140983137983161 99140400369830970 120 28 991404123983097983096
983117983137983160983145983149983157983149 983113983149983152983151983154983156 983107983137983152983137983139983145983156983161
(983117983113983107) 983107983144983137983154983143983141983155
983108983137983145983148983161 983107983144983137983154983143983141 983152983141983154 983147983126983105 983151983142 983117983113983107 9914040072 983147983126983105983140983137983161 991404007200 120 28 9914042419830972
983108983141983149983137983150983140 983107983144983137983154983143983141983155 983118983151983158983085983110983141983138
983117983151983150983156983144983148983161 983107983144983137983154983143983141 983152983141983154 983147983127 983151983142
983149983141983156983141983154983141983140 983117983108 983145983150 983137 983149983151983150983156983144983148983161
983152983141983154983145983151983140 (983123983157983138983146983141983139983156 983156983151 983137
983149983145983150983145983149983157983149 983139983144983137983154983143983141 983151983142 30983147983127) 991404189830975
983147983127983149983151983150983156
983144 99140418983097500 84 1 991404159830971983096 983108983137983161983156983145983149983141 983147983127983144 (983125983150983145983156) 983107983144983137983154983143983141
(0898309800 983085 19830979830985983097) 9914040148 983147983127983144 991404014800 11560 1
991404198309510983096
983096
983118983145983143983144983156983145983149983141 983147983127983144 (983125983150983145983156) 983118983145983143983144983156
983107983144983137983154983143983141983155 (2098309800983085079830985983097) 9914040064983097 983147983127983144 99140400649830970
983127983137983156983156983148983141983155983155 983125983150983145983156 983120983154983145983139983141 (983147983126983105983122983144
983145983150 983141983160983139983141983155983155 983151983142 983151983150983141 983156983144983145983154983140 983151983142 983156983151983156983137983148
983147983127983144) 991404000714 983147983126983137983154983144 991404000714 4787 1 991404341983096
983124983151983156983137983148 983106983145983148983148
983142983151983154
983110983141983138983154983157983137983154983161
9914042354983095
0
[16 Marks]
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2626
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
26
Potential Savings
bull Use Power Factor correction to eliminate Watt less unit charge and to reduce MIC
bull Shift Heating load to night time or weekend
bull Negotiate lower MIC
Potential savings from these measures
Eliminate Wattless charge save euro3417 month
Shift heating cycle to night time
Max demand it lowered by 20kW = saving of 20euro1895 = euro3790 month
Saving by moving 20kW to night rate = 20kW2 hours (euro0148-euro00649)= euro332
Negotiating a reduced MIC
If power factor correction is employed and the heating cycle is shifted to night time the
then maximum VA of the plant drops to close to its maximum kW demand of 64kW An
MIC of 80kVA should be sufficient to cover this Potential Saving = 40kVA
(euro0072+euro00369)28 = euro12196 per month
Total potential saving = euro3417+euro3790+euro332+euro12196=euro19735month
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 426
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 526
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
5
EX 5 A balanced three phase load of 10 MW with power factor 08 lagging is connected to a
three phase 50Hz supply via a line of impedance 02+j05Ω per phase The voltage at the load is
exactly 10kV line to line
bull Draw a per phase equivalent circuit of this systembull Calculate the line current in complex number form
bull Calculate the impedance of the load
bull Calculate the magnitude of the line t o line voltage at the supply
bull Calculate the amount of power factor VARS (correcting capacitance) that would
need to be fitted at the supply end to achieve 095 power factor at the supply
Soln (a)
(b) Pf = cos φ = 08 = PS
ZLine = 02 + j05
Vn Zload
IL
Vload
P=7008 (W)
j axis
Q =4659 (kVAr)
S = 8392(VA)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 626
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
6
S = Ppf = 10 106 08 = 12 5 MVA
P = radic3 VL IL cos φ = 10MW
A I L 69721=
IL = 72169 ang -3687o A
(c) Z2 = VL IL = 64 + j48
(d) Van = VL + IL Z1 = 610496 + j 2021 V
= 61083 ang 19o
|Van| = 61083 V
(e) pf orginal 08 new 095
P = 10 MW S = Ppf = 10 106 095 = 10526 MVA
Old Q = gt cos-108 = 3686o Old
Q = 10 106 Tan 3686 = 10 106 075 = 75 MVar
New Q = gt cos-1
095 = 18195o
Q = P tan 18195o = 10 10
6 032868 = 3287 MVar
Diff in MVar = 75 ndash 3287 = 4213 MVar required
Ex 6 (Past Exam q)
A balanced three phase load of 8 kVA with power factor 085 lagging is connected to a three
phase 50Hz supply via a line of impedance 05+j1Ω per phase The voltage at the load is exactly
400V line to line
bull Draw a per phase equivalent circuit of this system
bull Calculate the line current in complex number form
bull Calculate the impedance of the load
bull Calculate the magnitude of the line to line voltage at the supply
bull Calculate real power (P) and reactive power (Q) drawn from the supply
Soln (a)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 726
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
7
(b) Pf = cos φ = 085 = PS
P = Spf = 8 103 085 = 68 kW
A I L 5411=
IL = 981 ndash j607 = 1154 ang -3175o A
(c) Z2 = VL IL = Ωdegang 75319319 = 1695 + j1049
(d) Van = VL + IL Z1 = 2411ang 161o
|Van| = 241 V
(e) pf 085
S = 085 S = Ppf = 10526 MVA
P = S pf = 68 kW
S = P + jQ
Q = P tan 3179o = 4214 kVar
Ex 7
Delta to Star Transformation
ZLine = 05 + j1
Vn 8kVA
IL
Vload
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 826
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
8
C B A
B A
Z Z Z
Z Z
Z ++=1
C B A
C B
Z Z Z
Z Z Z
++=2
C B A
C A
Z Z Z
Z Z Z
++=3
Star ndash Delta Transformation
2
133221
Z
Z Z Z Z Z Z Z A
++=
3
133221
Z
Z Z Z Z Z Z Z B
++=
ZA ZB
ZC
1
2 3
Z1
Z2Z3
Delta
Star
ZA ZB
ZC
1
2 3
Z1
Z2Z3
Delta Star
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 926
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1026
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
10
RT = 2 [ 29(2+9)] = 3612 = 32727 Ω
Ind Machines
Ex 9
Why does the rotor of an induction motor turn slower than the revolving field
Ex 10
Describe briefly the key differences between a squirrel cage motor and wound-rotor motor
in terms of construction and method of operation
See lecture notes
Ex 11
Both the voltage and frequency induced in the rotor of an induction motor decrease as the
rotor speeds up Briefly explain the reason for this and include the relationships between
slip the applied stator angular velocity and rotor angular velocity
Ex 12A 3-phase 400V 6-pole 50Hz star-connected induction motor operates on full load at a speed of
1475 revmin Determine the slip s
Ex 13 a) Calculate the synchronous speed of a 3-phase 12-pole induction motor that is excited by a
50Hz source
b) What is the nominal speed (speed at full load) is the slip at full load is 5
9
9
9
2
2
2
a
b
c
RT
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1126
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
11
a) Synchronous speed rpmns = 120 f p revmin
500 revmin
(b) s = (ωs - ωr) ωs = (ns - nr) ns
nr = ns(1 - s) = 475 revmin
Ex 14
An 400 V 50 Hz six-pole star connected induction machine operates with full load slip
of 005 pu and has the following per-phase equivalent circuit parameters
stator resistance Rs = 012Ω stator leakage reactance Xs = 06 Ω and magnetising
reactance Xm = 2215 Ω The rotor referred resistance Rr` = 015 Ω and the rotor referred
leakage reactance Xr` = 0586 Ω The induction machine initially operates as a motoron full load Using an approximate per-phase equivalent circuit determine
(i) synchronous speed and the speed of the motor at rated slip in revmin and
radsec
(ii) The approximate equivalent circuit per-phase input impedance
(iii) The input current
(iv) The developed torque
(v) The starting torque
Soln
Motoring modeSynchronous speed ωs = 2 π fp = 1046 rads
ωm = ωs(1 ndash s) = 9937 rads = 9489 revmin
Rr ` s = 015005 = 3
Z = Rs + jXs+ Rr ` s +jXr
`)
= 3122 + j 1186
The total Z is Zin = ( jXm|| Z)
R`r s
j06ΩΩΩΩ
is i`r
Vs = 400 radic3 V
j0586ΩΩΩΩ 012ΩΩΩΩ
015005 j221515ΩΩΩΩ
A
B
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1226
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
12
= 7398ang110o 2354ang 8238
o
== 3143ang2762o
The input current
I i = 7318 ang-2762o A
The rotor current
I r = 6886 ang-208o A
6886 A at a pf of 0935 lagging
The developed torque
T = Pm ωm = ||3
r
2`r
s
R I
sω = 408 Nm
Starting Torque
T = ||3
r
2`r
s
R I
sω with s=1
= 204 Nm
Ex 15 A 4 pole induction motor running from a 3 phase 400V line to line 50 Hz
supply has the following parametersStator Winding Resistance (Rs) = 005Ω
Stator Leakage Inductance (Xs) = 01 Ω
Magnetising Inductance (Xm) = 20 Ω Rotor Winding Resistance referred to stator (Rrrsquo) = 01 Ω
Rotor Leakage Inductance referred to stator (Xrrsquo) = 02 Ω
Rotational losses may be assumed to be negligible
(i) Draw a per phase equivalent circuit of this machine
(ii) Calculate the rotor speed at which the machine will have a slip of 2
(iii) Calculate the Power over the gap (Pog) in this machine when it is running at a slip of 2
(iv) Calculate the mechanical power (Pm) and the torque (Tm) delivered by the machine
running at a slip of 2
Soln
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1326
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
13
Synchronous speed ωs = 2 π fp = 1500 revmin
ωm = ωs(1 ndash s) = 15386 rads = 146925 revmin
Rr` s = 01002 = 5
Z = Rs + jXs+ Rr ` s +jXr
`)
= 505 + j 03
= 5059ang34o
The rotor referred current
I r = 4546 ang-34o A
4546 A at a pf of 099 lagging
The air-gap power
Pg = ||3 r2`
r
s R I = 31kW
Prot = 0 ndash no rotational losses
Pm = Pg (1-s) = 3038 kW
T = 3038 103 15386
= 197 Nm
Ex 16 A 6 pole induction motor running from a 3 phase 400V line to line 50 Hz supply has
the following parameters
Stator Winding Resistance (Rs) = 015Ω
Stator Leakage Inductance (Xs) = 03 Ω
Magnetising Inductance (Xm) = 60 Ω
R`r s
j01ΩΩΩΩ
is i`r
Vs = 400 radic3 V
j02ΩΩΩΩ 005ΩΩΩΩ
01002 j20ΩΩΩΩ
A
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1426
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
14
Rotor Winding Resistance referred to stator (Rrrsquo) = 03 Ω
Rotor Leakage Inductance referred to stator (Xrrsquo) = 06 Ω
Rotational losses have previously been measured to be 1000W at 5 slip
bull Draw a per phase equivalent circuit of this machinebull Calculate the rotor speed at which the machine will have a slip of 5
bull Calculate the air-gap Power (Pg) in this machine when it is running at a slip of 5
bull Calculate the efficiency of the induction motor from electrical input to mechanical
output when the motor is running with a slip of 5
Soln(a)
(b) Synchronous speed ωs = 2 π fp = 1000 revmin
ωm = ωs(1 ndash s) = 9948 rads = 950 revmin
(c) Rr` s = 03005 = 6
Z = Rs + jXs+ Rr ` s +jXr
`)
= 615 + j 09
= 622ang832o
(d) The air-gap power
The rotor referred current
I r = 3698 ang-303o A
Pg = ||3
r
2`r
s
R I = 246 kW
(d) I i =inZ
3400
Zin = Xm || Z = j60 || 622ang832o
Zin = 6097 ang1409o
I = VZ = 3772 ang-1409oA
R`r s
j03ΩΩΩΩ
is i`r
Vs = 230V
j06ΩΩΩΩ 015ΩΩΩΩ
03005 j60ΩΩΩΩ
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1526
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
15
Pelec = radic3 VL IL cos φ = 25044 kW
Pm = Pg (1-s) ndash Prot = 2237 kW
η = Pelec Pm = 893s
Ex 17
A large 3-phase 4000V 8-pole 50Hz squirrel-cage induction motor draws a current of
380A and a total active power of 2344kW when operating at full load The speed at full load
is accurately measured and found to be 742rpm The stator is star-connected and the
resistance between two stator terminals is 01Ω The total iron losses are 234kW and the
windage and friction losses are 12kW
Calculate
a) The power factor at full loadb) The active power supplied to the rotor
c) The total i2R losses in the rotor
d) The mechanical power output to the load and torque delivered to the load
e) The motorrsquos net efficiency
a) The active power input to the motor is given as 2344kW The apparent power is
S = radic3 VL IL = 2633kVA
Therefore pf = PS 089 lagging
b) The active power supplied to the rotor is the active power input to the motor minus the ironlosses and copper losses in the stator The stator iron losses are given as 234kW To calculate
the copper losses note that the terminal resistance (01) is twice that of each phase for a wye
connection Therefore the copper losses are
Stator copper losses Pcu = 217kW
Therefore the active power supplied to the rotor is
Pr = Input power ndash stator copper loss ndash iron losses
Pr = 2344 kW ndash 234kW ndash 217kW
= 2298kW
c) The copper losses in the rotor are proportional to the slip (see lecture notes) Prcu sP
Therefore we must calculate the slip Neither the synchronous speed nor number of poles are
explicitly stated in the problem however by inspection we determine that the synchronous speed
must be 750rpm and the number of poles = 8 (For 50Hz supply synchronous speeds occur at
discrete intervals corresponding to pole pairs 2-pole 3000rpm 4-pole 1500rpm 6- pole
1000rpm 8-pole 750rpm etchellip We also know that induction motors operate near synchronous
speed ndash the speed is given as 742 rpm which is close to synchronous speed for an
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1626
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
16
8-pole machine)
Synchronous speed ωs = 2 π fp = 7854 rads
ωm = = 777 radss = 00107
Therefore the rotor copper losses are Pr cu = sPr = 2456 kw
(d)
The mechanical power (Pm) delivered to the load is the active power delivered to the rotor
minus rotor copper losses and windage and friction losses The windage and friction losses are
given as 12kW Therefore
Pm =22989 kW ndash 245kW-12 kW =22624 kW
kNmT L 129
60
2742
42262=
=
π
(e) the motors efficiency 5962344
42262===
kW
kW
P
P
in
out η
Ex 18
A 3-phase 440V 8-pole 50Hz star-connected induction motor has the following equivalent
circuit per-phase parameters
Rs = 01 Ω Rr = 015Ω xs = xr =06 Ω Rc = 100 Ω xm = 10 Ω
(a) Calculate using the equivalent circuit neglecting stator impedance at a slip of 5 the input
stator-current and power factor the rotor current referred to the stator the torque the
mechanical output power and the efficiency Calculate the starting torque Assume the
mechanical loss is 1kW for all speeds
Soln
(a) Synchronous speed ωs = 2 π fp 7854 rads
ωm = ωs(1 - s) = 7618 rads
Stator impedance neglected
Vs = 400 radic3 = 230
Ir` =
60050
10
230
j+ =
= 68 ang-1671 = 651 - j 1954
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1726
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
17
Io =230
100 + j230
10 = 23 ndash j 23
Is = Ir` + Io = 674 - j4294
= 799 A at cos φ = 084
Ir` = 68 A
T = 529 Nm
Pmech = 39365 kW
Pelect = 465 kW
Efficiency = 39365465 = 846
Ts = 39627 Nm
Ex19
Tests conducted at 50 Hz on a three-phase star connected 45 kW 2200V six-pole 50Hz squirrel
cage induction motor are listed below
The average dc resistance per stator phase Rs = 28 Ω (a) Briefly describe the tests required to determine the per-phase equivalent circuit
impedance parameters of an induction machine
(a) Determine the parameters of the per-phase equivalent circuit
(b) Determine the rotational losses of the machine Core losses can be neglected
load
0
No-loadV
R
I1
Ic
I
c X
Im
m
Figure 1 Equivalent Circuit during No-Load Test
torPower Fac0 times= I I C = 2025 A
22
0 cm I I I minus=
= 4019 A
Ω== 627c
c I
V R Ω== 5317
m
m I
V X
No Load Test Locked Rotor Test
Power Factor = 045 Power Factor = 09
Line Voltage = 22kV Line Voltage = 270V
Line Current = 45A Line Current = 25A
Input Power = 16kW Input Power = 9kW
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1826
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
18
For the locked rotor test slip = 1
Ω== 246s
s
LR I
V Z
Locked rotor reactance is
( )2
21
2
21 R R Z X X +minus=+
φ cos21 times=+= Z R R Rtotal = 5616Ω
R2rsquo = Rtot - R1 = 2816Ω
Ω=+ 72221 X X
(a) From the no-load test the no-load power isPNL = 16kW
The no-load rotational loss is
Prot = PNL ndash 3 I s2 Rs
= 14299W
Synch Machines
Ex 20
A 60 MVA 20 kV three-phase 50 Hz synchronous generator has a synchronous reactance
of 8 Ω per phase and negligible resistance The generator delivers rated power at a power
factor of 085 lagging at the rated terminal voltage to an infinite bus(a) Determine the excitation voltage per phase and the power angle δ
(b) With the excitation voltage Eg of the synchronous generator held constant at the
value determined in part (a) the input driving torque is reduced until the generator
delivers 28 MW Determine the armature current and the power factor
(c) If the generator operates at the excitation voltage of Eg obtained in part (a) what is
the steady state maximum power the machine can deliver before losing
jXsΩΩΩΩ jX`r ΩΩΩΩ Rs
R`r s
Is I`r
Vs ph
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1926
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
19
synchronism Determine the armature current corresponding to this maximum
power
(d) With real generated power at 42MW compute the power angle δ the stator
current Ia and power factor when the field excitation current If is adjusted so that
the excitation voltage Eg is decreased to 7919 of the value determined in part
(a)
Soln
(a) Apparent power S = 51 MW ndash j 316Mvar
Rated voltage per phase
V = 1154 ang 0deg kV
Rated current
Ia = S 3V
= 1733 ang -3178deg AExcitation voltage
Eg = 11540 + j9(1733) ang -3178deg)= 237922 ang 3386deg
Power angle = 3386o
(b) When delivering 28 MW
δ = sin-1
)5411)(79223(3
828 = 1578
o
P =
s
g
X
V E 3sinδ
Ia = 13069ang2967o j8
= 16336ang-6033o A
Power factor = cos(6033) = 0495 lagging
(c) The maximum power occurs at δ = 90o
Pmax =s
g
X
V E 3
= 102 MW
Ia = 324885ang25875o A
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2026
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
20
The power factor cos(25875) = 0899 leading
(d) Eg = 13463 V
The power angle δ = sin-1
)5411)(84318(3
842
= 31o
Ia = 86674ang0o A
-Power factor = 0
Ex 21 (Aug 13) A 3 phase synchronous machine with negligible stator winding resistance and a
synchronous reactance of 2 per phase is connected to a 400V 50 Hz 3-phase supply (infinite
bus) and operated as a generator It is currently generating 50kW with power factor 08 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)
(iv) What is the maximum power the machine can generate with this value of excitation
Soln(i)
(ii)P = 50kW
Apparent power S = 625 ang -3687
o
kVA
Rated voltage per phase V = 230ang 0deg VRated current
Ia = S 3V
= 9058ang -3687deg A
LoadGen
Xs =2
Eg Vt =400 radic3 V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2126
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
21
|Ia| = 9058 A
(iii)
Excitation voltage
Eg = 230 ang 0deg + j2(9058ang -3687deg)= 3684 ang 2318degV
|Eg| = 3684 V
Power angle = 2318o
(iv)
P = deg90sin2
23043683
(max occurs when δ = 90o)
= 1271 kW
Ex 22A 3 phase 4 pole synchronous machine with negligible stator winding resistance and a
synchronous reactance of 2 per phase is connected to a 480V 60 Hz 3phase supply
(infinite bus) and operated as a generator It is currently generating 80kW with power
factor 08 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)(iv) What is the maximum power the machine can generate with this value
excitation(v) What is the rotational speed of this generator
Soln (i)
(ii)P = 80kW Apparent power S 100 ang -3687
o kVA
Rated voltage per phase
V 277ang 0deg V
LoadGen
Xs =2
Eg Vt = 480V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2226
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
22
Rated current
Ia = S 3V
= 9058ang -3687deg A
|Ia| = 12028 A
(iii)
Excitation voltage
Eg = 277 ang 0deg + j2(12028ang -3687deg)
= 4632 ang 2455degV|Eg| = 4632 V
Power angle = 2455o
(iv)
P = deg90sin2
27724633
(max occurs when δ = 90o)
= 19238 kW
(v)Synchronous speed ωs = 2 π fp = 1885 rads
= 1800 revmin
Ex 23 (Jan 14)
A 3 phase synchronous machine with negligible stator winding resistance and a synchronous
reactance of 25 per phase is connected to a 20kV 50 Hz 3phase supply (infinite bus) and
operated as a generator It is currently generating 100MW with power factor 085 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)(iv) What minimum value of excitation voltage (E) would be required for this machine to
deliver 200MW into a 20kV bus assuming a 20deg stability margin must be maintained onpower angle (δ)
Soln
(i)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2326
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
23
(ii)
P = 100 MW Apparent power S = 100 085
= 11765 ang -3179o MVA
Rated voltage per phase
V= 11547ang 0deg VRated current
Ia = S 3V
= 3396 103ang -3179deg A
|Ia| = 3396 103A
(iii)
Excitation voltage
Eg = 11547 ang 0deg + j25(3396 103ang -3179deg)
= 17570 ang 2425degV|Eg| = 17570 V
Power angle = 2425o V
(iv)
P = )2090sin(52
3degminusdegt gV E
(max occurs when δ = 90o stability margin 20o)
200 MW = )2090sin(52
547113degminusdegg E
200 MW = 38130209397085613 gg E E =
|Eg| = 15360 V
LoadGen
Xs =25
Eg Vt =11547V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2426
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
24
Ex 24A manufacturing plant fed with a 400V 50Hz three phase runs from 0900 to 1800
Monday to Friday The factoryrsquos load is normally 80kVA with power factor 08 lagging
during these hours Once a month they need to run a special heating cycle which adds a
further 20kW to the factory load for a 2 hour period This heating cycle is timer
controlled and fully automated but the timer is currently set to run on the third Friday
afternoon of the month Outside of normal business hours and at weekends the factory
load is negligible The firm has a contractually agreed Maximum Import Capacity (MIC)
of 120kVA
bull Use the attached table of Low Voltage Maximum Demand charges to calculate the
firmrsquos electricity bill (exclusive of VAT) for the month of February (28 days)
bull Identify three steps that this firm could take to reduce their expenditure on
electricity assuming they stick with the same tariff Calculate the potential
savings
983116983151983159 983126983151983148983156983137983143983141 983117983137983160983145983149983157983149 983108983141983149983137983150983140 983107983144983137983154983143983141983155
983108983137983145983148983161 983123983156983137983150983140983145983150983143 983107983144983137983154983143983141 991404302983140983137983161
983120983123983119 983116983141983158983161
983120983123983119 983116983141983158983161 983152983141983154 983115983126983105 983151983142 983117983113983107 983152983141983154 983140983137983161 9914040036983097983147983126983105983140983137983161
983117983137983160983145983149983157983149 983113983149983152983151983154983156 983107983137983152983137983139983145983156983161 (983117983113983107) 983107983144983137983154983143983141983155
983108983137983145983148983161 983107983144983137983154983143983141 983152983141983154 983147983126983105 983151983142 983117983113983107 9914040072983147983126983105983140983137983161
983108983141983149983137983150983140 983107983144983137983154983143983141983155 983118983151983158983085983110983141983138 983117983137983154983085983119983139983156
983117983151983150983156983144983148983161 983107983144983137983154983143983141 983152983141983154 983147983127 983151983142 983149983141983156983141983154983141983140 983117983108 983145983150 983137
983149983151983150983156983144983148983161 983152983141983154983145983151983140 (983123983157983138983146983141983139983156 983156983151 983137 983149983145983150983145983149983157983149 983139983144983137983154983143983141 983151983142
30983147983127) 991404189830975983147983127983149983151983150983156983144 991404000
983108983137983161983156983145983149983141 983147983127983144 (983125983150983145983156) 983107983144983137983154983143983141 (0898309800 983085 19830979830985983097) 1480983139983147983127983144 1275983139983147983127983144
983118983145983143983144983156983145983149983141 983147983127983144 (983125983150983145983156) 983118983145983143983144983156 983107983144983137983154983143983141983155 (2098309800983085079830985983097) 64983097983139983147983127983144 64983097983139983147983127983144
983127983137983156983156983148983141983155983155 983125983150983145983156 983120983154983145983139983141 (983147983126983105983122983144 983145983150 983141983160983139983141983155983155 983151983142 983151983150983141 983156983144983145983154983140
983151983142 983156983151983156983137983148 983147983127983144) 991404000714 983147983126983105983122983144 991404000714 983147983126983105983122983144
bull Table Low Voltage Maximum Demand Charges (Q6)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2526
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
25
SolnNumber of days in February = 28
Number of working hours = 459 = 180
Normal kW = 80kVA08 = 64kW
Total kWh = 64kW180 hours+20kW2 hours = 11560 kWh
Normal kvar =80kVA sqrt(1-08^2) = 48kvar = sqrt(802-64
2) = 48 kVAr
Total kvarh = 48kvar180 hours = 8640 kvarh
Chargeable kvarh = 8640-(11560)3=4786 kvarh
983116983151983159 983126983151983148983156983137983143983141 983117983137983160983145983149983157983149 983108983141983149983137983150983140
983107983144983137983154983143983141983155 983152983141983154 983122983137983156983141 983121983156983161 983120983141983154983145983151983140 983107983144983137983154983143983141
983108983137983145983148983161 983123983156983137983150983140983145983150983143 983107983144983137983154983143983141 991404302 983140983137983161 991404302000 1 28 991404983096456
983120983123983119 983116983141983158983161
983120983123983119 983116983141983158983161 983152983141983154 983115983126983105 983152983141983154 983140983137983161 9914040036983097 983147983126983105983140983137983161 99140400369830970 120 28 991404123983097983096
983117983137983160983145983149983157983149 983113983149983152983151983154983156 983107983137983152983137983139983145983156983161
(983117983113983107) 983107983144983137983154983143983141983155
983108983137983145983148983161 983107983144983137983154983143983141 983152983141983154 983147983126983105 983151983142 983117983113983107 9914040072 983147983126983105983140983137983161 991404007200 120 28 9914042419830972
983108983141983149983137983150983140 983107983144983137983154983143983141983155 983118983151983158983085983110983141983138
983117983151983150983156983144983148983161 983107983144983137983154983143983141 983152983141983154 983147983127 983151983142
983149983141983156983141983154983141983140 983117983108 983145983150 983137 983149983151983150983156983144983148983161
983152983141983154983145983151983140 (983123983157983138983146983141983139983156 983156983151 983137
983149983145983150983145983149983157983149 983139983144983137983154983143983141 983151983142 30983147983127) 991404189830975
983147983127983149983151983150983156
983144 99140418983097500 84 1 991404159830971983096 983108983137983161983156983145983149983141 983147983127983144 (983125983150983145983156) 983107983144983137983154983143983141
(0898309800 983085 19830979830985983097) 9914040148 983147983127983144 991404014800 11560 1
991404198309510983096
983096
983118983145983143983144983156983145983149983141 983147983127983144 (983125983150983145983156) 983118983145983143983144983156
983107983144983137983154983143983141983155 (2098309800983085079830985983097) 9914040064983097 983147983127983144 99140400649830970
983127983137983156983156983148983141983155983155 983125983150983145983156 983120983154983145983139983141 (983147983126983105983122983144
983145983150 983141983160983139983141983155983155 983151983142 983151983150983141 983156983144983145983154983140 983151983142 983156983151983156983137983148
983147983127983144) 991404000714 983147983126983137983154983144 991404000714 4787 1 991404341983096
983124983151983156983137983148 983106983145983148983148
983142983151983154
983110983141983138983154983157983137983154983161
9914042354983095
0
[16 Marks]
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2626
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
26
Potential Savings
bull Use Power Factor correction to eliminate Watt less unit charge and to reduce MIC
bull Shift Heating load to night time or weekend
bull Negotiate lower MIC
Potential savings from these measures
Eliminate Wattless charge save euro3417 month
Shift heating cycle to night time
Max demand it lowered by 20kW = saving of 20euro1895 = euro3790 month
Saving by moving 20kW to night rate = 20kW2 hours (euro0148-euro00649)= euro332
Negotiating a reduced MIC
If power factor correction is employed and the heating cycle is shifted to night time the
then maximum VA of the plant drops to close to its maximum kW demand of 64kW An
MIC of 80kVA should be sufficient to cover this Potential Saving = 40kVA
(euro0072+euro00369)28 = euro12196 per month
Total potential saving = euro3417+euro3790+euro332+euro12196=euro19735month
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 526
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
5
EX 5 A balanced three phase load of 10 MW with power factor 08 lagging is connected to a
three phase 50Hz supply via a line of impedance 02+j05Ω per phase The voltage at the load is
exactly 10kV line to line
bull Draw a per phase equivalent circuit of this systembull Calculate the line current in complex number form
bull Calculate the impedance of the load
bull Calculate the magnitude of the line t o line voltage at the supply
bull Calculate the amount of power factor VARS (correcting capacitance) that would
need to be fitted at the supply end to achieve 095 power factor at the supply
Soln (a)
(b) Pf = cos φ = 08 = PS
ZLine = 02 + j05
Vn Zload
IL
Vload
P=7008 (W)
j axis
Q =4659 (kVAr)
S = 8392(VA)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 626
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
6
S = Ppf = 10 106 08 = 12 5 MVA
P = radic3 VL IL cos φ = 10MW
A I L 69721=
IL = 72169 ang -3687o A
(c) Z2 = VL IL = 64 + j48
(d) Van = VL + IL Z1 = 610496 + j 2021 V
= 61083 ang 19o
|Van| = 61083 V
(e) pf orginal 08 new 095
P = 10 MW S = Ppf = 10 106 095 = 10526 MVA
Old Q = gt cos-108 = 3686o Old
Q = 10 106 Tan 3686 = 10 106 075 = 75 MVar
New Q = gt cos-1
095 = 18195o
Q = P tan 18195o = 10 10
6 032868 = 3287 MVar
Diff in MVar = 75 ndash 3287 = 4213 MVar required
Ex 6 (Past Exam q)
A balanced three phase load of 8 kVA with power factor 085 lagging is connected to a three
phase 50Hz supply via a line of impedance 05+j1Ω per phase The voltage at the load is exactly
400V line to line
bull Draw a per phase equivalent circuit of this system
bull Calculate the line current in complex number form
bull Calculate the impedance of the load
bull Calculate the magnitude of the line to line voltage at the supply
bull Calculate real power (P) and reactive power (Q) drawn from the supply
Soln (a)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 726
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
7
(b) Pf = cos φ = 085 = PS
P = Spf = 8 103 085 = 68 kW
A I L 5411=
IL = 981 ndash j607 = 1154 ang -3175o A
(c) Z2 = VL IL = Ωdegang 75319319 = 1695 + j1049
(d) Van = VL + IL Z1 = 2411ang 161o
|Van| = 241 V
(e) pf 085
S = 085 S = Ppf = 10526 MVA
P = S pf = 68 kW
S = P + jQ
Q = P tan 3179o = 4214 kVar
Ex 7
Delta to Star Transformation
ZLine = 05 + j1
Vn 8kVA
IL
Vload
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 826
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
8
C B A
B A
Z Z Z
Z Z
Z ++=1
C B A
C B
Z Z Z
Z Z Z
++=2
C B A
C A
Z Z Z
Z Z Z
++=3
Star ndash Delta Transformation
2
133221
Z
Z Z Z Z Z Z Z A
++=
3
133221
Z
Z Z Z Z Z Z Z B
++=
ZA ZB
ZC
1
2 3
Z1
Z2Z3
Delta
Star
ZA ZB
ZC
1
2 3
Z1
Z2Z3
Delta Star
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 926
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1026
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
10
RT = 2 [ 29(2+9)] = 3612 = 32727 Ω
Ind Machines
Ex 9
Why does the rotor of an induction motor turn slower than the revolving field
Ex 10
Describe briefly the key differences between a squirrel cage motor and wound-rotor motor
in terms of construction and method of operation
See lecture notes
Ex 11
Both the voltage and frequency induced in the rotor of an induction motor decrease as the
rotor speeds up Briefly explain the reason for this and include the relationships between
slip the applied stator angular velocity and rotor angular velocity
Ex 12A 3-phase 400V 6-pole 50Hz star-connected induction motor operates on full load at a speed of
1475 revmin Determine the slip s
Ex 13 a) Calculate the synchronous speed of a 3-phase 12-pole induction motor that is excited by a
50Hz source
b) What is the nominal speed (speed at full load) is the slip at full load is 5
9
9
9
2
2
2
a
b
c
RT
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1126
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
11
a) Synchronous speed rpmns = 120 f p revmin
500 revmin
(b) s = (ωs - ωr) ωs = (ns - nr) ns
nr = ns(1 - s) = 475 revmin
Ex 14
An 400 V 50 Hz six-pole star connected induction machine operates with full load slip
of 005 pu and has the following per-phase equivalent circuit parameters
stator resistance Rs = 012Ω stator leakage reactance Xs = 06 Ω and magnetising
reactance Xm = 2215 Ω The rotor referred resistance Rr` = 015 Ω and the rotor referred
leakage reactance Xr` = 0586 Ω The induction machine initially operates as a motoron full load Using an approximate per-phase equivalent circuit determine
(i) synchronous speed and the speed of the motor at rated slip in revmin and
radsec
(ii) The approximate equivalent circuit per-phase input impedance
(iii) The input current
(iv) The developed torque
(v) The starting torque
Soln
Motoring modeSynchronous speed ωs = 2 π fp = 1046 rads
ωm = ωs(1 ndash s) = 9937 rads = 9489 revmin
Rr ` s = 015005 = 3
Z = Rs + jXs+ Rr ` s +jXr
`)
= 3122 + j 1186
The total Z is Zin = ( jXm|| Z)
R`r s
j06ΩΩΩΩ
is i`r
Vs = 400 radic3 V
j0586ΩΩΩΩ 012ΩΩΩΩ
015005 j221515ΩΩΩΩ
A
B
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1226
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
12
= 7398ang110o 2354ang 8238
o
== 3143ang2762o
The input current
I i = 7318 ang-2762o A
The rotor current
I r = 6886 ang-208o A
6886 A at a pf of 0935 lagging
The developed torque
T = Pm ωm = ||3
r
2`r
s
R I
sω = 408 Nm
Starting Torque
T = ||3
r
2`r
s
R I
sω with s=1
= 204 Nm
Ex 15 A 4 pole induction motor running from a 3 phase 400V line to line 50 Hz
supply has the following parametersStator Winding Resistance (Rs) = 005Ω
Stator Leakage Inductance (Xs) = 01 Ω
Magnetising Inductance (Xm) = 20 Ω Rotor Winding Resistance referred to stator (Rrrsquo) = 01 Ω
Rotor Leakage Inductance referred to stator (Xrrsquo) = 02 Ω
Rotational losses may be assumed to be negligible
(i) Draw a per phase equivalent circuit of this machine
(ii) Calculate the rotor speed at which the machine will have a slip of 2
(iii) Calculate the Power over the gap (Pog) in this machine when it is running at a slip of 2
(iv) Calculate the mechanical power (Pm) and the torque (Tm) delivered by the machine
running at a slip of 2
Soln
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1326
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
13
Synchronous speed ωs = 2 π fp = 1500 revmin
ωm = ωs(1 ndash s) = 15386 rads = 146925 revmin
Rr` s = 01002 = 5
Z = Rs + jXs+ Rr ` s +jXr
`)
= 505 + j 03
= 5059ang34o
The rotor referred current
I r = 4546 ang-34o A
4546 A at a pf of 099 lagging
The air-gap power
Pg = ||3 r2`
r
s R I = 31kW
Prot = 0 ndash no rotational losses
Pm = Pg (1-s) = 3038 kW
T = 3038 103 15386
= 197 Nm
Ex 16 A 6 pole induction motor running from a 3 phase 400V line to line 50 Hz supply has
the following parameters
Stator Winding Resistance (Rs) = 015Ω
Stator Leakage Inductance (Xs) = 03 Ω
Magnetising Inductance (Xm) = 60 Ω
R`r s
j01ΩΩΩΩ
is i`r
Vs = 400 radic3 V
j02ΩΩΩΩ 005ΩΩΩΩ
01002 j20ΩΩΩΩ
A
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1426
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
14
Rotor Winding Resistance referred to stator (Rrrsquo) = 03 Ω
Rotor Leakage Inductance referred to stator (Xrrsquo) = 06 Ω
Rotational losses have previously been measured to be 1000W at 5 slip
bull Draw a per phase equivalent circuit of this machinebull Calculate the rotor speed at which the machine will have a slip of 5
bull Calculate the air-gap Power (Pg) in this machine when it is running at a slip of 5
bull Calculate the efficiency of the induction motor from electrical input to mechanical
output when the motor is running with a slip of 5
Soln(a)
(b) Synchronous speed ωs = 2 π fp = 1000 revmin
ωm = ωs(1 ndash s) = 9948 rads = 950 revmin
(c) Rr` s = 03005 = 6
Z = Rs + jXs+ Rr ` s +jXr
`)
= 615 + j 09
= 622ang832o
(d) The air-gap power
The rotor referred current
I r = 3698 ang-303o A
Pg = ||3
r
2`r
s
R I = 246 kW
(d) I i =inZ
3400
Zin = Xm || Z = j60 || 622ang832o
Zin = 6097 ang1409o
I = VZ = 3772 ang-1409oA
R`r s
j03ΩΩΩΩ
is i`r
Vs = 230V
j06ΩΩΩΩ 015ΩΩΩΩ
03005 j60ΩΩΩΩ
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1526
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
15
Pelec = radic3 VL IL cos φ = 25044 kW
Pm = Pg (1-s) ndash Prot = 2237 kW
η = Pelec Pm = 893s
Ex 17
A large 3-phase 4000V 8-pole 50Hz squirrel-cage induction motor draws a current of
380A and a total active power of 2344kW when operating at full load The speed at full load
is accurately measured and found to be 742rpm The stator is star-connected and the
resistance between two stator terminals is 01Ω The total iron losses are 234kW and the
windage and friction losses are 12kW
Calculate
a) The power factor at full loadb) The active power supplied to the rotor
c) The total i2R losses in the rotor
d) The mechanical power output to the load and torque delivered to the load
e) The motorrsquos net efficiency
a) The active power input to the motor is given as 2344kW The apparent power is
S = radic3 VL IL = 2633kVA
Therefore pf = PS 089 lagging
b) The active power supplied to the rotor is the active power input to the motor minus the ironlosses and copper losses in the stator The stator iron losses are given as 234kW To calculate
the copper losses note that the terminal resistance (01) is twice that of each phase for a wye
connection Therefore the copper losses are
Stator copper losses Pcu = 217kW
Therefore the active power supplied to the rotor is
Pr = Input power ndash stator copper loss ndash iron losses
Pr = 2344 kW ndash 234kW ndash 217kW
= 2298kW
c) The copper losses in the rotor are proportional to the slip (see lecture notes) Prcu sP
Therefore we must calculate the slip Neither the synchronous speed nor number of poles are
explicitly stated in the problem however by inspection we determine that the synchronous speed
must be 750rpm and the number of poles = 8 (For 50Hz supply synchronous speeds occur at
discrete intervals corresponding to pole pairs 2-pole 3000rpm 4-pole 1500rpm 6- pole
1000rpm 8-pole 750rpm etchellip We also know that induction motors operate near synchronous
speed ndash the speed is given as 742 rpm which is close to synchronous speed for an
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1626
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
16
8-pole machine)
Synchronous speed ωs = 2 π fp = 7854 rads
ωm = = 777 radss = 00107
Therefore the rotor copper losses are Pr cu = sPr = 2456 kw
(d)
The mechanical power (Pm) delivered to the load is the active power delivered to the rotor
minus rotor copper losses and windage and friction losses The windage and friction losses are
given as 12kW Therefore
Pm =22989 kW ndash 245kW-12 kW =22624 kW
kNmT L 129
60
2742
42262=
=
π
(e) the motors efficiency 5962344
42262===
kW
kW
P
P
in
out η
Ex 18
A 3-phase 440V 8-pole 50Hz star-connected induction motor has the following equivalent
circuit per-phase parameters
Rs = 01 Ω Rr = 015Ω xs = xr =06 Ω Rc = 100 Ω xm = 10 Ω
(a) Calculate using the equivalent circuit neglecting stator impedance at a slip of 5 the input
stator-current and power factor the rotor current referred to the stator the torque the
mechanical output power and the efficiency Calculate the starting torque Assume the
mechanical loss is 1kW for all speeds
Soln
(a) Synchronous speed ωs = 2 π fp 7854 rads
ωm = ωs(1 - s) = 7618 rads
Stator impedance neglected
Vs = 400 radic3 = 230
Ir` =
60050
10
230
j+ =
= 68 ang-1671 = 651 - j 1954
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1726
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
17
Io =230
100 + j230
10 = 23 ndash j 23
Is = Ir` + Io = 674 - j4294
= 799 A at cos φ = 084
Ir` = 68 A
T = 529 Nm
Pmech = 39365 kW
Pelect = 465 kW
Efficiency = 39365465 = 846
Ts = 39627 Nm
Ex19
Tests conducted at 50 Hz on a three-phase star connected 45 kW 2200V six-pole 50Hz squirrel
cage induction motor are listed below
The average dc resistance per stator phase Rs = 28 Ω (a) Briefly describe the tests required to determine the per-phase equivalent circuit
impedance parameters of an induction machine
(a) Determine the parameters of the per-phase equivalent circuit
(b) Determine the rotational losses of the machine Core losses can be neglected
load
0
No-loadV
R
I1
Ic
I
c X
Im
m
Figure 1 Equivalent Circuit during No-Load Test
torPower Fac0 times= I I C = 2025 A
22
0 cm I I I minus=
= 4019 A
Ω== 627c
c I
V R Ω== 5317
m
m I
V X
No Load Test Locked Rotor Test
Power Factor = 045 Power Factor = 09
Line Voltage = 22kV Line Voltage = 270V
Line Current = 45A Line Current = 25A
Input Power = 16kW Input Power = 9kW
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1826
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
18
For the locked rotor test slip = 1
Ω== 246s
s
LR I
V Z
Locked rotor reactance is
( )2
21
2
21 R R Z X X +minus=+
φ cos21 times=+= Z R R Rtotal = 5616Ω
R2rsquo = Rtot - R1 = 2816Ω
Ω=+ 72221 X X
(a) From the no-load test the no-load power isPNL = 16kW
The no-load rotational loss is
Prot = PNL ndash 3 I s2 Rs
= 14299W
Synch Machines
Ex 20
A 60 MVA 20 kV three-phase 50 Hz synchronous generator has a synchronous reactance
of 8 Ω per phase and negligible resistance The generator delivers rated power at a power
factor of 085 lagging at the rated terminal voltage to an infinite bus(a) Determine the excitation voltage per phase and the power angle δ
(b) With the excitation voltage Eg of the synchronous generator held constant at the
value determined in part (a) the input driving torque is reduced until the generator
delivers 28 MW Determine the armature current and the power factor
(c) If the generator operates at the excitation voltage of Eg obtained in part (a) what is
the steady state maximum power the machine can deliver before losing
jXsΩΩΩΩ jX`r ΩΩΩΩ Rs
R`r s
Is I`r
Vs ph
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1926
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
19
synchronism Determine the armature current corresponding to this maximum
power
(d) With real generated power at 42MW compute the power angle δ the stator
current Ia and power factor when the field excitation current If is adjusted so that
the excitation voltage Eg is decreased to 7919 of the value determined in part
(a)
Soln
(a) Apparent power S = 51 MW ndash j 316Mvar
Rated voltage per phase
V = 1154 ang 0deg kV
Rated current
Ia = S 3V
= 1733 ang -3178deg AExcitation voltage
Eg = 11540 + j9(1733) ang -3178deg)= 237922 ang 3386deg
Power angle = 3386o
(b) When delivering 28 MW
δ = sin-1
)5411)(79223(3
828 = 1578
o
P =
s
g
X
V E 3sinδ
Ia = 13069ang2967o j8
= 16336ang-6033o A
Power factor = cos(6033) = 0495 lagging
(c) The maximum power occurs at δ = 90o
Pmax =s
g
X
V E 3
= 102 MW
Ia = 324885ang25875o A
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2026
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
20
The power factor cos(25875) = 0899 leading
(d) Eg = 13463 V
The power angle δ = sin-1
)5411)(84318(3
842
= 31o
Ia = 86674ang0o A
-Power factor = 0
Ex 21 (Aug 13) A 3 phase synchronous machine with negligible stator winding resistance and a
synchronous reactance of 2 per phase is connected to a 400V 50 Hz 3-phase supply (infinite
bus) and operated as a generator It is currently generating 50kW with power factor 08 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)
(iv) What is the maximum power the machine can generate with this value of excitation
Soln(i)
(ii)P = 50kW
Apparent power S = 625 ang -3687
o
kVA
Rated voltage per phase V = 230ang 0deg VRated current
Ia = S 3V
= 9058ang -3687deg A
LoadGen
Xs =2
Eg Vt =400 radic3 V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2126
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
21
|Ia| = 9058 A
(iii)
Excitation voltage
Eg = 230 ang 0deg + j2(9058ang -3687deg)= 3684 ang 2318degV
|Eg| = 3684 V
Power angle = 2318o
(iv)
P = deg90sin2
23043683
(max occurs when δ = 90o)
= 1271 kW
Ex 22A 3 phase 4 pole synchronous machine with negligible stator winding resistance and a
synchronous reactance of 2 per phase is connected to a 480V 60 Hz 3phase supply
(infinite bus) and operated as a generator It is currently generating 80kW with power
factor 08 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)(iv) What is the maximum power the machine can generate with this value
excitation(v) What is the rotational speed of this generator
Soln (i)
(ii)P = 80kW Apparent power S 100 ang -3687
o kVA
Rated voltage per phase
V 277ang 0deg V
LoadGen
Xs =2
Eg Vt = 480V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2226
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
22
Rated current
Ia = S 3V
= 9058ang -3687deg A
|Ia| = 12028 A
(iii)
Excitation voltage
Eg = 277 ang 0deg + j2(12028ang -3687deg)
= 4632 ang 2455degV|Eg| = 4632 V
Power angle = 2455o
(iv)
P = deg90sin2
27724633
(max occurs when δ = 90o)
= 19238 kW
(v)Synchronous speed ωs = 2 π fp = 1885 rads
= 1800 revmin
Ex 23 (Jan 14)
A 3 phase synchronous machine with negligible stator winding resistance and a synchronous
reactance of 25 per phase is connected to a 20kV 50 Hz 3phase supply (infinite bus) and
operated as a generator It is currently generating 100MW with power factor 085 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)(iv) What minimum value of excitation voltage (E) would be required for this machine to
deliver 200MW into a 20kV bus assuming a 20deg stability margin must be maintained onpower angle (δ)
Soln
(i)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2326
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
23
(ii)
P = 100 MW Apparent power S = 100 085
= 11765 ang -3179o MVA
Rated voltage per phase
V= 11547ang 0deg VRated current
Ia = S 3V
= 3396 103ang -3179deg A
|Ia| = 3396 103A
(iii)
Excitation voltage
Eg = 11547 ang 0deg + j25(3396 103ang -3179deg)
= 17570 ang 2425degV|Eg| = 17570 V
Power angle = 2425o V
(iv)
P = )2090sin(52
3degminusdegt gV E
(max occurs when δ = 90o stability margin 20o)
200 MW = )2090sin(52
547113degminusdegg E
200 MW = 38130209397085613 gg E E =
|Eg| = 15360 V
LoadGen
Xs =25
Eg Vt =11547V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2426
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
24
Ex 24A manufacturing plant fed with a 400V 50Hz three phase runs from 0900 to 1800
Monday to Friday The factoryrsquos load is normally 80kVA with power factor 08 lagging
during these hours Once a month they need to run a special heating cycle which adds a
further 20kW to the factory load for a 2 hour period This heating cycle is timer
controlled and fully automated but the timer is currently set to run on the third Friday
afternoon of the month Outside of normal business hours and at weekends the factory
