Power in Single Phase AC Circuitssiva/2020/ee381/Basic_Concepts.pdfthe load to improve the overall...
Transcript of Power in Single Phase AC Circuitssiva/2020/ee381/Basic_Concepts.pdfthe load to improve the overall...

Power in Single Phase AC Circuits
Let us consider the following circuit.
−+ v(t)
i(t)
LoadV
IθV θI
Letv(t) =
√2V sin(ωt + θV )
i(t) =√
2I sin(ωt + θI )

The instantaneous power delivered to the load is
p(t) = v(t)i(t)
p(t) = Vm sin(ωt + θV )Im sin(ωt + θI )
p(t) =VmIm
2(cos(θV − θI )− cos(2ωt + θV + θI ))
p(t) =VmIm
2cos(θV − θI )−
VmIm2
cos(2ωt + θV + θI )

ωt
p, v , i
v
i
Vm
Im
θIθV
φ
p
Figure: Voltage, current and power in RL circuit

Let θV − θI be φ.
p(t) = VI cosφ− VI cos(2ωt + θV + θI )
p(t) = VI cosφ− VI cos(2ωt + θV − θI + 2θI )
p(t) = VI cosφ− VI cos(2ωt + 2θI − φ)
p(t) = VI cosφ(1− cos(2ωt + 2θI ))︸ ︷︷ ︸pI
−VI sinφ sin(2ωt + 2θI )︸ ︷︷ ︸pII
pI has an average value of VI cosφ which is called the averagepower.
pII does not have an average. But it’s maximum value is VI sinφwhich is called reactive power.

ωt
pppI
pII
PQ
Figure: Power in RL circuit

PowerThe average power P is
P =VmIm
2cos(θV − θI ) = VI cos(φ)
where φ = θV − θI . Its unit is watts (W).The reactive power Q is
Q = VI sinφ
The apparent power S is
S  = VIIts unit is voltampere (VA).The ratio of real power (P) to apparent power is called as thepower factor (pf).
pf =VI cosφ
VI= cosφ
Since cosφ can never be greater than unity, P ≤ S .

Complex Power
Let us define voltage phasor and current phasor.
V = V∠θV , I = I∠θI
The complex power S isS = VI∗
S = V∠θV I∠− θI= VI∠(θV − θI )
S = VI cosφ+ VI sinφ
The real part of S is called the average power (P). The imaginarypart of S is called the reactive power (Q).
S = P + Q

Re
Im
V
IθV θI
φ
Re
Im
P
QS
φ
Figure: RL load
If V leads I (φ > 0), power factor is lagging.

Re
Im
V
I
θV = θI
Re
Im
P=S
Figure: Resisitive Load
If V and I are in phase (φ = 0), power factor is unity.

Re
Im
IV
θI θV
φ
Re
Im
P
QS
−φ
Figure: RC load
If I leads V (φ < 0), power factor is leading.

For two loads (inductive and capacitive) in parallel,
P1
Q1
S1 P2
Q2S2
ST QT
PT
PT = P1 + P2; QT = Q1 + Q2
ButST  6= S1+ S2

Power Factor Control
I If pf decreases, the current will increase to supply the samereal power.
I This will increase the line loss. (It is an additional cost to autility.)
I Capacitors which supply reactive power are connected inparallel to improve the power factor.
Load C
P
Q
S
−QcSnew
Qnewφnew

Example 1 : A singlephase inductive load draws to 1 kW at 0.6powerfactor lagging from a 230 V AC supply.
1. Find the current it draws.
2. Find the value of a capacitor to be connected in parallel withthe load to raise the power factor to 0.9 lagging. Determinethe current under this condition.
1.
I =1000
230× 0.6= 7.24 A
Q = 230× 7.24× 0.8 = 1.332 kVAr
I
Load
1 kW
1.332 kVAR
1.6652 kVA
53.13◦

2.pfnew = 0.9; φnew = 25.84
◦
Qnew = QL − QcQnew = P × tan 25.84◦ = 484.32 VAr
Qc = 847.68 VAr
Qc = V2ωC
C = 51 µF
I =1000
230× 0.9= 4.83 A
I
Load C
1 kW
QL
S
−QcSnewQnew
25.84◦

