AC Power Tut Qns PDF

26
7/21/2019 AC Power Tut Qns PDF http://slidepdf.com/reader/full/ac-power-tut-qns-pdf 1/26 DT022/3 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  resistor are connected across a 50 Hz, 1000 o  supply. (a) Calculate: (i) the impedance of the circuit (ii) the current in the circuit. (iii) the voltage drops, v r and v  L  across R and L. (b) Draw a phasor diagram showing e, v  , and v  L  .  (c) Det the active power, reactive powe and apparent power. Soln. (i) the impedance of the circuit is = 223.61 26.53 o   (ii) the current is I T  = e/Z T  = 1000 o  = 0.447 -26.56 o  A 223.6126.53 o  

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Transcript of AC Power Tut Qns PDF

Page 1: AC Power Tut Qns PDF

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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

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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

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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

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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)

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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)

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(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

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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

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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

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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

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= 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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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)

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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

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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

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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

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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

Page 2: AC Power Tut Qns PDF

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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

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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

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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)

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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)

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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

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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

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7212019 AC Power Tut Qns PDF

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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)

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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

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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)

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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]

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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

Page 3: AC Power Tut Qns PDF

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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

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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)

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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)

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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)

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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

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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

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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]

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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

Page 4: AC Power Tut Qns PDF

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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)

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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)

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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)

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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

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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

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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

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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

Page 5: AC Power Tut Qns PDF

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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)

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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)

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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

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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

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7212019 AC Power Tut Qns PDF

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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)

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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

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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

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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]

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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

Page 6: AC Power Tut Qns PDF

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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)

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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)

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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

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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)

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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]

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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

Page 7: AC Power Tut Qns PDF

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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)

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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

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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

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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]

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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

Page 8: AC Power Tut Qns PDF

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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)

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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

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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)

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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]

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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

Page 9: AC Power Tut Qns PDF

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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)

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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

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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

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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

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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

Page 10: AC Power Tut Qns PDF

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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)

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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

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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

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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

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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

Page 11: AC Power Tut Qns PDF

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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)

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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

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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)

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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]

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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

Page 12: AC Power Tut Qns PDF

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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)

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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

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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

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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]

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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

Page 13: AC Power Tut Qns PDF

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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)

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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

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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

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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]

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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

Page 14: AC Power Tut Qns PDF

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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

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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

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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

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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

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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

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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

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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

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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

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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)

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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

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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

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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

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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

Page 15: AC Power Tut Qns PDF

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

Page 16: AC Power Tut Qns PDF

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

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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

Page 17: AC Power Tut Qns PDF

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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

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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

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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

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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

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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

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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)

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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

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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

Page 18: AC Power Tut Qns PDF

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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

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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

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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

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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

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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

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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

Page 19: AC Power Tut Qns PDF

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

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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

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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

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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

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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

Page 20: AC Power Tut Qns PDF

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

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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

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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

Page 21: AC Power Tut Qns PDF

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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

Page 22: AC Power Tut Qns PDF

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

Page 23: AC Power Tut Qns PDF

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

Page 24: AC Power Tut Qns PDF

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

Page 25: AC Power Tut Qns PDF

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

Page 26: AC Power Tut Qns PDF

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