load is negligible The firm has a contractually agreed Maximum Import Capacity (MIC)
of 120kVA
bull Use the attached table of Low Voltage Maximum Demand charges to calculate the
firmrsquos electricity bill (exclusive of VAT) for the month of February (28 days)
bull Identify three steps that this firm could take to reduce their expenditure on
electricity assuming they stick with the same tariff Calculate the potential
savings
983116983151983159 983126983151983148983156983137983143983141 983117983137983160983145983149983157983149 983108983141983149983137983150983140 983107983144983137983154983143983141983155
983108983137983145983148983161 983123983156983137983150983140983145983150983143 983107983144983137983154983143983141 991404302983140983137983161
983120983123983119 983116983141983158983161
983120983123983119 983116983141983158983161 983152983141983154 983115983126983105 983151983142 983117983113983107 983152983141983154 983140983137983161 9914040036983097983147983126983105983140983137983161
983117983137983160983145983149983157983149 983113983149983152983151983154983156 983107983137983152983137983139983145983156983161 (983117983113983107) 983107983144983137983154983143983141983155
983108983137983145983148983161 983107983144983137983154983143983141 983152983141983154 983147983126983105 983151983142 983117983113983107 9914040072983147983126983105983140983137983161
983108983141983149983137983150983140 983107983144983137983154983143983141983155 983118983151983158983085983110983141983138 983117983137983154983085983119983139983156
983117983151983150983156983144983148983161 983107983144983137983154983143983141 983152983141983154 983147983127 983151983142 983149983141983156983141983154983141983140 983117983108 983145983150 983137
983149983151983150983156983144983148983161 983152983141983154983145983151983140 (983123983157983138983146983141983139983156 983156983151 983137 983149983145983150983145983149983157983149 983139983144983137983154983143983141 983151983142
30983147983127) 991404189830975983147983127983149983151983150983156983144 991404000
983108983137983161983156983145983149983141 983147983127983144 (983125983150983145983156) 983107983144983137983154983143983141 (0898309800 983085 19830979830985983097) 1480983139983147983127983144 1275983139983147983127983144
983118983145983143983144983156983145983149983141 983147983127983144 (983125983150983145983156) 983118983145983143983144983156 983107983144983137983154983143983141983155 (2098309800983085079830985983097) 64983097983139983147983127983144 64983097983139983147983127983144
983127983137983156983156983148983141983155983155 983125983150983145983156 983120983154983145983139983141 (983147983126983105983122983144 983145983150 983141983160983139983141983155983155 983151983142 983151983150983141 983156983144983145983154983140
983151983142 983156983151983156983137983148 983147983127983144) 991404000714 983147983126983105983122983144 991404000714 983147983126983105983122983144
bull Table Low Voltage Maximum Demand Charges (Q6)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2526
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
25
SolnNumber of days in February = 28
Number of working hours = 459 = 180
Normal kW = 80kVA08 = 64kW
Total kWh = 64kW180 hours+20kW2 hours = 11560 kWh
Normal kvar =80kVA sqrt(1-08^2) = 48kvar = sqrt(802-64
2) = 48 kVAr
Total kvarh = 48kvar180 hours = 8640 kvarh
Chargeable kvarh = 8640-(11560)3=4786 kvarh
983116983151983159 983126983151983148983156983137983143983141 983117983137983160983145983149983157983149 983108983141983149983137983150983140
983107983144983137983154983143983141983155 983152983141983154 983122983137983156983141 983121983156983161 983120983141983154983145983151983140 983107983144983137983154983143983141
983108983137983145983148983161 983123983156983137983150983140983145983150983143 983107983144983137983154983143983141 991404302 983140983137983161 991404302000 1 28 991404983096456
983120983123983119 983116983141983158983161
983120983123983119 983116983141983158983161 983152983141983154 983115983126983105 983152983141983154 983140983137983161 9914040036983097 983147983126983105983140983137983161 99140400369830970 120 28 991404123983097983096
983117983137983160983145983149983157983149 983113983149983152983151983154983156 983107983137983152983137983139983145983156983161
(983117983113983107) 983107983144983137983154983143983141983155
983108983137983145983148983161 983107983144983137983154983143983141 983152983141983154 983147983126983105 983151983142 983117983113983107 9914040072 983147983126983105983140983137983161 991404007200 120 28 9914042419830972
983108983141983149983137983150983140 983107983144983137983154983143983141983155 983118983151983158983085983110983141983138
983117983151983150983156983144983148983161 983107983144983137983154983143983141 983152983141983154 983147983127 983151983142
983149983141983156983141983154983141983140 983117983108 983145983150 983137 983149983151983150983156983144983148983161
983152983141983154983145983151983140 (983123983157983138983146983141983139983156 983156983151 983137
983149983145983150983145983149983157983149 983139983144983137983154983143983141 983151983142 30983147983127) 991404189830975
983147983127983149983151983150983156
983144 99140418983097500 84 1 991404159830971983096 983108983137983161983156983145983149983141 983147983127983144 (983125983150983145983156) 983107983144983137983154983143983141
(0898309800 983085 19830979830985983097) 9914040148 983147983127983144 991404014800 11560 1
991404198309510983096
983096
983118983145983143983144983156983145983149983141 983147983127983144 (983125983150983145983156) 983118983145983143983144983156
983107983144983137983154983143983141983155 (2098309800983085079830985983097) 9914040064983097 983147983127983144 99140400649830970
983127983137983156983156983148983141983155983155 983125983150983145983156 983120983154983145983139983141 (983147983126983105983122983144
983145983150 983141983160983139983141983155983155 983151983142 983151983150983141 983156983144983145983154983140 983151983142 983156983151983156983137983148
983147983127983144) 991404000714 983147983126983137983154983144 991404000714 4787 1 991404341983096
983124983151983156983137983148 983106983145983148983148
983142983151983154
983110983141983138983154983157983137983154983161
9914042354983095
0
[16 Marks]
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2626
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
26
Potential Savings
bull Use Power Factor correction to eliminate Watt less unit charge and to reduce MIC
bull Shift Heating load to night time or weekend
bull Negotiate lower MIC
Potential savings from these measures
Eliminate Wattless charge save euro3417 month
Shift heating cycle to night time
Max demand it lowered by 20kW = saving of 20euro1895 = euro3790 month
Saving by moving 20kW to night rate = 20kW2 hours (euro0148-euro00649)= euro332
Negotiating a reduced MIC
If power factor correction is employed and the heating cycle is shifted to night time the
then maximum VA of the plant drops to close to its maximum kW demand of 64kW An
MIC of 80kVA should be sufficient to cover this Potential Saving = 40kVA
(euro0072+euro00369)28 = euro12196 per month
Total potential saving = euro3417+euro3790+euro332+euro12196=euro19735month
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 626
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
6
S = Ppf = 10 106 08 = 12 5 MVA
P = radic3 VL IL cos φ = 10MW
A I L 69721=
IL = 72169 ang -3687o A
(c) Z2 = VL IL = 64 + j48
(d) Van = VL + IL Z1 = 610496 + j 2021 V
= 61083 ang 19o
|Van| = 61083 V
(e) pf orginal 08 new 095
P = 10 MW S = Ppf = 10 106 095 = 10526 MVA
Old Q = gt cos-108 = 3686o Old
Q = 10 106 Tan 3686 = 10 106 075 = 75 MVar
New Q = gt cos-1
095 = 18195o
Q = P tan 18195o = 10 10
6 032868 = 3287 MVar
Diff in MVar = 75 ndash 3287 = 4213 MVar required
Ex 6 (Past Exam q)
A balanced three phase load of 8 kVA with power factor 085 lagging is connected to a three
phase 50Hz supply via a line of impedance 05+j1Ω per phase The voltage at the load is exactly
400V line to line
bull Draw a per phase equivalent circuit of this system
bull Calculate the line current in complex number form
bull Calculate the impedance of the load
bull Calculate the magnitude of the line to line voltage at the supply
bull Calculate real power (P) and reactive power (Q) drawn from the supply
Soln (a)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 726
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
7
(b) Pf = cos φ = 085 = PS
P = Spf = 8 103 085 = 68 kW
A I L 5411=
IL = 981 ndash j607 = 1154 ang -3175o A
(c) Z2 = VL IL = Ωdegang 75319319 = 1695 + j1049
(d) Van = VL + IL Z1 = 2411ang 161o
|Van| = 241 V
(e) pf 085
S = 085 S = Ppf = 10526 MVA
P = S pf = 68 kW
S = P + jQ
Q = P tan 3179o = 4214 kVar
Ex 7
Delta to Star Transformation
ZLine = 05 + j1
Vn 8kVA
IL
Vload
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 826
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
8
C B A
B A
Z Z Z
Z Z
Z ++=1
C B A
C B
Z Z Z
Z Z Z
++=2
C B A
C A
Z Z Z
Z Z Z
++=3
Star ndash Delta Transformation
2
133221
Z
Z Z Z Z Z Z Z A
++=
3
133221
Z
Z Z Z Z Z Z Z B
++=
ZA ZB
ZC
1
2 3
Z1
Z2Z3
Delta
Star
ZA ZB
ZC
1
2 3
Z1
Z2Z3
Delta Star
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 926
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1026
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
10
RT = 2 [ 29(2+9)] = 3612 = 32727 Ω
Ind Machines
Ex 9
Why does the rotor of an induction motor turn slower than the revolving field
Ex 10
Describe briefly the key differences between a squirrel cage motor and wound-rotor motor
in terms of construction and method of operation
See lecture notes
Ex 11
Both the voltage and frequency induced in the rotor of an induction motor decrease as the
rotor speeds up Briefly explain the reason for this and include the relationships between
slip the applied stator angular velocity and rotor angular velocity
Ex 12A 3-phase 400V 6-pole 50Hz star-connected induction motor operates on full load at a speed of
1475 revmin Determine the slip s
Ex 13 a) Calculate the synchronous speed of a 3-phase 12-pole induction motor that is excited by a
50Hz source
b) What is the nominal speed (speed at full load) is the slip at full load is 5
9
9
9
2
2
2
a
b
c
RT
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1126
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
11
a) Synchronous speed rpmns = 120 f p revmin
500 revmin
(b) s = (ωs - ωr) ωs = (ns - nr) ns
nr = ns(1 - s) = 475 revmin
Ex 14
An 400 V 50 Hz six-pole star connected induction machine operates with full load slip
of 005 pu and has the following per-phase equivalent circuit parameters
stator resistance Rs = 012Ω stator leakage reactance Xs = 06 Ω and magnetising
reactance Xm = 2215 Ω The rotor referred resistance Rr` = 015 Ω and the rotor referred
leakage reactance Xr` = 0586 Ω The induction machine initially operates as a motoron full load Using an approximate per-phase equivalent circuit determine
(i) synchronous speed and the speed of the motor at rated slip in revmin and
radsec
(ii) The approximate equivalent circuit per-phase input impedance
(iii) The input current
(iv) The developed torque
(v) The starting torque
Soln
Motoring modeSynchronous speed ωs = 2 π fp = 1046 rads
ωm = ωs(1 ndash s) = 9937 rads = 9489 revmin
Rr ` s = 015005 = 3
Z = Rs + jXs+ Rr ` s +jXr
`)
= 3122 + j 1186
The total Z is Zin = ( jXm|| Z)
R`r s
j06ΩΩΩΩ
is i`r
Vs = 400 radic3 V
j0586ΩΩΩΩ 012ΩΩΩΩ
015005 j221515ΩΩΩΩ
A
B
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1226
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
12
= 7398ang110o 2354ang 8238
o
== 3143ang2762o
The input current
I i = 7318 ang-2762o A
The rotor current
I r = 6886 ang-208o A
6886 A at a pf of 0935 lagging
The developed torque
T = Pm ωm = ||3
r
2`r
s
R I
sω = 408 Nm
Starting Torque
T = ||3
r
2`r
s
R I
sω with s=1
= 204 Nm
Ex 15 A 4 pole induction motor running from a 3 phase 400V line to line 50 Hz
supply has the following parametersStator Winding Resistance (Rs) = 005Ω
Stator Leakage Inductance (Xs) = 01 Ω
Magnetising Inductance (Xm) = 20 Ω Rotor Winding Resistance referred to stator (Rrrsquo) = 01 Ω
Rotor Leakage Inductance referred to stator (Xrrsquo) = 02 Ω
Rotational losses may be assumed to be negligible
(i) Draw a per phase equivalent circuit of this machine
(ii) Calculate the rotor speed at which the machine will have a slip of 2
(iii) Calculate the Power over the gap (Pog) in this machine when it is running at a slip of 2
(iv) Calculate the mechanical power (Pm) and the torque (Tm) delivered by the machine
running at a slip of 2
Soln
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1326
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
13
Synchronous speed ωs = 2 π fp = 1500 revmin
ωm = ωs(1 ndash s) = 15386 rads = 146925 revmin
Rr` s = 01002 = 5
Z = Rs + jXs+ Rr ` s +jXr
`)
= 505 + j 03
= 5059ang34o
The rotor referred current
I r = 4546 ang-34o A
4546 A at a pf of 099 lagging
The air-gap power
Pg = ||3 r2`
r
s R I = 31kW
Prot = 0 ndash no rotational losses
Pm = Pg (1-s) = 3038 kW
T = 3038 103 15386
= 197 Nm
Ex 16 A 6 pole induction motor running from a 3 phase 400V line to line 50 Hz supply has
the following parameters
Stator Winding Resistance (Rs) = 015Ω
Stator Leakage Inductance (Xs) = 03 Ω
Magnetising Inductance (Xm) = 60 Ω
R`r s
j01ΩΩΩΩ
is i`r
Vs = 400 radic3 V
j02ΩΩΩΩ 005ΩΩΩΩ
01002 j20ΩΩΩΩ
A
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1426
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
14
Rotor Winding Resistance referred to stator (Rrrsquo) = 03 Ω
Rotor Leakage Inductance referred to stator (Xrrsquo) = 06 Ω
Rotational losses have previously been measured to be 1000W at 5 slip
bull Draw a per phase equivalent circuit of this machinebull Calculate the rotor speed at which the machine will have a slip of 5
bull Calculate the air-gap Power (Pg) in this machine when it is running at a slip of 5
bull Calculate the efficiency of the induction motor from electrical input to mechanical
output when the motor is running with a slip of 5
Soln(a)
(b) Synchronous speed ωs = 2 π fp = 1000 revmin
ωm = ωs(1 ndash s) = 9948 rads = 950 revmin
(c) Rr` s = 03005 = 6
Z = Rs + jXs+ Rr ` s +jXr
`)
= 615 + j 09
= 622ang832o
(d) The air-gap power
The rotor referred current
I r = 3698 ang-303o A
Pg = ||3
r
2`r
s
R I = 246 kW
(d) I i =inZ
3400
Zin = Xm || Z = j60 || 622ang832o
Zin = 6097 ang1409o
I = VZ = 3772 ang-1409oA
R`r s
j03ΩΩΩΩ
is i`r
Vs = 230V
j06ΩΩΩΩ 015ΩΩΩΩ
03005 j60ΩΩΩΩ
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1526
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
15
Pelec = radic3 VL IL cos φ = 25044 kW
Pm = Pg (1-s) ndash Prot = 2237 kW
η = Pelec Pm = 893s
Ex 17
A large 3-phase 4000V 8-pole 50Hz squirrel-cage induction motor draws a current of
380A and a total active power of 2344kW when operating at full load The speed at full load
is accurately measured and found to be 742rpm The stator is star-connected and the
resistance between two stator terminals is 01Ω The total iron losses are 234kW and the
windage and friction losses are 12kW
Calculate
a) The power factor at full loadb) The active power supplied to the rotor
c) The total i2R losses in the rotor
d) The mechanical power output to the load and torque delivered to the load
e) The motorrsquos net efficiency
a) The active power input to the motor is given as 2344kW The apparent power is
S = radic3 VL IL = 2633kVA
Therefore pf = PS 089 lagging
b) The active power supplied to the rotor is the active power input to the motor minus the ironlosses and copper losses in the stator The stator iron losses are given as 234kW To calculate
the copper losses note that the terminal resistance (01) is twice that of each phase for a wye
connection Therefore the copper losses are
Stator copper losses Pcu = 217kW
Therefore the active power supplied to the rotor is
Pr = Input power ndash stator copper loss ndash iron losses
Pr = 2344 kW ndash 234kW ndash 217kW
= 2298kW
c) The copper losses in the rotor are proportional to the slip (see lecture notes) Prcu sP
Therefore we must calculate the slip Neither the synchronous speed nor number of poles are
explicitly stated in the problem however by inspection we determine that the synchronous speed
must be 750rpm and the number of poles = 8 (For 50Hz supply synchronous speeds occur at
discrete intervals corresponding to pole pairs 2-pole 3000rpm 4-pole 1500rpm 6- pole
1000rpm 8-pole 750rpm etchellip We also know that induction motors operate near synchronous
speed ndash the speed is given as 742 rpm which is close to synchronous speed for an
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1626
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
16
8-pole machine)
Synchronous speed ωs = 2 π fp = 7854 rads
ωm = = 777 radss = 00107
Therefore the rotor copper losses are Pr cu = sPr = 2456 kw
(d)
The mechanical power (Pm) delivered to the load is the active power delivered to the rotor
minus rotor copper losses and windage and friction losses The windage and friction losses are
given as 12kW Therefore
Pm =22989 kW ndash 245kW-12 kW =22624 kW
kNmT L 129
60
2742
42262=
=
π
(e) the motors efficiency 5962344
42262===
kW
kW
P
P
in
out η
Ex 18
A 3-phase 440V 8-pole 50Hz star-connected induction motor has the following equivalent
circuit per-phase parameters
Rs = 01 Ω Rr = 015Ω xs = xr =06 Ω Rc = 100 Ω xm = 10 Ω
(a) Calculate using the equivalent circuit neglecting stator impedance at a slip of 5 the input
stator-current and power factor the rotor current referred to the stator the torque the
mechanical output power and the efficiency Calculate the starting torque Assume the
mechanical loss is 1kW for all speeds
Soln
(a) Synchronous speed ωs = 2 π fp 7854 rads
ωm = ωs(1 - s) = 7618 rads
Stator impedance neglected
Vs = 400 radic3 = 230
Ir` =
60050
10
230
j+ =
= 68 ang-1671 = 651 - j 1954
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1726
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
17
Io =230
100 + j230
10 = 23 ndash j 23
Is = Ir` + Io = 674 - j4294
= 799 A at cos φ = 084
Ir` = 68 A
T = 529 Nm
Pmech = 39365 kW
Pelect = 465 kW
Efficiency = 39365465 = 846
Ts = 39627 Nm
Ex19
Tests conducted at 50 Hz on a three-phase star connected 45 kW 2200V six-pole 50Hz squirrel
cage induction motor are listed below
The average dc resistance per stator phase Rs = 28 Ω (a) Briefly describe the tests required to determine the per-phase equivalent circuit
impedance parameters of an induction machine
(a) Determine the parameters of the per-phase equivalent circuit
(b) Determine the rotational losses of the machine Core losses can be neglected
load
0
No-loadV
R
I1
Ic
I
c X
Im
m
Figure 1 Equivalent Circuit during No-Load Test
torPower Fac0 times= I I C = 2025 A
22
0 cm I I I minus=
= 4019 A
Ω== 627c
c I
V R Ω== 5317
m
m I
V X
No Load Test Locked Rotor Test
Power Factor = 045 Power Factor = 09
Line Voltage = 22kV Line Voltage = 270V
Line Current = 45A Line Current = 25A
Input Power = 16kW Input Power = 9kW
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1826
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
18
For the locked rotor test slip = 1
Ω== 246s
s
LR I
V Z
Locked rotor reactance is
( )2
21
2
21 R R Z X X +minus=+
φ cos21 times=+= Z R R Rtotal = 5616Ω
R2rsquo = Rtot - R1 = 2816Ω
Ω=+ 72221 X X
(a) From the no-load test the no-load power isPNL = 16kW
The no-load rotational loss is
Prot = PNL ndash 3 I s2 Rs
= 14299W
Synch Machines
Ex 20
A 60 MVA 20 kV three-phase 50 Hz synchronous generator has a synchronous reactance
of 8 Ω per phase and negligible resistance The generator delivers rated power at a power
factor of 085 lagging at the rated terminal voltage to an infinite bus(a) Determine the excitation voltage per phase and the power angle δ
(b) With the excitation voltage Eg of the synchronous generator held constant at the
value determined in part (a) the input driving torque is reduced until the generator
delivers 28 MW Determine the armature current and the power factor
(c) If the generator operates at the excitation voltage of Eg obtained in part (a) what is
the steady state maximum power the machine can deliver before losing
jXsΩΩΩΩ jX`r ΩΩΩΩ Rs
R`r s
Is I`r
Vs ph
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1926
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
19
synchronism Determine the armature current corresponding to this maximum
power
(d) With real generated power at 42MW compute the power angle δ the stator
current Ia and power factor when the field excitation current If is adjusted so that
the excitation voltage Eg is decreased to 7919 of the value determined in part
(a)
Soln
(a) Apparent power S = 51 MW ndash j 316Mvar
Rated voltage per phase
V = 1154 ang 0deg kV
Rated current
Ia = S 3V
= 1733 ang -3178deg AExcitation voltage
Eg = 11540 + j9(1733) ang -3178deg)= 237922 ang 3386deg
Power angle = 3386o
(b) When delivering 28 MW
δ = sin-1
)5411)(79223(3
828 = 1578
o
P =
s
g
X
V E 3sinδ
Ia = 13069ang2967o j8
= 16336ang-6033o A
Power factor = cos(6033) = 0495 lagging
(c) The maximum power occurs at δ = 90o
Pmax =s
g
X
V E 3
= 102 MW
Ia = 324885ang25875o A
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2026
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
20
The power factor cos(25875) = 0899 leading
(d) Eg = 13463 V
The power angle δ = sin-1
)5411)(84318(3
842
= 31o
Ia = 86674ang0o A
-Power factor = 0
Ex 21 (Aug 13) A 3 phase synchronous machine with negligible stator winding resistance and a
synchronous reactance of 2 per phase is connected to a 400V 50 Hz 3-phase supply (infinite
bus) and operated as a generator It is currently generating 50kW with power factor 08 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)
(iv) What is the maximum power the machine can generate with this value of excitation
Soln(i)
(ii)P = 50kW
Apparent power S = 625 ang -3687
o
kVA
Rated voltage per phase V = 230ang 0deg VRated current
Ia = S 3V
= 9058ang -3687deg A
LoadGen
Xs =2
Eg Vt =400 radic3 V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2126
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
21
|Ia| = 9058 A
(iii)
Excitation voltage
Eg = 230 ang 0deg + j2(9058ang -3687deg)= 3684 ang 2318degV
|Eg| = 3684 V
Power angle = 2318o
(iv)
P = deg90sin2
23043683
(max occurs when δ = 90o)
= 1271 kW
Ex 22A 3 phase 4 pole synchronous machine with negligible stator winding resistance and a
synchronous reactance of 2 per phase is connected to a 480V 60 Hz 3phase supply
(infinite bus) and operated as a generator It is currently generating 80kW with power
factor 08 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)(iv) What is the maximum power the machine can generate with this value
excitation(v) What is the rotational speed of this generator
Soln (i)
(ii)P = 80kW Apparent power S 100 ang -3687
o kVA
Rated voltage per phase
V 277ang 0deg V
LoadGen
Xs =2
Eg Vt = 480V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2226
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
22
Rated current
Ia = S 3V
= 9058ang -3687deg A
|Ia| = 12028 A
(iii)
Excitation voltage
Eg = 277 ang 0deg + j2(12028ang -3687deg)
= 4632 ang 2455degV|Eg| = 4632 V
Power angle = 2455o
(iv)
P = deg90sin2
27724633
(max occurs when δ = 90o)
= 19238 kW
(v)Synchronous speed ωs = 2 π fp = 1885 rads
= 1800 revmin
Ex 23 (Jan 14)
A 3 phase synchronous machine with negligible stator winding resistance and a synchronous
reactance of 25 per phase is connected to a 20kV 50 Hz 3phase supply (infinite bus) and
operated as a generator It is currently generating 100MW with power factor 085 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)(iv) What minimum value of excitation voltage (E) would be required for this machine to
deliver 200MW into a 20kV bus assuming a 20deg stability margin must be maintained onpower angle (δ)
Soln
(i)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2326
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
23
(ii)
P = 100 MW Apparent power S = 100 085
= 11765 ang -3179o MVA
Rated voltage per phase
V= 11547ang 0deg VRated current
Ia = S 3V
= 3396 103ang -3179deg A
|Ia| = 3396 103A
(iii)
Excitation voltage
Eg = 11547 ang 0deg + j25(3396 103ang -3179deg)
= 17570 ang 2425degV|Eg| = 17570 V
Power angle = 2425o V
(iv)
P = )2090sin(52
3degminusdegt gV E
(max occurs when δ = 90o stability margin 20o)
200 MW = )2090sin(52
547113degminusdegg E
200 MW = 38130209397085613 gg E E =
|Eg| = 15360 V
LoadGen
Xs =25
Eg Vt =11547V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2426
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
24
Ex 24A manufacturing plant fed with a 400V 50Hz three phase runs from 0900 to 1800
Monday to Friday The factoryrsquos load is normally 80kVA with power factor 08 lagging
during these hours Once a month they need to run a special heating cycle which adds a
further 20kW to the factory load for a 2 hour period This heating cycle is timer
controlled and fully automated but the timer is currently set to run on the third Friday
afternoon of the month Outside of normal business hours and at weekends the factory
load is negligible The firm has a contractually agreed Maximum Import Capacity (MIC)
of 120kVA
bull Use the attached table of Low Voltage Maximum Demand charges to calculate the
firmrsquos electricity bill (exclusive of VAT) for the month of February (28 days)
bull Identify three steps that this firm could take to reduce their expenditure on
electricity assuming they stick with the same tariff Calculate the potential
savings
983116983151983159 983126983151983148983156983137983143983141 983117983137983160983145983149983157983149 983108983141983149983137983150983140 983107983144983137983154983143983141983155
983108983137983145983148983161 983123983156983137983150983140983145983150983143 983107983144983137983154983143983141 991404302983140983137983161
983120983123983119 983116983141983158983161
983120983123983119 983116983141983158983161 983152983141983154 983115983126983105 983151983142 983117983113983107 983152983141983154 983140983137983161 9914040036983097983147983126983105983140983137983161
983117983137983160983145983149983157983149 983113983149983152983151983154983156 983107983137983152983137983139983145983156983161 (983117983113983107) 983107983144983137983154983143983141983155
983108983137983145983148983161 983107983144983137983154983143983141 983152983141983154 983147983126983105 983151983142 983117983113983107 9914040072983147983126983105983140983137983161
983108983141983149983137983150983140 983107983144983137983154983143983141983155 983118983151983158983085983110983141983138 983117983137983154983085983119983139983156
983117983151983150983156983144983148983161 983107983144983137983154983143983141 983152983141983154 983147983127 983151983142 983149983141983156983141983154983141983140 983117983108 983145983150 983137
983149983151983150983156983144983148983161 983152983141983154983145983151983140 (983123983157983138983146983141983139983156 983156983151 983137 983149983145983150983145983149983157983149 983139983144983137983154983143983141 983151983142
30983147983127) 991404189830975983147983127983149983151983150983156983144 991404000
983108983137983161983156983145983149983141 983147983127983144 (983125983150983145983156) 983107983144983137983154983143983141 (0898309800 983085 19830979830985983097) 1480983139983147983127983144 1275983139983147983127983144
983118983145983143983144983156983145983149983141 983147983127983144 (983125983150983145983156) 983118983145983143983144983156 983107983144983137983154983143983141983155 (2098309800983085079830985983097) 64983097983139983147983127983144 64983097983139983147983127983144
983127983137983156983156983148983141983155983155 983125983150983145983156 983120983154983145983139983141 (983147983126983105983122983144 983145983150 983141983160983139983141983155983155 983151983142 983151983150983141 983156983144983145983154983140
983151983142 983156983151983156983137983148 983147983127983144) 991404000714 983147983126983105983122983144 991404000714 983147983126983105983122983144
bull Table Low Voltage Maximum Demand Charges (Q6)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2526
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
25
SolnNumber of days in February = 28
Number of working hours = 459 = 180
Normal kW = 80kVA08 = 64kW
Total kWh = 64kW180 hours+20kW2 hours = 11560 kWh
Normal kvar =80kVA sqrt(1-08^2) = 48kvar = sqrt(802-64
2) = 48 kVAr
Total kvarh = 48kvar180 hours = 8640 kvarh
Chargeable kvarh = 8640-(11560)3=4786 kvarh
983116983151983159 983126983151983148983156983137983143983141 983117983137983160983145983149983157983149 983108983141983149983137983150983140
983107983144983137983154983143983141983155 983152983141983154 983122983137983156983141 983121983156983161 983120983141983154983145983151983140 983107983144983137983154983143983141
983108983137983145983148983161 983123983156983137983150983140983145983150983143 983107983144983137983154983143983141 991404302 983140983137983161 991404302000 1 28 991404983096456
983120983123983119 983116983141983158983161
983120983123983119 983116983141983158983161 983152983141983154 983115983126983105 983152983141983154 983140983137983161 9914040036983097 983147983126983105983140983137983161 99140400369830970 120 28 991404123983097983096
983117983137983160983145983149983157983149 983113983149983152983151983154983156 983107983137983152983137983139983145983156983161
(983117983113983107) 983107983144983137983154983143983141983155
983108983137983145983148983161 983107983144983137983154983143983141 983152983141983154 983147983126983105 983151983142 983117983113983107 9914040072 983147983126983105983140983137983161 991404007200 120 28 9914042419830972
983108983141983149983137983150983140 983107983144983137983154983143983141983155 983118983151983158983085983110983141983138
983117983151983150983156983144983148983161 983107983144983137983154983143983141 983152983141983154 983147983127 983151983142
983149983141983156983141983154983141983140 983117983108 983145983150 983137 983149983151983150983156983144983148983161
983152983141983154983145983151983140 (983123983157983138983146983141983139983156 983156983151 983137
983149983145983150983145983149983157983149 983139983144983137983154983143983141 983151983142 30983147983127) 991404189830975
983147983127983149983151983150983156
983144 99140418983097500 84 1 991404159830971983096 983108983137983161983156983145983149983141 983147983127983144 (983125983150983145983156) 983107983144983137983154983143983141
(0898309800 983085 19830979830985983097) 9914040148 983147983127983144 991404014800 11560 1
991404198309510983096
983096
983118983145983143983144983156983145983149983141 983147983127983144 (983125983150983145983156) 983118983145983143983144983156
983107983144983137983154983143983141983155 (2098309800983085079830985983097) 9914040064983097 983147983127983144 99140400649830970
983127983137983156983156983148983141983155983155 983125983150983145983156 983120983154983145983139983141 (983147983126983105983122983144
983145983150 983141983160983139983141983155983155 983151983142 983151983150983141 983156983144983145983154983140 983151983142 983156983151983156983137983148
983147983127983144) 991404000714 983147983126983137983154983144 991404000714 4787 1 991404341983096
983124983151983156983137983148 983106983145983148983148
983142983151983154
983110983141983138983154983157983137983154983161
9914042354983095
0
[16 Marks]
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2626
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
26
Potential Savings
bull Use Power Factor correction to eliminate Watt less unit charge and to reduce MIC
bull Shift Heating load to night time or weekend
bull Negotiate lower MIC
Potential savings from these measures
Eliminate Wattless charge save euro3417 month
Shift heating cycle to night time
Max demand it lowered by 20kW = saving of 20euro1895 = euro3790 month
Saving by moving 20kW to night rate = 20kW2 hours (euro0148-euro00649)= euro332
Negotiating a reduced MIC
If power factor correction is employed and the heating cycle is shifted to night time the
then maximum VA of the plant drops to close to its maximum kW demand of 64kW An
MIC of 80kVA should be sufficient to cover this Potential Saving = 40kVA
(euro0072+euro00369)28 = euro12196 per month
Total potential saving = euro3417+euro3790+euro332+euro12196=euro19735month
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 726
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
7
(b) Pf = cos φ = 085 = PS
P = Spf = 8 103 085 = 68 kW
A I L 5411=
IL = 981 ndash j607 = 1154 ang -3175o A
(c) Z2 = VL IL = Ωdegang 75319319 = 1695 + j1049
(d) Van = VL + IL Z1 = 2411ang 161o
|Van| = 241 V
(e) pf 085
S = 085 S = Ppf = 10526 MVA
P = S pf = 68 kW
S = P + jQ
Q = P tan 3179o = 4214 kVar
Ex 7
Delta to Star Transformation
ZLine = 05 + j1
Vn 8kVA
IL
Vload
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 826
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
8
C B A
B A
Z Z Z
Z Z
Z ++=1
C B A
C B
Z Z Z
Z Z Z
++=2
C B A
C A
Z Z Z
Z Z Z
++=3
Star ndash Delta Transformation
2
133221
Z
Z Z Z Z Z Z Z A
++=
3
133221
Z
Z Z Z Z Z Z Z B
++=
ZA ZB
ZC
1
2 3
Z1
Z2Z3
Delta
Star
ZA ZB
ZC
1
2 3
Z1
Z2Z3
Delta Star
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 926
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1026
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
10
RT = 2 [ 29(2+9)] = 3612 = 32727 Ω
Ind Machines
Ex 9
Why does the rotor of an induction motor turn slower than the revolving field
Ex 10
Describe briefly the key differences between a squirrel cage motor and wound-rotor motor
in terms of construction and method of operation
See lecture notes
Ex 11
Both the voltage and frequency induced in the rotor of an induction motor decrease as the
rotor speeds up Briefly explain the reason for this and include the relationships between
slip the applied stator angular velocity and rotor angular velocity
Ex 12A 3-phase 400V 6-pole 50Hz star-connected induction motor operates on full load at a speed of
1475 revmin Determine the slip s
Ex 13 a) Calculate the synchronous speed of a 3-phase 12-pole induction motor that is excited by a
50Hz source
b) What is the nominal speed (speed at full load) is the slip at full load is 5
9
9
9
2
2
2
a
b
c
RT
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1126
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
11
a) Synchronous speed rpmns = 120 f p revmin
500 revmin
(b) s = (ωs - ωr) ωs = (ns - nr) ns
nr = ns(1 - s) = 475 revmin
Ex 14
An 400 V 50 Hz six-pole star connected induction machine operates with full load slip
of 005 pu and has the following per-phase equivalent circuit parameters
stator resistance Rs = 012Ω stator leakage reactance Xs = 06 Ω and magnetising
reactance Xm = 2215 Ω The rotor referred resistance Rr` = 015 Ω and the rotor referred
leakage reactance Xr` = 0586 Ω The induction machine initially operates as a motoron full load Using an approximate per-phase equivalent circuit determine
(i) synchronous speed and the speed of the motor at rated slip in revmin and
radsec
(ii) The approximate equivalent circuit per-phase input impedance
(iii) The input current
(iv) The developed torque
(v) The starting torque
Soln
Motoring modeSynchronous speed ωs = 2 π fp = 1046 rads
ωm = ωs(1 ndash s) = 9937 rads = 9489 revmin
Rr ` s = 015005 = 3
Z = Rs + jXs+ Rr ` s +jXr
`)
= 3122 + j 1186
The total Z is Zin = ( jXm|| Z)
R`r s
j06ΩΩΩΩ
is i`r
Vs = 400 radic3 V
j0586ΩΩΩΩ 012ΩΩΩΩ
015005 j221515ΩΩΩΩ
A
B
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1226
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
12
= 7398ang110o 2354ang 8238
o
== 3143ang2762o
The input current
I i = 7318 ang-2762o A
The rotor current
I r = 6886 ang-208o A
6886 A at a pf of 0935 lagging
The developed torque
T = Pm ωm = ||3
r
2`r
s
R I
sω = 408 Nm
Starting Torque
T = ||3
r
2`r
s
R I
sω with s=1
= 204 Nm
Ex 15 A 4 pole induction motor running from a 3 phase 400V line to line 50 Hz
supply has the following parametersStator Winding Resistance (Rs) = 005Ω
Stator Leakage Inductance (Xs) = 01 Ω
Magnetising Inductance (Xm) = 20 Ω Rotor Winding Resistance referred to stator (Rrrsquo) = 01 Ω
Rotor Leakage Inductance referred to stator (Xrrsquo) = 02 Ω
Rotational losses may be assumed to be negligible
(i) Draw a per phase equivalent circuit of this machine
(ii) Calculate the rotor speed at which the machine will have a slip of 2
(iii) Calculate the Power over the gap (Pog) in this machine when it is running at a slip of 2
(iv) Calculate the mechanical power (Pm) and the torque (Tm) delivered by the machine
running at a slip of 2
Soln
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1326
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
13
Synchronous speed ωs = 2 π fp = 1500 revmin
ωm = ωs(1 ndash s) = 15386 rads = 146925 revmin
Rr` s = 01002 = 5
Z = Rs + jXs+ Rr ` s +jXr
`)
= 505 + j 03
= 5059ang34o
The rotor referred current
I r = 4546 ang-34o A
4546 A at a pf of 099 lagging
The air-gap power
Pg = ||3 r2`
r
s R I = 31kW
Prot = 0 ndash no rotational losses
Pm = Pg (1-s) = 3038 kW
T = 3038 103 15386
= 197 Nm
Ex 16 A 6 pole induction motor running from a 3 phase 400V line to line 50 Hz supply has
the following parameters
Stator Winding Resistance (Rs) = 015Ω
Stator Leakage Inductance (Xs) = 03 Ω
Magnetising Inductance (Xm) = 60 Ω
R`r s
j01ΩΩΩΩ
is i`r
Vs = 400 radic3 V
j02ΩΩΩΩ 005ΩΩΩΩ
01002 j20ΩΩΩΩ
A
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1426
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
14
Rotor Winding Resistance referred to stator (Rrrsquo) = 03 Ω
Rotor Leakage Inductance referred to stator (Xrrsquo) = 06 Ω
Rotational losses have previously been measured to be 1000W at 5 slip
bull Draw a per phase equivalent circuit of this machinebull Calculate the rotor speed at which the machine will have a slip of 5
bull Calculate the air-gap Power (Pg) in this machine when it is running at a slip of 5
bull Calculate the efficiency of the induction motor from electrical input to mechanical
output when the motor is running with a slip of 5
Soln(a)
(b) Synchronous speed ωs = 2 π fp = 1000 revmin
ωm = ωs(1 ndash s) = 9948 rads = 950 revmin
(c) Rr` s = 03005 = 6
Z = Rs + jXs+ Rr ` s +jXr
`)
= 615 + j 09
= 622ang832o
(d) The air-gap power
The rotor referred current
I r = 3698 ang-303o A
Pg = ||3
r
2`r
s
R I = 246 kW
(d) I i =inZ
3400
Zin = Xm || Z = j60 || 622ang832o
Zin = 6097 ang1409o
I = VZ = 3772 ang-1409oA
R`r s
j03ΩΩΩΩ
is i`r
Vs = 230V
j06ΩΩΩΩ 015ΩΩΩΩ
03005 j60ΩΩΩΩ
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1526
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
15
Pelec = radic3 VL IL cos φ = 25044 kW
Pm = Pg (1-s) ndash Prot = 2237 kW
η = Pelec Pm = 893s
Ex 17
A large 3-phase 4000V 8-pole 50Hz squirrel-cage induction motor draws a current of
380A and a total active power of 2344kW when operating at full load The speed at full load
is accurately measured and found to be 742rpm The stator is star-connected and the
resistance between two stator terminals is 01Ω The total iron losses are 234kW and the
windage and friction losses are 12kW
Calculate
a) The power factor at full loadb) The active power supplied to the rotor
c) The total i2R losses in the rotor
d) The mechanical power output to the load and torque delivered to the load
e) The motorrsquos net efficiency
a) The active power input to the motor is given as 2344kW The apparent power is
S = radic3 VL IL = 2633kVA
Therefore pf = PS 089 lagging
b) The active power supplied to the rotor is the active power input to the motor minus the ironlosses and copper losses in the stator The stator iron losses are given as 234kW To calculate
the copper losses note that the terminal resistance (01) is twice that of each phase for a wye
connection Therefore the copper losses are
Stator copper losses Pcu = 217kW
Therefore the active power supplied to the rotor is
Pr = Input power ndash stator copper loss ndash iron losses
Pr = 2344 kW ndash 234kW ndash 217kW
= 2298kW
c) The copper losses in the rotor are proportional to the slip (see lecture notes) Prcu sP
Therefore we must calculate the slip Neither the synchronous speed nor number of poles are
explicitly stated in the problem however by inspection we determine that the synchronous speed
must be 750rpm and the number of poles = 8 (For 50Hz supply synchronous speeds occur at
discrete intervals corresponding to pole pairs 2-pole 3000rpm 4-pole 1500rpm 6- pole
1000rpm 8-pole 750rpm etchellip We also know that induction motors operate near synchronous
speed ndash the speed is given as 742 rpm which is close to synchronous speed for an
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1626
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
16
8-pole machine)
Synchronous speed ωs = 2 π fp = 7854 rads
ωm = = 777 radss = 00107
Therefore the rotor copper losses are Pr cu = sPr = 2456 kw
(d)
The mechanical power (Pm) delivered to the load is the active power delivered to the rotor
minus rotor copper losses and windage and friction losses The windage and friction losses are
given as 12kW Therefore
Pm =22989 kW ndash 245kW-12 kW =22624 kW
kNmT L 129
60
2742
42262=
=
π
(e) the motors efficiency 5962344
42262===
kW
kW
P
P
in
out η
Ex 18
A 3-phase 440V 8-pole 50Hz star-connected induction motor has the following equivalent
circuit per-phase parameters
Rs = 01 Ω Rr = 015Ω xs = xr =06 Ω Rc = 100 Ω xm = 10 Ω
(a) Calculate using the equivalent circuit neglecting stator impedance at a slip of 5 the input
stator-current and power factor the rotor current referred to the stator the torque the
mechanical output power and the efficiency Calculate the starting torque Assume the
mechanical loss is 1kW for all speeds
Soln
(a) Synchronous speed ωs = 2 π fp 7854 rads
ωm = ωs(1 - s) = 7618 rads
Stator impedance neglected
Vs = 400 radic3 = 230
Ir` =
60050
10
230
j+ =
= 68 ang-1671 = 651 - j 1954
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1726
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
17
Io =230
100 + j230
10 = 23 ndash j 23
Is = Ir` + Io = 674 - j4294
= 799 A at cos φ = 084
Ir` = 68 A
T = 529 Nm
Pmech = 39365 kW
Pelect = 465 kW
Efficiency = 39365465 = 846
Ts = 39627 Nm
Ex19
Tests conducted at 50 Hz on a three-phase star connected 45 kW 2200V six-pole 50Hz squirrel
cage induction motor are listed below
The average dc resistance per stator phase Rs = 28 Ω (a) Briefly describe the tests required to determine the per-phase equivalent circuit
impedance parameters of an induction machine
(a) Determine the parameters of the per-phase equivalent circuit
(b) Determine the rotational losses of the machine Core losses can be neglected
load
0
No-loadV
R
I1
Ic
I
c X
Im
m
Figure 1 Equivalent Circuit during No-Load Test
torPower Fac0 times= I I C = 2025 A
22
0 cm I I I minus=
= 4019 A
Ω== 627c
c I
V R Ω== 5317
m
m I
V X
No Load Test Locked Rotor Test
Power Factor = 045 Power Factor = 09
Line Voltage = 22kV Line Voltage = 270V
Line Current = 45A Line Current = 25A
Input Power = 16kW Input Power = 9kW
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1826
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
18
For the locked rotor test slip = 1
Ω== 246s
s
LR I
V Z
Locked rotor reactance is
( )2
21
2
21 R R Z X X +minus=+
φ cos21 times=+= Z R R Rtotal = 5616Ω
R2rsquo = Rtot - R1 = 2816Ω
Ω=+ 72221 X X
(a) From the no-load test the no-load power isPNL = 16kW
The no-load rotational loss is
Prot = PNL ndash 3 I s2 Rs
= 14299W
Synch Machines
Ex 20
A 60 MVA 20 kV three-phase 50 Hz synchronous generator has a synchronous reactance
of 8 Ω per phase and negligible resistance The generator delivers rated power at a power
factor of 085 lagging at the rated terminal voltage to an infinite bus(a) Determine the excitation voltage per phase and the power angle δ
(b) With the excitation voltage Eg of the synchronous generator held constant at the
value determined in part (a) the input driving torque is reduced until the generator
delivers 28 MW Determine the armature current and the power factor
(c) If the generator operates at the excitation voltage of Eg obtained in part (a) what is
the steady state maximum power the machine can deliver before losing
jXsΩΩΩΩ jX`r ΩΩΩΩ Rs
R`r s
Is I`r
Vs ph
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1926
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
19
synchronism Determine the armature current corresponding to this maximum
power
(d) With real generated power at 42MW compute the power angle δ the stator
current Ia and power factor when the field excitation current If is adjusted so that
the excitation voltage Eg is decreased to 7919 of the value determined in part
(a)
Soln
(a) Apparent power S = 51 MW ndash j 316Mvar
Rated voltage per phase
V = 1154 ang 0deg kV
Rated current
Ia = S 3V
= 1733 ang -3178deg AExcitation voltage
Eg = 11540 + j9(1733) ang -3178deg)= 237922 ang 3386deg
Power angle = 3386o
(b) When delivering 28 MW
δ = sin-1
)5411)(79223(3
828 = 1578
o
P =
s
g
X
V E 3sinδ
Ia = 13069ang2967o j8
= 16336ang-6033o A
Power factor = cos(6033) = 0495 lagging
(c) The maximum power occurs at δ = 90o
Pmax =s
g
X
V E 3
= 102 MW
Ia = 324885ang25875o A
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2026
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
20
The power factor cos(25875) = 0899 leading
(d) Eg = 13463 V
The power angle δ = sin-1
)5411)(84318(3
842
= 31o
Ia = 86674ang0o A
-Power factor = 0
Ex 21 (Aug 13) A 3 phase synchronous machine with negligible stator winding resistance and a
synchronous reactance of 2 per phase is connected to a 400V 50 Hz 3-phase supply (infinite
bus) and operated as a generator It is currently generating 50kW with power factor 08 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)
(iv) What is the maximum power the machine can generate with this value of excitation
Soln(i)
(ii)P = 50kW
Apparent power S = 625 ang -3687
o
kVA
Rated voltage per phase V = 230ang 0deg VRated current
Ia = S 3V
= 9058ang -3687deg A
LoadGen
Xs =2
Eg Vt =400 radic3 V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2126
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
21
|Ia| = 9058 A
(iii)
Excitation voltage
Eg = 230 ang 0deg + j2(9058ang -3687deg)= 3684 ang 2318degV
|Eg| = 3684 V
Power angle = 2318o
(iv)
P = deg90sin2
23043683
(max occurs when δ = 90o)
= 1271 kW
Ex 22A 3 phase 4 pole synchronous machine with negligible stator winding resistance and a
synchronous reactance of 2 per phase is connected to a 480V 60 Hz 3phase supply
(infinite bus) and operated as a generator It is currently generating 80kW with power
factor 08 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)(iv) What is the maximum power the machine can generate with this value
excitation(v) What is the rotational speed of this generator
Soln (i)
(ii)P = 80kW Apparent power S 100 ang -3687
o kVA
Rated voltage per phase
V 277ang 0deg V
LoadGen
Xs =2
Eg Vt = 480V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2226
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
22
Rated current
Ia = S 3V
= 9058ang -3687deg A
|Ia| = 12028 A
(iii)
Excitation voltage
Eg = 277 ang 0deg + j2(12028ang -3687deg)
= 4632 ang 2455degV|Eg| = 4632 V
Power angle = 2455o
(iv)
P = deg90sin2
27724633
(max occurs when δ = 90o)
= 19238 kW
(v)Synchronous speed ωs = 2 π fp = 1885 rads
= 1800 revmin
Ex 23 (Jan 14)
A 3 phase synchronous machine with negligible stator winding resistance and a synchronous
reactance of 25 per phase is connected to a 20kV 50 Hz 3phase supply (infinite bus) and
operated as a generator It is currently generating 100MW with power factor 085 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)(iv) What minimum value of excitation voltage (E) would be required for this machine to
deliver 200MW into a 20kV bus assuming a 20deg stability margin must be maintained onpower angle (δ)