Power in Balanced Three Phase Circuits
Let va, vb and vc be the instantaneous voltages of a balanced threephase source.
va =√
2V sin(ωt + θV )
vb =√
2V sin(ωt + θV − 120◦)vc =
√2V sin(ωt + θV − 240◦)
When it supplies a balanced load,
ia =√
2I sin(ωt + θI )
ib =√
2I sin(ωt + θI − 120◦)ic =
√2I sin(ωt + θI − 240◦)

The instantaneous power is
p = vaia + vbib + vc ic
p =√
2Vp sin(ωt + θV )×√
2Ip sin(ωt + θI )√2Vp sin(ωt + θV − 120◦)×
√2Ip sin(ωt + θI − 120◦)√
2Vp sin(ωt + θV − 240◦)×√
2Ip sin(ωt + θI − 240◦)
p =VpIp cos(θV − θI ) + VpIp cos(2ωt + θV + θI )VpIp cos(θV − θI ) + VpIp cos(2ωt + θV + θI − 120◦)VpIp cos(θV − θI ) + VpIp cos(2ωt + θV + θI − 240◦)
p = 3VpIp cosφ
where φ = θV − θI .The instantaneous power in a 3 phase balanced system is constant.

ωt
p, v , i
p
Figure: Voltage, current and power in a RL load

The average/real power in a 3phase system is
P = 3VpIp cosφ Watts
In a Y connected load, VL =√
3Vp and IL = Ip,
P =√
3VLIL cosφ
In a ∆ connected load, VL = Vp and IL√
3Ip,
P =√
3VLIL cosφ
Therefore, the three phase real power is
P = 3VpIp cosφ =√
3VLIL cosφ

Since the instantaneous power in a 3phase balanced system isconstant, it does not mean that there is no reactive power. Stillthe instantaneous power of individual phases is pulsating.
The 3phase reactive power is
Q = 3VpIP sinφ =√
3VLIL sinφ VAr
The apparent power is
S  =√
P2 + Q2 = 3VpIp =√
3VLIL VA

Per Phase Analysis
If a three phase system is balanced and there is no mutualinductance between phases, it is enough to analyze it on per phasebasis.
1. Convert all ∆ connected sources and loads into equivalent Yconnections.
2. Solve for phase a variables using the phase a circuit withneutrals connected.
3. Other phase variables can be found from the phase a variablesusing the symmetry.
4. If necessary, find lineline variables from the original circuit.

Synchronous Machine Model
The per phase equivalent circuit of a synchronous machine is
+
−E δ
Xs IaRa
+
−
Vt 0◦

Transformer Model
The per phase equivalent circuit of a transformer is
Req Xeq
Rc Xm
I Since the impedance of the shunt path is larger, Rc and Xmare neglected.
I Since Req is much smaller than Xeq, Req can be eliminated.
Xeq

Example 2: Consider a system where a three phase 440 V, 50 Hzsource is supplying power to two loads. Load 1 is a ∆ connectedload with a phase impedance of 10∠30◦ Ω and load 2 is a Yconnected load with a phase impedance of 5∠36.87◦ Ω.
1. Find the line current and the overall power factor of thesystem.
2. Determine the capacitance per phase in µF of a three phasebank of delta connected capacitors to be added in parallel tothe load to improve the overall power factor unity. Find theline current under this condition.