Soln
(i)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2326
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
23
(ii)
P = 100 MW Apparent power S = 100 085
= 11765 ang -3179o MVA
Rated voltage per phase
V= 11547ang 0deg VRated current
Ia = S 3V
= 3396 103ang -3179deg A
|Ia| = 3396 103A
(iii)
Excitation voltage
Eg = 11547 ang 0deg + j25(3396 103ang -3179deg)
= 17570 ang 2425degV|Eg| = 17570 V
Power angle = 2425o V
(iv)
P = )2090sin(52
3degminusdegt gV E
(max occurs when δ = 90o stability margin 20o)
200 MW = )2090sin(52
547113degminusdegg E
200 MW = 38130209397085613 gg E E =
|Eg| = 15360 V
LoadGen
Xs =25
Eg Vt =11547V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2426
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
24
Ex 24A manufacturing plant fed with a 400V 50Hz three phase runs from 0900 to 1800
Monday to Friday The factoryrsquos load is normally 80kVA with power factor 08 lagging
during these hours Once a month they need to run a special heating cycle which adds a
further 20kW to the factory load for a 2 hour period This heating cycle is timer
controlled and fully automated but the timer is currently set to run on the third Friday
afternoon of the month Outside of normal business hours and at weekends the factory
load is negligible The firm has a contractually agreed Maximum Import Capacity (MIC)
of 120kVA
bull Use the attached table of Low Voltage Maximum Demand charges to calculate the
firmrsquos electricity bill (exclusive of VAT) for the month of February (28 days)
bull Identify three steps that this firm could take to reduce their expenditure on
electricity assuming they stick with the same tariff Calculate the potential
savings
983116983151983159 983126983151983148983156983137983143983141 983117983137983160983145983149983157983149 983108983141983149983137983150983140 983107983144983137983154983143983141983155
983108983137983145983148983161 983123983156983137983150983140983145983150983143 983107983144983137983154983143983141 991404302983140983137983161
983120983123983119 983116983141983158983161
983120983123983119 983116983141983158983161 983152983141983154 983115983126983105 983151983142 983117983113983107 983152983141983154 983140983137983161 9914040036983097983147983126983105983140983137983161
983117983137983160983145983149983157983149 983113983149983152983151983154983156 983107983137983152983137983139983145983156983161 (983117983113983107) 983107983144983137983154983143983141983155
983108983137983145983148983161 983107983144983137983154983143983141 983152983141983154 983147983126983105 983151983142 983117983113983107 9914040072983147983126983105983140983137983161
983108983141983149983137983150983140 983107983144983137983154983143983141983155 983118983151983158983085983110983141983138 983117983137983154983085983119983139983156
983117983151983150983156983144983148983161 983107983144983137983154983143983141 983152983141983154 983147983127 983151983142 983149983141983156983141983154983141983140 983117983108 983145983150 983137
983149983151983150983156983144983148983161 983152983141983154983145983151983140 (983123983157983138983146983141983139983156 983156983151 983137 983149983145983150983145983149983157983149 983139983144983137983154983143983141 983151983142
30983147983127) 991404189830975983147983127983149983151983150983156983144 991404000
983108983137983161983156983145983149983141 983147983127983144 (983125983150983145983156) 983107983144983137983154983143983141 (0898309800 983085 19830979830985983097) 1480983139983147983127983144 1275983139983147983127983144
983118983145983143983144983156983145983149983141 983147983127983144 (983125983150983145983156) 983118983145983143983144983156 983107983144983137983154983143983141983155 (2098309800983085079830985983097) 64983097983139983147983127983144 64983097983139983147983127983144
983127983137983156983156983148983141983155983155 983125983150983145983156 983120983154983145983139983141 (983147983126983105983122983144 983145983150 983141983160983139983141983155983155 983151983142 983151983150983141 983156983144983145983154983140
983151983142 983156983151983156983137983148 983147983127983144) 991404000714 983147983126983105983122983144 991404000714 983147983126983105983122983144
bull Table Low Voltage Maximum Demand Charges (Q6)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2526
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
25
SolnNumber of days in February = 28
Number of working hours = 459 = 180
Normal kW = 80kVA08 = 64kW
Total kWh = 64kW180 hours+20kW2 hours = 11560 kWh
Normal kvar =80kVA sqrt(1-08^2) = 48kvar = sqrt(802-64
2) = 48 kVAr
Total kvarh = 48kvar180 hours = 8640 kvarh
Chargeable kvarh = 8640-(11560)3=4786 kvarh
983116983151983159 983126983151983148983156983137983143983141 983117983137983160983145983149983157983149 983108983141983149983137983150983140
983107983144983137983154983143983141983155 983152983141983154 983122983137983156983141 983121983156983161 983120983141983154983145983151983140 983107983144983137983154983143983141
983108983137983145983148983161 983123983156983137983150983140983145983150983143 983107983144983137983154983143983141 991404302 983140983137983161 991404302000 1 28 991404983096456
983120983123983119 983116983141983158983161
983120983123983119 983116983141983158983161 983152983141983154 983115983126983105 983152983141983154 983140983137983161 9914040036983097 983147983126983105983140983137983161 99140400369830970 120 28 991404123983097983096
983117983137983160983145983149983157983149 983113983149983152983151983154983156 983107983137983152983137983139983145983156983161
(983117983113983107) 983107983144983137983154983143983141983155
983108983137983145983148983161 983107983144983137983154983143983141 983152983141983154 983147983126983105 983151983142 983117983113983107 9914040072 983147983126983105983140983137983161 991404007200 120 28 9914042419830972
983108983141983149983137983150983140 983107983144983137983154983143983141983155 983118983151983158983085983110983141983138
983117983151983150983156983144983148983161 983107983144983137983154983143983141 983152983141983154 983147983127 983151983142
983149983141983156983141983154983141983140 983117983108 983145983150 983137 983149983151983150983156983144983148983161
983152983141983154983145983151983140 (983123983157983138983146983141983139983156 983156983151 983137
983149983145983150983145983149983157983149 983139983144983137983154983143983141 983151983142 30983147983127) 991404189830975
983147983127983149983151983150983156
983144 99140418983097500 84 1 991404159830971983096 983108983137983161983156983145983149983141 983147983127983144 (983125983150983145983156) 983107983144983137983154983143983141
(0898309800 983085 19830979830985983097) 9914040148 983147983127983144 991404014800 11560 1
991404198309510983096
983096
983118983145983143983144983156983145983149983141 983147983127983144 (983125983150983145983156) 983118983145983143983144983156
983107983144983137983154983143983141983155 (2098309800983085079830985983097) 9914040064983097 983147983127983144 99140400649830970
983127983137983156983156983148983141983155983155 983125983150983145983156 983120983154983145983139983141 (983147983126983105983122983144
983145983150 983141983160983139983141983155983155 983151983142 983151983150983141 983156983144983145983154983140 983151983142 983156983151983156983137983148
983147983127983144) 991404000714 983147983126983137983154983144 991404000714 4787 1 991404341983096
983124983151983156983137983148 983106983145983148983148
983142983151983154
983110983141983138983154983157983137983154983161
9914042354983095
0
[16 Marks]
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2626
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
26
Potential Savings
bull Use Power Factor correction to eliminate Watt less unit charge and to reduce MIC
bull Shift Heating load to night time or weekend
bull Negotiate lower MIC
Potential savings from these measures
Eliminate Wattless charge save euro3417 month
Shift heating cycle to night time
Max demand it lowered by 20kW = saving of 20euro1895 = euro3790 month
Saving by moving 20kW to night rate = 20kW2 hours (euro0148-euro00649)= euro332
Negotiating a reduced MIC
If power factor correction is employed and the heating cycle is shifted to night time the
then maximum VA of the plant drops to close to its maximum kW demand of 64kW An
MIC of 80kVA should be sufficient to cover this Potential Saving = 40kVA
(euro0072+euro00369)28 = euro12196 per month
Total potential saving = euro3417+euro3790+euro332+euro12196=euro19735month
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 826
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
8
C B A
B A
Z Z Z
Z Z
Z ++=1
C B A
C B
Z Z Z
Z Z Z
++=2
C B A
C A
Z Z Z
Z Z Z
++=3
Star ndash Delta Transformation
2
133221
Z
Z Z Z Z Z Z Z A
++=
3
133221
Z
Z Z Z Z Z Z Z B
++=
ZA ZB
ZC
1
2 3
Z1
Z2Z3
Delta
Star
ZA ZB
ZC
1
2 3
Z1
Z2Z3
Delta Star
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 926
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1026
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
10
RT = 2 [ 29(2+9)] = 3612 = 32727 Ω
Ind Machines
Ex 9
Why does the rotor of an induction motor turn slower than the revolving field
Ex 10
Describe briefly the key differences between a squirrel cage motor and wound-rotor motor
in terms of construction and method of operation
See lecture notes
Ex 11
Both the voltage and frequency induced in the rotor of an induction motor decrease as the
rotor speeds up Briefly explain the reason for this and include the relationships between
slip the applied stator angular velocity and rotor angular velocity
Ex 12A 3-phase 400V 6-pole 50Hz star-connected induction motor operates on full load at a speed of
1475 revmin Determine the slip s
Ex 13 a) Calculate the synchronous speed of a 3-phase 12-pole induction motor that is excited by a
50Hz source
b) What is the nominal speed (speed at full load) is the slip at full load is 5
9
9
9
2
2
2
a
b
c
RT
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1126
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
11
a) Synchronous speed rpmns = 120 f p revmin
500 revmin
(b) s = (ωs - ωr) ωs = (ns - nr) ns
nr = ns(1 - s) = 475 revmin
Ex 14
An 400 V 50 Hz six-pole star connected induction machine operates with full load slip
of 005 pu and has the following per-phase equivalent circuit parameters
stator resistance Rs = 012Ω stator leakage reactance Xs = 06 Ω and magnetising
reactance Xm = 2215 Ω The rotor referred resistance Rr` = 015 Ω and the rotor referred
leakage reactance Xr` = 0586 Ω The induction machine initially operates as a motoron full load Using an approximate per-phase equivalent circuit determine
(i) synchronous speed and the speed of the motor at rated slip in revmin and
radsec
(ii) The approximate equivalent circuit per-phase input impedance
(iii) The input current
(iv) The developed torque
(v) The starting torque
Soln
Motoring modeSynchronous speed ωs = 2 π fp = 1046 rads
ωm = ωs(1 ndash s) = 9937 rads = 9489 revmin
Rr ` s = 015005 = 3
Z = Rs + jXs+ Rr ` s +jXr
`)
= 3122 + j 1186
The total Z is Zin = ( jXm|| Z)
R`r s
j06ΩΩΩΩ
is i`r
Vs = 400 radic3 V
j0586ΩΩΩΩ 012ΩΩΩΩ
015005 j221515ΩΩΩΩ
A
B
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1226
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
12
= 7398ang110o 2354ang 8238
o
== 3143ang2762o
The input current
I i = 7318 ang-2762o A
The rotor current
I r = 6886 ang-208o A
6886 A at a pf of 0935 lagging
The developed torque
T = Pm ωm = ||3
r
2`r
s
R I
sω = 408 Nm
Starting Torque
T = ||3
r
2`r
s
R I
sω with s=1
= 204 Nm
Ex 15 A 4 pole induction motor running from a 3 phase 400V line to line 50 Hz
supply has the following parametersStator Winding Resistance (Rs) = 005Ω
Stator Leakage Inductance (Xs) = 01 Ω
Magnetising Inductance (Xm) = 20 Ω Rotor Winding Resistance referred to stator (Rrrsquo) = 01 Ω
Rotor Leakage Inductance referred to stator (Xrrsquo) = 02 Ω
Rotational losses may be assumed to be negligible
(i) Draw a per phase equivalent circuit of this machine
(ii) Calculate the rotor speed at which the machine will have a slip of 2
(iii) Calculate the Power over the gap (Pog) in this machine when it is running at a slip of 2
(iv) Calculate the mechanical power (Pm) and the torque (Tm) delivered by the machine
running at a slip of 2
Soln
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1326
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
13
Synchronous speed ωs = 2 π fp = 1500 revmin
ωm = ωs(1 ndash s) = 15386 rads = 146925 revmin
Rr` s = 01002 = 5
Z = Rs + jXs+ Rr ` s +jXr
`)
= 505 + j 03
= 5059ang34o
The rotor referred current
I r = 4546 ang-34o A
4546 A at a pf of 099 lagging
The air-gap power
Pg = ||3 r2`
r
s R I = 31kW
Prot = 0 ndash no rotational losses
Pm = Pg (1-s) = 3038 kW
T = 3038 103 15386
= 197 Nm
Ex 16 A 6 pole induction motor running from a 3 phase 400V line to line 50 Hz supply has
the following parameters
Stator Winding Resistance (Rs) = 015Ω
Stator Leakage Inductance (Xs) = 03 Ω
Magnetising Inductance (Xm) = 60 Ω
R`r s
j01ΩΩΩΩ
is i`r
Vs = 400 radic3 V
j02ΩΩΩΩ 005ΩΩΩΩ
01002 j20ΩΩΩΩ
A
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1426
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
14
Rotor Winding Resistance referred to stator (Rrrsquo) = 03 Ω
Rotor Leakage Inductance referred to stator (Xrrsquo) = 06 Ω
Rotational losses have previously been measured to be 1000W at 5 slip
bull Draw a per phase equivalent circuit of this machinebull Calculate the rotor speed at which the machine will have a slip of 5
bull Calculate the air-gap Power (Pg) in this machine when it is running at a slip of 5
bull Calculate the efficiency of the induction motor from electrical input to mechanical
output when the motor is running with a slip of 5
Soln(a)
(b) Synchronous speed ωs = 2 π fp = 1000 revmin
ωm = ωs(1 ndash s) = 9948 rads = 950 revmin
(c) Rr` s = 03005 = 6
Z = Rs + jXs+ Rr ` s +jXr
`)
= 615 + j 09
= 622ang832o
(d) The air-gap power
The rotor referred current
I r = 3698 ang-303o A
Pg = ||3
r
2`r
s
R I = 246 kW
(d) I i =inZ
3400
Zin = Xm || Z = j60 || 622ang832o
Zin = 6097 ang1409o
I = VZ = 3772 ang-1409oA
R`r s
j03ΩΩΩΩ
is i`r
Vs = 230V
j06ΩΩΩΩ 015ΩΩΩΩ
03005 j60ΩΩΩΩ
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1526
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
15
Pelec = radic3 VL IL cos φ = 25044 kW
Pm = Pg (1-s) ndash Prot = 2237 kW
η = Pelec Pm = 893s
Ex 17
A large 3-phase 4000V 8-pole 50Hz squirrel-cage induction motor draws a current of
380A and a total active power of 2344kW when operating at full load The speed at full load
is accurately measured and found to be 742rpm The stator is star-connected and the
resistance between two stator terminals is 01Ω The total iron losses are 234kW and the
windage and friction losses are 12kW
Calculate
a) The power factor at full loadb) The active power supplied to the rotor
c) The total i2R losses in the rotor
d) The mechanical power output to the load and torque delivered to the load
e) The motorrsquos net efficiency
a) The active power input to the motor is given as 2344kW The apparent power is
S = radic3 VL IL = 2633kVA
Therefore pf = PS 089 lagging
b) The active power supplied to the rotor is the active power input to the motor minus the ironlosses and copper losses in the stator The stator iron losses are given as 234kW To calculate
the copper losses note that the terminal resistance (01) is twice that of each phase for a wye
connection Therefore the copper losses are
Stator copper losses Pcu = 217kW
Therefore the active power supplied to the rotor is
Pr = Input power ndash stator copper loss ndash iron losses
Pr = 2344 kW ndash 234kW ndash 217kW
= 2298kW
c) The copper losses in the rotor are proportional to the slip (see lecture notes) Prcu sP
Therefore we must calculate the slip Neither the synchronous speed nor number of poles are
explicitly stated in the problem however by inspection we determine that the synchronous speed
must be 750rpm and the number of poles = 8 (For 50Hz supply synchronous speeds occur at
discrete intervals corresponding to pole pairs 2-pole 3000rpm 4-pole 1500rpm 6- pole
1000rpm 8-pole 750rpm etchellip We also know that induction motors operate near synchronous
speed ndash the speed is given as 742 rpm which is close to synchronous speed for an
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1626
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
16
8-pole machine)
Synchronous speed ωs = 2 π fp = 7854 rads
ωm = = 777 radss = 00107
Therefore the rotor copper losses are Pr cu = sPr = 2456 kw
(d)
The mechanical power (Pm) delivered to the load is the active power delivered to the rotor
minus rotor copper losses and windage and friction losses The windage and friction losses are
given as 12kW Therefore
Pm =22989 kW ndash 245kW-12 kW =22624 kW
kNmT L 129
60
2742
42262=
=
π
(e) the motors efficiency 5962344
42262===
kW
kW
P
P
in
out η
Ex 18
A 3-phase 440V 8-pole 50Hz star-connected induction motor has the following equivalent
circuit per-phase parameters
Rs = 01 Ω Rr = 015Ω xs = xr =06 Ω Rc = 100 Ω xm = 10 Ω
(a) Calculate using the equivalent circuit neglecting stator impedance at a slip of 5 the input
stator-current and power factor the rotor current referred to the stator the torque the
mechanical output power and the efficiency Calculate the starting torque Assume the
mechanical loss is 1kW for all speeds
Soln
(a) Synchronous speed ωs = 2 π fp 7854 rads
ωm = ωs(1 - s) = 7618 rads
Stator impedance neglected
Vs = 400 radic3 = 230
Ir` =
60050
10
230
j+ =
= 68 ang-1671 = 651 - j 1954
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1726
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
17
Io =230
100 + j230
10 = 23 ndash j 23
Is = Ir` + Io = 674 - j4294
= 799 A at cos φ = 084
Ir` = 68 A
T = 529 Nm
Pmech = 39365 kW
Pelect = 465 kW
Efficiency = 39365465 = 846
Ts = 39627 Nm
Ex19
Tests conducted at 50 Hz on a three-phase star connected 45 kW 2200V six-pole 50Hz squirrel
cage induction motor are listed below
The average dc resistance per stator phase Rs = 28 Ω (a) Briefly describe the tests required to determine the per-phase equivalent circuit
impedance parameters of an induction machine
(a) Determine the parameters of the per-phase equivalent circuit
(b) Determine the rotational losses of the machine Core losses can be neglected
load
0
No-loadV
R
I1
Ic
I
c X
Im
m
Figure 1 Equivalent Circuit during No-Load Test
torPower Fac0 times= I I C = 2025 A
22
0 cm I I I minus=
= 4019 A
Ω== 627c
c I
V R Ω== 5317
m
m I
V X
No Load Test Locked Rotor Test
Power Factor = 045 Power Factor = 09
Line Voltage = 22kV Line Voltage = 270V
Line Current = 45A Line Current = 25A
Input Power = 16kW Input Power = 9kW
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1826
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
18
For the locked rotor test slip = 1
Ω== 246s
s
LR I
V Z
Locked rotor reactance is
( )2
21
2
21 R R Z X X +minus=+
φ cos21 times=+= Z R R Rtotal = 5616Ω
R2rsquo = Rtot - R1 = 2816Ω
Ω=+ 72221 X X
(a) From the no-load test the no-load power isPNL = 16kW
The no-load rotational loss is
Prot = PNL ndash 3 I s2 Rs
= 14299W
Synch Machines
Ex 20
A 60 MVA 20 kV three-phase 50 Hz synchronous generator has a synchronous reactance
of 8 Ω per phase and negligible resistance The generator delivers rated power at a power
factor of 085 lagging at the rated terminal voltage to an infinite bus(a) Determine the excitation voltage per phase and the power angle δ
(b) With the excitation voltage Eg of the synchronous generator held constant at the
value determined in part (a) the input driving torque is reduced until the generator
delivers 28 MW Determine the armature current and the power factor
(c) If the generator operates at the excitation voltage of Eg obtained in part (a) what is
the steady state maximum power the machine can deliver before losing
jXsΩΩΩΩ jX`r ΩΩΩΩ Rs
R`r s
Is I`r
Vs ph
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1926
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
19
synchronism Determine the armature current corresponding to this maximum
power
(d) With real generated power at 42MW compute the power angle δ the stator
current Ia and power factor when the field excitation current If is adjusted so that
the excitation voltage Eg is decreased to 7919 of the value determined in part
(a)
Soln
(a) Apparent power S = 51 MW ndash j 316Mvar
Rated voltage per phase
V = 1154 ang 0deg kV
Rated current
Ia = S 3V
= 1733 ang -3178deg AExcitation voltage
Eg = 11540 + j9(1733) ang -3178deg)= 237922 ang 3386deg
Power angle = 3386o
(b) When delivering 28 MW
δ = sin-1
)5411)(79223(3
828 = 1578
o
P =
s
g
X
V E 3sinδ
Ia = 13069ang2967o j8
= 16336ang-6033o A
Power factor = cos(6033) = 0495 lagging
(c) The maximum power occurs at δ = 90o
Pmax =s
g
X
V E 3
= 102 MW
Ia = 324885ang25875o A
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2026
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
20
The power factor cos(25875) = 0899 leading
(d) Eg = 13463 V
The power angle δ = sin-1
)5411)(84318(3
842
= 31o
Ia = 86674ang0o A
-Power factor = 0
Ex 21 (Aug 13) A 3 phase synchronous machine with negligible stator winding resistance and a
synchronous reactance of 2 per phase is connected to a 400V 50 Hz 3-phase supply (infinite
bus) and operated as a generator It is currently generating 50kW with power factor 08 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)
(iv) What is the maximum power the machine can generate with this value of excitation
Soln(i)
(ii)P = 50kW
Apparent power S = 625 ang -3687
o
kVA
Rated voltage per phase V = 230ang 0deg VRated current
Ia = S 3V
= 9058ang -3687deg A
LoadGen
Xs =2
Eg Vt =400 radic3 V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2126
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
21
|Ia| = 9058 A
(iii)
Excitation voltage
Eg = 230 ang 0deg + j2(9058ang -3687deg)= 3684 ang 2318degV
|Eg| = 3684 V
Power angle = 2318o
(iv)
P = deg90sin2
23043683
(max occurs when δ = 90o)
= 1271 kW
Ex 22A 3 phase 4 pole synchronous machine with negligible stator winding resistance and a
synchronous reactance of 2 per phase is connected to a 480V 60 Hz 3phase supply
(infinite bus) and operated as a generator It is currently generating 80kW with power
factor 08 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)(iv) What is the maximum power the machine can generate with this value
excitation(v) What is the rotational speed of this generator
Soln (i)
(ii)P = 80kW Apparent power S 100 ang -3687
o kVA
Rated voltage per phase
V 277ang 0deg V
LoadGen
Xs =2
Eg Vt = 480V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2226
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
22
Rated current
Ia = S 3V
= 9058ang -3687deg A
|Ia| = 12028 A
(iii)
Excitation voltage
Eg = 277 ang 0deg + j2(12028ang -3687deg)
= 4632 ang 2455degV|Eg| = 4632 V
Power angle = 2455o
(iv)
P = deg90sin2
27724633
(max occurs when δ = 90o)
= 19238 kW
(v)Synchronous speed ωs = 2 π fp = 1885 rads
= 1800 revmin
Ex 23 (Jan 14)
A 3 phase synchronous machine with negligible stator winding resistance and a synchronous
reactance of 25 per phase is connected to a 20kV 50 Hz 3phase supply (infinite bus) and
operated as a generator It is currently generating 100MW with power factor 085 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)(iv) What minimum value of excitation voltage (E) would be required for this machine to
deliver 200MW into a 20kV bus assuming a 20deg stability margin must be maintained onpower angle (δ)
Soln
(i)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2326
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
23
(ii)
P = 100 MW Apparent power S = 100 085
= 11765 ang -3179o MVA
Rated voltage per phase
V= 11547ang 0deg VRated current
Ia = S 3V
= 3396 103ang -3179deg A
|Ia| = 3396 103A
(iii)
Excitation voltage
Eg = 11547 ang 0deg + j25(3396 103ang -3179deg)
= 17570 ang 2425degV|Eg| = 17570 V
Power angle = 2425o V
(iv)
P = )2090sin(52
3degminusdegt gV E
(max occurs when δ = 90o stability margin 20o)
200 MW = )2090sin(52
547113degminusdegg E
200 MW = 38130209397085613 gg E E =
|Eg| = 15360 V
LoadGen
Xs =25
Eg Vt =11547V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2426
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
24
Ex 24A manufacturing plant fed with a 400V 50Hz three phase runs from 0900 to 1800
Monday to Friday The factoryrsquos load is normally 80kVA with power factor 08 lagging
during these hours Once a month they need to run a special heating cycle which adds a
further 20kW to the factory load for a 2 hour period This heating cycle is timer
controlled and fully automated but the timer is currently set to run on the third Friday
afternoon of the month Outside of normal business hours and at weekends the factory
load is negligible The firm has a contractually agreed Maximum Import Capacity (MIC)
of 120kVA
bull Use the attached table of Low Voltage Maximum Demand charges to calculate the
firmrsquos electricity bill (exclusive of VAT) for the month of February (28 days)
bull Identify three steps that this firm could take to reduce their expenditure on
electricity assuming they stick with the same tariff Calculate the potential
savings
983116983151983159 983126983151983148983156983137983143983141 983117983137983160983145983149983157983149 983108983141983149983137983150983140 983107983144983137983154983143983141983155
983108983137983145983148983161 983123983156983137983150983140983145983150983143 983107983144983137983154983143983141 991404302983140983137983161
983120983123983119 983116983141983158983161
983120983123983119 983116983141983158983161 983152983141983154 983115983126983105 983151983142 983117983113983107 983152983141983154 983140983137983161 9914040036983097983147983126983105983140983137983161
983117983137983160983145983149983157983149 983113983149983152983151983154983156 983107983137983152983137983139983145983156983161 (983117983113983107) 983107983144983137983154983143983141983155
983108983137983145983148983161 983107983144983137983154983143983141 983152983141983154 983147983126983105 983151983142 983117983113983107 9914040072983147983126983105983140983137983161
983108983141983149983137983150983140 983107983144983137983154983143983141983155 983118983151983158983085983110983141983138 983117983137983154983085983119983139983156
983117983151983150983156983144983148983161 983107983144983137983154983143983141 983152983141983154 983147983127 983151983142 983149983141983156983141983154983141983140 983117983108 983145983150 983137
983149983151983150983156983144983148983161 983152983141983154983145983151983140 (983123983157983138983146983141983139983156 983156983151 983137 983149983145983150983145983149983157983149 983139983144983137983154983143983141 983151983142
30983147983127) 991404189830975983147983127983149983151983150983156983144 991404000
983108983137983161983156983145983149983141 983147983127983144 (983125983150983145983156) 983107983144983137983154983143983141 (0898309800 983085 19830979830985983097) 1480983139983147983127983144 1275983139983147983127983144
983118983145983143983144983156983145983149983141 983147983127983144 (983125983150983145983156) 983118983145983143983144983156 983107983144983137983154983143983141983155 (2098309800983085079830985983097) 64983097983139983147983127983144 64983097983139983147983127983144
983127983137983156983156983148983141983155983155 983125983150983145983156 983120983154983145983139983141 (983147983126983105983122983144 983145983150 983141983160983139983141983155983155 983151983142 983151983150983141 983156983144983145983154983140
983151983142 983156983151983156983137983148 983147983127983144) 991404000714 983147983126983105983122983144 991404000714 983147983126983105983122983144
bull Table Low Voltage Maximum Demand Charges (Q6)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2526
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
25
SolnNumber of days in February = 28
Number of working hours = 459 = 180
Normal kW = 80kVA08 = 64kW
Total kWh = 64kW180 hours+20kW2 hours = 11560 kWh
Normal kvar =80kVA sqrt(1-08^2) = 48kvar = sqrt(802-64
2) = 48 kVAr
Total kvarh = 48kvar180 hours = 8640 kvarh
Chargeable kvarh = 8640-(11560)3=4786 kvarh
983116983151983159 983126983151983148983156983137983143983141 983117983137983160983145983149983157983149 983108983141983149983137983150983140
983107983144983137983154983143983141983155 983152983141983154 983122983137983156983141 983121983156983161 983120983141983154983145983151983140 983107983144983137983154983143983141
983108983137983145983148983161 983123983156983137983150983140983145983150983143 983107983144983137983154983143983141 991404302 983140983137983161 991404302000 1 28 991404983096456
983120983123983119 983116983141983158983161
983120983123983119 983116983141983158983161 983152983141983154 983115983126983105 983152983141983154 983140983137983161 9914040036983097 983147983126983105983140983137983161 99140400369830970 120 28 991404123983097983096
983117983137983160983145983149983157983149 983113983149983152983151983154983156 983107983137983152983137983139983145983156983161
(983117983113983107) 983107983144983137983154983143983141983155
983108983137983145983148983161 983107983144983137983154983143983141 983152983141983154 983147983126983105 983151983142 983117983113983107 9914040072 983147983126983105983140983137983161 991404007200 120 28 9914042419830972
983108983141983149983137983150983140 983107983144983137983154983143983141983155 983118983151983158983085983110983141983138
983117983151983150983156983144983148983161 983107983144983137983154983143983141 983152983141983154 983147983127 983151983142
983149983141983156983141983154983141983140 983117983108 983145983150 983137 983149983151983150983156983144983148983161
983152983141983154983145983151983140 (983123983157983138983146983141983139983156 983156983151 983137
983149983145983150983145983149983157983149 983139983144983137983154983143983141 983151983142 30983147983127) 991404189830975
983147983127983149983151983150983156
983144 99140418983097500 84 1 991404159830971983096 983108983137983161983156983145983149983141 983147983127983144 (983125983150983145983156) 983107983144983137983154983143983141
(0898309800 983085 19830979830985983097) 9914040148 983147983127983144 991404014800 11560 1
991404198309510983096
983096
983118983145983143983144983156983145983149983141 983147983127983144 (983125983150983145983156) 983118983145983143983144983156
983107983144983137983154983143983141983155 (2098309800983085079830985983097) 9914040064983097 983147983127983144 99140400649830970
983127983137983156983156983148983141983155983155 983125983150983145983156 983120983154983145983139983141 (983147983126983105983122983144
983145983150 983141983160983139983141983155983155 983151983142 983151983150983141 983156983144983145983154983140 983151983142 983156983151983156983137983148
983147983127983144) 991404000714 983147983126983137983154983144 991404000714 4787 1 991404341983096
983124983151983156983137983148 983106983145983148983148
983142983151983154
983110983141983138983154983157983137983154983161
9914042354983095
0
[16 Marks]
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2626
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
26
Potential Savings
bull Use Power Factor correction to eliminate Watt less unit charge and to reduce MIC
bull Shift Heating load to night time or weekend
bull Negotiate lower MIC
Potential savings from these measures
Eliminate Wattless charge save euro3417 month
Shift heating cycle to night time
Max demand it lowered by 20kW = saving of 20euro1895 = euro3790 month
Saving by moving 20kW to night rate = 20kW2 hours (euro0148-euro00649)= euro332
Negotiating a reduced MIC
If power factor correction is employed and the heating cycle is shifted to night time the
then maximum VA of the plant drops to close to its maximum kW demand of 64kW An
MIC of 80kVA should be sufficient to cover this Potential Saving = 40kVA
(euro0072+euro00369)28 = euro12196 per month
Total potential saving = euro3417+euro3790+euro332+euro12196=euro19735month
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 926
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1026
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
10
RT = 2 [ 29(2+9)] = 3612 = 32727 Ω
Ind Machines
Ex 9
Why does the rotor of an induction motor turn slower than the revolving field
Ex 10
Describe briefly the key differences between a squirrel cage motor and wound-rotor motor
in terms of construction and method of operation
See lecture notes
Ex 11
Both the voltage and frequency induced in the rotor of an induction motor decrease as the
rotor speeds up Briefly explain the reason for this and include the relationships between
slip the applied stator angular velocity and rotor angular velocity
Ex 12A 3-phase 400V 6-pole 50Hz star-connected induction motor operates on full load at a speed of
1475 revmin Determine the slip s
Ex 13 a) Calculate the synchronous speed of a 3-phase 12-pole induction motor that is excited by a
50Hz source
b) What is the nominal speed (speed at full load) is the slip at full load is 5
9
9
9
2
2
2
a
b
c
RT
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1126
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
11
a) Synchronous speed rpmns = 120 f p revmin
500 revmin
(b) s = (ωs - ωr) ωs = (ns - nr) ns
nr = ns(1 - s) = 475 revmin
Ex 14
An 400 V 50 Hz six-pole star connected induction machine operates with full load slip
of 005 pu and has the following per-phase equivalent circuit parameters
stator resistance Rs = 012Ω stator leakage reactance Xs = 06 Ω and magnetising
reactance Xm = 2215 Ω The rotor referred resistance Rr` = 015 Ω and the rotor referred
leakage reactance Xr` = 0586 Ω The induction machine initially operates as a motoron full load Using an approximate per-phase equivalent circuit determine
(i) synchronous speed and the speed of the motor at rated slip in revmin and
radsec
(ii) The approximate equivalent circuit per-phase input impedance
(iii) The input current
(iv) The developed torque
(v) The starting torque
Soln
Motoring modeSynchronous speed ωs = 2 π fp = 1046 rads
ωm = ωs(1 ndash s) = 9937 rads = 9489 revmin
Rr ` s = 015005 = 3
Z = Rs + jXs+ Rr ` s +jXr
`)
= 3122 + j 1186
The total Z is Zin = ( jXm|| Z)
R`r s
j06ΩΩΩΩ
is i`r
Vs = 400 radic3 V
j0586ΩΩΩΩ 012ΩΩΩΩ
015005 j221515ΩΩΩΩ
A
B
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1226
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
12
= 7398ang110o 2354ang 8238
o
== 3143ang2762o
The input current
I i = 7318 ang-2762o A
The rotor current
I r = 6886 ang-208o A
6886 A at a pf of 0935 lagging
The developed torque
T = Pm ωm = ||3
r
2`r
s
R I
sω = 408 Nm
Starting Torque
T = ||3
r
2`r
s
R I
sω with s=1
= 204 Nm
Ex 15 A 4 pole induction motor running from a 3 phase 400V line to line 50 Hz
supply has the following parametersStator Winding Resistance (Rs) = 005Ω
Stator Leakage Inductance (Xs) = 01 Ω
Magnetising Inductance (Xm) = 20 Ω Rotor Winding Resistance referred to stator (Rrrsquo) = 01 Ω
Rotor Leakage Inductance referred to stator (Xrrsquo) = 02 Ω
Rotational losses may be assumed to be negligible
(i) Draw a per phase equivalent circuit of this machine
(ii) Calculate the rotor speed at which the machine will have a slip of 2
(iii) Calculate the Power over the gap (Pog) in this machine when it is running at a slip of 2
(iv) Calculate the mechanical power (Pm) and the torque (Tm) delivered by the machine
running at a slip of 2
Soln
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1326
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
13
Synchronous speed ωs = 2 π fp = 1500 revmin
ωm = ωs(1 ndash s) = 15386 rads = 146925 revmin
Rr` s = 01002 = 5
Z = Rs + jXs+ Rr ` s +jXr
`)
= 505 + j 03
= 5059ang34o
The rotor referred current
I r = 4546 ang-34o A
4546 A at a pf of 099 lagging
The air-gap power
Pg = ||3 r2`
r
s R I = 31kW
Prot = 0 ndash no rotational losses
Pm = Pg (1-s) = 3038 kW
T = 3038 103 15386
= 197 Nm
Ex 16 A 6 pole induction motor running from a 3 phase 400V line to line 50 Hz supply has
the following parameters
Stator Winding Resistance (Rs) = 015Ω
Stator Leakage Inductance (Xs) = 03 Ω
Magnetising Inductance (Xm) = 60 Ω
R`r s
j01ΩΩΩΩ
is i`r
Vs = 400 radic3 V
j02ΩΩΩΩ 005ΩΩΩΩ
01002 j20ΩΩΩΩ
A
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1426
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
14
Rotor Winding Resistance referred to stator (Rrrsquo) = 03 Ω
Rotor Leakage Inductance referred to stator (Xrrsquo) = 06 Ω
Rotational losses have previously been measured to be 1000W at 5 slip
bull Draw a per phase equivalent circuit of this machinebull Calculate the rotor speed at which the machine will have a slip of 5
bull Calculate the air-gap Power (Pg) in this machine when it is running at a slip of 5
bull Calculate the efficiency of the induction motor from electrical input to mechanical
output when the motor is running with a slip of 5
Soln(a)
(b) Synchronous speed ωs = 2 π fp = 1000 revmin
ωm = ωs(1 ndash s) = 9948 rads = 950 revmin
(c) Rr` s = 03005 = 6
Z = Rs + jXs+ Rr ` s +jXr
`)
= 615 + j 09
= 622ang832o
(d) The air-gap power
The rotor referred current
I r = 3698 ang-303o A
Pg = ||3
r
2`r
s
R I = 246 kW
(d) I i =inZ
3400
Zin = Xm || Z = j60 || 622ang832o
Zin = 6097 ang1409o
I = VZ = 3772 ang-1409oA
R`r s
j03ΩΩΩΩ
is i`r
Vs = 230V
j06ΩΩΩΩ 015ΩΩΩΩ
03005 j60ΩΩΩΩ
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1526
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
15
Pelec = radic3 VL IL cos φ = 25044 kW
Pm = Pg (1-s) ndash Prot = 2237 kW
η = Pelec Pm = 893s
Ex 17
A large 3-phase 4000V 8-pole 50Hz squirrel-cage induction motor draws a current of
380A and a total active power of 2344kW when operating at full load The speed at full load
is accurately measured and found to be 742rpm The stator is star-connected and the
resistance between two stator terminals is 01Ω The total iron losses are 234kW and the
windage and friction losses are 12kW
Calculate
a) The power factor at full loadb) The active power supplied to the rotor
c) The total i2R losses in the rotor
d) The mechanical power output to the load and torque delivered to the load
e) The motorrsquos net efficiency
a) The active power input to the motor is given as 2344kW The apparent power is
S = radic3 VL IL = 2633kVA
Therefore pf = PS 089 lagging
b) The active power supplied to the rotor is the active power input to the motor minus the ironlosses and copper losses in the stator The stator iron losses are given as 234kW To calculate
the copper losses note that the terminal resistance (01) is twice that of each phase for a wye
connection Therefore the copper losses are
Stator copper losses Pcu = 217kW
Therefore the active power supplied to the rotor is
Pr = Input power ndash stator copper loss ndash iron losses
Pr = 2344 kW ndash 234kW ndash 217kW
= 2298kW
c) The copper losses in the rotor are proportional to the slip (see lecture notes) Prcu sP
Therefore we must calculate the slip Neither the synchronous speed nor number of poles are
explicitly stated in the problem however by inspection we determine that the synchronous speed
must be 750rpm and the number of poles = 8 (For 50Hz supply synchronous speeds occur at
discrete intervals corresponding to pole pairs 2-pole 3000rpm 4-pole 1500rpm 6- pole
1000rpm 8-pole 750rpm etchellip We also know that induction motors operate near synchronous
speed ndash the speed is given as 742 rpm which is close to synchronous speed for an
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1626
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
16
8-pole machine)
Synchronous speed ωs = 2 π fp = 7854 rads
ωm = = 777 radss = 00107
Therefore the rotor copper losses are Pr cu = sPr = 2456 kw
(d)
The mechanical power (Pm) delivered to the load is the active power delivered to the rotor
minus rotor copper losses and windage and friction losses The windage and friction losses are
given as 12kW Therefore
Pm =22989 kW ndash 245kW-12 kW =22624 kW
kNmT L 129
60
2742
42262=
=
π
(e) the motors efficiency 5962344
42262===
kW
kW
P
P
in
out η
Ex 18
A 3-phase 440V 8-pole 50Hz star-connected induction motor has the following equivalent
circuit per-phase parameters
Rs = 01 Ω Rr = 015Ω xs = xr =06 Ω Rc = 100 Ω xm = 10 Ω
(a) Calculate using the equivalent circuit neglecting stator impedance at a slip of 5 the input
stator-current and power factor the rotor current referred to the stator the torque the
mechanical output power and the efficiency Calculate the starting torque Assume the
mechanical loss is 1kW for all speeds
Soln
(a) Synchronous speed ωs = 2 π fp 7854 rads
ωm = ωs(1 - s) = 7618 rads
Stator impedance neglected
Vs = 400 radic3 = 230
Ir` =
60050
10
230
j+ =
= 68 ang-1671 = 651 - j 1954
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1726
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
17
Io =230
100 + j230
10 = 23 ndash j 23
Is = Ir` + Io = 674 - j4294
= 799 A at cos φ = 084
Ir` = 68 A
T = 529 Nm
Pmech = 39365 kW
Pelect = 465 kW
Efficiency = 39365465 = 846
Ts = 39627 Nm
Ex19
Tests conducted at 50 Hz on a three-phase star connected 45 kW 2200V six-pole 50Hz squirrel
cage induction motor are listed below
The average dc resistance per stator phase Rs = 28 Ω (a) Briefly describe the tests required to determine the per-phase equivalent circuit
impedance parameters of an induction machine
(a) Determine the parameters of the per-phase equivalent circuit
(b) Determine the rotational losses of the machine Core losses can be neglected
load
0
No-loadV
R
I1
Ic
I
c X
Im
m
Figure 1 Equivalent Circuit during No-Load Test
torPower Fac0 times= I I C = 2025 A
22
0 cm I I I minus=
= 4019 A
Ω== 627c
c I
V R Ω== 5317
m
m I
V X
No Load Test Locked Rotor Test
Power Factor = 045 Power Factor = 09
Line Voltage = 22kV Line Voltage = 270V
Line Current = 45A Line Current = 25A
Input Power = 16kW Input Power = 9kW
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1826
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
18
For the locked rotor test slip = 1
Ω== 246s
s
LR I
V Z
Locked rotor reactance is
( )2
21
2
21 R R Z X X +minus=+
φ cos21 times=+= Z R R Rtotal = 5616Ω
R2rsquo = Rtot - R1 = 2816Ω
Ω=+ 72221 X X
(a) From the no-load test the no-load power isPNL = 16kW
The no-load rotational loss is
Prot = PNL ndash 3 I s2 Rs
= 14299W
Synch Machines
Ex 20
A 60 MVA 20 kV three-phase 50 Hz synchronous generator has a synchronous reactance
of 8 Ω per phase and negligible resistance The generator delivers rated power at a power
factor of 085 lagging at the rated terminal voltage to an infinite bus(a) Determine the excitation voltage per phase and the power angle δ
(b) With the excitation voltage Eg of the synchronous generator held constant at the
value determined in part (a) the input driving torque is reduced until the generator
delivers 28 MW Determine the armature current and the power factor
(c) If the generator operates at the excitation voltage of Eg obtained in part (a) what is
the steady state maximum power the machine can deliver before losing
jXsΩΩΩΩ jX`r ΩΩΩΩ Rs
R`r s
Is I`r
Vs ph
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1926
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
19
synchronism Determine the armature current corresponding to this maximum
power
(d) With real generated power at 42MW compute the power angle δ the stator
current Ia and power factor when the field excitation current If is adjusted so that
the excitation voltage Eg is decreased to 7919 of the value determined in part
(a)
Soln
(a) Apparent power S = 51 MW ndash j 316Mvar
Rated voltage per phase
V = 1154 ang 0deg kV
Rated current
Ia = S 3V
= 1733 ang -3178deg AExcitation voltage
Eg = 11540 + j9(1733) ang -3178deg)= 237922 ang 3386deg
Power angle = 3386o
(b) When delivering 28 MW
δ = sin-1
)5411)(79223(3
828 = 1578
o
P =
s
g
X
V E 3sinδ
Ia = 13069ang2967o j8
= 16336ang-6033o A
Power factor = cos(6033) = 0495 lagging
(c) The maximum power occurs at δ = 90o
Pmax =s
g
X
V E 3
= 102 MW
Ia = 324885ang25875o A
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2026
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
20
The power factor cos(25875) = 0899 leading
(d) Eg = 13463 V
The power angle δ = sin-1
)5411)(84318(3
842
= 31o
Ia = 86674ang0o A
-Power factor = 0
Ex 21 (Aug 13) A 3 phase synchronous machine with negligible stator winding resistance and a
synchronous reactance of 2 per phase is connected to a 400V 50 Hz 3-phase supply (infinite
bus) and operated as a generator It is currently generating 50kW with power factor 08 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)
(iv) What is the maximum power the machine can generate with this value of excitation
Soln(i)
(ii)P = 50kW
Apparent power S = 625 ang -3687
o
kVA
Rated voltage per phase V = 230ang 0deg VRated current
Ia = S 3V
= 9058ang -3687deg A
LoadGen
Xs =2
Eg Vt =400 radic3 V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2126
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
21
|Ia| = 9058 A
(iii)
Excitation voltage
Eg = 230 ang 0deg + j2(9058ang -3687deg)= 3684 ang 2318degV
|Eg| = 3684 V
Power angle = 2318o
(iv)
P = deg90sin2
23043683
(max occurs when δ = 90o)
= 1271 kW
Ex 22A 3 phase 4 pole synchronous machine with negligible stator winding resistance and a
synchronous reactance of 2 per phase is connected to a 480V 60 Hz 3phase supply
(infinite bus) and operated as a generator It is currently generating 80kW with power
factor 08 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)(iv) What is the maximum power the machine can generate with this value
excitation(v) What is the rotational speed of this generator
Soln (i)
(ii)P = 80kW Apparent power S 100 ang -3687
o kVA
Rated voltage per phase
V 277ang 0deg V
LoadGen
Xs =2
Eg Vt = 480V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2226
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
22
Rated current
Ia = S 3V
= 9058ang -3687deg A
|Ia| = 12028 A
(iii)
Excitation voltage
Eg = 277 ang 0deg + j2(12028ang -3687deg)
= 4632 ang 2455degV|Eg| = 4632 V
Power angle = 2455o
(iv)
P = deg90sin2
27724633
(max occurs when δ = 90o)
= 19238 kW
(v)Synchronous speed ωs = 2 π fp = 1885 rads
= 1800 revmin
Ex 23 (Jan 14)
A 3 phase synchronous machine with negligible stator winding resistance and a synchronous
reactance of 25 per phase is connected to a 20kV 50 Hz 3phase supply (infinite bus) and
operated as a generator It is currently generating 100MW with power factor 085 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)(iv) What minimum value of excitation voltage (E) would be required for this machine to
deliver 200MW into a 20kV bus assuming a 20deg stability margin must be maintained onpower angle (δ)
Soln
(i)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2326
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
23
(ii)
P = 100 MW Apparent power S = 100 085
= 11765 ang -3179o MVA
Rated voltage per phase
V= 11547ang 0deg VRated current
Ia = S 3V
= 3396 103ang -3179deg A
|Ia| = 3396 103A
(iii)
Excitation voltage
Eg = 11547 ang 0deg + j25(3396 103ang -3179deg)
= 17570 ang 2425degV|Eg| = 17570 V
Power angle = 2425o V
(iv)
P = )2090sin(52
3degminusdegt gV E
(max occurs when δ = 90o stability margin 20o)
200 MW = )2090sin(52
547113degminusdegg E
200 MW = 38130209397085613 gg E E =
|Eg| = 15360 V
LoadGen
Xs =25
Eg Vt =11547V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2426
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
24
Ex 24A manufacturing plant fed with a 400V 50Hz three phase runs from 0900 to 1800
Monday to Friday The factoryrsquos load is normally 80kVA with power factor 08 lagging
during these hours Once a month they need to run a special heating cycle which adds a
further 20kW to the factory load for a 2 hour period This heating cycle is timer
controlled and fully automated but the timer is currently set to run on the third Friday
afternoon of the month Outside of normal business hours and at weekends the factory
load is negligible The firm has a contractually agreed Maximum Import Capacity (MIC)
of 120kVA
bull Use the attached table of Low Voltage Maximum Demand charges to calculate the
firmrsquos electricity bill (exclusive of VAT) for the month of February (28 days)
bull Identify three steps that this firm could take to reduce their expenditure on
electricity assuming they stick with the same tariff Calculate the potential
savings
983116983151983159 983126983151983148983156983137983143983141 983117983137983160983145983149983157983149 983108983141983149983137983150983140 983107983144983137983154983143983141983155
983108983137983145983148983161 983123983156983137983150983140983145983150983143 983107983144983137983154983143983141 991404302983140983137983161
983120983123983119 983116983141983158983161
983120983123983119 983116983141983158983161 983152983141983154 983115983126983105 983151983142 983117983113983107 983152983141983154 983140983137983161 9914040036983097983147983126983105983140983137983161
983117983137983160983145983149983157983149 983113983149983152983151983154983156 983107983137983152983137983139983145983156983161 (983117983113983107) 983107983144983137983154983143983141983155
983108983137983145983148983161 983107983144983137983154983143983141 983152983141983154 983147983126983105 983151983142 983117983113983107 9914040072983147983126983105983140983137983161
983108983141983149983137983150983140 983107983144983137983154983143983141983155 983118983151983158983085983110983141983138 983117983137983154983085983119983139983156
983117983151983150983156983144983148983161 983107983144983137983154983143983141 983152983141983154 983147983127 983151983142 983149983141983156983141983154983141983140 983117983108 983145983150 983137
983149983151983150983156983144983148983161 983152983141983154983145983151983140 (983123983157983138983146983141983139983156 983156983151 983137 983149983145983150983145983149983157983149 983139983144983137983154983143983141 983151983142
30983147983127) 991404189830975983147983127983149983151983150983156983144 991404000
983108983137983161983156983145983149983141 983147983127983144 (983125983150983145983156) 983107983144983137983154983143983141 (0898309800 983085 19830979830985983097) 1480983139983147983127983144 1275983139983147983127983144
983118983145983143983144983156983145983149983141 983147983127983144 (983125983150983145983156) 983118983145983143983144983156 983107983144983137983154983143983141983155 (2098309800983085079830985983097) 64983097983139983147983127983144 64983097983139983147983127983144
983127983137983156983156983148983141983155983155 983125983150983145983156 983120983154983145983139983141 (983147983126983105983122983144 983145983150 983141983160983139983141983155983155 983151983142 983151983150983141 983156983144983145983154983140
983151983142 983156983151983156983137983148 983147983127983144) 991404000714 983147983126983105983122983144 991404000714 983147983126983105983122983144
bull Table Low Voltage Maximum Demand Charges (Q6)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2526
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
25
SolnNumber of days in February = 28
Number of working hours = 459 = 180
Normal kW = 80kVA08 = 64kW
Total kWh = 64kW180 hours+20kW2 hours = 11560 kWh
Normal kvar =80kVA sqrt(1-08^2) = 48kvar = sqrt(802-64
2) = 48 kVAr
Total kvarh = 48kvar180 hours = 8640 kvarh
Chargeable kvarh = 8640-(11560)3=4786 kvarh
983116983151983159 983126983151983148983156983137983143983141 983117983137983160983145983149983157983149 983108983141983149983137983150983140
983107983144983137983154983143983141983155 983152983141983154 983122983137983156983141 983121983156983161 983120983141983154983145983151983140 983107983144983137983154983143983141
983108983137983145983148983161 983123983156983137983150983140983145983150983143 983107983144983137983154983143983141 991404302 983140983137983161 991404302000 1 28 991404983096456
983120983123983119 983116983141983158983161
983120983123983119 983116983141983158983161 983152983141983154 983115983126983105 983152983141983154 983140983137983161 9914040036983097 983147983126983105983140983137983161 99140400369830970 120 28 991404123983097983096
983117983137983160983145983149983157983149 983113983149983152983151983154983156 983107983137983152983137983139983145983156983161
(983117983113983107) 983107983144983137983154983143983141983155
983108983137983145983148983161 983107983144983137983154983143983141 983152983141983154 983147983126983105 983151983142 983117983113983107 9914040072 983147983126983105983140983137983161 991404007200 120 28 9914042419830972
983108983141983149983137983150983140 983107983144983137983154983143983141983155 983118983151983158983085983110983141983138
983117983151983150983156983144983148983161 983107983144983137983154983143983141 983152983141983154 983147983127 983151983142
983149983141983156983141983154983141983140 983117983108 983145983150 983137 983149983151983150983156983144983148983161
983152983141983154983145983151983140 (983123983157983138983146983141983139983156 983156983151 983137
983149983145983150983145983149983157983149 983139983144983137983154983143983141 983151983142 30983147983127) 991404189830975
983147983127983149983151983150983156
983144 99140418983097500 84 1 991404159830971983096 983108983137983161983156983145983149983141 983147983127983144 (983125983150983145983156) 983107983144983137983154983143983141
(0898309800 983085 19830979830985983097) 9914040148 983147983127983144 991404014800 11560 1
991404198309510983096
983096
983118983145983143983144983156983145983149983141 983147983127983144 (983125983150983145983156) 983118983145983143983144983156
983107983144983137983154983143983141983155 (2098309800983085079830985983097) 9914040064983097 983147983127983144 99140400649830970
983127983137983156983156983148983141983155983155 983125983150983145983156 983120983154983145983139983141 (983147983126983105983122983144
983145983150 983141983160983139983141983155983155 983151983142 983151983150983141 983156983144983145983154983140 983151983142 983156983151983156983137983148
983147983127983144) 991404000714 983147983126983137983154983144 991404000714 4787 1 991404341983096
983124983151983156983137983148 983106983145983148983148
983142983151983154
983110983141983138983154983157983137983154983161
9914042354983095
0
[16 Marks]
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2626
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
26
Potential Savings
bull Use Power Factor correction to eliminate Watt less unit charge and to reduce MIC
bull Shift Heating load to night time or weekend
bull Negotiate lower MIC
Potential savings from these measures
Eliminate Wattless charge save euro3417 month
Shift heating cycle to night time
Max demand it lowered by 20kW = saving of 20euro1895 = euro3790 month
Saving by moving 20kW to night rate = 20kW2 hours (euro0148-euro00649)= euro332
Negotiating a reduced MIC
If power factor correction is employed and the heating cycle is shifted to night time the
then maximum VA of the plant drops to close to its maximum kW demand of 64kW An
MIC of 80kVA should be sufficient to cover this Potential Saving = 40kVA
(euro0072+euro00369)28 = euro12196 per month
Total potential saving = euro3417+euro3790+euro332+euro12196=euro19735month
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1026
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
10
RT = 2 [ 29(2+9)] = 3612 = 32727 Ω
Ind Machines
Ex 9
Why does the rotor of an induction motor turn slower than the revolving field
Ex 10
Describe briefly the key differences between a squirrel cage motor and wound-rotor motor
in terms of construction and method of operation
See lecture notes
Ex 11
Both the voltage and frequency induced in the rotor of an induction motor decrease as the
rotor speeds up Briefly explain the reason for this and include the relationships between
slip the applied stator angular velocity and rotor angular velocity
Ex 12A 3-phase 400V 6-pole 50Hz star-connected induction motor operates on full load at a speed of
1475 revmin Determine the slip s
Ex 13 a) Calculate the synchronous speed of a 3-phase 12-pole induction motor that is excited by a
50Hz source
b) What is the nominal speed (speed at full load) is the slip at full load is 5
9
9
9
2
2
2
a
b
c
RT
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1126
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
11
a) Synchronous speed rpmns = 120 f p revmin
500 revmin
(b) s = (ωs - ωr) ωs = (ns - nr) ns
nr = ns(1 - s) = 475 revmin
Ex 14
An 400 V 50 Hz six-pole star connected induction machine operates with full load slip
of 005 pu and has the following per-phase equivalent circuit parameters
stator resistance Rs = 012Ω stator leakage reactance Xs = 06 Ω and magnetising
reactance Xm = 2215 Ω The rotor referred resistance Rr` = 015 Ω and the rotor referred
leakage reactance Xr` = 0586 Ω The induction machine initially operates as a motoron full load Using an approximate per-phase equivalent circuit determine
(i) synchronous speed and the speed of the motor at rated slip in revmin and
radsec
(ii) The approximate equivalent circuit per-phase input impedance
(iii) The input current
(iv) The developed torque
(v) The starting torque
Soln
Motoring modeSynchronous speed ωs = 2 π fp = 1046 rads
ωm = ωs(1 ndash s) = 9937 rads = 9489 revmin
Rr ` s = 015005 = 3
Z = Rs + jXs+ Rr ` s +jXr
`)
= 3122 + j 1186
The total Z is Zin = ( jXm|| Z)
R`r s
j06ΩΩΩΩ
is i`r
Vs = 400 radic3 V
j0586ΩΩΩΩ 012ΩΩΩΩ
015005 j221515ΩΩΩΩ
A
B
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1226
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
12
= 7398ang110o 2354ang 8238
o
== 3143ang2762o
The input current
I i = 7318 ang-2762o A
The rotor current
I r = 6886 ang-208o A
6886 A at a pf of 0935 lagging
The developed torque
T = Pm ωm = ||3
r
2`r
s
R I
sω = 408 Nm
Starting Torque
T = ||3
r
2`r
s
R I
sω with s=1
= 204 Nm
Ex 15 A 4 pole induction motor running from a 3 phase 400V line to line 50 Hz
supply has the following parametersStator Winding Resistance (Rs) = 005Ω
Stator Leakage Inductance (Xs) = 01 Ω
Magnetising Inductance (Xm) = 20 Ω Rotor Winding Resistance referred to stator (Rrrsquo) = 01 Ω
Rotor Leakage Inductance referred to stator (Xrrsquo) = 02 Ω
Rotational losses may be assumed to be negligible
(i) Draw a per phase equivalent circuit of this machine
(ii) Calculate the rotor speed at which the machine will have a slip of 2
(iii) Calculate the Power over the gap (Pog) in this machine when it is running at a slip of 2
(iv) Calculate the mechanical power (Pm) and the torque (Tm) delivered by the machine
running at a slip of 2
Soln
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1326
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
13
Synchronous speed ωs = 2 π fp = 1500 revmin
ωm = ωs(1 ndash s) = 15386 rads = 146925 revmin
Rr` s = 01002 = 5
Z = Rs + jXs+ Rr ` s +jXr
`)
= 505 + j 03
= 5059ang34o
The rotor referred current
I r = 4546 ang-34o A
4546 A at a pf of 099 lagging
The air-gap power
Pg = ||3 r2`
r
s R I = 31kW
Prot = 0 ndash no rotational losses
Pm = Pg (1-s) = 3038 kW
T = 3038 103 15386
= 197 Nm
Ex 16 A 6 pole induction motor running from a 3 phase 400V line to line 50 Hz supply has
the following parameters
Stator Winding Resistance (Rs) = 015Ω
Stator Leakage Inductance (Xs) = 03 Ω
Magnetising Inductance (Xm) = 60 Ω
R`r s
j01ΩΩΩΩ
is i`r
Vs = 400 radic3 V
j02ΩΩΩΩ 005ΩΩΩΩ
01002 j20ΩΩΩΩ
A
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1426
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
14
Rotor Winding Resistance referred to stator (Rrrsquo) = 03 Ω
Rotor Leakage Inductance referred to stator (Xrrsquo) = 06 Ω
Rotational losses have previously been measured to be 1000W at 5 slip
bull Draw a per phase equivalent circuit of this machinebull Calculate the rotor speed at which the machine will have a slip of 5
bull Calculate the air-gap Power (Pg) in this machine when it is running at a slip of 5
bull Calculate the efficiency of the induction motor from electrical input to mechanical
output when the motor is running with a slip of 5
Soln(a)
(b) Synchronous speed ωs = 2 π fp = 1000 revmin
ωm = ωs(1 ndash s) = 9948 rads = 950 revmin
(c) Rr` s = 03005 = 6
Z = Rs + jXs+ Rr ` s +jXr
`)
= 615 + j 09
= 622ang832o
(d) The air-gap power
The rotor referred current
I r = 3698 ang-303o A
Pg = ||3
r
2`r
s
R I = 246 kW
(d) I i =inZ
3400
Zin = Xm || Z = j60 || 622ang832o
Zin = 6097 ang1409o
I = VZ = 3772 ang-1409oA
R`r s
j03ΩΩΩΩ
is i`r
Vs = 230V
j06ΩΩΩΩ 015ΩΩΩΩ
03005 j60ΩΩΩΩ
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1526
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
15
Pelec = radic3 VL IL cos φ = 25044 kW
Pm = Pg (1-s) ndash Prot = 2237 kW
η = Pelec Pm = 893s
Ex 17
A large 3-phase 4000V 8-pole 50Hz squirrel-cage induction motor draws a current of
380A and a total active power of 2344kW when operating at full load The speed at full load
is accurately measured and found to be 742rpm The stator is star-connected and the
resistance between two stator terminals is 01Ω The total iron losses are 234kW and the
windage and friction losses are 12kW
Calculate
a) The power factor at full loadb) The active power supplied to the rotor
c) The total i2R losses in the rotor
d) The mechanical power output to the load and torque delivered to the load
e) The motorrsquos net efficiency
a) The active power input to the motor is given as 2344kW The apparent power is
S = radic3 VL IL = 2633kVA
Therefore pf = PS 089 lagging
b) The active power supplied to the rotor is the active power input to the motor minus the ironlosses and copper losses in the stator The stator iron losses are given as 234kW To calculate
the copper losses note that the terminal resistance (01) is twice that of each phase for a wye
connection Therefore the copper losses are
Stator copper losses Pcu = 217kW
Therefore the active power supplied to the rotor is
Pr = Input power ndash stator copper loss ndash iron losses
Pr = 2344 kW ndash 234kW ndash 217kW
= 2298kW
c) The copper losses in the rotor are proportional to the slip (see lecture notes) Prcu sP
Therefore we must calculate the slip Neither the synchronous speed nor number of poles are
explicitly stated in the problem however by inspection we determine that the synchronous speed
must be 750rpm and the number of poles = 8 (For 50Hz supply synchronous speeds occur at
discrete intervals corresponding to pole pairs 2-pole 3000rpm 4-pole 1500rpm 6- pole
1000rpm 8-pole 750rpm etchellip We also know that induction motors operate near synchronous
speed ndash the speed is given as 742 rpm which is close to synchronous speed for an
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1626
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
16
8-pole machine)
Synchronous speed ωs = 2 π fp = 7854 rads
ωm = = 777 radss = 00107
Therefore the rotor copper losses are Pr cu = sPr = 2456 kw
(d)
The mechanical power (Pm) delivered to the load is the active power delivered to the rotor
minus rotor copper losses and windage and friction losses The windage and friction losses are
given as 12kW Therefore
Pm =22989 kW ndash 245kW-12 kW =22624 kW
kNmT L 129
60
2742
42262=
=
π
(e) the motors efficiency 5962344
42262===
kW
kW
P
P
in
out η
Ex 18
A 3-phase 440V 8-pole 50Hz star-connected induction motor has the following equivalent
circuit per-phase parameters
Rs = 01 Ω Rr = 015Ω xs = xr =06 Ω Rc = 100 Ω xm = 10 Ω
(a) Calculate using the equivalent circuit neglecting stator impedance at a slip of 5 the input
stator-current and power factor the rotor current referred to the stator the torque the
mechanical output power and the efficiency Calculate the starting torque Assume the
mechanical loss is 1kW for all speeds
Soln
(a) Synchronous speed ωs = 2 π fp 7854 rads
ωm = ωs(1 - s) = 7618 rads
Stator impedance neglected
Vs = 400 radic3 = 230
Ir` =
60050
10
230
j+ =
= 68 ang-1671 = 651 - j 1954
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1726
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
17
Io =230
100 + j230
10 = 23 ndash j 23
Is = Ir` + Io = 674 - j4294
= 799 A at cos φ = 084
Ir` = 68 A
T = 529 Nm
Pmech = 39365 kW
Pelect = 465 kW
Efficiency = 39365465 = 846
Ts = 39627 Nm
Ex19
Tests conducted at 50 Hz on a three-phase star connected 45 kW 2200V six-pole 50Hz squirrel
cage induction motor are listed below
The average dc resistance per stator phase Rs = 28 Ω (a) Briefly describe the tests required to determine the per-phase equivalent circuit
impedance parameters of an induction machine
(a) Determine the parameters of the per-phase equivalent circuit
(b) Determine the rotational losses of the machine Core losses can be neglected
load
0
No-loadV
R
I1
Ic
I
c X
Im
m
Figure 1 Equivalent Circuit during No-Load Test
torPower Fac0 times= I I C = 2025 A
22
0 cm I I I minus=
= 4019 A
Ω== 627c
c I
V R Ω== 5317
m
m I
V X
No Load Test Locked Rotor Test
Power Factor = 045 Power Factor = 09
Line Voltage = 22kV Line Voltage = 270V
Line Current = 45A Line Current = 25A
Input Power = 16kW Input Power = 9kW
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1826
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
18
For the locked rotor test slip = 1
Ω== 246s
s
LR I
V Z
Locked rotor reactance is
( )2
21
2
21 R R Z X X +minus=+
φ cos21 times=+= Z R R Rtotal = 5616Ω
R2rsquo = Rtot - R1 = 2816Ω
Ω=+ 72221 X X
(a) From the no-load test the no-load power isPNL = 16kW
The no-load rotational loss is
Prot = PNL ndash 3 I s2 Rs
= 14299W
Synch Machines
Ex 20
A 60 MVA 20 kV three-phase 50 Hz synchronous generator has a synchronous reactance
of 8 Ω per phase and negligible resistance The generator delivers rated power at a power
factor of 085 lagging at the rated terminal voltage to an infinite bus(a) Determine the excitation voltage per phase and the power angle δ
(b) With the excitation voltage Eg of the synchronous generator held constant at the
value determined in part (a) the input driving torque is reduced until the generator
delivers 28 MW Determine the armature current and the power factor
(c) If the generator operates at the excitation voltage of Eg obtained in part (a) what is
the steady state maximum power the machine can deliver before losing
jXsΩΩΩΩ jX`r ΩΩΩΩ Rs
R`r s
Is I`r
Vs ph
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1926
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
19
synchronism Determine the armature current corresponding to this maximum
power
(d) With real generated power at 42MW compute the power angle δ the stator
current Ia and power factor when the field excitation current If is adjusted so that
the excitation voltage Eg is decreased to 7919 of the value determined in part
(a)
Soln
(a) Apparent power S = 51 MW ndash j 316Mvar
Rated voltage per phase
V = 1154 ang 0deg kV
Rated current
Ia = S 3V
= 1733 ang -3178deg AExcitation voltage
Eg = 11540 + j9(1733) ang -3178deg)= 237922 ang 3386deg
Power angle = 3386o
(b) When delivering 28 MW
δ = sin-1
)5411)(79223(3
828 = 1578
o
P =
s
g
X
V E 3sinδ
Ia = 13069ang2967o j8
= 16336ang-6033o A
Power factor = cos(6033) = 0495 lagging
(c) The maximum power occurs at δ = 90o
Pmax =s
g
X
V E 3
= 102 MW
Ia = 324885ang25875o A
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2026
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
20
The power factor cos(25875) = 0899 leading
(d) Eg = 13463 V
The power angle δ = sin-1
)5411)(84318(3
842
= 31o
Ia = 86674ang0o A
-Power factor = 0
Ex 21 (Aug 13) A 3 phase synchronous machine with negligible stator winding resistance and a
synchronous reactance of 2 per phase is connected to a 400V 50 Hz 3-phase supply (infinite
bus) and operated as a generator It is currently generating 50kW with power factor 08 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)
(iv) What is the maximum power the machine can generate with this value of excitation
Soln(i)
(ii)P = 50kW
Apparent power S = 625 ang -3687
o
kVA
Rated voltage per phase V = 230ang 0deg VRated current
Ia = S 3V
= 9058ang -3687deg A
LoadGen
Xs =2
Eg Vt =400 radic3 V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2126
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
21
|Ia| = 9058 A
(iii)
Excitation voltage
Eg = 230 ang 0deg + j2(9058ang -3687deg)= 3684 ang 2318degV
|Eg| = 3684 V
Power angle = 2318o
(iv)
P = deg90sin2
23043683
(max occurs when δ = 90o)
= 1271 kW
Ex 22A 3 phase 4 pole synchronous machine with negligible stator winding resistance and a
synchronous reactance of 2 per phase is connected to a 480V 60 Hz 3phase supply
(infinite bus) and operated as a generator It is currently generating 80kW with power
factor 08 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)(iv) What is the maximum power the machine can generate with this value
excitation(v) What is the rotational speed of this generator
Soln (i)
(ii)P = 80kW Apparent power S 100 ang -3687
o kVA
Rated voltage per phase
V 277ang 0deg V
LoadGen
Xs =2
Eg Vt = 480V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2226
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
22
Rated current
Ia = S 3V
= 9058ang -3687deg A
|Ia| = 12028 A
(iii)
Excitation voltage
Eg = 277 ang 0deg + j2(12028ang -3687deg)
= 4632 ang 2455degV|Eg| = 4632 V
Power angle = 2455o
(iv)
P = deg90sin2
27724633
(max occurs when δ = 90o)
= 19238 kW
(v)Synchronous speed ωs = 2 π fp = 1885 rads
= 1800 revmin
Ex 23 (Jan 14)
A 3 phase synchronous machine with negligible stator winding resistance and a synchronous
reactance of 25 per phase is connected to a 20kV 50 Hz 3phase supply (infinite bus) and
operated as a generator It is currently generating 100MW with power factor 085 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)(iv) What minimum value of excitation voltage (E) would be required for this machine to
deliver 200MW into a 20kV bus assuming a 20deg stability margin must be maintained onpower angle (δ)
Soln
(i)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2326
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
23
(ii)
P = 100 MW Apparent power S = 100 085
= 11765 ang -3179o MVA
Rated voltage per phase
V= 11547ang 0deg VRated current
Ia = S 3V
= 3396 103ang -3179deg A
|Ia| = 3396 103A
(iii)
Excitation voltage
Eg = 11547 ang 0deg + j25(3396 103ang -3179deg)
= 17570 ang 2425degV|Eg| = 17570 V
Power angle = 2425o V
(iv)
P = )2090sin(52
3degminusdegt gV E
(max occurs when δ = 90o stability margin 20o)
200 MW = )2090sin(52
547113degminusdegg E
200 MW = 38130209397085613 gg E E =
|Eg| = 15360 V
LoadGen
Xs =25
Eg Vt =11547V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2426
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
24
Ex 24A manufacturing plant fed with a 400V 50Hz three phase runs from 0900 to 1800
Monday to Friday The factoryrsquos load is normally 80kVA with power factor 08 lagging
during these hours Once a month they need to run a special heating cycle which adds a
further 20kW to the factory load for a 2 hour period This heating cycle is timer
controlled and fully automated but the timer is currently set to run on the third Friday
afternoon of the month Outside of normal business hours and at weekends the factory
load is negligible The firm has a contractually agreed Maximum Import Capacity (MIC)
of 120kVA
bull Use the attached table of Low Voltage Maximum Demand charges to calculate the
firmrsquos electricity bill (exclusive of VAT) for the month of February (28 days)
bull Identify three steps that this firm could take to reduce their expenditure on
electricity assuming they stick with the same tariff Calculate the potential
savings
983116983151983159 983126983151983148983156983137983143983141 983117983137983160983145983149983157983149 983108983141983149983137983150983140 983107983144983137983154983143983141983155
983108983137983145983148983161 983123983156983137983150983140983145983150983143 983107983144983137983154983143983141 991404302983140983137983161
983120983123983119 983116983141983158983161
983120983123983119 983116983141983158983161 983152983141983154 983115983126983105 983151983142 983117983113983107 983152983141983154 983140983137983161 9914040036983097983147983126983105983140983137983161
983117983137983160983145983149983157983149 983113983149983152983151983154983156 983107983137983152983137983139983145983156983161 (983117983113983107) 983107983144983137983154983143983141983155
983108983137983145983148983161 983107983144983137983154983143983141 983152983141983154 983147983126983105 983151983142 983117983113983107 9914040072983147983126983105983140983137983161
983108983141983149983137983150983140 983107983144983137983154983143983141983155 983118983151983158983085983110983141983138 983117983137983154983085983119983139983156
983117983151983150983156983144983148983161 983107983144983137983154983143983141 983152983141983154 983147983127 983151983142 983149983141983156983141983154983141983140 983117983108 983145983150 983137
983149983151983150983156983144983148983161 983152983141983154983145983151983140 (983123983157983138983146983141983139983156 983156983151 983137 983149983145983150983145983149983157983149 983139983144983137983154983143983141 983151983142
30983147983127) 991404189830975983147983127983149983151983150983156983144 991404000
983108983137983161983156983145983149983141 983147983127983144 (983125983150983145983156) 983107983144983137983154983143983141 (0898309800 983085 19830979830985983097) 1480983139983147983127983144 1275983139983147983127983144
983118983145983143983144983156983145983149983141 983147983127983144 (983125983150983145983156) 983118983145983143983144983156 983107983144983137983154983143983141983155 (2098309800983085079830985983097) 64983097983139983147983127983144 64983097983139983147983127983144
983127983137983156983156983148983141983155983155 983125983150983145983156 983120983154983145983139983141 (983147983126983105983122983144 983145983150 983141983160983139983141983155983155 983151983142 983151983150983141 983156983144983145983154983140
983151983142 983156983151983156983137983148 983147983127983144) 991404000714 983147983126983105983122983144 991404000714 983147983126983105983122983144
bull Table Low Voltage Maximum Demand Charges (Q6)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2526
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
25
SolnNumber of days in February = 28
Number of working hours = 459 = 180
Normal kW = 80kVA08 = 64kW
Total kWh = 64kW180 hours+20kW2 hours = 11560 kWh
Normal kvar =80kVA sqrt(1-08^2) = 48kvar = sqrt(802-64
2) = 48 kVAr
Total kvarh = 48kvar180 hours = 8640 kvarh
Chargeable kvarh = 8640-(11560)3=4786 kvarh
983116983151983159 983126983151983148983156983137983143983141 983117983137983160983145983149983157983149 983108983141983149983137983150983140
983107983144983137983154983143983141983155 983152983141983154 983122983137983156983141 983121983156983161 983120983141983154983145983151983140 983107983144983137983154983143983141
983108983137983145983148983161 983123983156983137983150983140983145983150983143 983107983144983137983154983143983141 991404302 983140983137983161 991404302000 1 28 991404983096456
983120983123983119 983116983141983158983161
983120983123983119 983116983141983158983161 983152983141983154 983115983126983105 983152983141983154 983140983137983161 9914040036983097 983147983126983105983140983137983161 99140400369830970 120 28 991404123983097983096
983117983137983160983145983149983157983149 983113983149983152983151983154983156 983107983137983152983137983139983145983156983161
(983117983113983107) 983107983144983137983154983143983141983155
983108983137983145983148983161 983107983144983137983154983143983141 983152983141983154 983147983126983105 983151983142 983117983113983107 9914040072 983147983126983105983140983137983161 991404007200 120 28 9914042419830972
983108983141983149983137983150983140 983107983144983137983154983143983141983155 983118983151983158983085983110983141983138
983117983151983150983156983144983148983161 983107983144983137983154983143983141 983152983141983154 983147983127 983151983142
983149983141983156983141983154983141983140 983117983108 983145983150 983137 983149983151983150983156983144983148983161
983152983141983154983145983151983140 (983123983157983138983146983141983139983156 983156983151 983137
983149983145983150983145983149983157983149 983139983144983137983154983143983141 983151983142 30983147983127) 991404189830975
983147983127983149983151983150983156
983144 99140418983097500 84 1 991404159830971983096 983108983137983161983156983145983149983141 983147983127983144 (983125983150983145983156) 983107983144983137983154983143983141
(0898309800 983085 19830979830985983097) 9914040148 983147983127983144 991404014800 11560 1
991404198309510983096
983096
983118983145983143983144983156983145983149983141 983147983127983144 (983125983150983145983156) 983118983145983143983144983156
983107983144983137983154983143983141983155 (2098309800983085079830985983097) 9914040064983097 983147983127983144 99140400649830970
983127983137983156983156983148983141983155983155 983125983150983145983156 983120983154983145983139983141 (983147983126983105983122983144
983145983150 983141983160983139983141983155983155 983151983142 983151983150983141 983156983144983145983154983140 983151983142 983156983151983156983137983148
983147983127983144) 991404000714 983147983126983137983154983144 991404000714 4787 1 991404341983096
983124983151983156983137983148 983106983145983148983148
983142983151983154
983110983141983138983154983157983137983154983161
9914042354983095
0
[16 Marks]
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2626
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
26
Potential Savings
bull Use Power Factor correction to eliminate Watt less unit charge and to reduce MIC
bull Shift Heating load to night time or weekend
bull Negotiate lower MIC
Potential savings from these measures
Eliminate Wattless charge save euro3417 month
Shift heating cycle to night time
Max demand it lowered by 20kW = saving of 20euro1895 = euro3790 month
Saving by moving 20kW to night rate = 20kW2 hours (euro0148-euro00649)= euro332
Negotiating a reduced MIC
If power factor correction is employed and the heating cycle is shifted to night time the
then maximum VA of the plant drops to close to its maximum kW demand of 64kW An
MIC of 80kVA should be sufficient to cover this Potential Saving = 40kVA
(euro0072+euro00369)28 = euro12196 per month
Total potential saving = euro3417+euro3790+euro332+euro12196=euro19735month
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1126
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
11
a) Synchronous speed rpmns = 120 f p revmin
500 revmin
(b) s = (ωs - ωr) ωs = (ns - nr) ns
nr = ns(1 - s) = 475 revmin
Ex 14
An 400 V 50 Hz six-pole star connected induction machine operates with full load slip
of 005 pu and has the following per-phase equivalent circuit parameters
stator resistance Rs = 012Ω stator leakage reactance Xs = 06 Ω and magnetising
reactance Xm = 2215 Ω The rotor referred resistance Rr` = 015 Ω and the rotor referred
leakage reactance Xr` = 0586 Ω The induction machine initially operates as a motoron full load Using an approximate per-phase equivalent circuit determine
(i) synchronous speed and the speed of the motor at rated slip in revmin and
radsec
(ii) The approximate equivalent circuit per-phase input impedance
(iii) The input current
(iv) The developed torque
(v) The starting torque
Soln
Motoring modeSynchronous speed ωs = 2 π fp = 1046 rads
ωm = ωs(1 ndash s) = 9937 rads = 9489 revmin
Rr ` s = 015005 = 3
Z = Rs + jXs+ Rr ` s +jXr
`)
= 3122 + j 1186
The total Z is Zin = ( jXm|| Z)
R`r s
j06ΩΩΩΩ
is i`r
Vs = 400 radic3 V
j0586ΩΩΩΩ 012ΩΩΩΩ
015005 j221515ΩΩΩΩ
A
B
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1226
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
12
= 7398ang110o 2354ang 8238
o
== 3143ang2762o
The input current
I i = 7318 ang-2762o A
The rotor current
I r = 6886 ang-208o A
6886 A at a pf of 0935 lagging
The developed torque
T = Pm ωm = ||3
r
2`r
s
R I
sω = 408 Nm
Starting Torque
T = ||3
r
2`r
s
R I
sω with s=1
= 204 Nm
Ex 15 A 4 pole induction motor running from a 3 phase 400V line to line 50 Hz
supply has the following parametersStator Winding Resistance (Rs) = 005Ω
Stator Leakage Inductance (Xs) = 01 Ω
Magnetising Inductance (Xm) = 20 Ω Rotor Winding Resistance referred to stator (Rrrsquo) = 01 Ω
Rotor Leakage Inductance referred to stator (Xrrsquo) = 02 Ω
Rotational losses may be assumed to be negligible
(i) Draw a per phase equivalent circuit of this machine
(ii) Calculate the rotor speed at which the machine will have a slip of 2
(iii) Calculate the Power over the gap (Pog) in this machine when it is running at a slip of 2
(iv) Calculate the mechanical power (Pm) and the torque (Tm) delivered by the machine
running at a slip of 2
Soln
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1326
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
13
Synchronous speed ωs = 2 π fp = 1500 revmin
ωm = ωs(1 ndash s) = 15386 rads = 146925 revmin
Rr` s = 01002 = 5
Z = Rs + jXs+ Rr ` s +jXr
`)
= 505 + j 03
= 5059ang34o
The rotor referred current
I r = 4546 ang-34o A
4546 A at a pf of 099 lagging
The air-gap power
Pg = ||3 r2`
r
s R I = 31kW
Prot = 0 ndash no rotational losses
Pm = Pg (1-s) = 3038 kW
T = 3038 103 15386
= 197 Nm
Ex 16 A 6 pole induction motor running from a 3 phase 400V line to line 50 Hz supply has
the following parameters
Stator Winding Resistance (Rs) = 015Ω
Stator Leakage Inductance (Xs) = 03 Ω
Magnetising Inductance (Xm) = 60 Ω
R`r s
j01ΩΩΩΩ
is i`r
Vs = 400 radic3 V
j02ΩΩΩΩ 005ΩΩΩΩ
01002 j20ΩΩΩΩ
A
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1426
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
14
Rotor Winding Resistance referred to stator (Rrrsquo) = 03 Ω
Rotor Leakage Inductance referred to stator (Xrrsquo) = 06 Ω
Rotational losses have previously been measured to be 1000W at 5 slip
bull Draw a per phase equivalent circuit of this machinebull Calculate the rotor speed at which the machine will have a slip of 5
bull Calculate the air-gap Power (Pg) in this machine when it is running at a slip of 5
bull Calculate the efficiency of the induction motor from electrical input to mechanical
output when the motor is running with a slip of 5
Soln(a)
(b) Synchronous speed ωs = 2 π fp = 1000 revmin
ωm = ωs(1 ndash s) = 9948 rads = 950 revmin
(c) Rr` s = 03005 = 6
Z = Rs + jXs+ Rr ` s +jXr
`)
= 615 + j 09
= 622ang832o
(d) The air-gap power
The rotor referred current
I r = 3698 ang-303o A
Pg = ||3
r
2`r
s
R I = 246 kW
(d) I i =inZ
3400
Zin = Xm || Z = j60 || 622ang832o
Zin = 6097 ang1409o
I = VZ = 3772 ang-1409oA
R`r s
j03ΩΩΩΩ
is i`r
Vs = 230V
j06ΩΩΩΩ 015ΩΩΩΩ
03005 j60ΩΩΩΩ
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1526
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
15
Pelec = radic3 VL IL cos φ = 25044 kW
Pm = Pg (1-s) ndash Prot = 2237 kW
η = Pelec Pm = 893s
Ex 17
A large 3-phase 4000V 8-pole 50Hz squirrel-cage induction motor draws a current of
380A and a total active power of 2344kW when operating at full load The speed at full load
is accurately measured and found to be 742rpm The stator is star-connected and the
resistance between two stator terminals is 01Ω The total iron losses are 234kW and the
windage and friction losses are 12kW
Calculate
a) The power factor at full loadb) The active power supplied to the rotor
c) The total i2R losses in the rotor
d) The mechanical power output to the load and torque delivered to the load
e) The motorrsquos net efficiency
a) The active power input to the motor is given as 2344kW The apparent power is
S = radic3 VL IL = 2633kVA
Therefore pf = PS 089 lagging
b) The active power supplied to the rotor is the active power input to the motor minus the ironlosses and copper losses in the stator The stator iron losses are given as 234kW To calculate
the copper losses note that the terminal resistance (01) is twice that of each phase for a wye
connection Therefore the copper losses are
Stator copper losses Pcu = 217kW
Therefore the active power supplied to the rotor is
Pr = Input power ndash stator copper loss ndash iron losses
Pr = 2344 kW ndash 234kW ndash 217kW
= 2298kW
c) The copper losses in the rotor are proportional to the slip (see lecture notes) Prcu sP
Therefore we must calculate the slip Neither the synchronous speed nor number of poles are
explicitly stated in the problem however by inspection we determine that the synchronous speed
must be 750rpm and the number of poles = 8 (For 50Hz supply synchronous speeds occur at
discrete intervals corresponding to pole pairs 2-pole 3000rpm 4-pole 1500rpm 6- pole
1000rpm 8-pole 750rpm etchellip We also know that induction motors operate near synchronous
speed ndash the speed is given as 742 rpm which is close to synchronous speed for an
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1626
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
16
8-pole machine)
Synchronous speed ωs = 2 π fp = 7854 rads
ωm = = 777 radss = 00107
Therefore the rotor copper losses are Pr cu = sPr = 2456 kw
(d)
The mechanical power (Pm) delivered to the load is the active power delivered to the rotor
minus rotor copper losses and windage and friction losses The windage and friction losses are
given as 12kW Therefore
Pm =22989 kW ndash 245kW-12 kW =22624 kW
kNmT L 129
60
2742
42262=
=
π
(e) the motors efficiency 5962344
42262===
kW
kW
P
P
in
out η
Ex 18
A 3-phase 440V 8-pole 50Hz star-connected induction motor has the following equivalent
circuit per-phase parameters
Rs = 01 Ω Rr = 015Ω xs = xr =06 Ω Rc = 100 Ω xm = 10 Ω
(a) Calculate using the equivalent circuit neglecting stator impedance at a slip of 5 the input
stator-current and power factor the rotor current referred to the stator the torque the
mechanical output power and the efficiency Calculate the starting torque Assume the
mechanical loss is 1kW for all speeds
Soln
(a) Synchronous speed ωs = 2 π fp 7854 rads
ωm = ωs(1 - s) = 7618 rads
Stator impedance neglected
Vs = 400 radic3 = 230
Ir` =
60050
10
230
j+ =
= 68 ang-1671 = 651 - j 1954
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1726
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
17
Io =230
100 + j230
10 = 23 ndash j 23
Is = Ir` + Io = 674 - j4294
= 799 A at cos φ = 084
Ir` = 68 A
T = 529 Nm
Pmech = 39365 kW
Pelect = 465 kW
Efficiency = 39365465 = 846
Ts = 39627 Nm
Ex19
Tests conducted at 50 Hz on a three-phase star connected 45 kW 2200V six-pole 50Hz squirrel
cage induction motor are listed below
The average dc resistance per stator phase Rs = 28 Ω (a) Briefly describe the tests required to determine the per-phase equivalent circuit
impedance parameters of an induction machine
(a) Determine the parameters of the per-phase equivalent circuit
(b) Determine the rotational losses of the machine Core losses can be neglected
load
0
No-loadV
R
I1
Ic
I
c X
Im
m
Figure 1 Equivalent Circuit during No-Load Test
torPower Fac0 times= I I C = 2025 A
22
0 cm I I I minus=
= 4019 A
Ω== 627c
c I
V R Ω== 5317
m
m I
V X
No Load Test Locked Rotor Test
Power Factor = 045 Power Factor = 09
Line Voltage = 22kV Line Voltage = 270V
Line Current = 45A Line Current = 25A
Input Power = 16kW Input Power = 9kW
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1826
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
18
For the locked rotor test slip = 1
Ω== 246s
s
LR I
V Z
Locked rotor reactance is
( )2
21
2
21 R R Z X X +minus=+
φ cos21 times=+= Z R R Rtotal = 5616Ω
R2rsquo = Rtot - R1 = 2816Ω
Ω=+ 72221 X X
(a) From the no-load test the no-load power isPNL = 16kW
The no-load rotational loss is
Prot = PNL ndash 3 I s2 Rs
= 14299W
Synch Machines
Ex 20
A 60 MVA 20 kV three-phase 50 Hz synchronous generator has a synchronous reactance
of 8 Ω per phase and negligible resistance The generator delivers rated power at a power
factor of 085 lagging at the rated terminal voltage to an infinite bus(a) Determine the excitation voltage per phase and the power angle δ
(b) With the excitation voltage Eg of the synchronous generator held constant at the
value determined in part (a) the input driving torque is reduced until the generator
delivers 28 MW Determine the armature current and the power factor
(c) If the generator operates at the excitation voltage of Eg obtained in part (a) what is
the steady state maximum power the machine can deliver before losing
jXsΩΩΩΩ jX`r ΩΩΩΩ Rs
R`r s
Is I`r
Vs ph
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1926
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
19
synchronism Determine the armature current corresponding to this maximum
power
(d) With real generated power at 42MW compute the power angle δ the stator
current Ia and power factor when the field excitation current If is adjusted so that
the excitation voltage Eg is decreased to 7919 of the value determined in part
(a)
Soln
(a) Apparent power S = 51 MW ndash j 316Mvar
Rated voltage per phase
V = 1154 ang 0deg kV
Rated current
Ia = S 3V
= 1733 ang -3178deg AExcitation voltage
Eg = 11540 + j9(1733) ang -3178deg)= 237922 ang 3386deg
Power angle = 3386o
(b) When delivering 28 MW
δ = sin-1
)5411)(79223(3
828 = 1578
o
P =
s
g
X
V E 3sinδ
Ia = 13069ang2967o j8
= 16336ang-6033o A
Power factor = cos(6033) = 0495 lagging
(c) The maximum power occurs at δ = 90o
Pmax =s
g
X
V E 3
= 102 MW
Ia = 324885ang25875o A
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2026
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
20
The power factor cos(25875) = 0899 leading
(d) Eg = 13463 V
The power angle δ = sin-1
)5411)(84318(3
842
= 31o
Ia = 86674ang0o A
-Power factor = 0
Ex 21 (Aug 13) A 3 phase synchronous machine with negligible stator winding resistance and a
synchronous reactance of 2 per phase is connected to a 400V 50 Hz 3-phase supply (infinite
bus) and operated as a generator It is currently generating 50kW with power factor 08 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)
(iv) What is the maximum power the machine can generate with this value of excitation
Soln(i)
(ii)P = 50kW
Apparent power S = 625 ang -3687
o
kVA
Rated voltage per phase V = 230ang 0deg VRated current
Ia = S 3V
= 9058ang -3687deg A
LoadGen
Xs =2
Eg Vt =400 radic3 V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2126
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
21
|Ia| = 9058 A
(iii)
Excitation voltage
Eg = 230 ang 0deg + j2(9058ang -3687deg)= 3684 ang 2318degV
|Eg| = 3684 V
Power angle = 2318o
(iv)
P = deg90sin2
23043683
(max occurs when δ = 90o)
= 1271 kW
Ex 22A 3 phase 4 pole synchronous machine with negligible stator winding resistance and a
synchronous reactance of 2 per phase is connected to a 480V 60 Hz 3phase supply
(infinite bus) and operated as a generator It is currently generating 80kW with power
factor 08 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)(iv) What is the maximum power the machine can generate with this value
excitation(v) What is the rotational speed of this generator
Soln (i)
(ii)P = 80kW Apparent power S 100 ang -3687
o kVA
Rated voltage per phase
V 277ang 0deg V
LoadGen
Xs =2
Eg Vt = 480V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2226
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
22
Rated current
Ia = S 3V
= 9058ang -3687deg A
|Ia| = 12028 A
(iii)
Excitation voltage
Eg = 277 ang 0deg + j2(12028ang -3687deg)
= 4632 ang 2455degV|Eg| = 4632 V
Power angle = 2455o
(iv)
P = deg90sin2
27724633
(max occurs when δ = 90o)
= 19238 kW
(v)Synchronous speed ωs = 2 π fp = 1885 rads
= 1800 revmin
Ex 23 (Jan 14)
A 3 phase synchronous machine with negligible stator winding resistance and a synchronous
reactance of 25 per phase is connected to a 20kV 50 Hz 3phase supply (infinite bus) and
operated as a generator It is currently generating 100MW with power factor 085 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)(iv) What minimum value of excitation voltage (E) would be required for this machine to
deliver 200MW into a 20kV bus assuming a 20deg stability margin must be maintained onpower angle (δ)
Soln
(i)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2326
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
23
(ii)
P = 100 MW Apparent power S = 100 085
= 11765 ang -3179o MVA
Rated voltage per phase
V= 11547ang 0deg VRated current
Ia = S 3V
= 3396 103ang -3179deg A
|Ia| = 3396 103A
(iii)
Excitation voltage
Eg = 11547 ang 0deg + j25(3396 103ang -3179deg)
= 17570 ang 2425degV|Eg| = 17570 V
Power angle = 2425o V
(iv)
P = )2090sin(52
3degminusdegt gV E
(max occurs when δ = 90o stability margin 20o)
200 MW = )2090sin(52
547113degminusdegg E
200 MW = 38130209397085613 gg E E =
|Eg| = 15360 V
LoadGen
Xs =25
Eg Vt =11547V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2426
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
24
Ex 24A manufacturing plant fed with a 400V 50Hz three phase runs from 0900 to 1800
Monday to Friday The factoryrsquos load is normally 80kVA with power factor 08 lagging
during these hours Once a month they need to run a special heating cycle which adds a
further 20kW to the factory load for a 2 hour period This heating cycle is timer
controlled and fully automated but the timer is currently set to run on the third Friday
afternoon of the month Outside of normal business hours and at weekends the factory
load is negligible The firm has a contractually agreed Maximum Import Capacity (MIC)
of 120kVA
bull Use the attached table of Low Voltage Maximum Demand charges to calculate the
firmrsquos electricity bill (exclusive of VAT) for the month of February (28 days)
bull Identify three steps that this firm could take to reduce their expenditure on
electricity assuming they stick with the same tariff Calculate the potential
savings
983116983151983159 983126983151983148983156983137983143983141 983117983137983160983145983149983157983149 983108983141983149983137983150983140 983107983144983137983154983143983141983155
983108983137983145983148983161 983123983156983137983150983140983145983150983143 983107983144983137983154983143983141 991404302983140983137983161
983120983123983119 983116983141983158983161
983120983123983119 983116983141983158983161 983152983141983154 983115983126983105 983151983142 983117983113983107 983152983141983154 983140983137983161 9914040036983097983147983126983105983140983137983161
983117983137983160983145983149983157983149 983113983149983152983151983154983156 983107983137983152983137983139983145983156983161 (983117983113983107) 983107983144983137983154983143983141983155
983108983137983145983148983161 983107983144983137983154983143983141 983152983141983154 983147983126983105 983151983142 983117983113983107 9914040072983147983126983105983140983137983161
983108983141983149983137983150983140 983107983144983137983154983143983141983155 983118983151983158983085983110983141983138 983117983137983154983085983119983139983156
983117983151983150983156983144983148983161 983107983144983137983154983143983141 983152983141983154 983147983127 983151983142 983149983141983156983141983154983141983140 983117983108 983145983150 983137
983149983151983150983156983144983148983161 983152983141983154983145983151983140 (983123983157983138983146983141983139983156 983156983151 983137 983149983145983150983145983149983157983149 983139983144983137983154983143983141 983151983142
30983147983127) 991404189830975983147983127983149983151983150983156983144 991404000
983108983137983161983156983145983149983141 983147983127983144 (983125983150983145983156) 983107983144983137983154983143983141 (0898309800 983085 19830979830985983097) 1480983139983147983127983144 1275983139983147983127983144
983118983145983143983144983156983145983149983141 983147983127983144 (983125983150983145983156) 983118983145983143983144983156 983107983144983137983154983143983141983155 (2098309800983085079830985983097) 64983097983139983147983127983144 64983097983139983147983127983144
983127983137983156983156983148983141983155983155 983125983150983145983156 983120983154983145983139983141 (983147983126983105983122983144 983145983150 983141983160983139983141983155983155 983151983142 983151983150983141 983156983144983145983154983140
983151983142 983156983151983156983137983148 983147983127983144) 991404000714 983147983126983105983122983144 991404000714 983147983126983105983122983144
bull Table Low Voltage Maximum Demand Charges (Q6)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2526
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
25
SolnNumber of days in February = 28
Number of working hours = 459 = 180
Normal kW = 80kVA08 = 64kW
Total kWh = 64kW180 hours+20kW2 hours = 11560 kWh
Normal kvar =80kVA sqrt(1-08^2) = 48kvar = sqrt(802-64
2) = 48 kVAr
Total kvarh = 48kvar180 hours = 8640 kvarh
Chargeable kvarh = 8640-(11560)3=4786 kvarh
983116983151983159 983126983151983148983156983137983143983141 983117983137983160983145983149983157983149 983108983141983149983137983150983140
983107983144983137983154983143983141983155 983152983141983154 983122983137983156983141 983121983156983161 983120983141983154983145983151983140 983107983144983137983154983143983141
983108983137983145983148983161 983123983156983137983150983140983145983150983143 983107983144983137983154983143983141 991404302 983140983137983161 991404302000 1 28 991404983096456
983120983123983119 983116983141983158983161
983120983123983119 983116983141983158983161 983152983141983154 983115983126983105 983152983141983154 983140983137983161 9914040036983097 983147983126983105983140983137983161 99140400369830970 120 28 991404123983097983096
983117983137983160983145983149983157983149 983113983149983152983151983154983156 983107983137983152983137983139983145983156983161
(983117983113983107) 983107983144983137983154983143983141983155
983108983137983145983148983161 983107983144983137983154983143983141 983152983141983154 983147983126983105 983151983142 983117983113983107 9914040072 983147983126983105983140983137983161 991404007200 120 28 9914042419830972
983108983141983149983137983150983140 983107983144983137983154983143983141983155 983118983151983158983085983110983141983138
983117983151983150983156983144983148983161 983107983144983137983154983143983141 983152983141983154 983147983127 983151983142
983149983141983156983141983154983141983140 983117983108 983145983150 983137 983149983151983150983156983144983148983161
983152983141983154983145983151983140 (983123983157983138983146983141983139983156 983156983151 983137
983149983145983150983145983149983157983149 983139983144983137983154983143983141 983151983142 30983147983127) 991404189830975
983147983127983149983151983150983156
983144 99140418983097500 84 1 991404159830971983096 983108983137983161983156983145983149983141 983147983127983144 (983125983150983145983156) 983107983144983137983154983143983141
(0898309800 983085 19830979830985983097) 9914040148 983147983127983144 991404014800 11560 1
991404198309510983096
983096
983118983145983143983144983156983145983149983141 983147983127983144 (983125983150983145983156) 983118983145983143983144983156
983107983144983137983154983143983141983155 (2098309800983085079830985983097) 9914040064983097 983147983127983144 99140400649830970
983127983137983156983156983148983141983155983155 983125983150983145983156 983120983154983145983139983141 (983147983126983105983122983144
983145983150 983141983160983139983141983155983155 983151983142 983151983150983141 983156983144983145983154983140 983151983142 983156983151983156983137983148
983147983127983144) 991404000714 983147983126983137983154983144 991404000714 4787 1 991404341983096
983124983151983156983137983148 983106983145983148983148
983142983151983154
983110983141983138983154983157983137983154983161
9914042354983095
0
[16 Marks]
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2626
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
26
Potential Savings
bull Use Power Factor correction to eliminate Watt less unit charge and to reduce MIC
bull Shift Heating load to night time or weekend
bull Negotiate lower MIC
Potential savings from these measures
Eliminate Wattless charge save euro3417 month
Shift heating cycle to night time
Max demand it lowered by 20kW = saving of 20euro1895 = euro3790 month
Saving by moving 20kW to night rate = 20kW2 hours (euro0148-euro00649)= euro332
Negotiating a reduced MIC
If power factor correction is employed and the heating cycle is shifted to night time the
then maximum VA of the plant drops to close to its maximum kW demand of 64kW An
MIC of 80kVA should be sufficient to cover this Potential Saving = 40kVA
(euro0072+euro00369)28 = euro12196 per month
Total potential saving = euro3417+euro3790+euro332+euro12196=euro19735month
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1226
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
12
= 7398ang110o 2354ang 8238
o
== 3143ang2762o
The input current
I i = 7318 ang-2762o A
The rotor current
I r = 6886 ang-208o A
6886 A at a pf of 0935 lagging
The developed torque
T = Pm ωm = ||3
r
2`r
s
R I
sω = 408 Nm
Starting Torque
T = ||3
r
2`r
s
R I
sω with s=1
= 204 Nm
Ex 15 A 4 pole induction motor running from a 3 phase 400V line to line 50 Hz
supply has the following parametersStator Winding Resistance (Rs) = 005Ω
Stator Leakage Inductance (Xs) = 01 Ω
Magnetising Inductance (Xm) = 20 Ω Rotor Winding Resistance referred to stator (Rrrsquo) = 01 Ω
Rotor Leakage Inductance referred to stator (Xrrsquo) = 02 Ω
Rotational losses may be assumed to be negligible
(i) Draw a per phase equivalent circuit of this machine
(ii) Calculate the rotor speed at which the machine will have a slip of 2
(iii) Calculate the Power over the gap (Pog) in this machine when it is running at a slip of 2
(iv) Calculate the mechanical power (Pm) and the torque (Tm) delivered by the machine
running at a slip of 2
Soln
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1326
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
13
Synchronous speed ωs = 2 π fp = 1500 revmin
ωm = ωs(1 ndash s) = 15386 rads = 146925 revmin
Rr` s = 01002 = 5
Z = Rs + jXs+ Rr ` s +jXr
`)
= 505 + j 03
= 5059ang34o
The rotor referred current
I r = 4546 ang-34o A
4546 A at a pf of 099 lagging
The air-gap power
Pg = ||3 r2`
r
s R I = 31kW
Prot = 0 ndash no rotational losses
Pm = Pg (1-s) = 3038 kW
T = 3038 103 15386
= 197 Nm
Ex 16 A 6 pole induction motor running from a 3 phase 400V line to line 50 Hz supply has
the following parameters
Stator Winding Resistance (Rs) = 015Ω
Stator Leakage Inductance (Xs) = 03 Ω
Magnetising Inductance (Xm) = 60 Ω
R`r s
j01ΩΩΩΩ
is i`r
Vs = 400 radic3 V
j02ΩΩΩΩ 005ΩΩΩΩ
01002 j20ΩΩΩΩ
A
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1426
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
14
Rotor Winding Resistance referred to stator (Rrrsquo) = 03 Ω
Rotor Leakage Inductance referred to stator (Xrrsquo) = 06 Ω
Rotational losses have previously been measured to be 1000W at 5 slip
bull Draw a per phase equivalent circuit of this machinebull Calculate the rotor speed at which the machine will have a slip of 5
bull Calculate the air-gap Power (Pg) in this machine when it is running at a slip of 5
bull Calculate the efficiency of the induction motor from electrical input to mechanical
output when the motor is running with a slip of 5
Soln(a)
(b) Synchronous speed ωs = 2 π fp = 1000 revmin
ωm = ωs(1 ndash s) = 9948 rads = 950 revmin
(c) Rr` s = 03005 = 6
Z = Rs + jXs+ Rr ` s +jXr
`)
= 615 + j 09
= 622ang832o
(d) The air-gap power
The rotor referred current
I r = 3698 ang-303o A
Pg = ||3
r
2`r
s
R I = 246 kW
(d) I i =inZ
3400
Zin = Xm || Z = j60 || 622ang832o
Zin = 6097 ang1409o
I = VZ = 3772 ang-1409oA
R`r s
j03ΩΩΩΩ
is i`r
Vs = 230V
j06ΩΩΩΩ 015ΩΩΩΩ
03005 j60ΩΩΩΩ
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1526
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
15
Pelec = radic3 VL IL cos φ = 25044 kW
Pm = Pg (1-s) ndash Prot = 2237 kW
η = Pelec Pm = 893s
Ex 17
A large 3-phase 4000V 8-pole 50Hz squirrel-cage induction motor draws a current of
380A and a total active power of 2344kW when operating at full load The speed at full load
is accurately measured and found to be 742rpm The stator is star-connected and the
resistance between two stator terminals is 01Ω The total iron losses are 234kW and the
windage and friction losses are 12kW
Calculate
a) The power factor at full loadb) The active power supplied to the rotor
c) The total i2R losses in the rotor
d) The mechanical power output to the load and torque delivered to the load
e) The motorrsquos net efficiency
a) The active power input to the motor is given as 2344kW The apparent power is
S = radic3 VL IL = 2633kVA
Therefore pf = PS 089 lagging
b) The active power supplied to the rotor is the active power input to the motor minus the ironlosses and copper losses in the stator The stator iron losses are given as 234kW To calculate
the copper losses note that the terminal resistance (01) is twice that of each phase for a wye
connection Therefore the copper losses are
Stator copper losses Pcu = 217kW
Therefore the active power supplied to the rotor is
Pr = Input power ndash stator copper loss ndash iron losses
Pr = 2344 kW ndash 234kW ndash 217kW
= 2298kW
c) The copper losses in the rotor are proportional to the slip (see lecture notes) Prcu sP
Therefore we must calculate the slip Neither the synchronous speed nor number of poles are
explicitly stated in the problem however by inspection we determine that the synchronous speed
must be 750rpm and the number of poles = 8 (For 50Hz supply synchronous speeds occur at
discrete intervals corresponding to pole pairs 2-pole 3000rpm 4-pole 1500rpm 6- pole
1000rpm 8-pole 750rpm etchellip We also know that induction motors operate near synchronous
speed ndash the speed is given as 742 rpm which is close to synchronous speed for an
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1626
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
16
8-pole machine)
Synchronous speed ωs = 2 π fp = 7854 rads
ωm = = 777 radss = 00107
Therefore the rotor copper losses are Pr cu = sPr = 2456 kw
(d)
The mechanical power (Pm) delivered to the load is the active power delivered to the rotor
minus rotor copper losses and windage and friction losses The windage and friction losses are
given as 12kW Therefore
Pm =22989 kW ndash 245kW-12 kW =22624 kW
kNmT L 129
60
2742
42262=
=
π
(e) the motors efficiency 5962344
42262===
kW
kW
P
P
in
out η
Ex 18
A 3-phase 440V 8-pole 50Hz star-connected induction motor has the following equivalent
circuit per-phase parameters
Rs = 01 Ω Rr = 015Ω xs = xr =06 Ω Rc = 100 Ω xm = 10 Ω
(a) Calculate using the equivalent circuit neglecting stator impedance at a slip of 5 the input
stator-current and power factor the rotor current referred to the stator the torque the
mechanical output power and the efficiency Calculate the starting torque Assume the
mechanical loss is 1kW for all speeds
Soln
(a) Synchronous speed ωs = 2 π fp 7854 rads
ωm = ωs(1 - s) = 7618 rads
Stator impedance neglected
Vs = 400 radic3 = 230
Ir` =
60050
10
230
j+ =
= 68 ang-1671 = 651 - j 1954
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1726
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
17
Io =230
100 + j230
10 = 23 ndash j 23
Is = Ir` + Io = 674 - j4294
= 799 A at cos φ = 084
Ir` = 68 A
T = 529 Nm
Pmech = 39365 kW
Pelect = 465 kW
Efficiency = 39365465 = 846
Ts = 39627 Nm
Ex19
Tests conducted at 50 Hz on a three-phase star connected 45 kW 2200V six-pole 50Hz squirrel
cage induction motor are listed below
The average dc resistance per stator phase Rs = 28 Ω (a) Briefly describe the tests required to determine the per-phase equivalent circuit
impedance parameters of an induction machine
(a) Determine the parameters of the per-phase equivalent circuit
(b) Determine the rotational losses of the machine Core losses can be neglected
load
0
No-loadV
R
I1
Ic
I
c X
Im
m
Figure 1 Equivalent Circuit during No-Load Test
torPower Fac0 times= I I C = 2025 A
22
0 cm I I I minus=
= 4019 A
Ω== 627c
c I
V R Ω== 5317
m
m I
V X
No Load Test Locked Rotor Test
Power Factor = 045 Power Factor = 09
Line Voltage = 22kV Line Voltage = 270V
Line Current = 45A Line Current = 25A
Input Power = 16kW Input Power = 9kW
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1826
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
18
For the locked rotor test slip = 1
Ω== 246s
s
LR I
V Z
Locked rotor reactance is
( )2
21
2
21 R R Z X X +minus=+
φ cos21 times=+= Z R R Rtotal = 5616Ω
R2rsquo = Rtot - R1 = 2816Ω
Ω=+ 72221 X X
(a) From the no-load test the no-load power isPNL = 16kW
The no-load rotational loss is
Prot = PNL ndash 3 I s2 Rs
= 14299W
Synch Machines
Ex 20
A 60 MVA 20 kV three-phase 50 Hz synchronous generator has a synchronous reactance
of 8 Ω per phase and negligible resistance The generator delivers rated power at a power
factor of 085 lagging at the rated terminal voltage to an infinite bus(a) Determine the excitation voltage per phase and the power angle δ
(b) With the excitation voltage Eg of the synchronous generator held constant at the
value determined in part (a) the input driving torque is reduced until the generator
delivers 28 MW Determine the armature current and the power factor
(c) If the generator operates at the excitation voltage of Eg obtained in part (a) what is
the steady state maximum power the machine can deliver before losing
jXsΩΩΩΩ jX`r ΩΩΩΩ Rs
R`r s
Is I`r
Vs ph
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1926
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
19
synchronism Determine the armature current corresponding to this maximum
power
(d) With real generated power at 42MW compute the power angle δ the stator
current Ia and power factor when the field excitation current If is adjusted so that
the excitation voltage Eg is decreased to 7919 of the value determined in part
(a)
Soln
(a) Apparent power S = 51 MW ndash j 316Mvar
Rated voltage per phase
V = 1154 ang 0deg kV
Rated current
Ia = S 3V
= 1733 ang -3178deg AExcitation voltage
Eg = 11540 + j9(1733) ang -3178deg)= 237922 ang 3386deg
Power angle = 3386o
(b) When delivering 28 MW
δ = sin-1
)5411)(79223(3
828 = 1578
o
P =
s
g
X
V E 3sinδ
Ia = 13069ang2967o j8
= 16336ang-6033o A
Power factor = cos(6033) = 0495 lagging
(c) The maximum power occurs at δ = 90o
Pmax =s
g
X
V E 3
= 102 MW
Ia = 324885ang25875o A
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2026
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
20
The power factor cos(25875) = 0899 leading
(d) Eg = 13463 V
The power angle δ = sin-1
)5411)(84318(3
842
= 31o
Ia = 86674ang0o A
-Power factor = 0
Ex 21 (Aug 13) A 3 phase synchronous machine with negligible stator winding resistance and a
synchronous reactance of 2 per phase is connected to a 400V 50 Hz 3-phase supply (infinite
bus) and operated as a generator It is currently generating 50kW with power factor 08 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)
(iv) What is the maximum power the machine can generate with this value of excitation
Soln(i)
(ii)P = 50kW
Apparent power S = 625 ang -3687
o
kVA
Rated voltage per phase V = 230ang 0deg VRated current
Ia = S 3V
= 9058ang -3687deg A
LoadGen
Xs =2
Eg Vt =400 radic3 V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2126
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
21
|Ia| = 9058 A
(iii)
Excitation voltage
Eg = 230 ang 0deg + j2(9058ang -3687deg)= 3684 ang 2318degV
|Eg| = 3684 V
Power angle = 2318o
(iv)
P = deg90sin2
23043683
(max occurs when δ = 90o)
= 1271 kW
Ex 22A 3 phase 4 pole synchronous machine with negligible stator winding resistance and a
synchronous reactance of 2 per phase is connected to a 480V 60 Hz 3phase supply
(infinite bus) and operated as a generator It is currently generating 80kW with power
factor 08 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)(iv) What is the maximum power the machine can generate with this value
excitation(v) What is the rotational speed of this generator
Soln (i)
(ii)P = 80kW Apparent power S 100 ang -3687
o kVA
Rated voltage per phase
V 277ang 0deg V
LoadGen
Xs =2
Eg Vt = 480V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2226
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
22
Rated current
Ia = S 3V
= 9058ang -3687deg A
|Ia| = 12028 A
(iii)
Excitation voltage
Eg = 277 ang 0deg + j2(12028ang -3687deg)
= 4632 ang 2455degV|Eg| = 4632 V
Power angle = 2455o
(iv)
P = deg90sin2
27724633
(max occurs when δ = 90o)
= 19238 kW
(v)Synchronous speed ωs = 2 π fp = 1885 rads
= 1800 revmin
Ex 23 (Jan 14)
A 3 phase synchronous machine with negligible stator winding resistance and a synchronous
reactance of 25 per phase is connected to a 20kV 50 Hz 3phase supply (infinite bus) and
operated as a generator It is currently generating 100MW with power factor 085 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)(iv) What minimum value of excitation voltage (E) would be required for this machine to
deliver 200MW into a 20kV bus assuming a 20deg stability margin must be maintained onpower angle (δ)
Soln
(i)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2326
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
23
(ii)
P = 100 MW Apparent power S = 100 085
= 11765 ang -3179o MVA
Rated voltage per phase
V= 11547ang 0deg VRated current
Ia = S 3V
= 3396 103ang -3179deg A
|Ia| = 3396 103A
(iii)
Excitation voltage
Eg = 11547 ang 0deg + j25(3396 103ang -3179deg)
= 17570 ang 2425degV|Eg| = 17570 V
Power angle = 2425o V
(iv)
P = )2090sin(52
3degminusdegt gV E
(max occurs when δ = 90o stability margin 20o)
200 MW = )2090sin(52
547113degminusdegg E
200 MW = 38130209397085613 gg E E =
|Eg| = 15360 V
LoadGen
Xs =25
Eg Vt =11547V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2426
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
24
Ex 24A manufacturing plant fed with a 400V 50Hz three phase runs from 0900 to 1800
Monday to Friday The factoryrsquos load is normally 80kVA with power factor 08 lagging
during these hours Once a month they need to run a special heating cycle which adds a
further 20kW to the factory load for a 2 hour period This heating cycle is timer
controlled and fully automated but the timer is currently set to run on the third Friday
afternoon of the month Outside of normal business hours and at weekends the factory
load is negligible The firm has a contractually agreed Maximum Import Capacity (MIC)
of 120kVA
bull Use the attached table of Low Voltage Maximum Demand charges to calculate the
firmrsquos electricity bill (exclusive of VAT) for the month of February (28 days)
bull Identify three steps that this firm could take to reduce their expenditure on
electricity assuming they stick with the same tariff Calculate the potential
savings
983116983151983159 983126983151983148983156983137983143983141 983117983137983160983145983149983157983149 983108983141983149983137983150983140 983107983144983137983154983143983141983155
983108983137983145983148983161 983123983156983137983150983140983145983150983143 983107983144983137983154983143983141 991404302983140983137983161
983120983123983119 983116983141983158983161
983120983123983119 983116983141983158983161 983152983141983154 983115983126983105 983151983142 983117983113983107 983152983141983154 983140983137983161 9914040036983097983147983126983105983140983137983161
983117983137983160983145983149983157983149 983113983149983152983151983154983156 983107983137983152983137983139983145983156983161 (983117983113983107) 983107983144983137983154983143983141983155
983108983137983145983148983161 983107983144983137983154983143983141 983152983141983154 983147983126983105 983151983142 983117983113983107 9914040072983147983126983105983140983137983161
983108983141983149983137983150983140 983107983144983137983154983143983141983155 983118983151983158983085983110983141983138 983117983137983154983085983119983139983156
983117983151983150983156983144983148983161 983107983144983137983154983143983141 983152983141983154 983147983127 983151983142 983149983141983156983141983154983141983140 983117983108 983145983150 983137
983149983151983150983156983144983148983161 983152983141983154983145983151983140 (983123983157983138983146983141983139983156 983156983151 983137 983149983145983150983145983149983157983149 983139983144983137983154983143983141 983151983142
30983147983127) 991404189830975983147983127983149983151983150983156983144 991404000
983108983137983161983156983145983149983141 983147983127983144 (983125983150983145983156) 983107983144983137983154983143983141 (0898309800 983085 19830979830985983097) 1480983139983147983127983144 1275983139983147983127983144
983118983145983143983144983156983145983149983141 983147983127983144 (983125983150983145983156) 983118983145983143983144983156 983107983144983137983154983143983141983155 (2098309800983085079830985983097) 64983097983139983147983127983144 64983097983139983147983127983144
983127983137983156983156983148983141983155983155 983125983150983145983156 983120983154983145983139983141 (983147983126983105983122983144 983145983150 983141983160983139983141983155983155 983151983142 983151983150983141 983156983144983145983154983140
983151983142 983156983151983156983137983148 983147983127983144) 991404000714 983147983126983105983122983144 991404000714 983147983126983105983122983144
bull Table Low Voltage Maximum Demand Charges (Q6)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2526
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
25
SolnNumber of days in February = 28
Number of working hours = 459 = 180
Normal kW = 80kVA08 = 64kW
Total kWh = 64kW180 hours+20kW2 hours = 11560 kWh
Normal kvar =80kVA sqrt(1-08^2) = 48kvar = sqrt(802-64
2) = 48 kVAr
Total kvarh = 48kvar180 hours = 8640 kvarh
Chargeable kvarh = 8640-(11560)3=4786 kvarh
983116983151983159 983126983151983148983156983137983143983141 983117983137983160983145983149983157983149 983108983141983149983137983150983140
983107983144983137983154983143983141983155 983152983141983154 983122983137983156983141 983121983156983161 983120983141983154983145983151983140 983107983144983137983154983143983141
983108983137983145983148983161 983123983156983137983150983140983145983150983143 983107983144983137983154983143983141 991404302 983140983137983161 991404302000 1 28 991404983096456
983120983123983119 983116983141983158983161
983120983123983119 983116983141983158983161 983152983141983154 983115983126983105 983152983141983154 983140983137983161 9914040036983097 983147983126983105983140983137983161 99140400369830970 120 28 991404123983097983096
983117983137983160983145983149983157983149 983113983149983152983151983154983156 983107983137983152983137983139983145983156983161
(983117983113983107) 983107983144983137983154983143983141983155
983108983137983145983148983161 983107983144983137983154983143983141 983152983141983154 983147983126983105 983151983142 983117983113983107 9914040072 983147983126983105983140983137983161 991404007200 120 28 9914042419830972
983108983141983149983137983150983140 983107983144983137983154983143983141983155 983118983151983158983085983110983141983138
983117983151983150983156983144983148983161 983107983144983137983154983143983141 983152983141983154 983147983127 983151983142
983149983141983156983141983154983141983140 983117983108 983145983150 983137 983149983151983150983156983144983148983161
983152983141983154983145983151983140 (983123983157983138983146983141983139983156 983156983151 983137
983149983145983150983145983149983157983149 983139983144983137983154983143983141 983151983142 30983147983127) 991404189830975
983147983127983149983151983150983156
983144 99140418983097500 84 1 991404159830971983096 983108983137983161983156983145983149983141 983147983127983144 (983125983150983145983156) 983107983144983137983154983143983141
(0898309800 983085 19830979830985983097) 9914040148 983147983127983144 991404014800 11560 1
991404198309510983096
983096
983118983145983143983144983156983145983149983141 983147983127983144 (983125983150983145983156) 983118983145983143983144983156
983107983144983137983154983143983141983155 (2098309800983085079830985983097) 9914040064983097 983147983127983144 99140400649830970
983127983137983156983156983148983141983155983155 983125983150983145983156 983120983154983145983139983141 (983147983126983105983122983144
983145983150 983141983160983139983141983155983155 983151983142 983151983150983141 983156983144983145983154983140 983151983142 983156983151983156983137983148
983147983127983144) 991404000714 983147983126983137983154983144 991404000714 4787 1 991404341983096
983124983151983156983137983148 983106983145983148983148
983142983151983154
983110983141983138983154983157983137983154983161
9914042354983095
0
[16 Marks]
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2626
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
26
Potential Savings
bull Use Power Factor correction to eliminate Watt less unit charge and to reduce MIC
bull Shift Heating load to night time or weekend
bull Negotiate lower MIC
Potential savings from these measures
Eliminate Wattless charge save euro3417 month
Shift heating cycle to night time
Max demand it lowered by 20kW = saving of 20euro1895 = euro3790 month
Saving by moving 20kW to night rate = 20kW2 hours (euro0148-euro00649)= euro332
Negotiating a reduced MIC
If power factor correction is employed and the heating cycle is shifted to night time the
then maximum VA of the plant drops to close to its maximum kW demand of 64kW An
MIC of 80kVA should be sufficient to cover this Potential Saving = 40kVA
(euro0072+euro00369)28 = euro12196 per month
Total potential saving = euro3417+euro3790+euro332+euro12196=euro19735month
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1326
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
13
Synchronous speed ωs = 2 π fp = 1500 revmin
ωm = ωs(1 ndash s) = 15386 rads = 146925 revmin
Rr` s = 01002 = 5
Z = Rs + jXs+ Rr ` s +jXr
`)
= 505 + j 03
= 5059ang34o
The rotor referred current
I r = 4546 ang-34o A
4546 A at a pf of 099 lagging
The air-gap power
Pg = ||3 r2`
r
s R I = 31kW
Prot = 0 ndash no rotational losses
Pm = Pg (1-s) = 3038 kW
T = 3038 103 15386
= 197 Nm
Ex 16 A 6 pole induction motor running from a 3 phase 400V line to line 50 Hz supply has
the following parameters
Stator Winding Resistance (Rs) = 015Ω
Stator Leakage Inductance (Xs) = 03 Ω
Magnetising Inductance (Xm) = 60 Ω
R`r s
j01ΩΩΩΩ
is i`r
Vs = 400 radic3 V
j02ΩΩΩΩ 005ΩΩΩΩ
01002 j20ΩΩΩΩ
A
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1426
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
14
Rotor Winding Resistance referred to stator (Rrrsquo) = 03 Ω
Rotor Leakage Inductance referred to stator (Xrrsquo) = 06 Ω
Rotational losses have previously been measured to be 1000W at 5 slip
bull Draw a per phase equivalent circuit of this machinebull Calculate the rotor speed at which the machine will have a slip of 5
bull Calculate the air-gap Power (Pg) in this machine when it is running at a slip of 5
bull Calculate the efficiency of the induction motor from electrical input to mechanical
output when the motor is running with a slip of 5
Soln(a)
(b) Synchronous speed ωs = 2 π fp = 1000 revmin
ωm = ωs(1 ndash s) = 9948 rads = 950 revmin
(c) Rr` s = 03005 = 6
Z = Rs + jXs+ Rr ` s +jXr
`)
= 615 + j 09
= 622ang832o
(d) The air-gap power
The rotor referred current
I r = 3698 ang-303o A
Pg = ||3
r
2`r
s
R I = 246 kW
(d) I i =inZ
3400
Zin = Xm || Z = j60 || 622ang832o
Zin = 6097 ang1409o
I = VZ = 3772 ang-1409oA
R`r s
j03ΩΩΩΩ
is i`r
Vs = 230V
j06ΩΩΩΩ 015ΩΩΩΩ
03005 j60ΩΩΩΩ
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1526
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
15
Pelec = radic3 VL IL cos φ = 25044 kW
Pm = Pg (1-s) ndash Prot = 2237 kW
η = Pelec Pm = 893s
Ex 17
A large 3-phase 4000V 8-pole 50Hz squirrel-cage induction motor draws a current of
380A and a total active power of 2344kW when operating at full load The speed at full load
is accurately measured and found to be 742rpm The stator is star-connected and the
resistance between two stator terminals is 01Ω The total iron losses are 234kW and the
windage and friction losses are 12kW
Calculate
a) The power factor at full loadb) The active power supplied to the rotor
c) The total i2R losses in the rotor
d) The mechanical power output to the load and torque delivered to the load
e) The motorrsquos net efficiency
a) The active power input to the motor is given as 2344kW The apparent power is
S = radic3 VL IL = 2633kVA
Therefore pf = PS 089 lagging
b) The active power supplied to the rotor is the active power input to the motor minus the ironlosses and copper losses in the stator The stator iron losses are given as 234kW To calculate
the copper losses note that the terminal resistance (01) is twice that of each phase for a wye
connection Therefore the copper losses are
Stator copper losses Pcu = 217kW
Therefore the active power supplied to the rotor is
Pr = Input power ndash stator copper loss ndash iron losses
Pr = 2344 kW ndash 234kW ndash 217kW
= 2298kW
c) The copper losses in the rotor are proportional to the slip (see lecture notes) Prcu sP
Therefore we must calculate the slip Neither the synchronous speed nor number of poles are
explicitly stated in the problem however by inspection we determine that the synchronous speed
must be 750rpm and the number of poles = 8 (For 50Hz supply synchronous speeds occur at
discrete intervals corresponding to pole pairs 2-pole 3000rpm 4-pole 1500rpm 6- pole
1000rpm 8-pole 750rpm etchellip We also know that induction motors operate near synchronous
speed ndash the speed is given as 742 rpm which is close to synchronous speed for an
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1626
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
16
8-pole machine)
Synchronous speed ωs = 2 π fp = 7854 rads
ωm = = 777 radss = 00107
Therefore the rotor copper losses are Pr cu = sPr = 2456 kw
(d)
The mechanical power (Pm) delivered to the load is the active power delivered to the rotor
minus rotor copper losses and windage and friction losses The windage and friction losses are
given as 12kW Therefore
Pm =22989 kW ndash 245kW-12 kW =22624 kW
kNmT L 129
60
2742
42262=
=
π
(e) the motors efficiency 5962344
42262===
kW
kW
P
P
in
out η
Ex 18
A 3-phase 440V 8-pole 50Hz star-connected induction motor has the following equivalent
circuit per-phase parameters
Rs = 01 Ω Rr = 015Ω xs = xr =06 Ω Rc = 100 Ω xm = 10 Ω
(a) Calculate using the equivalent circuit neglecting stator impedance at a slip of 5 the input
stator-current and power factor the rotor current referred to the stator the torque the
mechanical output power and the efficiency Calculate the starting torque Assume the
mechanical loss is 1kW for all speeds
Soln
(a) Synchronous speed ωs = 2 π fp 7854 rads
ωm = ωs(1 - s) = 7618 rads
Stator impedance neglected
Vs = 400 radic3 = 230
Ir` =
60050
10
230
j+ =
= 68 ang-1671 = 651 - j 1954
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1726
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
17
Io =230
100 + j230
10 = 23 ndash j 23
Is = Ir` + Io = 674 - j4294
= 799 A at cos φ = 084
Ir` = 68 A
T = 529 Nm
Pmech = 39365 kW
Pelect = 465 kW
Efficiency = 39365465 = 846
Ts = 39627 Nm
Ex19
Tests conducted at 50 Hz on a three-phase star connected 45 kW 2200V six-pole 50Hz squirrel
cage induction motor are listed below
The average dc resistance per stator phase Rs = 28 Ω (a) Briefly describe the tests required to determine the per-phase equivalent circuit
impedance parameters of an induction machine
(a) Determine the parameters of the per-phase equivalent circuit
(b) Determine the rotational losses of the machine Core losses can be neglected
load
0
No-loadV
R
I1
Ic
I
c X
Im
m
Figure 1 Equivalent Circuit during No-Load Test
torPower Fac0 times= I I C = 2025 A
22
0 cm I I I minus=
= 4019 A
Ω== 627c
c I
V R Ω== 5317
m
m I
V X
No Load Test Locked Rotor Test
Power Factor = 045 Power Factor = 09
Line Voltage = 22kV Line Voltage = 270V
Line Current = 45A Line Current = 25A
Input Power = 16kW Input Power = 9kW
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1826
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
18
For the locked rotor test slip = 1
Ω== 246s
s
LR I
V Z
Locked rotor reactance is
( )2
21
2
21 R R Z X X +minus=+
φ cos21 times=+= Z R R Rtotal = 5616Ω
R2rsquo = Rtot - R1 = 2816Ω
Ω=+ 72221 X X
(a) From the no-load test the no-load power isPNL = 16kW
The no-load rotational loss is
Prot = PNL ndash 3 I s2 Rs
= 14299W
Synch Machines
Ex 20
A 60 MVA 20 kV three-phase 50 Hz synchronous generator has a synchronous reactance
of 8 Ω per phase and negligible resistance The generator delivers rated power at a power
factor of 085 lagging at the rated terminal voltage to an infinite bus(a) Determine the excitation voltage per phase and the power angle δ
(b) With the excitation voltage Eg of the synchronous generator held constant at the
value determined in part (a) the input driving torque is reduced until the generator
delivers 28 MW Determine the armature current and the power factor
(c) If the generator operates at the excitation voltage of Eg obtained in part (a) what is
the steady state maximum power the machine can deliver before losing
jXsΩΩΩΩ jX`r ΩΩΩΩ Rs
R`r s
Is I`r
Vs ph
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1926
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
19
synchronism Determine the armature current corresponding to this maximum
power
(d) With real generated power at 42MW compute the power angle δ the stator
current Ia and power factor when the field excitation current If is adjusted so that
the excitation voltage Eg is decreased to 7919 of the value determined in part
(a)
Soln
(a) Apparent power S = 51 MW ndash j 316Mvar
Rated voltage per phase
V = 1154 ang 0deg kV
Rated current
Ia = S 3V
= 1733 ang -3178deg AExcitation voltage
Eg = 11540 + j9(1733) ang -3178deg)= 237922 ang 3386deg
Power angle = 3386o
(b) When delivering 28 MW
δ = sin-1
)5411)(79223(3
828 = 1578
o
P =
s
g
X
V E 3sinδ
Ia = 13069ang2967o j8
= 16336ang-6033o A
Power factor = cos(6033) = 0495 lagging
(c) The maximum power occurs at δ = 90o
Pmax =s
g
X
V E 3
= 102 MW
Ia = 324885ang25875o A
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2026
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
20
The power factor cos(25875) = 0899 leading
(d) Eg = 13463 V
The power angle δ = sin-1
)5411)(84318(3
842
= 31o
Ia = 86674ang0o A
-Power factor = 0
Ex 21 (Aug 13) A 3 phase synchronous machine with negligible stator winding resistance and a
synchronous reactance of 2 per phase is connected to a 400V 50 Hz 3-phase supply (infinite
bus) and operated as a generator It is currently generating 50kW with power factor 08 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)
(iv) What is the maximum power the machine can generate with this value of excitation
Soln(i)
(ii)P = 50kW
Apparent power S = 625 ang -3687
o
kVA
Rated voltage per phase V = 230ang 0deg VRated current
Ia = S 3V
= 9058ang -3687deg A
LoadGen
Xs =2
Eg Vt =400 radic3 V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2126
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
21
|Ia| = 9058 A
(iii)
Excitation voltage
Eg = 230 ang 0deg + j2(9058ang -3687deg)= 3684 ang 2318degV
|Eg| = 3684 V
Power angle = 2318o
(iv)
P = deg90sin2
23043683
(max occurs when δ = 90o)
= 1271 kW
Ex 22A 3 phase 4 pole synchronous machine with negligible stator winding resistance and a
synchronous reactance of 2 per phase is connected to a 480V 60 Hz 3phase supply
(infinite bus) and operated as a generator It is currently generating 80kW with power
factor 08 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)(iv) What is the maximum power the machine can generate with this value
excitation(v) What is the rotational speed of this generator
Soln (i)
(ii)P = 80kW Apparent power S 100 ang -3687
o kVA
Rated voltage per phase
V 277ang 0deg V
LoadGen
Xs =2
Eg Vt = 480V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2226
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
22
Rated current
Ia = S 3V
= 9058ang -3687deg A
|Ia| = 12028 A
(iii)
Excitation voltage
Eg = 277 ang 0deg + j2(12028ang -3687deg)
= 4632 ang 2455degV|Eg| = 4632 V
Power angle = 2455o
(iv)
P = deg90sin2
27724633
(max occurs when δ = 90o)
= 19238 kW
(v)Synchronous speed ωs = 2 π fp = 1885 rads
= 1800 revmin
Ex 23 (Jan 14)
A 3 phase synchronous machine with negligible stator winding resistance and a synchronous
reactance of 25 per phase is connected to a 20kV 50 Hz 3phase supply (infinite bus) and
operated as a generator It is currently generating 100MW with power factor 085 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)(iv) What minimum value of excitation voltage (E) would be required for this machine to
deliver 200MW into a 20kV bus assuming a 20deg stability margin must be maintained onpower angle (δ)
Soln
(i)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2326
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
23
(ii)
P = 100 MW Apparent power S = 100 085
= 11765 ang -3179o MVA
Rated voltage per phase
V= 11547ang 0deg VRated current
Ia = S 3V
= 3396 103ang -3179deg A
|Ia| = 3396 103A
(iii)
Excitation voltage
Eg = 11547 ang 0deg + j25(3396 103ang -3179deg)
= 17570 ang 2425degV|Eg| = 17570 V
Power angle = 2425o V
(iv)
P = )2090sin(52
3degminusdegt gV E
(max occurs when δ = 90o stability margin 20o)
200 MW = )2090sin(52
547113degminusdegg E
200 MW = 38130209397085613 gg E E =
|Eg| = 15360 V
LoadGen
Xs =25
Eg Vt =11547V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2426
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
24
Ex 24A manufacturing plant fed with a 400V 50Hz three phase runs from 0900 to 1800
Monday to Friday The factoryrsquos load is normally 80kVA with power factor 08 lagging
during these hours Once a month they need to run a special heating cycle which adds a
further 20kW to the factory load for a 2 hour period This heating cycle is timer
controlled and fully automated but the timer is currently set to run on the third Friday
afternoon of the month Outside of normal business hours and at weekends the factory
load is negligible The firm has a contractually agreed Maximum Import Capacity (MIC)
of 120kVA
bull Use the attached table of Low Voltage Maximum Demand charges to calculate the
firmrsquos electricity bill (exclusive of VAT) for the month of February (28 days)
bull Identify three steps that this firm could take to reduce their expenditure on
electricity assuming they stick with the same tariff Calculate the potential
savings
983116983151983159 983126983151983148983156983137983143983141 983117983137983160983145983149983157983149 983108983141983149983137983150983140 983107983144983137983154983143983141983155
983108983137983145983148983161 983123983156983137983150983140983145983150983143 983107983144983137983154983143983141 991404302983140983137983161
983120983123983119 983116983141983158983161
983120983123983119 983116983141983158983161 983152983141983154 983115983126983105 983151983142 983117983113983107 983152983141983154 983140983137983161 9914040036983097983147983126983105983140983137983161
983117983137983160983145983149983157983149 983113983149983152983151983154983156 983107983137983152983137983139983145983156983161 (983117983113983107) 983107983144983137983154983143983141983155
983108983137983145983148983161 983107983144983137983154983143983141 983152983141983154 983147983126983105 983151983142 983117983113983107 9914040072983147983126983105983140983137983161
983108983141983149983137983150983140 983107983144983137983154983143983141983155 983118983151983158983085983110983141983138 983117983137983154983085983119983139983156
983117983151983150983156983144983148983161 983107983144983137983154983143983141 983152983141983154 983147983127 983151983142 983149983141983156983141983154983141983140 983117983108 983145983150 983137
983149983151983150983156983144983148983161 983152983141983154983145983151983140 (983123983157983138983146983141983139983156 983156983151 983137 983149983145983150983145983149983157983149 983139983144983137983154983143983141 983151983142
30983147983127) 991404189830975983147983127983149983151983150983156983144 991404000
983108983137983161983156983145983149983141 983147983127983144 (983125983150983145983156) 983107983144983137983154983143983141 (0898309800 983085 19830979830985983097) 1480983139983147983127983144 1275983139983147983127983144
983118983145983143983144983156983145983149983141 983147983127983144 (983125983150983145983156) 983118983145983143983144983156 983107983144983137983154983143983141983155 (2098309800983085079830985983097) 64983097983139983147983127983144 64983097983139983147983127983144
983127983137983156983156983148983141983155983155 983125983150983145983156 983120983154983145983139983141 (983147983126983105983122983144 983145983150 983141983160983139983141983155983155 983151983142 983151983150983141 983156983144983145983154983140
983151983142 983156983151983156983137983148 983147983127983144) 991404000714 983147983126983105983122983144 991404000714 983147983126983105983122983144
bull Table Low Voltage Maximum Demand Charges (Q6)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2526
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
25
SolnNumber of days in February = 28
Number of working hours = 459 = 180
Normal kW = 80kVA08 = 64kW
Total kWh = 64kW180 hours+20kW2 hours = 11560 kWh
Normal kvar =80kVA sqrt(1-08^2) = 48kvar = sqrt(802-64
2) = 48 kVAr
Total kvarh = 48kvar180 hours = 8640 kvarh
Chargeable kvarh = 8640-(11560)3=4786 kvarh
983116983151983159 983126983151983148983156983137983143983141 983117983137983160983145983149983157983149 983108983141983149983137983150983140
983107983144983137983154983143983141983155 983152983141983154 983122983137983156983141 983121983156983161 983120983141983154983145983151983140 983107983144983137983154983143983141
983108983137983145983148983161 983123983156983137983150983140983145983150983143 983107983144983137983154983143983141 991404302 983140983137983161 991404302000 1 28 991404983096456
983120983123983119 983116983141983158983161
983120983123983119 983116983141983158983161 983152983141983154 983115983126983105 983152983141983154 983140983137983161 9914040036983097 983147983126983105983140983137983161 99140400369830970 120 28 991404123983097983096
983117983137983160983145983149983157983149 983113983149983152983151983154983156 983107983137983152983137983139983145983156983161
(983117983113983107) 983107983144983137983154983143983141983155
983108983137983145983148983161 983107983144983137983154983143983141 983152983141983154 983147983126983105 983151983142 983117983113983107 9914040072 983147983126983105983140983137983161 991404007200 120 28 9914042419830972
983108983141983149983137983150983140 983107983144983137983154983143983141983155 983118983151983158983085983110983141983138
983117983151983150983156983144983148983161 983107983144983137983154983143983141 983152983141983154 983147983127 983151983142
983149983141983156983141983154983141983140 983117983108 983145983150 983137 983149983151983150983156983144983148983161
983152983141983154983145983151983140 (983123983157983138983146983141983139983156 983156983151 983137
983149983145983150983145983149983157983149 983139983144983137983154983143983141 983151983142 30983147983127) 991404189830975
983147983127983149983151983150983156
983144 99140418983097500 84 1 991404159830971983096 983108983137983161983156983145983149983141 983147983127983144 (983125983150983145983156) 983107983144983137983154983143983141
(0898309800 983085 19830979830985983097) 9914040148 983147983127983144 991404014800 11560 1
991404198309510983096
983096
983118983145983143983144983156983145983149983141 983147983127983144 (983125983150983145983156) 983118983145983143983144983156
983107983144983137983154983143983141983155 (2098309800983085079830985983097) 9914040064983097 983147983127983144 99140400649830970
983127983137983156983156983148983141983155983155 983125983150983145983156 983120983154983145983139983141 (983147983126983105983122983144
983145983150 983141983160983139983141983155983155 983151983142 983151983150983141 983156983144983145983154983140 983151983142 983156983151983156983137983148
983147983127983144) 991404000714 983147983126983137983154983144 991404000714 4787 1 991404341983096
983124983151983156983137983148 983106983145983148983148
983142983151983154
983110983141983138983154983157983137983154983161
9914042354983095
0
[16 Marks]
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2626
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
26
Potential Savings
bull Use Power Factor correction to eliminate Watt less unit charge and to reduce MIC
bull Shift Heating load to night time or weekend
bull Negotiate lower MIC
Potential savings from these measures
Eliminate Wattless charge save euro3417 month
Shift heating cycle to night time
Max demand it lowered by 20kW = saving of 20euro1895 = euro3790 month
Saving by moving 20kW to night rate = 20kW2 hours (euro0148-euro00649)= euro332
Negotiating a reduced MIC
If power factor correction is employed and the heating cycle is shifted to night time the
then maximum VA of the plant drops to close to its maximum kW demand of 64kW An
MIC of 80kVA should be sufficient to cover this Potential Saving = 40kVA
(euro0072+euro00369)28 = euro12196 per month
Total potential saving = euro3417+euro3790+euro332+euro12196=euro19735month
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1426
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
14
Rotor Winding Resistance referred to stator (Rrrsquo) = 03 Ω
Rotor Leakage Inductance referred to stator (Xrrsquo) = 06 Ω
Rotational losses have previously been measured to be 1000W at 5 slip
bull Draw a per phase equivalent circuit of this machinebull Calculate the rotor speed at which the machine will have a slip of 5
bull Calculate the air-gap Power (Pg) in this machine when it is running at a slip of 5
bull Calculate the efficiency of the induction motor from electrical input to mechanical
output when the motor is running with a slip of 5
Soln(a)
(b) Synchronous speed ωs = 2 π fp = 1000 revmin
ωm = ωs(1 ndash s) = 9948 rads = 950 revmin
(c) Rr` s = 03005 = 6
Z = Rs + jXs+ Rr ` s +jXr
`)
= 615 + j 09
= 622ang832o
(d) The air-gap power
The rotor referred current
I r = 3698 ang-303o A
Pg = ||3
r
2`r
s
R I = 246 kW
(d) I i =inZ
3400
Zin = Xm || Z = j60 || 622ang832o
Zin = 6097 ang1409o
I = VZ = 3772 ang-1409oA
R`r s
j03ΩΩΩΩ
is i`r
Vs = 230V
j06ΩΩΩΩ 015ΩΩΩΩ
03005 j60ΩΩΩΩ
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1526
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
15
Pelec = radic3 VL IL cos φ = 25044 kW
Pm = Pg (1-s) ndash Prot = 2237 kW
η = Pelec Pm = 893s
Ex 17
A large 3-phase 4000V 8-pole 50Hz squirrel-cage induction motor draws a current of
380A and a total active power of 2344kW when operating at full load The speed at full load
is accurately measured and found to be 742rpm The stator is star-connected and the
resistance between two stator terminals is 01Ω The total iron losses are 234kW and the
windage and friction losses are 12kW
Calculate
a) The power factor at full loadb) The active power supplied to the rotor
c) The total i2R losses in the rotor
d) The mechanical power output to the load and torque delivered to the load
e) The motorrsquos net efficiency
a) The active power input to the motor is given as 2344kW The apparent power is
S = radic3 VL IL = 2633kVA
Therefore pf = PS 089 lagging
b) The active power supplied to the rotor is the active power input to the motor minus the ironlosses and copper losses in the stator The stator iron losses are given as 234kW To calculate
the copper losses note that the terminal resistance (01) is twice that of each phase for a wye
connection Therefore the copper losses are
Stator copper losses Pcu = 217kW
Therefore the active power supplied to the rotor is
Pr = Input power ndash stator copper loss ndash iron losses
Pr = 2344 kW ndash 234kW ndash 217kW
= 2298kW
c) The copper losses in the rotor are proportional to the slip (see lecture notes) Prcu sP
Therefore we must calculate the slip Neither the synchronous speed nor number of poles are
explicitly stated in the problem however by inspection we determine that the synchronous speed
must be 750rpm and the number of poles = 8 (For 50Hz supply synchronous speeds occur at
discrete intervals corresponding to pole pairs 2-pole 3000rpm 4-pole 1500rpm 6- pole
1000rpm 8-pole 750rpm etchellip We also know that induction motors operate near synchronous
speed ndash the speed is given as 742 rpm which is close to synchronous speed for an
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1626
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
16
8-pole machine)
Synchronous speed ωs = 2 π fp = 7854 rads
ωm = = 777 radss = 00107
Therefore the rotor copper losses are Pr cu = sPr = 2456 kw
(d)
The mechanical power (Pm) delivered to the load is the active power delivered to the rotor
minus rotor copper losses and windage and friction losses The windage and friction losses are
given as 12kW Therefore
Pm =22989 kW ndash 245kW-12 kW =22624 kW
kNmT L 129
60
2742
42262=
=
π
(e) the motors efficiency 5962344
42262===
kW
kW
P
P
in
out η
Ex 18
A 3-phase 440V 8-pole 50Hz star-connected induction motor has the following equivalent
circuit per-phase parameters
Rs = 01 Ω Rr = 015Ω xs = xr =06 Ω Rc = 100 Ω xm = 10 Ω
(a) Calculate using the equivalent circuit neglecting stator impedance at a slip of 5 the input
stator-current and power factor the rotor current referred to the stator the torque the
mechanical output power and the efficiency Calculate the starting torque Assume the
mechanical loss is 1kW for all speeds
Soln
(a) Synchronous speed ωs = 2 π fp 7854 rads
ωm = ωs(1 - s) = 7618 rads
Stator impedance neglected
Vs = 400 radic3 = 230
Ir` =
60050
10
230
j+ =
= 68 ang-1671 = 651 - j 1954
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1726
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
17
Io =230
100 + j230
10 = 23 ndash j 23
Is = Ir` + Io = 674 - j4294
= 799 A at cos φ = 084
Ir` = 68 A
T = 529 Nm
Pmech = 39365 kW
Pelect = 465 kW
Efficiency = 39365465 = 846
Ts = 39627 Nm
Ex19
Tests conducted at 50 Hz on a three-phase star connected 45 kW 2200V six-pole 50Hz squirrel
cage induction motor are listed below
The average dc resistance per stator phase Rs = 28 Ω (a) Briefly describe the tests required to determine the per-phase equivalent circuit
impedance parameters of an induction machine
(a) Determine the parameters of the per-phase equivalent circuit
(b) Determine the rotational losses of the machine Core losses can be neglected
load
0
No-loadV
R
I1
Ic
I
c X
Im
m
Figure 1 Equivalent Circuit during No-Load Test
torPower Fac0 times= I I C = 2025 A
22
0 cm I I I minus=
= 4019 A
Ω== 627c
c I
V R Ω== 5317
m
m I
V X
No Load Test Locked Rotor Test
Power Factor = 045 Power Factor = 09
Line Voltage = 22kV Line Voltage = 270V
Line Current = 45A Line Current = 25A
Input Power = 16kW Input Power = 9kW
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1826
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
18
For the locked rotor test slip = 1
Ω== 246s
s
LR I
V Z
Locked rotor reactance is
( )2
21
2
21 R R Z X X +minus=+
φ cos21 times=+= Z R R Rtotal = 5616Ω
R2rsquo = Rtot - R1 = 2816Ω
Ω=+ 72221 X X
(a) From the no-load test the no-load power isPNL = 16kW
The no-load rotational loss is
Prot = PNL ndash 3 I s2 Rs
= 14299W
Synch Machines
Ex 20
A 60 MVA 20 kV three-phase 50 Hz synchronous generator has a synchronous reactance
of 8 Ω per phase and negligible resistance The generator delivers rated power at a power
factor of 085 lagging at the rated terminal voltage to an infinite bus(a) Determine the excitation voltage per phase and the power angle δ
(b) With the excitation voltage Eg of the synchronous generator held constant at the
value determined in part (a) the input driving torque is reduced until the generator
delivers 28 MW Determine the armature current and the power factor
(c) If the generator operates at the excitation voltage of Eg obtained in part (a) what is
the steady state maximum power the machine can deliver before losing
jXsΩΩΩΩ jX`r ΩΩΩΩ Rs
R`r s
Is I`r
Vs ph
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1926
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
19
synchronism Determine the armature current corresponding to this maximum
power
(d) With real generated power at 42MW compute the power angle δ the stator
current Ia and power factor when the field excitation current If is adjusted so that
the excitation voltage Eg is decreased to 7919 of the value determined in part
(a)
Soln
(a) Apparent power S = 51 MW ndash j 316Mvar
Rated voltage per phase
V = 1154 ang 0deg kV
Rated current
Ia = S 3V
= 1733 ang -3178deg AExcitation voltage
Eg = 11540 + j9(1733) ang -3178deg)= 237922 ang 3386deg
Power angle = 3386o
(b) When delivering 28 MW
δ = sin-1
)5411)(79223(3
828 = 1578
o
P =
s
g
X
V E 3sinδ
Ia = 13069ang2967o j8
= 16336ang-6033o A
Power factor = cos(6033) = 0495 lagging
(c) The maximum power occurs at δ = 90o
Pmax =s
g
X
V E 3
= 102 MW
Ia = 324885ang25875o A
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2026
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
20
The power factor cos(25875) = 0899 leading
(d) Eg = 13463 V
The power angle δ = sin-1
)5411)(84318(3
842
= 31o
Ia = 86674ang0o A
-Power factor = 0
Ex 21 (Aug 13) A 3 phase synchronous machine with negligible stator winding resistance and a
synchronous reactance of 2 per phase is connected to a 400V 50 Hz 3-phase supply (infinite
bus) and operated as a generator It is currently generating 50kW with power factor 08 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)
(iv) What is the maximum power the machine can generate with this value of excitation
Soln(i)
(ii)P = 50kW
Apparent power S = 625 ang -3687
o
kVA
Rated voltage per phase V = 230ang 0deg VRated current
Ia = S 3V
= 9058ang -3687deg A
LoadGen
Xs =2
Eg Vt =400 radic3 V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2126
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
21
|Ia| = 9058 A
(iii)
Excitation voltage
Eg = 230 ang 0deg + j2(9058ang -3687deg)= 3684 ang 2318degV
|Eg| = 3684 V
Power angle = 2318o
(iv)
P = deg90sin2
23043683
(max occurs when δ = 90o)
= 1271 kW
Ex 22A 3 phase 4 pole synchronous machine with negligible stator winding resistance and a
synchronous reactance of 2 per phase is connected to a 480V 60 Hz 3phase supply
(infinite bus) and operated as a generator It is currently generating 80kW with power
factor 08 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)(iv) What is the maximum power the machine can generate with this value
excitation(v) What is the rotational speed of this generator
Soln (i)
(ii)P = 80kW Apparent power S 100 ang -3687
o kVA
Rated voltage per phase
V 277ang 0deg V
LoadGen
Xs =2
Eg Vt = 480V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2226
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
22
Rated current
Ia = S 3V
= 9058ang -3687deg A
|Ia| = 12028 A
(iii)
Excitation voltage
Eg = 277 ang 0deg + j2(12028ang -3687deg)
= 4632 ang 2455degV|Eg| = 4632 V
Power angle = 2455o
(iv)
P = deg90sin2
27724633
(max occurs when δ = 90o)
= 19238 kW
(v)Synchronous speed ωs = 2 π fp = 1885 rads
= 1800 revmin
Ex 23 (Jan 14)
A 3 phase synchronous machine with negligible stator winding resistance and a synchronous
reactance of 25 per phase is connected to a 20kV 50 Hz 3phase supply (infinite bus) and
operated as a generator It is currently generating 100MW with power factor 085 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)(iv) What minimum value of excitation voltage (E) would be required for this machine to
deliver 200MW into a 20kV bus assuming a 20deg stability margin must be maintained onpower angle (δ)
Soln
(i)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2326
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
23
(ii)
P = 100 MW Apparent power S = 100 085
= 11765 ang -3179o MVA
Rated voltage per phase
V= 11547ang 0deg VRated current
Ia = S 3V
= 3396 103ang -3179deg A
|Ia| = 3396 103A
(iii)
Excitation voltage
Eg = 11547 ang 0deg + j25(3396 103ang -3179deg)
= 17570 ang 2425degV|Eg| = 17570 V
Power angle = 2425o V
(iv)
P = )2090sin(52
3degminusdegt gV E
(max occurs when δ = 90o stability margin 20o)
200 MW = )2090sin(52
547113degminusdegg E
200 MW = 38130209397085613 gg E E =
|Eg| = 15360 V
LoadGen
Xs =25
Eg Vt =11547V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2426
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
24
Ex 24A manufacturing plant fed with a 400V 50Hz three phase runs from 0900 to 1800
Monday to Friday The factoryrsquos load is normally 80kVA with power factor 08 lagging
during these hours Once a month they need to run a special heating cycle which adds a
further 20kW to the factory load for a 2 hour period This heating cycle is timer
controlled and fully automated but the timer is currently set to run on the third Friday
afternoon of the month Outside of normal business hours and at weekends the factory
load is negligible The firm has a contractually agreed Maximum Import Capacity (MIC)
of 120kVA
bull Use the attached table of Low Voltage Maximum Demand charges to calculate the
firmrsquos electricity bill (exclusive of VAT) for the month of February (28 days)
bull Identify three steps that this firm could take to reduce their expenditure on
electricity assuming they stick with the same tariff Calculate the potential
savings
983116983151983159 983126983151983148983156983137983143983141 983117983137983160983145983149983157983149 983108983141983149983137983150983140 983107983144983137983154983143983141983155
983108983137983145983148983161 983123983156983137983150983140983145983150983143 983107983144983137983154983143983141 991404302983140983137983161
983120983123983119 983116983141983158983161
983120983123983119 983116983141983158983161 983152983141983154 983115983126983105 983151983142 983117983113983107 983152983141983154 983140983137983161 9914040036983097983147983126983105983140983137983161
983117983137983160983145983149983157983149 983113983149983152983151983154983156 983107983137983152983137983139983145983156983161 (983117983113983107) 983107983144983137983154983143983141983155
983108983137983145983148983161 983107983144983137983154983143983141 983152983141983154 983147983126983105 983151983142 983117983113983107 9914040072983147983126983105983140983137983161
983108983141983149983137983150983140 983107983144983137983154983143983141983155 983118983151983158983085983110983141983138 983117983137983154983085983119983139983156
983117983151983150983156983144983148983161 983107983144983137983154983143983141 983152983141983154 983147983127 983151983142 983149983141983156983141983154983141983140 983117983108 983145983150 983137
983149983151983150983156983144983148983161 983152983141983154983145983151983140 (983123983157983138983146983141983139983156 983156983151 983137 983149983145983150983145983149983157983149 983139983144983137983154983143983141 983151983142
30983147983127) 991404189830975983147983127983149983151983150983156983144 991404000
983108983137983161983156983145983149983141 983147983127983144 (983125983150983145983156) 983107983144983137983154983143983141 (0898309800 983085 19830979830985983097) 1480983139983147983127983144 1275983139983147983127983144
983118983145983143983144983156983145983149983141 983147983127983144 (983125983150983145983156) 983118983145983143983144983156 983107983144983137983154983143983141983155 (2098309800983085079830985983097) 64983097983139983147983127983144 64983097983139983147983127983144
983127983137983156983156983148983141983155983155 983125983150983145983156 983120983154983145983139983141 (983147983126983105983122983144 983145983150 983141983160983139983141983155983155 983151983142 983151983150983141 983156983144983145983154983140
983151983142 983156983151983156983137983148 983147983127983144) 991404000714 983147983126983105983122983144 991404000714 983147983126983105983122983144
bull Table Low Voltage Maximum Demand Charges (Q6)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2526
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
25
SolnNumber of days in February = 28
Number of working hours = 459 = 180
Normal kW = 80kVA08 = 64kW
Total kWh = 64kW180 hours+20kW2 hours = 11560 kWh
Normal kvar =80kVA sqrt(1-08^2) = 48kvar = sqrt(802-64
2) = 48 kVAr
Total kvarh = 48kvar180 hours = 8640 kvarh
Chargeable kvarh = 8640-(11560)3=4786 kvarh
983116983151983159 983126983151983148983156983137983143983141 983117983137983160983145983149983157983149 983108983141983149983137983150983140
983107983144983137983154983143983141983155 983152983141983154 983122983137983156983141 983121983156983161 983120983141983154983145983151983140 983107983144983137983154983143983141
983108983137983145983148983161 983123983156983137983150983140983145983150983143 983107983144983137983154983143983141 991404302 983140983137983161 991404302000 1 28 991404983096456
983120983123983119 983116983141983158983161
983120983123983119 983116983141983158983161 983152983141983154 983115983126983105 983152983141983154 983140983137983161 9914040036983097 983147983126983105983140983137983161 99140400369830970 120 28 991404123983097983096
983117983137983160983145983149983157983149 983113983149983152983151983154983156 983107983137983152983137983139983145983156983161
(983117983113983107) 983107983144983137983154983143983141983155
983108983137983145983148983161 983107983144983137983154983143983141 983152983141983154 983147983126983105 983151983142 983117983113983107 9914040072 983147983126983105983140983137983161 991404007200 120 28 9914042419830972
983108983141983149983137983150983140 983107983144983137983154983143983141983155 983118983151983158983085983110983141983138
983117983151983150983156983144983148983161 983107983144983137983154983143983141 983152983141983154 983147983127 983151983142
983149983141983156983141983154983141983140 983117983108 983145983150 983137 983149983151983150983156983144983148983161
983152983141983154983145983151983140 (983123983157983138983146983141983139983156 983156983151 983137
983149983145983150983145983149983157983149 983139983144983137983154983143983141 983151983142 30983147983127) 991404189830975
983147983127983149983151983150983156
983144 99140418983097500 84 1 991404159830971983096 983108983137983161983156983145983149983141 983147983127983144 (983125983150983145983156) 983107983144983137983154983143983141
(0898309800 983085 19830979830985983097) 9914040148 983147983127983144 991404014800 11560 1
991404198309510983096
983096
983118983145983143983144983156983145983149983141 983147983127983144 (983125983150983145983156) 983118983145983143983144983156
983107983144983137983154983143983141983155 (2098309800983085079830985983097) 9914040064983097 983147983127983144 99140400649830970
983127983137983156983156983148983141983155983155 983125983150983145983156 983120983154983145983139983141 (983147983126983105983122983144
983145983150 983141983160983139983141983155983155 983151983142 983151983150983141 983156983144983145983154983140 983151983142 983156983151983156983137983148
983147983127983144) 991404000714 983147983126983137983154983144 991404000714 4787 1 991404341983096
983124983151983156983137983148 983106983145983148983148
983142983151983154
983110983141983138983154983157983137983154983161
9914042354983095
0
[16 Marks]
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2626
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
26
Potential Savings
bull Use Power Factor correction to eliminate Watt less unit charge and to reduce MIC
bull Shift Heating load to night time or weekend
bull Negotiate lower MIC
Potential savings from these measures
Eliminate Wattless charge save euro3417 month
Shift heating cycle to night time
Max demand it lowered by 20kW = saving of 20euro1895 = euro3790 month
Saving by moving 20kW to night rate = 20kW2 hours (euro0148-euro00649)= euro332
Negotiating a reduced MIC
If power factor correction is employed and the heating cycle is shifted to night time the
then maximum VA of the plant drops to close to its maximum kW demand of 64kW An
MIC of 80kVA should be sufficient to cover this Potential Saving = 40kVA
(euro0072+euro00369)28 = euro12196 per month
Total potential saving = euro3417+euro3790+euro332+euro12196=euro19735month
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1526
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
15
Pelec = radic3 VL IL cos φ = 25044 kW
Pm = Pg (1-s) ndash Prot = 2237 kW
η = Pelec Pm = 893s
Ex 17
A large 3-phase 4000V 8-pole 50Hz squirrel-cage induction motor draws a current of
380A and a total active power of 2344kW when operating at full load The speed at full load
is accurately measured and found to be 742rpm The stator is star-connected and the
resistance between two stator terminals is 01Ω The total iron losses are 234kW and the
windage and friction losses are 12kW
Calculate
a) The power factor at full loadb) The active power supplied to the rotor
c) The total i2R losses in the rotor
d) The mechanical power output to the load and torque delivered to the load
e) The motorrsquos net efficiency
a) The active power input to the motor is given as 2344kW The apparent power is
S = radic3 VL IL = 2633kVA
Therefore pf = PS 089 lagging
b) The active power supplied to the rotor is the active power input to the motor minus the ironlosses and copper losses in the stator The stator iron losses are given as 234kW To calculate
the copper losses note that the terminal resistance (01) is twice that of each phase for a wye
connection Therefore the copper losses are
Stator copper losses Pcu = 217kW
Therefore the active power supplied to the rotor is
Pr = Input power ndash stator copper loss ndash iron losses
Pr = 2344 kW ndash 234kW ndash 217kW
= 2298kW
c) The copper losses in the rotor are proportional to the slip (see lecture notes) Prcu sP
Therefore we must calculate the slip Neither the synchronous speed nor number of poles are
explicitly stated in the problem however by inspection we determine that the synchronous speed
must be 750rpm and the number of poles = 8 (For 50Hz supply synchronous speeds occur at
discrete intervals corresponding to pole pairs 2-pole 3000rpm 4-pole 1500rpm 6- pole
1000rpm 8-pole 750rpm etchellip We also know that induction motors operate near synchronous
speed ndash the speed is given as 742 rpm which is close to synchronous speed for an
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1626
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
16
8-pole machine)
Synchronous speed ωs = 2 π fp = 7854 rads
ωm = = 777 radss = 00107
Therefore the rotor copper losses are Pr cu = sPr = 2456 kw
(d)
The mechanical power (Pm) delivered to the load is the active power delivered to the rotor
minus rotor copper losses and windage and friction losses The windage and friction losses are
given as 12kW Therefore
Pm =22989 kW ndash 245kW-12 kW =22624 kW
kNmT L 129
60
2742
42262=
=
π
(e) the motors efficiency 5962344
42262===
kW
kW
P
P
in
out η
Ex 18
A 3-phase 440V 8-pole 50Hz star-connected induction motor has the following equivalent
circuit per-phase parameters
Rs = 01 Ω Rr = 015Ω xs = xr =06 Ω Rc = 100 Ω xm = 10 Ω
(a) Calculate using the equivalent circuit neglecting stator impedance at a slip of 5 the input
stator-current and power factor the rotor current referred to the stator the torque the
mechanical output power and the efficiency Calculate the starting torque Assume the
mechanical loss is 1kW for all speeds
Soln
(a) Synchronous speed ωs = 2 π fp 7854 rads
ωm = ωs(1 - s) = 7618 rads
Stator impedance neglected
Vs = 400 radic3 = 230
Ir` =
60050
10
230
j+ =
= 68 ang-1671 = 651 - j 1954
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1726
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
17
Io =230
100 + j230
10 = 23 ndash j 23
Is = Ir` + Io = 674 - j4294
= 799 A at cos φ = 084
Ir` = 68 A
T = 529 Nm
Pmech = 39365 kW
Pelect = 465 kW
Efficiency = 39365465 = 846
Ts = 39627 Nm
Ex19
Tests conducted at 50 Hz on a three-phase star connected 45 kW 2200V six-pole 50Hz squirrel
cage induction motor are listed below
The average dc resistance per stator phase Rs = 28 Ω (a) Briefly describe the tests required to determine the per-phase equivalent circuit
impedance parameters of an induction machine
(a) Determine the parameters of the per-phase equivalent circuit
(b) Determine the rotational losses of the machine Core losses can be neglected
load
0
No-loadV
R
I1
Ic
I
c X
Im
m
Figure 1 Equivalent Circuit during No-Load Test
torPower Fac0 times= I I C = 2025 A
22
0 cm I I I minus=
= 4019 A
Ω== 627c
c I
V R Ω== 5317
m
m I
V X
No Load Test Locked Rotor Test
Power Factor = 045 Power Factor = 09
Line Voltage = 22kV Line Voltage = 270V
Line Current = 45A Line Current = 25A
Input Power = 16kW Input Power = 9kW
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1826
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
18
For the locked rotor test slip = 1
Ω== 246s
s
LR I
V Z
Locked rotor reactance is
( )2
21
2
21 R R Z X X +minus=+
φ cos21 times=+= Z R R Rtotal = 5616Ω
R2rsquo = Rtot - R1 = 2816Ω
Ω=+ 72221 X X
(a) From the no-load test the no-load power isPNL = 16kW
The no-load rotational loss is
Prot = PNL ndash 3 I s2 Rs
= 14299W
Synch Machines
Ex 20
A 60 MVA 20 kV three-phase 50 Hz synchronous generator has a synchronous reactance
of 8 Ω per phase and negligible resistance The generator delivers rated power at a power
factor of 085 lagging at the rated terminal voltage to an infinite bus(a) Determine the excitation voltage per phase and the power angle δ
(b) With the excitation voltage Eg of the synchronous generator held constant at the
value determined in part (a) the input driving torque is reduced until the generator
delivers 28 MW Determine the armature current and the power factor
(c) If the generator operates at the excitation voltage of Eg obtained in part (a) what is
the steady state maximum power the machine can deliver before losing
jXsΩΩΩΩ jX`r ΩΩΩΩ Rs
R`r s
Is I`r
Vs ph
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1926
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
19
synchronism Determine the armature current corresponding to this maximum
power
(d) With real generated power at 42MW compute the power angle δ the stator
current Ia and power factor when the field excitation current If is adjusted so that
the excitation voltage Eg is decreased to 7919 of the value determined in part
(a)
Soln
(a) Apparent power S = 51 MW ndash j 316Mvar
Rated voltage per phase
V = 1154 ang 0deg kV
Rated current
Ia = S 3V
= 1733 ang -3178deg AExcitation voltage
Eg = 11540 + j9(1733) ang -3178deg)= 237922 ang 3386deg
Power angle = 3386o
(b) When delivering 28 MW
δ = sin-1
)5411)(79223(3
828 = 1578
o
P =
s
g
X
V E 3sinδ
Ia = 13069ang2967o j8
= 16336ang-6033o A
Power factor = cos(6033) = 0495 lagging
(c) The maximum power occurs at δ = 90o
Pmax =s
g
X
V E 3
= 102 MW
Ia = 324885ang25875o A
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2026
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
20
The power factor cos(25875) = 0899 leading
(d) Eg = 13463 V
The power angle δ = sin-1
)5411)(84318(3
842
= 31o
Ia = 86674ang0o A
-Power factor = 0
Ex 21 (Aug 13) A 3 phase synchronous machine with negligible stator winding resistance and a
synchronous reactance of 2 per phase is connected to a 400V 50 Hz 3-phase supply (infinite
bus) and operated as a generator It is currently generating 50kW with power factor 08 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)
(iv) What is the maximum power the machine can generate with this value of excitation
Soln(i)
(ii)P = 50kW
Apparent power S = 625 ang -3687
o
kVA
Rated voltage per phase V = 230ang 0deg VRated current
Ia = S 3V
= 9058ang -3687deg A
LoadGen
Xs =2
Eg Vt =400 radic3 V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2126
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
21
|Ia| = 9058 A
(iii)
Excitation voltage
Eg = 230 ang 0deg + j2(9058ang -3687deg)= 3684 ang 2318degV
|Eg| = 3684 V
Power angle = 2318o
(iv)
P = deg90sin2
23043683
(max occurs when δ = 90o)
= 1271 kW
Ex 22A 3 phase 4 pole synchronous machine with negligible stator winding resistance and a
synchronous reactance of 2 per phase is connected to a 480V 60 Hz 3phase supply
(infinite bus) and operated as a generator It is currently generating 80kW with power
factor 08 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)(iv) What is the maximum power the machine can generate with this value
excitation(v) What is the rotational speed of this generator
Soln (i)
(ii)P = 80kW Apparent power S 100 ang -3687
o kVA
Rated voltage per phase
V 277ang 0deg V
LoadGen
Xs =2
Eg Vt = 480V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2226
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
22
Rated current
Ia = S 3V
= 9058ang -3687deg A
|Ia| = 12028 A
(iii)
Excitation voltage
Eg = 277 ang 0deg + j2(12028ang -3687deg)
= 4632 ang 2455degV|Eg| = 4632 V
Power angle = 2455o
(iv)
P = deg90sin2
27724633
(max occurs when δ = 90o)
= 19238 kW
(v)Synchronous speed ωs = 2 π fp = 1885 rads
= 1800 revmin
Ex 23 (Jan 14)
A 3 phase synchronous machine with negligible stator winding resistance and a synchronous
reactance of 25 per phase is connected to a 20kV 50 Hz 3phase supply (infinite bus) and
operated as a generator It is currently generating 100MW with power factor 085 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)(iv) What minimum value of excitation voltage (E) would be required for this machine to
deliver 200MW into a 20kV bus assuming a 20deg stability margin must be maintained onpower angle (δ)
Soln
(i)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2326
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
23
(ii)
P = 100 MW Apparent power S = 100 085
= 11765 ang -3179o MVA
Rated voltage per phase
V= 11547ang 0deg VRated current
Ia = S 3V
= 3396 103ang -3179deg A
|Ia| = 3396 103A
(iii)
Excitation voltage
Eg = 11547 ang 0deg + j25(3396 103ang -3179deg)
= 17570 ang 2425degV|Eg| = 17570 V
Power angle = 2425o V
(iv)
P = )2090sin(52
3degminusdegt gV E
(max occurs when δ = 90o stability margin 20o)
200 MW = )2090sin(52
547113degminusdegg E
200 MW = 38130209397085613 gg E E =
|Eg| = 15360 V
LoadGen
Xs =25
Eg Vt =11547V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2426
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
24
Ex 24A manufacturing plant fed with a 400V 50Hz three phase runs from 0900 to 1800
Monday to Friday The factoryrsquos load is normally 80kVA with power factor 08 lagging
during these hours Once a month they need to run a special heating cycle which adds a
further 20kW to the factory load for a 2 hour period This heating cycle is timer
controlled and fully automated but the timer is currently set to run on the third Friday
afternoon of the month Outside of normal business hours and at weekends the factory
load is negligible The firm has a contractually agreed Maximum Import Capacity (MIC)
of 120kVA
bull Use the attached table of Low Voltage Maximum Demand charges to calculate the
firmrsquos electricity bill (exclusive of VAT) for the month of February (28 days)
bull Identify three steps that this firm could take to reduce their expenditure on
electricity assuming they stick with the same tariff Calculate the potential
savings
983116983151983159 983126983151983148983156983137983143983141 983117983137983160983145983149983157983149 983108983141983149983137983150983140 983107983144983137983154983143983141983155
983108983137983145983148983161 983123983156983137983150983140983145983150983143 983107983144983137983154983143983141 991404302983140983137983161
983120983123983119 983116983141983158983161
983120983123983119 983116983141983158983161 983152983141983154 983115983126983105 983151983142 983117983113983107 983152983141983154 983140983137983161 9914040036983097983147983126983105983140983137983161
983117983137983160983145983149983157983149 983113983149983152983151983154983156 983107983137983152983137983139983145983156983161 (983117983113983107) 983107983144983137983154983143983141983155
983108983137983145983148983161 983107983144983137983154983143983141 983152983141983154 983147983126983105 983151983142 983117983113983107 9914040072983147983126983105983140983137983161
983108983141983149983137983150983140 983107983144983137983154983143983141983155 983118983151983158983085983110983141983138 983117983137983154983085983119983139983156
983117983151983150983156983144983148983161 983107983144983137983154983143983141 983152983141983154 983147983127 983151983142 983149983141983156983141983154983141983140 983117983108 983145983150 983137
983149983151983150983156983144983148983161 983152983141983154983145983151983140 (983123983157983138983146983141983139983156 983156983151 983137 983149983145983150983145983149983157983149 983139983144983137983154983143983141 983151983142
30983147983127) 991404189830975983147983127983149983151983150983156983144 991404000
983108983137983161983156983145983149983141 983147983127983144 (983125983150983145983156) 983107983144983137983154983143983141 (0898309800 983085 19830979830985983097) 1480983139983147983127983144 1275983139983147983127983144
983118983145983143983144983156983145983149983141 983147983127983144 (983125983150983145983156) 983118983145983143983144983156 983107983144983137983154983143983141983155 (2098309800983085079830985983097) 64983097983139983147983127983144 64983097983139983147983127983144
983127983137983156983156983148983141983155983155 983125983150983145983156 983120983154983145983139983141 (983147983126983105983122983144 983145983150 983141983160983139983141983155983155 983151983142 983151983150983141 983156983144983145983154983140
983151983142 983156983151983156983137983148 983147983127983144) 991404000714 983147983126983105983122983144 991404000714 983147983126983105983122983144
bull Table Low Voltage Maximum Demand Charges (Q6)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2526
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
25
SolnNumber of days in February = 28
Number of working hours = 459 = 180
Normal kW = 80kVA08 = 64kW
Total kWh = 64kW180 hours+20kW2 hours = 11560 kWh
Normal kvar =80kVA sqrt(1-08^2) = 48kvar = sqrt(802-64
2) = 48 kVAr
Total kvarh = 48kvar180 hours = 8640 kvarh
Chargeable kvarh = 8640-(11560)3=4786 kvarh
983116983151983159 983126983151983148983156983137983143983141 983117983137983160983145983149983157983149 983108983141983149983137983150983140
983107983144983137983154983143983141983155 983152983141983154 983122983137983156983141 983121983156983161 983120983141983154983145983151983140 983107983144983137983154983143983141
983108983137983145983148983161 983123983156983137983150983140983145983150983143 983107983144983137983154983143983141 991404302 983140983137983161 991404302000 1 28 991404983096456
983120983123983119 983116983141983158983161
983120983123983119 983116983141983158983161 983152983141983154 983115983126983105 983152983141983154 983140983137983161 9914040036983097 983147983126983105983140983137983161 99140400369830970 120 28 991404123983097983096
983117983137983160983145983149983157983149 983113983149983152983151983154983156 983107983137983152983137983139983145983156983161
(983117983113983107) 983107983144983137983154983143983141983155
983108983137983145983148983161 983107983144983137983154983143983141 983152983141983154 983147983126983105 983151983142 983117983113983107 9914040072 983147983126983105983140983137983161 991404007200 120 28 9914042419830972
983108983141983149983137983150983140 983107983144983137983154983143983141983155 983118983151983158983085983110983141983138
983117983151983150983156983144983148983161 983107983144983137983154983143983141 983152983141983154 983147983127 983151983142
983149983141983156983141983154983141983140 983117983108 983145983150 983137 983149983151983150983156983144983148983161
983152983141983154983145983151983140 (983123983157983138983146983141983139983156 983156983151 983137
983149983145983150983145983149983157983149 983139983144983137983154983143983141 983151983142 30983147983127) 991404189830975
983147983127983149983151983150983156
983144 99140418983097500 84 1 991404159830971983096 983108983137983161983156983145983149983141 983147983127983144 (983125983150983145983156) 983107983144983137983154983143983141
(0898309800 983085 19830979830985983097) 9914040148 983147983127983144 991404014800 11560 1
991404198309510983096
983096
983118983145983143983144983156983145983149983141 983147983127983144 (983125983150983145983156) 983118983145983143983144983156
983107983144983137983154983143983141983155 (2098309800983085079830985983097) 9914040064983097 983147983127983144 99140400649830970
983127983137983156983156983148983141983155983155 983125983150983145983156 983120983154983145983139983141 (983147983126983105983122983144
983145983150 983141983160983139983141983155983155 983151983142 983151983150983141 983156983144983145983154983140 983151983142 983156983151983156983137983148
983147983127983144) 991404000714 983147983126983137983154983144 991404000714 4787 1 991404341983096
983124983151983156983137983148 983106983145983148983148
983142983151983154
983110983141983138983154983157983137983154983161
9914042354983095
0
[16 Marks]
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2626
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
26
Potential Savings
bull Use Power Factor correction to eliminate Watt less unit charge and to reduce MIC
bull Shift Heating load to night time or weekend
bull Negotiate lower MIC
Potential savings from these measures
Eliminate Wattless charge save euro3417 month
Shift heating cycle to night time
Max demand it lowered by 20kW = saving of 20euro1895 = euro3790 month
Saving by moving 20kW to night rate = 20kW2 hours (euro0148-euro00649)= euro332
Negotiating a reduced MIC
If power factor correction is employed and the heating cycle is shifted to night time the
then maximum VA of the plant drops to close to its maximum kW demand of 64kW An
MIC of 80kVA should be sufficient to cover this Potential Saving = 40kVA
(euro0072+euro00369)28 = euro12196 per month
Total potential saving = euro3417+euro3790+euro332+euro12196=euro19735month
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1626
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
16
8-pole machine)
Synchronous speed ωs = 2 π fp = 7854 rads
ωm = = 777 radss = 00107
Therefore the rotor copper losses are Pr cu = sPr = 2456 kw
(d)
The mechanical power (Pm) delivered to the load is the active power delivered to the rotor
minus rotor copper losses and windage and friction losses The windage and friction losses are
given as 12kW Therefore
Pm =22989 kW ndash 245kW-12 kW =22624 kW
kNmT L 129
60
2742
42262=
=
π
(e) the motors efficiency 5962344
42262===
kW
kW
P
P
in
out η
Ex 18
A 3-phase 440V 8-pole 50Hz star-connected induction motor has the following equivalent
circuit per-phase parameters
Rs = 01 Ω Rr = 015Ω xs = xr =06 Ω Rc = 100 Ω xm = 10 Ω
(a) Calculate using the equivalent circuit neglecting stator impedance at a slip of 5 the input
stator-current and power factor the rotor current referred to the stator the torque the
mechanical output power and the efficiency Calculate the starting torque Assume the
mechanical loss is 1kW for all speeds
Soln
(a) Synchronous speed ωs = 2 π fp 7854 rads
ωm = ωs(1 - s) = 7618 rads
Stator impedance neglected
Vs = 400 radic3 = 230
Ir` =
60050
10
230
j+ =
= 68 ang-1671 = 651 - j 1954
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1726
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
17
Io =230
100 + j230
10 = 23 ndash j 23
Is = Ir` + Io = 674 - j4294
= 799 A at cos φ = 084
Ir` = 68 A
T = 529 Nm
Pmech = 39365 kW
Pelect = 465 kW
Efficiency = 39365465 = 846
Ts = 39627 Nm
Ex19
Tests conducted at 50 Hz on a three-phase star connected 45 kW 2200V six-pole 50Hz squirrel
cage induction motor are listed below
The average dc resistance per stator phase Rs = 28 Ω (a) Briefly describe the tests required to determine the per-phase equivalent circuit
impedance parameters of an induction machine
(a) Determine the parameters of the per-phase equivalent circuit
(b) Determine the rotational losses of the machine Core losses can be neglected
load
0
No-loadV
R
I1
Ic
I
c X
Im
m
Figure 1 Equivalent Circuit during No-Load Test
torPower Fac0 times= I I C = 2025 A
22
0 cm I I I minus=
= 4019 A
Ω== 627c
c I
V R Ω== 5317
m
m I
V X
No Load Test Locked Rotor Test
Power Factor = 045 Power Factor = 09
Line Voltage = 22kV Line Voltage = 270V
Line Current = 45A Line Current = 25A
Input Power = 16kW Input Power = 9kW
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1826
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
18
For the locked rotor test slip = 1
Ω== 246s
s
LR I
V Z
Locked rotor reactance is
( )2
21
2
21 R R Z X X +minus=+
φ cos21 times=+= Z R R Rtotal = 5616Ω
R2rsquo = Rtot - R1 = 2816Ω
Ω=+ 72221 X X
(a) From the no-load test the no-load power isPNL = 16kW
The no-load rotational loss is
Prot = PNL ndash 3 I s2 Rs
= 14299W
Synch Machines
Ex 20
A 60 MVA 20 kV three-phase 50 Hz synchronous generator has a synchronous reactance
of 8 Ω per phase and negligible resistance The generator delivers rated power at a power
factor of 085 lagging at the rated terminal voltage to an infinite bus(a) Determine the excitation voltage per phase and the power angle δ
(b) With the excitation voltage Eg of the synchronous generator held constant at the
value determined in part (a) the input driving torque is reduced until the generator
delivers 28 MW Determine the armature current and the power factor
(c) If the generator operates at the excitation voltage of Eg obtained in part (a) what is
the steady state maximum power the machine can deliver before losing
jXsΩΩΩΩ jX`r ΩΩΩΩ Rs
R`r s
Is I`r
Vs ph
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1926
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
19
synchronism Determine the armature current corresponding to this maximum
power
(d) With real generated power at 42MW compute the power angle δ the stator
current Ia and power factor when the field excitation current If is adjusted so that
the excitation voltage Eg is decreased to 7919 of the value determined in part
(a)
Soln
(a) Apparent power S = 51 MW ndash j 316Mvar
Rated voltage per phase
V = 1154 ang 0deg kV
Rated current
Ia = S 3V
= 1733 ang -3178deg AExcitation voltage
Eg = 11540 + j9(1733) ang -3178deg)= 237922 ang 3386deg
Power angle = 3386o
(b) When delivering 28 MW
δ = sin-1
)5411)(79223(3
828 = 1578
o
P =
s
g
X
V E 3sinδ
Ia = 13069ang2967o j8
= 16336ang-6033o A
Power factor = cos(6033) = 0495 lagging
(c) The maximum power occurs at δ = 90o
Pmax =s
g
X
V E 3
= 102 MW
Ia = 324885ang25875o A
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2026
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
20
The power factor cos(25875) = 0899 leading
(d) Eg = 13463 V
The power angle δ = sin-1
)5411)(84318(3
842
= 31o
Ia = 86674ang0o A
-Power factor = 0
Ex 21 (Aug 13) A 3 phase synchronous machine with negligible stator winding resistance and a
synchronous reactance of 2 per phase is connected to a 400V 50 Hz 3-phase supply (infinite
bus) and operated as a generator It is currently generating 50kW with power factor 08 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)
(iv) What is the maximum power the machine can generate with this value of excitation
Soln(i)
(ii)P = 50kW
Apparent power S = 625 ang -3687
o
kVA
Rated voltage per phase V = 230ang 0deg VRated current
Ia = S 3V
= 9058ang -3687deg A
LoadGen
Xs =2
Eg Vt =400 radic3 V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2126
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
21
|Ia| = 9058 A
(iii)
Excitation voltage
Eg = 230 ang 0deg + j2(9058ang -3687deg)= 3684 ang 2318degV
|Eg| = 3684 V
Power angle = 2318o
(iv)
P = deg90sin2
23043683
(max occurs when δ = 90o)
= 1271 kW
Ex 22A 3 phase 4 pole synchronous machine with negligible stator winding resistance and a
synchronous reactance of 2 per phase is connected to a 480V 60 Hz 3phase supply
(infinite bus) and operated as a generator It is currently generating 80kW with power
factor 08 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)(iv) What is the maximum power the machine can generate with this value
excitation(v) What is the rotational speed of this generator
Soln (i)
(ii)P = 80kW Apparent power S 100 ang -3687
o kVA
Rated voltage per phase
V 277ang 0deg V
LoadGen
Xs =2
Eg Vt = 480V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2226
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
22
Rated current
Ia = S 3V
= 9058ang -3687deg A
|Ia| = 12028 A
(iii)
Excitation voltage
Eg = 277 ang 0deg + j2(12028ang -3687deg)
= 4632 ang 2455degV|Eg| = 4632 V
Power angle = 2455o
(iv)
P = deg90sin2
27724633
(max occurs when δ = 90o)
= 19238 kW
(v)Synchronous speed ωs = 2 π fp = 1885 rads
= 1800 revmin
Ex 23 (Jan 14)
A 3 phase synchronous machine with negligible stator winding resistance and a synchronous
reactance of 25 per phase is connected to a 20kV 50 Hz 3phase supply (infinite bus) and
operated as a generator It is currently generating 100MW with power factor 085 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)(iv) What minimum value of excitation voltage (E) would be required for this machine to
deliver 200MW into a 20kV bus assuming a 20deg stability margin must be maintained onpower angle (δ)
Soln
(i)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2326
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
23
(ii)
P = 100 MW Apparent power S = 100 085
= 11765 ang -3179o MVA
Rated voltage per phase
V= 11547ang 0deg VRated current
Ia = S 3V
= 3396 103ang -3179deg A
|Ia| = 3396 103A
(iii)
Excitation voltage
Eg = 11547 ang 0deg + j25(3396 103ang -3179deg)
= 17570 ang 2425degV|Eg| = 17570 V
Power angle = 2425o V
(iv)
P = )2090sin(52
3degminusdegt gV E
(max occurs when δ = 90o stability margin 20o)
200 MW = )2090sin(52
547113degminusdegg E
200 MW = 38130209397085613 gg E E =
|Eg| = 15360 V
LoadGen
Xs =25
Eg Vt =11547V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2426
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
24
Ex 24A manufacturing plant fed with a 400V 50Hz three phase runs from 0900 to 1800
Monday to Friday The factoryrsquos load is normally 80kVA with power factor 08 lagging
during these hours Once a month they need to run a special heating cycle which adds a
further 20kW to the factory load for a 2 hour period This heating cycle is timer
controlled and fully automated but the timer is currently set to run on the third Friday
afternoon of the month Outside of normal business hours and at weekends the factory
load is negligible The firm has a contractually agreed Maximum Import Capacity (MIC)
of 120kVA
bull Use the attached table of Low Voltage Maximum Demand charges to calculate the
firmrsquos electricity bill (exclusive of VAT) for the month of February (28 days)
bull Identify three steps that this firm could take to reduce their expenditure on
electricity assuming they stick with the same tariff Calculate the potential
savings
983116983151983159 983126983151983148983156983137983143983141 983117983137983160983145983149983157983149 983108983141983149983137983150983140 983107983144983137983154983143983141983155
983108983137983145983148983161 983123983156983137983150983140983145983150983143 983107983144983137983154983143983141 991404302983140983137983161
983120983123983119 983116983141983158983161
983120983123983119 983116983141983158983161 983152983141983154 983115983126983105 983151983142 983117983113983107 983152983141983154 983140983137983161 9914040036983097983147983126983105983140983137983161
983117983137983160983145983149983157983149 983113983149983152983151983154983156 983107983137983152983137983139983145983156983161 (983117983113983107) 983107983144983137983154983143983141983155
983108983137983145983148983161 983107983144983137983154983143983141 983152983141983154 983147983126983105 983151983142 983117983113983107 9914040072983147983126983105983140983137983161
983108983141983149983137983150983140 983107983144983137983154983143983141983155 983118983151983158983085983110983141983138 983117983137983154983085983119983139983156
983117983151983150983156983144983148983161 983107983144983137983154983143983141 983152983141983154 983147983127 983151983142 983149983141983156983141983154983141983140 983117983108 983145983150 983137
983149983151983150983156983144983148983161 983152983141983154983145983151983140 (983123983157983138983146983141983139983156 983156983151 983137 983149983145983150983145983149983157983149 983139983144983137983154983143983141 983151983142
30983147983127) 991404189830975983147983127983149983151983150983156983144 991404000
983108983137983161983156983145983149983141 983147983127983144 (983125983150983145983156) 983107983144983137983154983143983141 (0898309800 983085 19830979830985983097) 1480983139983147983127983144 1275983139983147983127983144
983118983145983143983144983156983145983149983141 983147983127983144 (983125983150983145983156) 983118983145983143983144983156 983107983144983137983154983143983141983155 (2098309800983085079830985983097) 64983097983139983147983127983144 64983097983139983147983127983144
983127983137983156983156983148983141983155983155 983125983150983145983156 983120983154983145983139983141 (983147983126983105983122983144 983145983150 983141983160983139983141983155983155 983151983142 983151983150983141 983156983144983145983154983140
983151983142 983156983151983156983137983148 983147983127983144) 991404000714 983147983126983105983122983144 991404000714 983147983126983105983122983144
bull Table Low Voltage Maximum Demand Charges (Q6)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2526
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
25
SolnNumber of days in February = 28
Number of working hours = 459 = 180
Normal kW = 80kVA08 = 64kW
Total kWh = 64kW180 hours+20kW2 hours = 11560 kWh
Normal kvar =80kVA sqrt(1-08^2) = 48kvar = sqrt(802-64
2) = 48 kVAr
Total kvarh = 48kvar180 hours = 8640 kvarh
Chargeable kvarh = 8640-(11560)3=4786 kvarh
983116983151983159 983126983151983148983156983137983143983141 983117983137983160983145983149983157983149 983108983141983149983137983150983140
983107983144983137983154983143983141983155 983152983141983154 983122983137983156983141 983121983156983161 983120983141983154983145983151983140 983107983144983137983154983143983141
983108983137983145983148983161 983123983156983137983150983140983145983150983143 983107983144983137983154983143983141 991404302 983140983137983161 991404302000 1 28 991404983096456
983120983123983119 983116983141983158983161
983120983123983119 983116983141983158983161 983152983141983154 983115983126983105 983152983141983154 983140983137983161 9914040036983097 983147983126983105983140983137983161 99140400369830970 120 28 991404123983097983096
983117983137983160983145983149983157983149 983113983149983152983151983154983156 983107983137983152983137983139983145983156983161
(983117983113983107) 983107983144983137983154983143983141983155
983108983137983145983148983161 983107983144983137983154983143983141 983152983141983154 983147983126983105 983151983142 983117983113983107 9914040072 983147983126983105983140983137983161 991404007200 120 28 9914042419830972
983108983141983149983137983150983140 983107983144983137983154983143983141983155 983118983151983158983085983110983141983138
983117983151983150983156983144983148983161 983107983144983137983154983143983141 983152983141983154 983147983127 983151983142
983149983141983156983141983154983141983140 983117983108 983145983150 983137 983149983151983150983156983144983148983161
983152983141983154983145983151983140 (983123983157983138983146983141983139983156 983156983151 983137
983149983145983150983145983149983157983149 983139983144983137983154983143983141 983151983142 30983147983127) 991404189830975
983147983127983149983151983150983156
983144 99140418983097500 84 1 991404159830971983096 983108983137983161983156983145983149983141 983147983127983144 (983125983150983145983156) 983107983144983137983154983143983141
(0898309800 983085 19830979830985983097) 9914040148 983147983127983144 991404014800 11560 1
991404198309510983096
983096
983118983145983143983144983156983145983149983141 983147983127983144 (983125983150983145983156) 983118983145983143983144983156
983107983144983137983154983143983141983155 (2098309800983085079830985983097) 9914040064983097 983147983127983144 99140400649830970
983127983137983156983156983148983141983155983155 983125983150983145983156 983120983154983145983139983141 (983147983126983105983122983144
983145983150 983141983160983139983141983155983155 983151983142 983151983150983141 983156983144983145983154983140 983151983142 983156983151983156983137983148
983147983127983144) 991404000714 983147983126983137983154983144 991404000714 4787 1 991404341983096
983124983151983156983137983148 983106983145983148983148
983142983151983154
983110983141983138983154983157983137983154983161
9914042354983095
0
[16 Marks]
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2626
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
26
Potential Savings
bull Use Power Factor correction to eliminate Watt less unit charge and to reduce MIC
bull Shift Heating load to night time or weekend
bull Negotiate lower MIC
Potential savings from these measures
Eliminate Wattless charge save euro3417 month
Shift heating cycle to night time
Max demand it lowered by 20kW = saving of 20euro1895 = euro3790 month
Saving by moving 20kW to night rate = 20kW2 hours (euro0148-euro00649)= euro332
Negotiating a reduced MIC
If power factor correction is employed and the heating cycle is shifted to night time the
then maximum VA of the plant drops to close to its maximum kW demand of 64kW An
MIC of 80kVA should be sufficient to cover this Potential Saving = 40kVA
(euro0072+euro00369)28 = euro12196 per month
Total potential saving = euro3417+euro3790+euro332+euro12196=euro19735month
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1726
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
17
Io =230
100 + j230
10 = 23 ndash j 23
Is = Ir` + Io = 674 - j4294
= 799 A at cos φ = 084
Ir` = 68 A
T = 529 Nm
Pmech = 39365 kW
Pelect = 465 kW
Efficiency = 39365465 = 846
Ts = 39627 Nm
Ex19
Tests conducted at 50 Hz on a three-phase star connected 45 kW 2200V six-pole 50Hz squirrel
cage induction motor are listed below
The average dc resistance per stator phase Rs = 28 Ω (a) Briefly describe the tests required to determine the per-phase equivalent circuit
impedance parameters of an induction machine
(a) Determine the parameters of the per-phase equivalent circuit
(b) Determine the rotational losses of the machine Core losses can be neglected
load
0
No-loadV
R
I1
Ic
I
c X
Im
m
Figure 1 Equivalent Circuit during No-Load Test
torPower Fac0 times= I I C = 2025 A
22
0 cm I I I minus=
= 4019 A
Ω== 627c
c I
V R Ω== 5317
m
m I
V X
No Load Test Locked Rotor Test
Power Factor = 045 Power Factor = 09
Line Voltage = 22kV Line Voltage = 270V
Line Current = 45A Line Current = 25A
Input Power = 16kW Input Power = 9kW
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1826
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
18
For the locked rotor test slip = 1
Ω== 246s
s
LR I
V Z
Locked rotor reactance is
( )2
21
2
21 R R Z X X +minus=+
φ cos21 times=+= Z R R Rtotal = 5616Ω
R2rsquo = Rtot - R1 = 2816Ω
Ω=+ 72221 X X
(a) From the no-load test the no-load power isPNL = 16kW
The no-load rotational loss is
Prot = PNL ndash 3 I s2 Rs
= 14299W
Synch Machines
Ex 20
A 60 MVA 20 kV three-phase 50 Hz synchronous generator has a synchronous reactance
of 8 Ω per phase and negligible resistance The generator delivers rated power at a power
factor of 085 lagging at the rated terminal voltage to an infinite bus(a) Determine the excitation voltage per phase and the power angle δ
(b) With the excitation voltage Eg of the synchronous generator held constant at the
value determined in part (a) the input driving torque is reduced until the generator
delivers 28 MW Determine the armature current and the power factor
(c) If the generator operates at the excitation voltage of Eg obtained in part (a) what is
the steady state maximum power the machine can deliver before losing
jXsΩΩΩΩ jX`r ΩΩΩΩ Rs
R`r s
Is I`r
Vs ph
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1926
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
19
synchronism Determine the armature current corresponding to this maximum
power
(d) With real generated power at 42MW compute the power angle δ the stator
current Ia and power factor when the field excitation current If is adjusted so that
the excitation voltage Eg is decreased to 7919 of the value determined in part
(a)
Soln
(a) Apparent power S = 51 MW ndash j 316Mvar
Rated voltage per phase
V = 1154 ang 0deg kV
Rated current
Ia = S 3V
= 1733 ang -3178deg AExcitation voltage
Eg = 11540 + j9(1733) ang -3178deg)= 237922 ang 3386deg
Power angle = 3386o
(b) When delivering 28 MW
δ = sin-1
)5411)(79223(3
828 = 1578
o
P =
s
g
X
V E 3sinδ
Ia = 13069ang2967o j8
= 16336ang-6033o A
Power factor = cos(6033) = 0495 lagging
(c) The maximum power occurs at δ = 90o
Pmax =s
g
X
V E 3
= 102 MW
Ia = 324885ang25875o A
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2026
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
20
The power factor cos(25875) = 0899 leading
(d) Eg = 13463 V
The power angle δ = sin-1
)5411)(84318(3
842
= 31o
Ia = 86674ang0o A
-Power factor = 0
Ex 21 (Aug 13) A 3 phase synchronous machine with negligible stator winding resistance and a
synchronous reactance of 2 per phase is connected to a 400V 50 Hz 3-phase supply (infinite
bus) and operated as a generator It is currently generating 50kW with power factor 08 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)
(iv) What is the maximum power the machine can generate with this value of excitation
Soln(i)
(ii)P = 50kW
Apparent power S = 625 ang -3687
o
kVA
Rated voltage per phase V = 230ang 0deg VRated current
Ia = S 3V
= 9058ang -3687deg A
LoadGen
Xs =2
Eg Vt =400 radic3 V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2126
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
21
|Ia| = 9058 A
(iii)
Excitation voltage
Eg = 230 ang 0deg + j2(9058ang -3687deg)= 3684 ang 2318degV
|Eg| = 3684 V
Power angle = 2318o
(iv)
P = deg90sin2
23043683
(max occurs when δ = 90o)
= 1271 kW
Ex 22A 3 phase 4 pole synchronous machine with negligible stator winding resistance and a
synchronous reactance of 2 per phase is connected to a 480V 60 Hz 3phase supply
(infinite bus) and operated as a generator It is currently generating 80kW with power
factor 08 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)(iv) What is the maximum power the machine can generate with this value
excitation(v) What is the rotational speed of this generator
Soln (i)
(ii)P = 80kW Apparent power S 100 ang -3687
o kVA
Rated voltage per phase
V 277ang 0deg V
LoadGen
Xs =2
Eg Vt = 480V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2226
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
22
Rated current
Ia = S 3V
= 9058ang -3687deg A
|Ia| = 12028 A
(iii)
Excitation voltage
Eg = 277 ang 0deg + j2(12028ang -3687deg)
= 4632 ang 2455degV|Eg| = 4632 V
Power angle = 2455o
(iv)
P = deg90sin2
27724633
(max occurs when δ = 90o)
= 19238 kW
(v)Synchronous speed ωs = 2 π fp = 1885 rads
= 1800 revmin
Ex 23 (Jan 14)
A 3 phase synchronous machine with negligible stator winding resistance and a synchronous
reactance of 25 per phase is connected to a 20kV 50 Hz 3phase supply (infinite bus) and
operated as a generator It is currently generating 100MW with power factor 085 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)(iv) What minimum value of excitation voltage (E) would be required for this machine to
deliver 200MW into a 20kV bus assuming a 20deg stability margin must be maintained onpower angle (δ)
Soln
(i)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2326
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
23
(ii)
P = 100 MW Apparent power S = 100 085
= 11765 ang -3179o MVA
Rated voltage per phase
V= 11547ang 0deg VRated current
Ia = S 3V
= 3396 103ang -3179deg A
|Ia| = 3396 103A
(iii)
Excitation voltage
Eg = 11547 ang 0deg + j25(3396 103ang -3179deg)
= 17570 ang 2425degV|Eg| = 17570 V
Power angle = 2425o V
(iv)
P = )2090sin(52
3degminusdegt gV E
(max occurs when δ = 90o stability margin 20o)
200 MW = )2090sin(52
547113degminusdegg E
200 MW = 38130209397085613 gg E E =
|Eg| = 15360 V
LoadGen
Xs =25
Eg Vt =11547V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2426
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
24
Ex 24A manufacturing plant fed with a 400V 50Hz three phase runs from 0900 to 1800
Monday to Friday The factoryrsquos load is normally 80kVA with power factor 08 lagging
during these hours Once a month they need to run a special heating cycle which adds a
further 20kW to the factory load for a 2 hour period This heating cycle is timer
controlled and fully automated but the timer is currently set to run on the third Friday
afternoon of the month Outside of normal business hours and at weekends the factory
load is negligible The firm has a contractually agreed Maximum Import Capacity (MIC)
of 120kVA
bull Use the attached table of Low Voltage Maximum Demand charges to calculate the
firmrsquos electricity bill (exclusive of VAT) for the month of February (28 days)
bull Identify three steps that this firm could take to reduce their expenditure on
electricity assuming they stick with the same tariff Calculate the potential
savings
983116983151983159 983126983151983148983156983137983143983141 983117983137983160983145983149983157983149 983108983141983149983137983150983140 983107983144983137983154983143983141983155
983108983137983145983148983161 983123983156983137983150983140983145983150983143 983107983144983137983154983143983141 991404302983140983137983161
983120983123983119 983116983141983158983161
983120983123983119 983116983141983158983161 983152983141983154 983115983126983105 983151983142 983117983113983107 983152983141983154 983140983137983161 9914040036983097983147983126983105983140983137983161
983117983137983160983145983149983157983149 983113983149983152983151983154983156 983107983137983152983137983139983145983156983161 (983117983113983107) 983107983144983137983154983143983141983155
983108983137983145983148983161 983107983144983137983154983143983141 983152983141983154 983147983126983105 983151983142 983117983113983107 9914040072983147983126983105983140983137983161
983108983141983149983137983150983140 983107983144983137983154983143983141983155 983118983151983158983085983110983141983138 983117983137983154983085983119983139983156
983117983151983150983156983144983148983161 983107983144983137983154983143983141 983152983141983154 983147983127 983151983142 983149983141983156983141983154983141983140 983117983108 983145983150 983137
983149983151983150983156983144983148983161 983152983141983154983145983151983140 (983123983157983138983146983141983139983156 983156983151 983137 983149983145983150983145983149983157983149 983139983144983137983154983143983141 983151983142
30983147983127) 991404189830975983147983127983149983151983150983156983144 991404000
983108983137983161983156983145983149983141 983147983127983144 (983125983150983145983156) 983107983144983137983154983143983141 (0898309800 983085 19830979830985983097) 1480983139983147983127983144 1275983139983147983127983144
983118983145983143983144983156983145983149983141 983147983127983144 (983125983150983145983156) 983118983145983143983144983156 983107983144983137983154983143983141983155 (2098309800983085079830985983097) 64983097983139983147983127983144 64983097983139983147983127983144
983127983137983156983156983148983141983155983155 983125983150983145983156 983120983154983145983139983141 (983147983126983105983122983144 983145983150 983141983160983139983141983155983155 983151983142 983151983150983141 983156983144983145983154983140
983151983142 983156983151983156983137983148 983147983127983144) 991404000714 983147983126983105983122983144 991404000714 983147983126983105983122983144
bull Table Low Voltage Maximum Demand Charges (Q6)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2526
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
25
SolnNumber of days in February = 28
Number of working hours = 459 = 180
Normal kW = 80kVA08 = 64kW
Total kWh = 64kW180 hours+20kW2 hours = 11560 kWh
Normal kvar =80kVA sqrt(1-08^2) = 48kvar = sqrt(802-64
2) = 48 kVAr
Total kvarh = 48kvar180 hours = 8640 kvarh
Chargeable kvarh = 8640-(11560)3=4786 kvarh
983116983151983159 983126983151983148983156983137983143983141 983117983137983160983145983149983157983149 983108983141983149983137983150983140
983107983144983137983154983143983141983155 983152983141983154 983122983137983156983141 983121983156983161 983120983141983154983145983151983140 983107983144983137983154983143983141
983108983137983145983148983161 983123983156983137983150983140983145983150983143 983107983144983137983154983143983141 991404302 983140983137983161 991404302000 1 28 991404983096456
983120983123983119 983116983141983158983161
983120983123983119 983116983141983158983161 983152983141983154 983115983126983105 983152983141983154 983140983137983161 9914040036983097 983147983126983105983140983137983161 99140400369830970 120 28 991404123983097983096
983117983137983160983145983149983157983149 983113983149983152983151983154983156 983107983137983152983137983139983145983156983161
(983117983113983107) 983107983144983137983154983143983141983155
983108983137983145983148983161 983107983144983137983154983143983141 983152983141983154 983147983126983105 983151983142 983117983113983107 9914040072 983147983126983105983140983137983161 991404007200 120 28 9914042419830972
983108983141983149983137983150983140 983107983144983137983154983143983141983155 983118983151983158983085983110983141983138
983117983151983150983156983144983148983161 983107983144983137983154983143983141 983152983141983154 983147983127 983151983142
983149983141983156983141983154983141983140 983117983108 983145983150 983137 983149983151983150983156983144983148983161
983152983141983154983145983151983140 (983123983157983138983146983141983139983156 983156983151 983137
983149983145983150983145983149983157983149 983139983144983137983154983143983141 983151983142 30983147983127) 991404189830975
983147983127983149983151983150983156
983144 99140418983097500 84 1 991404159830971983096 983108983137983161983156983145983149983141 983147983127983144 (983125983150983145983156) 983107983144983137983154983143983141
(0898309800 983085 19830979830985983097) 9914040148 983147983127983144 991404014800 11560 1
991404198309510983096
983096
983118983145983143983144983156983145983149983141 983147983127983144 (983125983150983145983156) 983118983145983143983144983156
983107983144983137983154983143983141983155 (2098309800983085079830985983097) 9914040064983097 983147983127983144 99140400649830970
983127983137983156983156983148983141983155983155 983125983150983145983156 983120983154983145983139983141 (983147983126983105983122983144
983145983150 983141983160983139983141983155983155 983151983142 983151983150983141 983156983144983145983154983140 983151983142 983156983151983156983137983148
983147983127983144) 991404000714 983147983126983137983154983144 991404000714 4787 1 991404341983096
983124983151983156983137983148 983106983145983148983148
983142983151983154
983110983141983138983154983157983137983154983161
9914042354983095
0
[16 Marks]
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2626
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
26
Potential Savings
bull Use Power Factor correction to eliminate Watt less unit charge and to reduce MIC
bull Shift Heating load to night time or weekend
bull Negotiate lower MIC
Potential savings from these measures
Eliminate Wattless charge save euro3417 month
Shift heating cycle to night time
Max demand it lowered by 20kW = saving of 20euro1895 = euro3790 month
Saving by moving 20kW to night rate = 20kW2 hours (euro0148-euro00649)= euro332
Negotiating a reduced MIC
If power factor correction is employed and the heating cycle is shifted to night time the
then maximum VA of the plant drops to close to its maximum kW demand of 64kW An
MIC of 80kVA should be sufficient to cover this Potential Saving = 40kVA
(euro0072+euro00369)28 = euro12196 per month
Total potential saving = euro3417+euro3790+euro332+euro12196=euro19735month
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1826
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
18
For the locked rotor test slip = 1
Ω== 246s
s
LR I
V Z
Locked rotor reactance is
( )2
21
2
21 R R Z X X +minus=+
φ cos21 times=+= Z R R Rtotal = 5616Ω
R2rsquo = Rtot - R1 = 2816Ω
Ω=+ 72221 X X
(a) From the no-load test the no-load power isPNL = 16kW
The no-load rotational loss is
Prot = PNL ndash 3 I s2 Rs
= 14299W
Synch Machines
Ex 20
A 60 MVA 20 kV three-phase 50 Hz synchronous generator has a synchronous reactance
of 8 Ω per phase and negligible resistance The generator delivers rated power at a power
factor of 085 lagging at the rated terminal voltage to an infinite bus(a) Determine the excitation voltage per phase and the power angle δ
(b) With the excitation voltage Eg of the synchronous generator held constant at the
value determined in part (a) the input driving torque is reduced until the generator
delivers 28 MW Determine the armature current and the power factor
(c) If the generator operates at the excitation voltage of Eg obtained in part (a) what is
the steady state maximum power the machine can deliver before losing
jXsΩΩΩΩ jX`r ΩΩΩΩ Rs
R`r s
Is I`r
Vs ph
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1926
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
19
synchronism Determine the armature current corresponding to this maximum
power
(d) With real generated power at 42MW compute the power angle δ the stator
current Ia and power factor when the field excitation current If is adjusted so that
the excitation voltage Eg is decreased to 7919 of the value determined in part
(a)
Soln
(a) Apparent power S = 51 MW ndash j 316Mvar
Rated voltage per phase
V = 1154 ang 0deg kV
Rated current
Ia = S 3V
= 1733 ang -3178deg AExcitation voltage
Eg = 11540 + j9(1733) ang -3178deg)= 237922 ang 3386deg
Power angle = 3386o
(b) When delivering 28 MW
δ = sin-1
)5411)(79223(3
828 = 1578
o
P =
s
g
X
V E 3sinδ
Ia = 13069ang2967o j8
= 16336ang-6033o A
Power factor = cos(6033) = 0495 lagging
(c) The maximum power occurs at δ = 90o
Pmax =s
g
X
V E 3
= 102 MW
Ia = 324885ang25875o A
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2026
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
20
The power factor cos(25875) = 0899 leading
(d) Eg = 13463 V
The power angle δ = sin-1
)5411)(84318(3
842
= 31o
Ia = 86674ang0o A
-Power factor = 0
Ex 21 (Aug 13) A 3 phase synchronous machine with negligible stator winding resistance and a
synchronous reactance of 2 per phase is connected to a 400V 50 Hz 3-phase supply (infinite
bus) and operated as a generator It is currently generating 50kW with power factor 08 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)
(iv) What is the maximum power the machine can generate with this value of excitation
Soln(i)
(ii)P = 50kW
Apparent power S = 625 ang -3687
o
kVA
Rated voltage per phase V = 230ang 0deg VRated current
Ia = S 3V
= 9058ang -3687deg A
LoadGen
Xs =2
Eg Vt =400 radic3 V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2126
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
21
|Ia| = 9058 A
(iii)
Excitation voltage
Eg = 230 ang 0deg + j2(9058ang -3687deg)= 3684 ang 2318degV
|Eg| = 3684 V
Power angle = 2318o
(iv)
P = deg90sin2
23043683
(max occurs when δ = 90o)
= 1271 kW
Ex 22A 3 phase 4 pole synchronous machine with negligible stator winding resistance and a
synchronous reactance of 2 per phase is connected to a 480V 60 Hz 3phase supply
(infinite bus) and operated as a generator It is currently generating 80kW with power
factor 08 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)(iv) What is the maximum power the machine can generate with this value
excitation(v) What is the rotational speed of this generator
Soln (i)
(ii)P = 80kW Apparent power S 100 ang -3687
o kVA
Rated voltage per phase
V 277ang 0deg V
LoadGen
Xs =2
Eg Vt = 480V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2226
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
22
Rated current
Ia = S 3V
= 9058ang -3687deg A
|Ia| = 12028 A
(iii)
Excitation voltage
Eg = 277 ang 0deg + j2(12028ang -3687deg)
= 4632 ang 2455degV|Eg| = 4632 V
Power angle = 2455o
(iv)
P = deg90sin2
27724633
(max occurs when δ = 90o)
= 19238 kW
(v)Synchronous speed ωs = 2 π fp = 1885 rads
= 1800 revmin
Ex 23 (Jan 14)
A 3 phase synchronous machine with negligible stator winding resistance and a synchronous
reactance of 25 per phase is connected to a 20kV 50 Hz 3phase supply (infinite bus) and
operated as a generator It is currently generating 100MW with power factor 085 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)(iv) What minimum value of excitation voltage (E) would be required for this machine to
deliver 200MW into a 20kV bus assuming a 20deg stability margin must be maintained onpower angle (δ)
Soln
(i)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2326
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
23
(ii)
P = 100 MW Apparent power S = 100 085
= 11765 ang -3179o MVA
Rated voltage per phase
V= 11547ang 0deg VRated current
Ia = S 3V
= 3396 103ang -3179deg A
|Ia| = 3396 103A
(iii)
Excitation voltage
Eg = 11547 ang 0deg + j25(3396 103ang -3179deg)
= 17570 ang 2425degV|Eg| = 17570 V
Power angle = 2425o V
(iv)
P = )2090sin(52
3degminusdegt gV E
(max occurs when δ = 90o stability margin 20o)
200 MW = )2090sin(52
547113degminusdegg E
200 MW = 38130209397085613 gg E E =
|Eg| = 15360 V
LoadGen
Xs =25
Eg Vt =11547V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2426
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
24
Ex 24A manufacturing plant fed with a 400V 50Hz three phase runs from 0900 to 1800
Monday to Friday The factoryrsquos load is normally 80kVA with power factor 08 lagging
during these hours Once a month they need to run a special heating cycle which adds a
further 20kW to the factory load for a 2 hour period This heating cycle is timer
controlled and fully automated but the timer is currently set to run on the third Friday
afternoon of the month Outside of normal business hours and at weekends the factory
load is negligible The firm has a contractually agreed Maximum Import Capacity (MIC)
of 120kVA
bull Use the attached table of Low Voltage Maximum Demand charges to calculate the
firmrsquos electricity bill (exclusive of VAT) for the month of February (28 days)
bull Identify three steps that this firm could take to reduce their expenditure on
electricity assuming they stick with the same tariff Calculate the potential
savings
983116983151983159 983126983151983148983156983137983143983141 983117983137983160983145983149983157983149 983108983141983149983137983150983140 983107983144983137983154983143983141983155
983108983137983145983148983161 983123983156983137983150983140983145983150983143 983107983144983137983154983143983141 991404302983140983137983161
983120983123983119 983116983141983158983161
983120983123983119 983116983141983158983161 983152983141983154 983115983126983105 983151983142 983117983113983107 983152983141983154 983140983137983161 9914040036983097983147983126983105983140983137983161
983117983137983160983145983149983157983149 983113983149983152983151983154983156 983107983137983152983137983139983145983156983161 (983117983113983107) 983107983144983137983154983143983141983155
983108983137983145983148983161 983107983144983137983154983143983141 983152983141983154 983147983126983105 983151983142 983117983113983107 9914040072983147983126983105983140983137983161
983108983141983149983137983150983140 983107983144983137983154983143983141983155 983118983151983158983085983110983141983138 983117983137983154983085983119983139983156
983117983151983150983156983144983148983161 983107983144983137983154983143983141 983152983141983154 983147983127 983151983142 983149983141983156983141983154983141983140 983117983108 983145983150 983137
983149983151983150983156983144983148983161 983152983141983154983145983151983140 (983123983157983138983146983141983139983156 983156983151 983137 983149983145983150983145983149983157983149 983139983144983137983154983143983141 983151983142
30983147983127) 991404189830975983147983127983149983151983150983156983144 991404000
983108983137983161983156983145983149983141 983147983127983144 (983125983150983145983156) 983107983144983137983154983143983141 (0898309800 983085 19830979830985983097) 1480983139983147983127983144 1275983139983147983127983144
983118983145983143983144983156983145983149983141 983147983127983144 (983125983150983145983156) 983118983145983143983144983156 983107983144983137983154983143983141983155 (2098309800983085079830985983097) 64983097983139983147983127983144 64983097983139983147983127983144
983127983137983156983156983148983141983155983155 983125983150983145983156 983120983154983145983139983141 (983147983126983105983122983144 983145983150 983141983160983139983141983155983155 983151983142 983151983150983141 983156983144983145983154983140
983151983142 983156983151983156983137983148 983147983127983144) 991404000714 983147983126983105983122983144 991404000714 983147983126983105983122983144
bull Table Low Voltage Maximum Demand Charges (Q6)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2526
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
25
SolnNumber of days in February = 28
Number of working hours = 459 = 180
Normal kW = 80kVA08 = 64kW
Total kWh = 64kW180 hours+20kW2 hours = 11560 kWh
Normal kvar =80kVA sqrt(1-08^2) = 48kvar = sqrt(802-64
2) = 48 kVAr
Total kvarh = 48kvar180 hours = 8640 kvarh
Chargeable kvarh = 8640-(11560)3=4786 kvarh
983116983151983159 983126983151983148983156983137983143983141 983117983137983160983145983149983157983149 983108983141983149983137983150983140
983107983144983137983154983143983141983155 983152983141983154 983122983137983156983141 983121983156983161 983120983141983154983145983151983140 983107983144983137983154983143983141
983108983137983145983148983161 983123983156983137983150983140983145983150983143 983107983144983137983154983143983141 991404302 983140983137983161 991404302000 1 28 991404983096456
983120983123983119 983116983141983158983161
983120983123983119 983116983141983158983161 983152983141983154 983115983126983105 983152983141983154 983140983137983161 9914040036983097 983147983126983105983140983137983161 99140400369830970 120 28 991404123983097983096
983117983137983160983145983149983157983149 983113983149983152983151983154983156 983107983137983152983137983139983145983156983161
(983117983113983107) 983107983144983137983154983143983141983155
983108983137983145983148983161 983107983144983137983154983143983141 983152983141983154 983147983126983105 983151983142 983117983113983107 9914040072 983147983126983105983140983137983161 991404007200 120 28 9914042419830972
983108983141983149983137983150983140 983107983144983137983154983143983141983155 983118983151983158983085983110983141983138
983117983151983150983156983144983148983161 983107983144983137983154983143983141 983152983141983154 983147983127 983151983142
983149983141983156983141983154983141983140 983117983108 983145983150 983137 983149983151983150983156983144983148983161
983152983141983154983145983151983140 (983123983157983138983146983141983139983156 983156983151 983137
983149983145983150983145983149983157983149 983139983144983137983154983143983141 983151983142 30983147983127) 991404189830975
983147983127983149983151983150983156
983144 99140418983097500 84 1 991404159830971983096 983108983137983161983156983145983149983141 983147983127983144 (983125983150983145983156) 983107983144983137983154983143983141
(0898309800 983085 19830979830985983097) 9914040148 983147983127983144 991404014800 11560 1
991404198309510983096
983096
983118983145983143983144983156983145983149983141 983147983127983144 (983125983150983145983156) 983118983145983143983144983156
983107983144983137983154983143983141983155 (2098309800983085079830985983097) 9914040064983097 983147983127983144 99140400649830970
983127983137983156983156983148983141983155983155 983125983150983145983156 983120983154983145983139983141 (983147983126983105983122983144
983145983150 983141983160983139983141983155983155 983151983142 983151983150983141 983156983144983145983154983140 983151983142 983156983151983156983137983148
983147983127983144) 991404000714 983147983126983137983154983144 991404000714 4787 1 991404341983096
983124983151983156983137983148 983106983145983148983148
983142983151983154
983110983141983138983154983157983137983154983161
9914042354983095
0
[16 Marks]
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2626
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
26
Potential Savings
bull Use Power Factor correction to eliminate Watt less unit charge and to reduce MIC
bull Shift Heating load to night time or weekend
bull Negotiate lower MIC
Potential savings from these measures
Eliminate Wattless charge save euro3417 month
Shift heating cycle to night time
Max demand it lowered by 20kW = saving of 20euro1895 = euro3790 month
Saving by moving 20kW to night rate = 20kW2 hours (euro0148-euro00649)= euro332
Negotiating a reduced MIC
If power factor correction is employed and the heating cycle is shifted to night time the
then maximum VA of the plant drops to close to its maximum kW demand of 64kW An
MIC of 80kVA should be sufficient to cover this Potential Saving = 40kVA
(euro0072+euro00369)28 = euro12196 per month
Total potential saving = euro3417+euro3790+euro332+euro12196=euro19735month
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 1926
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
19
synchronism Determine the armature current corresponding to this maximum
power
(d) With real generated power at 42MW compute the power angle δ the stator
current Ia and power factor when the field excitation current If is adjusted so that
the excitation voltage Eg is decreased to 7919 of the value determined in part
(a)
Soln
(a) Apparent power S = 51 MW ndash j 316Mvar
Rated voltage per phase
V = 1154 ang 0deg kV
Rated current
Ia = S 3V
= 1733 ang -3178deg AExcitation voltage
Eg = 11540 + j9(1733) ang -3178deg)= 237922 ang 3386deg
Power angle = 3386o
(b) When delivering 28 MW
δ = sin-1
)5411)(79223(3
828 = 1578
o
P =
s
g
X
V E 3sinδ
Ia = 13069ang2967o j8
= 16336ang-6033o A
Power factor = cos(6033) = 0495 lagging
(c) The maximum power occurs at δ = 90o
Pmax =s
g
X
V E 3
= 102 MW
Ia = 324885ang25875o A
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2026
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
20
The power factor cos(25875) = 0899 leading
(d) Eg = 13463 V
The power angle δ = sin-1
)5411)(84318(3
842
= 31o
Ia = 86674ang0o A
-Power factor = 0
Ex 21 (Aug 13) A 3 phase synchronous machine with negligible stator winding resistance and a
synchronous reactance of 2 per phase is connected to a 400V 50 Hz 3-phase supply (infinite
bus) and operated as a generator It is currently generating 50kW with power factor 08 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)
(iv) What is the maximum power the machine can generate with this value of excitation
Soln(i)
(ii)P = 50kW
Apparent power S = 625 ang -3687
o
kVA
Rated voltage per phase V = 230ang 0deg VRated current
Ia = S 3V
= 9058ang -3687deg A
LoadGen
Xs =2
Eg Vt =400 radic3 V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2126
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
21
|Ia| = 9058 A
(iii)
Excitation voltage
Eg = 230 ang 0deg + j2(9058ang -3687deg)= 3684 ang 2318degV
|Eg| = 3684 V
Power angle = 2318o
(iv)
P = deg90sin2
23043683
(max occurs when δ = 90o)
= 1271 kW
Ex 22A 3 phase 4 pole synchronous machine with negligible stator winding resistance and a
synchronous reactance of 2 per phase is connected to a 480V 60 Hz 3phase supply
(infinite bus) and operated as a generator It is currently generating 80kW with power
factor 08 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)(iv) What is the maximum power the machine can generate with this value
excitation(v) What is the rotational speed of this generator
Soln (i)
(ii)P = 80kW Apparent power S 100 ang -3687
o kVA
Rated voltage per phase
V 277ang 0deg V
LoadGen
Xs =2
Eg Vt = 480V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2226
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
22
Rated current
Ia = S 3V
= 9058ang -3687deg A
|Ia| = 12028 A
(iii)
Excitation voltage
Eg = 277 ang 0deg + j2(12028ang -3687deg)
= 4632 ang 2455degV|Eg| = 4632 V
Power angle = 2455o
(iv)
P = deg90sin2
27724633
(max occurs when δ = 90o)
= 19238 kW
(v)Synchronous speed ωs = 2 π fp = 1885 rads
= 1800 revmin
Ex 23 (Jan 14)
A 3 phase synchronous machine with negligible stator winding resistance and a synchronous
reactance of 25 per phase is connected to a 20kV 50 Hz 3phase supply (infinite bus) and
operated as a generator It is currently generating 100MW with power factor 085 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)(iv) What minimum value of excitation voltage (E) would be required for this machine to
deliver 200MW into a 20kV bus assuming a 20deg stability margin must be maintained onpower angle (δ)
Soln
(i)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2326
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
23
(ii)
P = 100 MW Apparent power S = 100 085
= 11765 ang -3179o MVA
Rated voltage per phase
V= 11547ang 0deg VRated current
Ia = S 3V
= 3396 103ang -3179deg A
|Ia| = 3396 103A
(iii)
Excitation voltage
Eg = 11547 ang 0deg + j25(3396 103ang -3179deg)
= 17570 ang 2425degV|Eg| = 17570 V
Power angle = 2425o V
(iv)
P = )2090sin(52
3degminusdegt gV E
(max occurs when δ = 90o stability margin 20o)
200 MW = )2090sin(52
547113degminusdegg E
200 MW = 38130209397085613 gg E E =
|Eg| = 15360 V
LoadGen
Xs =25
Eg Vt =11547V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2426
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
24
Ex 24A manufacturing plant fed with a 400V 50Hz three phase runs from 0900 to 1800
Monday to Friday The factoryrsquos load is normally 80kVA with power factor 08 lagging
during these hours Once a month they need to run a special heating cycle which adds a
further 20kW to the factory load for a 2 hour period This heating cycle is timer
controlled and fully automated but the timer is currently set to run on the third Friday
afternoon of the month Outside of normal business hours and at weekends the factory
load is negligible The firm has a contractually agreed Maximum Import Capacity (MIC)
of 120kVA
bull Use the attached table of Low Voltage Maximum Demand charges to calculate the
firmrsquos electricity bill (exclusive of VAT) for the month of February (28 days)
bull Identify three steps that this firm could take to reduce their expenditure on
electricity assuming they stick with the same tariff Calculate the potential
savings
983116983151983159 983126983151983148983156983137983143983141 983117983137983160983145983149983157983149 983108983141983149983137983150983140 983107983144983137983154983143983141983155
983108983137983145983148983161 983123983156983137983150983140983145983150983143 983107983144983137983154983143983141 991404302983140983137983161
983120983123983119 983116983141983158983161
983120983123983119 983116983141983158983161 983152983141983154 983115983126983105 983151983142 983117983113983107 983152983141983154 983140983137983161 9914040036983097983147983126983105983140983137983161
983117983137983160983145983149983157983149 983113983149983152983151983154983156 983107983137983152983137983139983145983156983161 (983117983113983107) 983107983144983137983154983143983141983155
983108983137983145983148983161 983107983144983137983154983143983141 983152983141983154 983147983126983105 983151983142 983117983113983107 9914040072983147983126983105983140983137983161
983108983141983149983137983150983140 983107983144983137983154983143983141983155 983118983151983158983085983110983141983138 983117983137983154983085983119983139983156
983117983151983150983156983144983148983161 983107983144983137983154983143983141 983152983141983154 983147983127 983151983142 983149983141983156983141983154983141983140 983117983108 983145983150 983137
983149983151983150983156983144983148983161 983152983141983154983145983151983140 (983123983157983138983146983141983139983156 983156983151 983137 983149983145983150983145983149983157983149 983139983144983137983154983143983141 983151983142
30983147983127) 991404189830975983147983127983149983151983150983156983144 991404000
983108983137983161983156983145983149983141 983147983127983144 (983125983150983145983156) 983107983144983137983154983143983141 (0898309800 983085 19830979830985983097) 1480983139983147983127983144 1275983139983147983127983144
983118983145983143983144983156983145983149983141 983147983127983144 (983125983150983145983156) 983118983145983143983144983156 983107983144983137983154983143983141983155 (2098309800983085079830985983097) 64983097983139983147983127983144 64983097983139983147983127983144
983127983137983156983156983148983141983155983155 983125983150983145983156 983120983154983145983139983141 (983147983126983105983122983144 983145983150 983141983160983139983141983155983155 983151983142 983151983150983141 983156983144983145983154983140
983151983142 983156983151983156983137983148 983147983127983144) 991404000714 983147983126983105983122983144 991404000714 983147983126983105983122983144
bull Table Low Voltage Maximum Demand Charges (Q6)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2526
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
25
SolnNumber of days in February = 28
Number of working hours = 459 = 180
Normal kW = 80kVA08 = 64kW
Total kWh = 64kW180 hours+20kW2 hours = 11560 kWh
Normal kvar =80kVA sqrt(1-08^2) = 48kvar = sqrt(802-64
2) = 48 kVAr
Total kvarh = 48kvar180 hours = 8640 kvarh
Chargeable kvarh = 8640-(11560)3=4786 kvarh
983116983151983159 983126983151983148983156983137983143983141 983117983137983160983145983149983157983149 983108983141983149983137983150983140
983107983144983137983154983143983141983155 983152983141983154 983122983137983156983141 983121983156983161 983120983141983154983145983151983140 983107983144983137983154983143983141
983108983137983145983148983161 983123983156983137983150983140983145983150983143 983107983144983137983154983143983141 991404302 983140983137983161 991404302000 1 28 991404983096456
983120983123983119 983116983141983158983161
983120983123983119 983116983141983158983161 983152983141983154 983115983126983105 983152983141983154 983140983137983161 9914040036983097 983147983126983105983140983137983161 99140400369830970 120 28 991404123983097983096
983117983137983160983145983149983157983149 983113983149983152983151983154983156 983107983137983152983137983139983145983156983161
(983117983113983107) 983107983144983137983154983143983141983155
983108983137983145983148983161 983107983144983137983154983143983141 983152983141983154 983147983126983105 983151983142 983117983113983107 9914040072 983147983126983105983140983137983161 991404007200 120 28 9914042419830972
983108983141983149983137983150983140 983107983144983137983154983143983141983155 983118983151983158983085983110983141983138
983117983151983150983156983144983148983161 983107983144983137983154983143983141 983152983141983154 983147983127 983151983142
983149983141983156983141983154983141983140 983117983108 983145983150 983137 983149983151983150983156983144983148983161
983152983141983154983145983151983140 (983123983157983138983146983141983139983156 983156983151 983137
983149983145983150983145983149983157983149 983139983144983137983154983143983141 983151983142 30983147983127) 991404189830975
983147983127983149983151983150983156
983144 99140418983097500 84 1 991404159830971983096 983108983137983161983156983145983149983141 983147983127983144 (983125983150983145983156) 983107983144983137983154983143983141
(0898309800 983085 19830979830985983097) 9914040148 983147983127983144 991404014800 11560 1
991404198309510983096
983096
983118983145983143983144983156983145983149983141 983147983127983144 (983125983150983145983156) 983118983145983143983144983156
983107983144983137983154983143983141983155 (2098309800983085079830985983097) 9914040064983097 983147983127983144 99140400649830970
983127983137983156983156983148983141983155983155 983125983150983145983156 983120983154983145983139983141 (983147983126983105983122983144
983145983150 983141983160983139983141983155983155 983151983142 983151983150983141 983156983144983145983154983140 983151983142 983156983151983156983137983148
983147983127983144) 991404000714 983147983126983137983154983144 991404000714 4787 1 991404341983096
983124983151983156983137983148 983106983145983148983148
983142983151983154
983110983141983138983154983157983137983154983161
9914042354983095
0
[16 Marks]
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2626
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
26
Potential Savings
bull Use Power Factor correction to eliminate Watt less unit charge and to reduce MIC
bull Shift Heating load to night time or weekend
bull Negotiate lower MIC
Potential savings from these measures
Eliminate Wattless charge save euro3417 month
Shift heating cycle to night time
Max demand it lowered by 20kW = saving of 20euro1895 = euro3790 month
Saving by moving 20kW to night rate = 20kW2 hours (euro0148-euro00649)= euro332
Negotiating a reduced MIC
If power factor correction is employed and the heating cycle is shifted to night time the
then maximum VA of the plant drops to close to its maximum kW demand of 64kW An
MIC of 80kVA should be sufficient to cover this Potential Saving = 40kVA
(euro0072+euro00369)28 = euro12196 per month
Total potential saving = euro3417+euro3790+euro332+euro12196=euro19735month
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2026
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
20
The power factor cos(25875) = 0899 leading
(d) Eg = 13463 V
The power angle δ = sin-1
)5411)(84318(3
842
= 31o
Ia = 86674ang0o A
-Power factor = 0
Ex 21 (Aug 13) A 3 phase synchronous machine with negligible stator winding resistance and a
synchronous reactance of 2 per phase is connected to a 400V 50 Hz 3-phase supply (infinite
bus) and operated as a generator It is currently generating 50kW with power factor 08 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)
(iv) What is the maximum power the machine can generate with this value of excitation
Soln(i)
(ii)P = 50kW
Apparent power S = 625 ang -3687
o
kVA
Rated voltage per phase V = 230ang 0deg VRated current
Ia = S 3V
= 9058ang -3687deg A
LoadGen
Xs =2
Eg Vt =400 radic3 V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2126
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
21
|Ia| = 9058 A
(iii)
Excitation voltage
Eg = 230 ang 0deg + j2(9058ang -3687deg)= 3684 ang 2318degV
|Eg| = 3684 V
Power angle = 2318o
(iv)
P = deg90sin2
23043683
(max occurs when δ = 90o)
= 1271 kW
Ex 22A 3 phase 4 pole synchronous machine with negligible stator winding resistance and a
synchronous reactance of 2 per phase is connected to a 480V 60 Hz 3phase supply
(infinite bus) and operated as a generator It is currently generating 80kW with power
factor 08 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)(iv) What is the maximum power the machine can generate with this value
excitation(v) What is the rotational speed of this generator
Soln (i)
(ii)P = 80kW Apparent power S 100 ang -3687
o kVA
Rated voltage per phase
V 277ang 0deg V
LoadGen
Xs =2
Eg Vt = 480V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2226
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
22
Rated current
Ia = S 3V
= 9058ang -3687deg A
|Ia| = 12028 A
(iii)
Excitation voltage
Eg = 277 ang 0deg + j2(12028ang -3687deg)
= 4632 ang 2455degV|Eg| = 4632 V
Power angle = 2455o
(iv)
P = deg90sin2
27724633
(max occurs when δ = 90o)
= 19238 kW
(v)Synchronous speed ωs = 2 π fp = 1885 rads
= 1800 revmin
Ex 23 (Jan 14)
A 3 phase synchronous machine with negligible stator winding resistance and a synchronous
reactance of 25 per phase is connected to a 20kV 50 Hz 3phase supply (infinite bus) and
operated as a generator It is currently generating 100MW with power factor 085 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)(iv) What minimum value of excitation voltage (E) would be required for this machine to
deliver 200MW into a 20kV bus assuming a 20deg stability margin must be maintained onpower angle (δ)
Soln
(i)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2326
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
23
(ii)
P = 100 MW Apparent power S = 100 085
= 11765 ang -3179o MVA
Rated voltage per phase
V= 11547ang 0deg VRated current
Ia = S 3V
= 3396 103ang -3179deg A
|Ia| = 3396 103A
(iii)
Excitation voltage
Eg = 11547 ang 0deg + j25(3396 103ang -3179deg)
= 17570 ang 2425degV|Eg| = 17570 V
Power angle = 2425o V
(iv)
P = )2090sin(52
3degminusdegt gV E
(max occurs when δ = 90o stability margin 20o)
200 MW = )2090sin(52
547113degminusdegg E
200 MW = 38130209397085613 gg E E =
|Eg| = 15360 V
LoadGen
Xs =25
Eg Vt =11547V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2426
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
24
Ex 24A manufacturing plant fed with a 400V 50Hz three phase runs from 0900 to 1800
Monday to Friday The factoryrsquos load is normally 80kVA with power factor 08 lagging
during these hours Once a month they need to run a special heating cycle which adds a
further 20kW to the factory load for a 2 hour period This heating cycle is timer
controlled and fully automated but the timer is currently set to run on the third Friday
afternoon of the month Outside of normal business hours and at weekends the factory
load is negligible The firm has a contractually agreed Maximum Import Capacity (MIC)
of 120kVA
bull Use the attached table of Low Voltage Maximum Demand charges to calculate the
firmrsquos electricity bill (exclusive of VAT) for the month of February (28 days)
bull Identify three steps that this firm could take to reduce their expenditure on
electricity assuming they stick with the same tariff Calculate the potential
savings
983116983151983159 983126983151983148983156983137983143983141 983117983137983160983145983149983157983149 983108983141983149983137983150983140 983107983144983137983154983143983141983155
983108983137983145983148983161 983123983156983137983150983140983145983150983143 983107983144983137983154983143983141 991404302983140983137983161
983120983123983119 983116983141983158983161
983120983123983119 983116983141983158983161 983152983141983154 983115983126983105 983151983142 983117983113983107 983152983141983154 983140983137983161 9914040036983097983147983126983105983140983137983161
983117983137983160983145983149983157983149 983113983149983152983151983154983156 983107983137983152983137983139983145983156983161 (983117983113983107) 983107983144983137983154983143983141983155
983108983137983145983148983161 983107983144983137983154983143983141 983152983141983154 983147983126983105 983151983142 983117983113983107 9914040072983147983126983105983140983137983161
983108983141983149983137983150983140 983107983144983137983154983143983141983155 983118983151983158983085983110983141983138 983117983137983154983085983119983139983156
983117983151983150983156983144983148983161 983107983144983137983154983143983141 983152983141983154 983147983127 983151983142 983149983141983156983141983154983141983140 983117983108 983145983150 983137
983149983151983150983156983144983148983161 983152983141983154983145983151983140 (983123983157983138983146983141983139983156 983156983151 983137 983149983145983150983145983149983157983149 983139983144983137983154983143983141 983151983142
30983147983127) 991404189830975983147983127983149983151983150983156983144 991404000
983108983137983161983156983145983149983141 983147983127983144 (983125983150983145983156) 983107983144983137983154983143983141 (0898309800 983085 19830979830985983097) 1480983139983147983127983144 1275983139983147983127983144
983118983145983143983144983156983145983149983141 983147983127983144 (983125983150983145983156) 983118983145983143983144983156 983107983144983137983154983143983141983155 (2098309800983085079830985983097) 64983097983139983147983127983144 64983097983139983147983127983144
983127983137983156983156983148983141983155983155 983125983150983145983156 983120983154983145983139983141 (983147983126983105983122983144 983145983150 983141983160983139983141983155983155 983151983142 983151983150983141 983156983144983145983154983140
983151983142 983156983151983156983137983148 983147983127983144) 991404000714 983147983126983105983122983144 991404000714 983147983126983105983122983144
bull Table Low Voltage Maximum Demand Charges (Q6)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2526
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
25
SolnNumber of days in February = 28
Number of working hours = 459 = 180
Normal kW = 80kVA08 = 64kW
Total kWh = 64kW180 hours+20kW2 hours = 11560 kWh
Normal kvar =80kVA sqrt(1-08^2) = 48kvar = sqrt(802-64
2) = 48 kVAr
Total kvarh = 48kvar180 hours = 8640 kvarh
Chargeable kvarh = 8640-(11560)3=4786 kvarh
983116983151983159 983126983151983148983156983137983143983141 983117983137983160983145983149983157983149 983108983141983149983137983150983140
983107983144983137983154983143983141983155 983152983141983154 983122983137983156983141 983121983156983161 983120983141983154983145983151983140 983107983144983137983154983143983141
983108983137983145983148983161 983123983156983137983150983140983145983150983143 983107983144983137983154983143983141 991404302 983140983137983161 991404302000 1 28 991404983096456
983120983123983119 983116983141983158983161
983120983123983119 983116983141983158983161 983152983141983154 983115983126983105 983152983141983154 983140983137983161 9914040036983097 983147983126983105983140983137983161 99140400369830970 120 28 991404123983097983096
983117983137983160983145983149983157983149 983113983149983152983151983154983156 983107983137983152983137983139983145983156983161
(983117983113983107) 983107983144983137983154983143983141983155
983108983137983145983148983161 983107983144983137983154983143983141 983152983141983154 983147983126983105 983151983142 983117983113983107 9914040072 983147983126983105983140983137983161 991404007200 120 28 9914042419830972
983108983141983149983137983150983140 983107983144983137983154983143983141983155 983118983151983158983085983110983141983138
983117983151983150983156983144983148983161 983107983144983137983154983143983141 983152983141983154 983147983127 983151983142
983149983141983156983141983154983141983140 983117983108 983145983150 983137 983149983151983150983156983144983148983161
983152983141983154983145983151983140 (983123983157983138983146983141983139983156 983156983151 983137
983149983145983150983145983149983157983149 983139983144983137983154983143983141 983151983142 30983147983127) 991404189830975
983147983127983149983151983150983156
983144 99140418983097500 84 1 991404159830971983096 983108983137983161983156983145983149983141 983147983127983144 (983125983150983145983156) 983107983144983137983154983143983141
(0898309800 983085 19830979830985983097) 9914040148 983147983127983144 991404014800 11560 1
991404198309510983096
983096
983118983145983143983144983156983145983149983141 983147983127983144 (983125983150983145983156) 983118983145983143983144983156
983107983144983137983154983143983141983155 (2098309800983085079830985983097) 9914040064983097 983147983127983144 99140400649830970
983127983137983156983156983148983141983155983155 983125983150983145983156 983120983154983145983139983141 (983147983126983105983122983144
983145983150 983141983160983139983141983155983155 983151983142 983151983150983141 983156983144983145983154983140 983151983142 983156983151983156983137983148
983147983127983144) 991404000714 983147983126983137983154983144 991404000714 4787 1 991404341983096
983124983151983156983137983148 983106983145983148983148
983142983151983154
983110983141983138983154983157983137983154983161
9914042354983095
0
[16 Marks]
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2626
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
26
Potential Savings
bull Use Power Factor correction to eliminate Watt less unit charge and to reduce MIC
bull Shift Heating load to night time or weekend
bull Negotiate lower MIC
Potential savings from these measures
Eliminate Wattless charge save euro3417 month
Shift heating cycle to night time
Max demand it lowered by 20kW = saving of 20euro1895 = euro3790 month
Saving by moving 20kW to night rate = 20kW2 hours (euro0148-euro00649)= euro332
Negotiating a reduced MIC
If power factor correction is employed and the heating cycle is shifted to night time the
then maximum VA of the plant drops to close to its maximum kW demand of 64kW An
MIC of 80kVA should be sufficient to cover this Potential Saving = 40kVA
(euro0072+euro00369)28 = euro12196 per month
Total potential saving = euro3417+euro3790+euro332+euro12196=euro19735month
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2126
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
21
|Ia| = 9058 A
(iii)
Excitation voltage
Eg = 230 ang 0deg + j2(9058ang -3687deg)= 3684 ang 2318degV
|Eg| = 3684 V
Power angle = 2318o
(iv)
P = deg90sin2
23043683
(max occurs when δ = 90o)
= 1271 kW
Ex 22A 3 phase 4 pole synchronous machine with negligible stator winding resistance and a
synchronous reactance of 2 per phase is connected to a 480V 60 Hz 3phase supply
(infinite bus) and operated as a generator It is currently generating 80kW with power
factor 08 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)(iv) What is the maximum power the machine can generate with this value
excitation(v) What is the rotational speed of this generator
Soln (i)
(ii)P = 80kW Apparent power S 100 ang -3687
o kVA
Rated voltage per phase
V 277ang 0deg V
LoadGen
Xs =2
Eg Vt = 480V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2226
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
22
Rated current
Ia = S 3V
= 9058ang -3687deg A
|Ia| = 12028 A
(iii)
Excitation voltage
Eg = 277 ang 0deg + j2(12028ang -3687deg)
= 4632 ang 2455degV|Eg| = 4632 V
Power angle = 2455o
(iv)
P = deg90sin2
27724633
(max occurs when δ = 90o)
= 19238 kW
(v)Synchronous speed ωs = 2 π fp = 1885 rads
= 1800 revmin
Ex 23 (Jan 14)
A 3 phase synchronous machine with negligible stator winding resistance and a synchronous
reactance of 25 per phase is connected to a 20kV 50 Hz 3phase supply (infinite bus) and
operated as a generator It is currently generating 100MW with power factor 085 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)(iv) What minimum value of excitation voltage (E) would be required for this machine to
deliver 200MW into a 20kV bus assuming a 20deg stability margin must be maintained onpower angle (δ)
Soln
(i)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2326
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
23
(ii)
P = 100 MW Apparent power S = 100 085
= 11765 ang -3179o MVA
Rated voltage per phase
V= 11547ang 0deg VRated current
Ia = S 3V
= 3396 103ang -3179deg A
|Ia| = 3396 103A
(iii)
Excitation voltage
Eg = 11547 ang 0deg + j25(3396 103ang -3179deg)
= 17570 ang 2425degV|Eg| = 17570 V
Power angle = 2425o V
(iv)
P = )2090sin(52
3degminusdegt gV E
(max occurs when δ = 90o stability margin 20o)
200 MW = )2090sin(52
547113degminusdegg E
200 MW = 38130209397085613 gg E E =
|Eg| = 15360 V
LoadGen
Xs =25
Eg Vt =11547V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2426
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
24
Ex 24A manufacturing plant fed with a 400V 50Hz three phase runs from 0900 to 1800
Monday to Friday The factoryrsquos load is normally 80kVA with power factor 08 lagging
during these hours Once a month they need to run a special heating cycle which adds a
further 20kW to the factory load for a 2 hour period This heating cycle is timer
controlled and fully automated but the timer is currently set to run on the third Friday
afternoon of the month Outside of normal business hours and at weekends the factory
load is negligible The firm has a contractually agreed Maximum Import Capacity (MIC)
of 120kVA
bull Use the attached table of Low Voltage Maximum Demand charges to calculate the
firmrsquos electricity bill (exclusive of VAT) for the month of February (28 days)
bull Identify three steps that this firm could take to reduce their expenditure on
electricity assuming they stick with the same tariff Calculate the potential
savings
983116983151983159 983126983151983148983156983137983143983141 983117983137983160983145983149983157983149 983108983141983149983137983150983140 983107983144983137983154983143983141983155
983108983137983145983148983161 983123983156983137983150983140983145983150983143 983107983144983137983154983143983141 991404302983140983137983161
983120983123983119 983116983141983158983161
983120983123983119 983116983141983158983161 983152983141983154 983115983126983105 983151983142 983117983113983107 983152983141983154 983140983137983161 9914040036983097983147983126983105983140983137983161
983117983137983160983145983149983157983149 983113983149983152983151983154983156 983107983137983152983137983139983145983156983161 (983117983113983107) 983107983144983137983154983143983141983155
983108983137983145983148983161 983107983144983137983154983143983141 983152983141983154 983147983126983105 983151983142 983117983113983107 9914040072983147983126983105983140983137983161
983108983141983149983137983150983140 983107983144983137983154983143983141983155 983118983151983158983085983110983141983138 983117983137983154983085983119983139983156
983117983151983150983156983144983148983161 983107983144983137983154983143983141 983152983141983154 983147983127 983151983142 983149983141983156983141983154983141983140 983117983108 983145983150 983137
983149983151983150983156983144983148983161 983152983141983154983145983151983140 (983123983157983138983146983141983139983156 983156983151 983137 983149983145983150983145983149983157983149 983139983144983137983154983143983141 983151983142
30983147983127) 991404189830975983147983127983149983151983150983156983144 991404000
983108983137983161983156983145983149983141 983147983127983144 (983125983150983145983156) 983107983144983137983154983143983141 (0898309800 983085 19830979830985983097) 1480983139983147983127983144 1275983139983147983127983144
983118983145983143983144983156983145983149983141 983147983127983144 (983125983150983145983156) 983118983145983143983144983156 983107983144983137983154983143983141983155 (2098309800983085079830985983097) 64983097983139983147983127983144 64983097983139983147983127983144
983127983137983156983156983148983141983155983155 983125983150983145983156 983120983154983145983139983141 (983147983126983105983122983144 983145983150 983141983160983139983141983155983155 983151983142 983151983150983141 983156983144983145983154983140
983151983142 983156983151983156983137983148 983147983127983144) 991404000714 983147983126983105983122983144 991404000714 983147983126983105983122983144
bull Table Low Voltage Maximum Demand Charges (Q6)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2526
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
25
SolnNumber of days in February = 28
Number of working hours = 459 = 180
Normal kW = 80kVA08 = 64kW
Total kWh = 64kW180 hours+20kW2 hours = 11560 kWh
Normal kvar =80kVA sqrt(1-08^2) = 48kvar = sqrt(802-64
2) = 48 kVAr
Total kvarh = 48kvar180 hours = 8640 kvarh
Chargeable kvarh = 8640-(11560)3=4786 kvarh
983116983151983159 983126983151983148983156983137983143983141 983117983137983160983145983149983157983149 983108983141983149983137983150983140
983107983144983137983154983143983141983155 983152983141983154 983122983137983156983141 983121983156983161 983120983141983154983145983151983140 983107983144983137983154983143983141
983108983137983145983148983161 983123983156983137983150983140983145983150983143 983107983144983137983154983143983141 991404302 983140983137983161 991404302000 1 28 991404983096456
983120983123983119 983116983141983158983161
983120983123983119 983116983141983158983161 983152983141983154 983115983126983105 983152983141983154 983140983137983161 9914040036983097 983147983126983105983140983137983161 99140400369830970 120 28 991404123983097983096
983117983137983160983145983149983157983149 983113983149983152983151983154983156 983107983137983152983137983139983145983156983161
(983117983113983107) 983107983144983137983154983143983141983155
983108983137983145983148983161 983107983144983137983154983143983141 983152983141983154 983147983126983105 983151983142 983117983113983107 9914040072 983147983126983105983140983137983161 991404007200 120 28 9914042419830972
983108983141983149983137983150983140 983107983144983137983154983143983141983155 983118983151983158983085983110983141983138
983117983151983150983156983144983148983161 983107983144983137983154983143983141 983152983141983154 983147983127 983151983142
983149983141983156983141983154983141983140 983117983108 983145983150 983137 983149983151983150983156983144983148983161
983152983141983154983145983151983140 (983123983157983138983146983141983139983156 983156983151 983137
983149983145983150983145983149983157983149 983139983144983137983154983143983141 983151983142 30983147983127) 991404189830975
983147983127983149983151983150983156
983144 99140418983097500 84 1 991404159830971983096 983108983137983161983156983145983149983141 983147983127983144 (983125983150983145983156) 983107983144983137983154983143983141
(0898309800 983085 19830979830985983097) 9914040148 983147983127983144 991404014800 11560 1
991404198309510983096
983096
983118983145983143983144983156983145983149983141 983147983127983144 (983125983150983145983156) 983118983145983143983144983156
983107983144983137983154983143983141983155 (2098309800983085079830985983097) 9914040064983097 983147983127983144 99140400649830970
983127983137983156983156983148983141983155983155 983125983150983145983156 983120983154983145983139983141 (983147983126983105983122983144
983145983150 983141983160983139983141983155983155 983151983142 983151983150983141 983156983144983145983154983140 983151983142 983156983151983156983137983148
983147983127983144) 991404000714 983147983126983137983154983144 991404000714 4787 1 991404341983096
983124983151983156983137983148 983106983145983148983148
983142983151983154
983110983141983138983154983157983137983154983161
9914042354983095
0
[16 Marks]
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2626
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
26
Potential Savings
bull Use Power Factor correction to eliminate Watt less unit charge and to reduce MIC
bull Shift Heating load to night time or weekend
bull Negotiate lower MIC
Potential savings from these measures
Eliminate Wattless charge save euro3417 month
Shift heating cycle to night time
Max demand it lowered by 20kW = saving of 20euro1895 = euro3790 month
Saving by moving 20kW to night rate = 20kW2 hours (euro0148-euro00649)= euro332
Negotiating a reduced MIC
If power factor correction is employed and the heating cycle is shifted to night time the
then maximum VA of the plant drops to close to its maximum kW demand of 64kW An
MIC of 80kVA should be sufficient to cover this Potential Saving = 40kVA
(euro0072+euro00369)28 = euro12196 per month
Total potential saving = euro3417+euro3790+euro332+euro12196=euro19735month
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2226
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
22
Rated current
Ia = S 3V
= 9058ang -3687deg A
|Ia| = 12028 A
(iii)
Excitation voltage
Eg = 277 ang 0deg + j2(12028ang -3687deg)
= 4632 ang 2455degV|Eg| = 4632 V
Power angle = 2455o
(iv)
P = deg90sin2
27724633
(max occurs when δ = 90o)
= 19238 kW
(v)Synchronous speed ωs = 2 π fp = 1885 rads
= 1800 revmin
Ex 23 (Jan 14)
A 3 phase synchronous machine with negligible stator winding resistance and a synchronous
reactance of 25 per phase is connected to a 20kV 50 Hz 3phase supply (infinite bus) and
operated as a generator It is currently generating 100MW with power factor 085 lagging
(i) Draw a single phase equivalent circuit of the machine connected as a generator
(ii) What is the magnitude of the current being delivered
(iii) Determine the excitation voltage (E) of the machine and the power angle (δ)(iv) What minimum value of excitation voltage (E) would be required for this machine to
deliver 200MW into a 20kV bus assuming a 20deg stability margin must be maintained onpower angle (δ)
Soln
(i)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2326
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
23
(ii)
P = 100 MW Apparent power S = 100 085
= 11765 ang -3179o MVA
Rated voltage per phase
V= 11547ang 0deg VRated current
Ia = S 3V
= 3396 103ang -3179deg A
|Ia| = 3396 103A
(iii)
Excitation voltage
Eg = 11547 ang 0deg + j25(3396 103ang -3179deg)
= 17570 ang 2425degV|Eg| = 17570 V
Power angle = 2425o V
(iv)
P = )2090sin(52
3degminusdegt gV E
(max occurs when δ = 90o stability margin 20o)
200 MW = )2090sin(52
547113degminusdegg E
200 MW = 38130209397085613 gg E E =
|Eg| = 15360 V
LoadGen
Xs =25
Eg Vt =11547V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2426
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
24
Ex 24A manufacturing plant fed with a 400V 50Hz three phase runs from 0900 to 1800
Monday to Friday The factoryrsquos load is normally 80kVA with power factor 08 lagging
during these hours Once a month they need to run a special heating cycle which adds a
further 20kW to the factory load for a 2 hour period This heating cycle is timer
controlled and fully automated but the timer is currently set to run on the third Friday
afternoon of the month Outside of normal business hours and at weekends the factory
load is negligible The firm has a contractually agreed Maximum Import Capacity (MIC)
of 120kVA
bull Use the attached table of Low Voltage Maximum Demand charges to calculate the
firmrsquos electricity bill (exclusive of VAT) for the month of February (28 days)
bull Identify three steps that this firm could take to reduce their expenditure on
electricity assuming they stick with the same tariff Calculate the potential
savings
983116983151983159 983126983151983148983156983137983143983141 983117983137983160983145983149983157983149 983108983141983149983137983150983140 983107983144983137983154983143983141983155
983108983137983145983148983161 983123983156983137983150983140983145983150983143 983107983144983137983154983143983141 991404302983140983137983161
983120983123983119 983116983141983158983161
983120983123983119 983116983141983158983161 983152983141983154 983115983126983105 983151983142 983117983113983107 983152983141983154 983140983137983161 9914040036983097983147983126983105983140983137983161
983117983137983160983145983149983157983149 983113983149983152983151983154983156 983107983137983152983137983139983145983156983161 (983117983113983107) 983107983144983137983154983143983141983155
983108983137983145983148983161 983107983144983137983154983143983141 983152983141983154 983147983126983105 983151983142 983117983113983107 9914040072983147983126983105983140983137983161
983108983141983149983137983150983140 983107983144983137983154983143983141983155 983118983151983158983085983110983141983138 983117983137983154983085983119983139983156
983117983151983150983156983144983148983161 983107983144983137983154983143983141 983152983141983154 983147983127 983151983142 983149983141983156983141983154983141983140 983117983108 983145983150 983137
983149983151983150983156983144983148983161 983152983141983154983145983151983140 (983123983157983138983146983141983139983156 983156983151 983137 983149983145983150983145983149983157983149 983139983144983137983154983143983141 983151983142
30983147983127) 991404189830975983147983127983149983151983150983156983144 991404000
983108983137983161983156983145983149983141 983147983127983144 (983125983150983145983156) 983107983144983137983154983143983141 (0898309800 983085 19830979830985983097) 1480983139983147983127983144 1275983139983147983127983144
983118983145983143983144983156983145983149983141 983147983127983144 (983125983150983145983156) 983118983145983143983144983156 983107983144983137983154983143983141983155 (2098309800983085079830985983097) 64983097983139983147983127983144 64983097983139983147983127983144
983127983137983156983156983148983141983155983155 983125983150983145983156 983120983154983145983139983141 (983147983126983105983122983144 983145983150 983141983160983139983141983155983155 983151983142 983151983150983141 983156983144983145983154983140
983151983142 983156983151983156983137983148 983147983127983144) 991404000714 983147983126983105983122983144 991404000714 983147983126983105983122983144
bull Table Low Voltage Maximum Demand Charges (Q6)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2526
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
25
SolnNumber of days in February = 28
Number of working hours = 459 = 180
Normal kW = 80kVA08 = 64kW
Total kWh = 64kW180 hours+20kW2 hours = 11560 kWh
Normal kvar =80kVA sqrt(1-08^2) = 48kvar = sqrt(802-64
2) = 48 kVAr
Total kvarh = 48kvar180 hours = 8640 kvarh
Chargeable kvarh = 8640-(11560)3=4786 kvarh
983116983151983159 983126983151983148983156983137983143983141 983117983137983160983145983149983157983149 983108983141983149983137983150983140
983107983144983137983154983143983141983155 983152983141983154 983122983137983156983141 983121983156983161 983120983141983154983145983151983140 983107983144983137983154983143983141
983108983137983145983148983161 983123983156983137983150983140983145983150983143 983107983144983137983154983143983141 991404302 983140983137983161 991404302000 1 28 991404983096456
983120983123983119 983116983141983158983161
983120983123983119 983116983141983158983161 983152983141983154 983115983126983105 983152983141983154 983140983137983161 9914040036983097 983147983126983105983140983137983161 99140400369830970 120 28 991404123983097983096
983117983137983160983145983149983157983149 983113983149983152983151983154983156 983107983137983152983137983139983145983156983161
(983117983113983107) 983107983144983137983154983143983141983155
983108983137983145983148983161 983107983144983137983154983143983141 983152983141983154 983147983126983105 983151983142 983117983113983107 9914040072 983147983126983105983140983137983161 991404007200 120 28 9914042419830972
983108983141983149983137983150983140 983107983144983137983154983143983141983155 983118983151983158983085983110983141983138
983117983151983150983156983144983148983161 983107983144983137983154983143983141 983152983141983154 983147983127 983151983142
983149983141983156983141983154983141983140 983117983108 983145983150 983137 983149983151983150983156983144983148983161
983152983141983154983145983151983140 (983123983157983138983146983141983139983156 983156983151 983137
983149983145983150983145983149983157983149 983139983144983137983154983143983141 983151983142 30983147983127) 991404189830975
983147983127983149983151983150983156
983144 99140418983097500 84 1 991404159830971983096 983108983137983161983156983145983149983141 983147983127983144 (983125983150983145983156) 983107983144983137983154983143983141
(0898309800 983085 19830979830985983097) 9914040148 983147983127983144 991404014800 11560 1
991404198309510983096
983096
983118983145983143983144983156983145983149983141 983147983127983144 (983125983150983145983156) 983118983145983143983144983156
983107983144983137983154983143983141983155 (2098309800983085079830985983097) 9914040064983097 983147983127983144 99140400649830970
983127983137983156983156983148983141983155983155 983125983150983145983156 983120983154983145983139983141 (983147983126983105983122983144
983145983150 983141983160983139983141983155983155 983151983142 983151983150983141 983156983144983145983154983140 983151983142 983156983151983156983137983148
983147983127983144) 991404000714 983147983126983137983154983144 991404000714 4787 1 991404341983096
983124983151983156983137983148 983106983145983148983148
983142983151983154
983110983141983138983154983157983137983154983161
9914042354983095
0
[16 Marks]
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2626
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
26
Potential Savings
bull Use Power Factor correction to eliminate Watt less unit charge and to reduce MIC
bull Shift Heating load to night time or weekend
bull Negotiate lower MIC
Potential savings from these measures
Eliminate Wattless charge save euro3417 month
Shift heating cycle to night time
Max demand it lowered by 20kW = saving of 20euro1895 = euro3790 month
Saving by moving 20kW to night rate = 20kW2 hours (euro0148-euro00649)= euro332
Negotiating a reduced MIC
If power factor correction is employed and the heating cycle is shifted to night time the
then maximum VA of the plant drops to close to its maximum kW demand of 64kW An
MIC of 80kVA should be sufficient to cover this Potential Saving = 40kVA
(euro0072+euro00369)28 = euro12196 per month
Total potential saving = euro3417+euro3790+euro332+euro12196=euro19735month
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2326
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
23
(ii)
P = 100 MW Apparent power S = 100 085
= 11765 ang -3179o MVA
Rated voltage per phase
V= 11547ang 0deg VRated current
Ia = S 3V
= 3396 103ang -3179deg A
|Ia| = 3396 103A
(iii)
Excitation voltage
Eg = 11547 ang 0deg + j25(3396 103ang -3179deg)
= 17570 ang 2425degV|Eg| = 17570 V
Power angle = 2425o V
(iv)
P = )2090sin(52
3degminusdegt gV E
(max occurs when δ = 90o stability margin 20o)
200 MW = )2090sin(52
547113degminusdegg E
200 MW = 38130209397085613 gg E E =
|Eg| = 15360 V
LoadGen
Xs =25
Eg Vt =11547V
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2426
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
24
Ex 24A manufacturing plant fed with a 400V 50Hz three phase runs from 0900 to 1800
Monday to Friday The factoryrsquos load is normally 80kVA with power factor 08 lagging
during these hours Once a month they need to run a special heating cycle which adds a
further 20kW to the factory load for a 2 hour period This heating cycle is timer
controlled and fully automated but the timer is currently set to run on the third Friday
afternoon of the month Outside of normal business hours and at weekends the factory
load is negligible The firm has a contractually agreed Maximum Import Capacity (MIC)
of 120kVA
bull Use the attached table of Low Voltage Maximum Demand charges to calculate the
firmrsquos electricity bill (exclusive of VAT) for the month of February (28 days)
bull Identify three steps that this firm could take to reduce their expenditure on
electricity assuming they stick with the same tariff Calculate the potential
savings
983116983151983159 983126983151983148983156983137983143983141 983117983137983160983145983149983157983149 983108983141983149983137983150983140 983107983144983137983154983143983141983155
983108983137983145983148983161 983123983156983137983150983140983145983150983143 983107983144983137983154983143983141 991404302983140983137983161
983120983123983119 983116983141983158983161
983120983123983119 983116983141983158983161 983152983141983154 983115983126983105 983151983142 983117983113983107 983152983141983154 983140983137983161 9914040036983097983147983126983105983140983137983161
983117983137983160983145983149983157983149 983113983149983152983151983154983156 983107983137983152983137983139983145983156983161 (983117983113983107) 983107983144983137983154983143983141983155
983108983137983145983148983161 983107983144983137983154983143983141 983152983141983154 983147983126983105 983151983142 983117983113983107 9914040072983147983126983105983140983137983161
983108983141983149983137983150983140 983107983144983137983154983143983141983155 983118983151983158983085983110983141983138 983117983137983154983085983119983139983156
983117983151983150983156983144983148983161 983107983144983137983154983143983141 983152983141983154 983147983127 983151983142 983149983141983156983141983154983141983140 983117983108 983145983150 983137
983149983151983150983156983144983148983161 983152983141983154983145983151983140 (983123983157983138983146983141983139983156 983156983151 983137 983149983145983150983145983149983157983149 983139983144983137983154983143983141 983151983142
30983147983127) 991404189830975983147983127983149983151983150983156983144 991404000
983108983137983161983156983145983149983141 983147983127983144 (983125983150983145983156) 983107983144983137983154983143983141 (0898309800 983085 19830979830985983097) 1480983139983147983127983144 1275983139983147983127983144
983118983145983143983144983156983145983149983141 983147983127983144 (983125983150983145983156) 983118983145983143983144983156 983107983144983137983154983143983141983155 (2098309800983085079830985983097) 64983097983139983147983127983144 64983097983139983147983127983144
983127983137983156983156983148983141983155983155 983125983150983145983156 983120983154983145983139983141 (983147983126983105983122983144 983145983150 983141983160983139983141983155983155 983151983142 983151983150983141 983156983144983145983154983140
983151983142 983156983151983156983137983148 983147983127983144) 991404000714 983147983126983105983122983144 991404000714 983147983126983105983122983144
bull Table Low Voltage Maximum Demand Charges (Q6)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2526
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
25
SolnNumber of days in February = 28
Number of working hours = 459 = 180
Normal kW = 80kVA08 = 64kW
Total kWh = 64kW180 hours+20kW2 hours = 11560 kWh
Normal kvar =80kVA sqrt(1-08^2) = 48kvar = sqrt(802-64
2) = 48 kVAr
Total kvarh = 48kvar180 hours = 8640 kvarh
Chargeable kvarh = 8640-(11560)3=4786 kvarh
983116983151983159 983126983151983148983156983137983143983141 983117983137983160983145983149983157983149 983108983141983149983137983150983140
983107983144983137983154983143983141983155 983152983141983154 983122983137983156983141 983121983156983161 983120983141983154983145983151983140 983107983144983137983154983143983141
983108983137983145983148983161 983123983156983137983150983140983145983150983143 983107983144983137983154983143983141 991404302 983140983137983161 991404302000 1 28 991404983096456
983120983123983119 983116983141983158983161
983120983123983119 983116983141983158983161 983152983141983154 983115983126983105 983152983141983154 983140983137983161 9914040036983097 983147983126983105983140983137983161 99140400369830970 120 28 991404123983097983096
983117983137983160983145983149983157983149 983113983149983152983151983154983156 983107983137983152983137983139983145983156983161
(983117983113983107) 983107983144983137983154983143983141983155
983108983137983145983148983161 983107983144983137983154983143983141 983152983141983154 983147983126983105 983151983142 983117983113983107 9914040072 983147983126983105983140983137983161 991404007200 120 28 9914042419830972
983108983141983149983137983150983140 983107983144983137983154983143983141983155 983118983151983158983085983110983141983138
983117983151983150983156983144983148983161 983107983144983137983154983143983141 983152983141983154 983147983127 983151983142
983149983141983156983141983154983141983140 983117983108 983145983150 983137 983149983151983150983156983144983148983161
983152983141983154983145983151983140 (983123983157983138983146983141983139983156 983156983151 983137
983149983145983150983145983149983157983149 983139983144983137983154983143983141 983151983142 30983147983127) 991404189830975
983147983127983149983151983150983156
983144 99140418983097500 84 1 991404159830971983096 983108983137983161983156983145983149983141 983147983127983144 (983125983150983145983156) 983107983144983137983154983143983141
(0898309800 983085 19830979830985983097) 9914040148 983147983127983144 991404014800 11560 1
991404198309510983096
983096
983118983145983143983144983156983145983149983141 983147983127983144 (983125983150983145983156) 983118983145983143983144983156
983107983144983137983154983143983141983155 (2098309800983085079830985983097) 9914040064983097 983147983127983144 99140400649830970
983127983137983156983156983148983141983155983155 983125983150983145983156 983120983154983145983139983141 (983147983126983105983122983144
983145983150 983141983160983139983141983155983155 983151983142 983151983150983141 983156983144983145983154983140 983151983142 983156983151983156983137983148
983147983127983144) 991404000714 983147983126983137983154983144 991404000714 4787 1 991404341983096
983124983151983156983137983148 983106983145983148983148
983142983151983154
983110983141983138983154983157983137983154983161
9914042354983095
0
[16 Marks]
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2626
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
26
Potential Savings
bull Use Power Factor correction to eliminate Watt less unit charge and to reduce MIC
bull Shift Heating load to night time or weekend
bull Negotiate lower MIC
Potential savings from these measures
Eliminate Wattless charge save euro3417 month
Shift heating cycle to night time
Max demand it lowered by 20kW = saving of 20euro1895 = euro3790 month
Saving by moving 20kW to night rate = 20kW2 hours (euro0148-euro00649)= euro332
Negotiating a reduced MIC
If power factor correction is employed and the heating cycle is shifted to night time the
then maximum VA of the plant drops to close to its maximum kW demand of 64kW An
MIC of 80kVA should be sufficient to cover this Potential Saving = 40kVA
(euro0072+euro00369)28 = euro12196 per month
Total potential saving = euro3417+euro3790+euro332+euro12196=euro19735month
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2426
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
24
Ex 24A manufacturing plant fed with a 400V 50Hz three phase runs from 0900 to 1800
Monday to Friday The factoryrsquos load is normally 80kVA with power factor 08 lagging
during these hours Once a month they need to run a special heating cycle which adds a
further 20kW to the factory load for a 2 hour period This heating cycle is timer
controlled and fully automated but the timer is currently set to run on the third Friday
afternoon of the month Outside of normal business hours and at weekends the factory
load is negligible The firm has a contractually agreed Maximum Import Capacity (MIC)
of 120kVA
bull Use the attached table of Low Voltage Maximum Demand charges to calculate the
firmrsquos electricity bill (exclusive of VAT) for the month of February (28 days)
bull Identify three steps that this firm could take to reduce their expenditure on
electricity assuming they stick with the same tariff Calculate the potential
savings
983116983151983159 983126983151983148983156983137983143983141 983117983137983160983145983149983157983149 983108983141983149983137983150983140 983107983144983137983154983143983141983155
983108983137983145983148983161 983123983156983137983150983140983145983150983143 983107983144983137983154983143983141 991404302983140983137983161
983120983123983119 983116983141983158983161
983120983123983119 983116983141983158983161 983152983141983154 983115983126983105 983151983142 983117983113983107 983152983141983154 983140983137983161 9914040036983097983147983126983105983140983137983161
983117983137983160983145983149983157983149 983113983149983152983151983154983156 983107983137983152983137983139983145983156983161 (983117983113983107) 983107983144983137983154983143983141983155
983108983137983145983148983161 983107983144983137983154983143983141 983152983141983154 983147983126983105 983151983142 983117983113983107 9914040072983147983126983105983140983137983161
983108983141983149983137983150983140 983107983144983137983154983143983141983155 983118983151983158983085983110983141983138 983117983137983154983085983119983139983156
983117983151983150983156983144983148983161 983107983144983137983154983143983141 983152983141983154 983147983127 983151983142 983149983141983156983141983154983141983140 983117983108 983145983150 983137
983149983151983150983156983144983148983161 983152983141983154983145983151983140 (983123983157983138983146983141983139983156 983156983151 983137 983149983145983150983145983149983157983149 983139983144983137983154983143983141 983151983142
30983147983127) 991404189830975983147983127983149983151983150983156983144 991404000
983108983137983161983156983145983149983141 983147983127983144 (983125983150983145983156) 983107983144983137983154983143983141 (0898309800 983085 19830979830985983097) 1480983139983147983127983144 1275983139983147983127983144
983118983145983143983144983156983145983149983141 983147983127983144 (983125983150983145983156) 983118983145983143983144983156 983107983144983137983154983143983141983155 (2098309800983085079830985983097) 64983097983139983147983127983144 64983097983139983147983127983144
983127983137983156983156983148983141983155983155 983125983150983145983156 983120983154983145983139983141 (983147983126983105983122983144 983145983150 983141983160983139983141983155983155 983151983142 983151983150983141 983156983144983145983154983140
983151983142 983156983151983156983137983148 983147983127983144) 991404000714 983147983126983105983122983144 991404000714 983147983126983105983122983144
bull Table Low Voltage Maximum Demand Charges (Q6)
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2526
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
25
SolnNumber of days in February = 28
Number of working hours = 459 = 180
Normal kW = 80kVA08 = 64kW
Total kWh = 64kW180 hours+20kW2 hours = 11560 kWh
Normal kvar =80kVA sqrt(1-08^2) = 48kvar = sqrt(802-64
2) = 48 kVAr
Total kvarh = 48kvar180 hours = 8640 kvarh
Chargeable kvarh = 8640-(11560)3=4786 kvarh
983116983151983159 983126983151983148983156983137983143983141 983117983137983160983145983149983157983149 983108983141983149983137983150983140
983107983144983137983154983143983141983155 983152983141983154 983122983137983156983141 983121983156983161 983120983141983154983145983151983140 983107983144983137983154983143983141
983108983137983145983148983161 983123983156983137983150983140983145983150983143 983107983144983137983154983143983141 991404302 983140983137983161 991404302000 1 28 991404983096456
983120983123983119 983116983141983158983161
983120983123983119 983116983141983158983161 983152983141983154 983115983126983105 983152983141983154 983140983137983161 9914040036983097 983147983126983105983140983137983161 99140400369830970 120 28 991404123983097983096
983117983137983160983145983149983157983149 983113983149983152983151983154983156 983107983137983152983137983139983145983156983161
(983117983113983107) 983107983144983137983154983143983141983155
983108983137983145983148983161 983107983144983137983154983143983141 983152983141983154 983147983126983105 983151983142 983117983113983107 9914040072 983147983126983105983140983137983161 991404007200 120 28 9914042419830972
983108983141983149983137983150983140 983107983144983137983154983143983141983155 983118983151983158983085983110983141983138
983117983151983150983156983144983148983161 983107983144983137983154983143983141 983152983141983154 983147983127 983151983142
983149983141983156983141983154983141983140 983117983108 983145983150 983137 983149983151983150983156983144983148983161
983152983141983154983145983151983140 (983123983157983138983146983141983139983156 983156983151 983137
983149983145983150983145983149983157983149 983139983144983137983154983143983141 983151983142 30983147983127) 991404189830975
983147983127983149983151983150983156
983144 99140418983097500 84 1 991404159830971983096 983108983137983161983156983145983149983141 983147983127983144 (983125983150983145983156) 983107983144983137983154983143983141
(0898309800 983085 19830979830985983097) 9914040148 983147983127983144 991404014800 11560 1
991404198309510983096
983096
983118983145983143983144983156983145983149983141 983147983127983144 (983125983150983145983156) 983118983145983143983144983156
983107983144983137983154983143983141983155 (2098309800983085079830985983097) 9914040064983097 983147983127983144 99140400649830970
983127983137983156983156983148983141983155983155 983125983150983145983156 983120983154983145983139983141 (983147983126983105983122983144
983145983150 983141983160983139983141983155983155 983151983142 983151983150983141 983156983144983145983154983140 983151983142 983156983151983156983137983148
983147983127983144) 991404000714 983147983126983137983154983144 991404000714 4787 1 991404341983096
983124983151983156983137983148 983106983145983148983148
983142983151983154
983110983141983138983154983157983137983154983161
9914042354983095
0
[16 Marks]
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2626
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
26
Potential Savings
bull Use Power Factor correction to eliminate Watt less unit charge and to reduce MIC
bull Shift Heating load to night time or weekend
bull Negotiate lower MIC
Potential savings from these measures
Eliminate Wattless charge save euro3417 month
Shift heating cycle to night time
Max demand it lowered by 20kW = saving of 20euro1895 = euro3790 month
Saving by moving 20kW to night rate = 20kW2 hours (euro0148-euro00649)= euro332
Negotiating a reduced MIC
If power factor correction is employed and the heating cycle is shifted to night time the
then maximum VA of the plant drops to close to its maximum kW demand of 64kW An
MIC of 80kVA should be sufficient to cover this Potential Saving = 40kVA
(euro0072+euro00369)28 = euro12196 per month
Total potential saving = euro3417+euro3790+euro332+euro12196=euro19735month
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2526
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
25
SolnNumber of days in February = 28
Number of working hours = 459 = 180
Normal kW = 80kVA08 = 64kW
Total kWh = 64kW180 hours+20kW2 hours = 11560 kWh
Normal kvar =80kVA sqrt(1-08^2) = 48kvar = sqrt(802-64
2) = 48 kVAr
Total kvarh = 48kvar180 hours = 8640 kvarh
Chargeable kvarh = 8640-(11560)3=4786 kvarh
983116983151983159 983126983151983148983156983137983143983141 983117983137983160983145983149983157983149 983108983141983149983137983150983140
983107983144983137983154983143983141983155 983152983141983154 983122983137983156983141 983121983156983161 983120983141983154983145983151983140 983107983144983137983154983143983141
983108983137983145983148983161 983123983156983137983150983140983145983150983143 983107983144983137983154983143983141 991404302 983140983137983161 991404302000 1 28 991404983096456
983120983123983119 983116983141983158983161
983120983123983119 983116983141983158983161 983152983141983154 983115983126983105 983152983141983154 983140983137983161 9914040036983097 983147983126983105983140983137983161 99140400369830970 120 28 991404123983097983096
983117983137983160983145983149983157983149 983113983149983152983151983154983156 983107983137983152983137983139983145983156983161
(983117983113983107) 983107983144983137983154983143983141983155
983108983137983145983148983161 983107983144983137983154983143983141 983152983141983154 983147983126983105 983151983142 983117983113983107 9914040072 983147983126983105983140983137983161 991404007200 120 28 9914042419830972
983108983141983149983137983150983140 983107983144983137983154983143983141983155 983118983151983158983085983110983141983138
983117983151983150983156983144983148983161 983107983144983137983154983143983141 983152983141983154 983147983127 983151983142
983149983141983156983141983154983141983140 983117983108 983145983150 983137 983149983151983150983156983144983148983161
983152983141983154983145983151983140 (983123983157983138983146983141983139983156 983156983151 983137
983149983145983150983145983149983157983149 983139983144983137983154983143983141 983151983142 30983147983127) 991404189830975
983147983127983149983151983150983156
983144 99140418983097500 84 1 991404159830971983096 983108983137983161983156983145983149983141 983147983127983144 (983125983150983145983156) 983107983144983137983154983143983141
(0898309800 983085 19830979830985983097) 9914040148 983147983127983144 991404014800 11560 1
991404198309510983096
983096
983118983145983143983144983156983145983149983141 983147983127983144 (983125983150983145983156) 983118983145983143983144983156
983107983144983137983154983143983141983155 (2098309800983085079830985983097) 9914040064983097 983147983127983144 99140400649830970
983127983137983156983156983148983141983155983155 983125983150983145983156 983120983154983145983139983141 (983147983126983105983122983144
983145983150 983141983160983139983141983155983155 983151983142 983151983150983141 983156983144983145983154983140 983151983142 983156983151983156983137983148
983147983127983144) 991404000714 983147983126983137983154983144 991404000714 4787 1 991404341983096
983124983151983156983137983148 983106983145983148983148
983142983151983154
983110983141983138983154983157983137983154983161
9914042354983095
0
[16 Marks]
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2626
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
26
Potential Savings
bull Use Power Factor correction to eliminate Watt less unit charge and to reduce MIC
bull Shift Heating load to night time or weekend
bull Negotiate lower MIC
Potential savings from these measures
Eliminate Wattless charge save euro3417 month
Shift heating cycle to night time
Max demand it lowered by 20kW = saving of 20euro1895 = euro3790 month
Saving by moving 20kW to night rate = 20kW2 hours (euro0148-euro00649)= euro332
Negotiating a reduced MIC
If power factor correction is employed and the heating cycle is shifted to night time the
then maximum VA of the plant drops to close to its maximum kW demand of 64kW An
MIC of 80kVA should be sufficient to cover this Potential Saving = 40kVA
(euro0072+euro00369)28 = euro12196 per month
Total potential saving = euro3417+euro3790+euro332+euro12196=euro19735month
7212019 AC Power Tut Qns PDF
httpslidepdfcomreaderfullac-power-tut-qns-pdf 2626
DT0223 Electrical Engeneering Tut Qns Dr J Kearney
26
Potential Savings
bull Use Power Factor correction to eliminate Watt less unit charge and to reduce MIC
bull Shift Heating load to night time or weekend
bull Negotiate lower MIC
Potential savings from these measures
Eliminate Wattless charge save euro3417 month
Shift heating cycle to night time
Max demand it lowered by 20kW = saving of 20euro1895 = euro3790 month
Saving by moving 20kW to night rate = 20kW2 hours (euro0148-euro00649)= euro332
Negotiating a reduced MIC
If power factor correction is employed and the heating cycle is shifted to night time the
then maximum VA of the plant drops to close to its maximum kW demand of 64kW An
MIC of 80kVA should be sufficient to cover this Potential Saving = 40kVA
(euro0072+euro00369)28 = euro12196 per month
Total potential saving = euro3417+euro3790+euro332+euro12196=euro19735month