1. The per phase equivalent circuit after using ∆ to Ytransformation,
−
+
440√3
IL
10 30◦
35 36.87◦
IL =440/
√3
10/3 30◦+
440/√
3
5 36.87◦
IL = 106.64− 68.6A
IL = 126.8 −32.75◦ A
pf = cos(−32.75◦) = 0.84 lag

2. The per phase equivalent circuit with a capacitor bank
−
+
440√3
IL
10 30◦
35 36.87◦ −
Xc3
To make overall power factor unity, IL must be in phase withthe voltage.
∴ Ic = 68.6A
Xc =3× 440√3× 68.6
= 11.11 Ω
C = 286.52 µF

Example 3: Consider a system where a three phase 400 V, 50 Hzsource is supplying power to two loads. Load 1 draws 5 kW at 0.8pf lagging and load 2 draws 5 kW at unity power factor. Thevoltage across the loads is 400 V.
1. Find the line current and the overall power factor.
2. Find the value of kVAR required from a bank of capacitorsconnected across the loads to improve the overall power factorto unity. Determine the line current under this condition.
1.
IL1 =5000√
3× 400× 0.8= 9 A
IL2 =5000√
3× 400× 1= 7.2; A
IL = 9 −36.87◦ + 7.2 0◦ = 15.38 −20.56◦ A
pf = 0.9363 lag
QT = Q1 + Q2 = 3.75 + 0 = 3.75 kVAR

2. To make overall power factor unity,
QC + QT = 0
QC = −3.75 kVAR
(ve indicates that the capacitor supplies reactive power.)
∴ QC = 3.75 kVAR
When pf is unity, S = P.
IL =PT√3VL
=10× 103√
3× 400= 14.4 A
IL = 14.4 0◦ A

per unit
The per unit is defined as
per unit =actual value in any unit
base value in the same unit
There are normally four quantities associated with a power system.
S ,V , I ,Z
How to find base quantities?
I Choose any two. Normally Sbase and Vbase are chosen.
I Find the remaining two using their relations.

Let us start with single phase.
Sb = S1φ MVA; Vb = V1φ kV
Ib =Sb(MVA)
Vb(kV)kA
Zb =Vb(kV)
Ib(kA)Ω
Substituting Ib in Zb,
Zb =V 2b ( in kV)
Sb( in MVA)Ω
Zp.u. =Zactual(Ω)
Zb(Ω)
∴ Zp.u. =Zactual(Ω)× Sb( 1φ MVA)
V 2b ( L N in kV)

For three phase.
Sb = S3φ MVA; Vb = VL−L kV
Ib =Sb(MVA)√
3Vb(kV)kA
Zb =Vb(kV)√
3× Ib(kA)Ω
Substituting Ib in Zb,
Zb =V 2b ( in kV)
Sb( in MVA)Ω
Zp.u. =Zactual(Ω)
Zb(Ω)
∴ Zp.u. =Zactual(Ω)× Sb( 3φ MVA)
V 2b (LL in kV)

S3φb = 3S1φb ; VbL−L =
√3VbL−N
S3φp.u. =S3φ
S3φb=
3× S1φ
3× S1φb= Sp.u.
Vp.u. =VL−LVbL−L
=
√3VL−N√3VbL−N
= Vp.u.
I If the voltage magnitude is 1 p.u., the lineline voltage is 1p.u. and the lineneutral voltage is also 1 p.u.
I Similarly, three phase power in p.u. and the single phasepower in p.u. are the same.
Zb =(VbL−L)
2
S3φb=
(√
3VbL−N )2
3× S1φb=
(VbL−N )2
S1φb

Impedance values of a component when given in per unit withoutspecified bases are generally understood to be based on the MVAand kV ratings of the component.
To change p.u. from one base to new base:
Zp.u. ∝SbV 2b
Zp.u. (new) = Zp.u. (given) ×Sb(new in MVA)
Sb(given in MVA)×
V 2b (given in kV)
V 2b (new in kV)
Advantages:
I The per unit values of impedance, voltage and current of atransformer are the same regardless of whether they arereferred to the HV side or LV side. This is possible bychoosing base voltages on either side of the transformer usingthe voltage ratio of the transformer.
I The factors√
3 and 3 get eliminated in the per unit powerand voltage the equations.

Example 3 : Let us do the same example in p.u.
1.Sb = 5 kVA; Vb = 400 V
P1p.u. =5
5= 1; P2p.u. =
5
5= 1
Since Ip.u. =Pp.u.
V p.u.× pf
IL1p.u. =P1
V × pf=
1
1× 0.8= 1.25
IL2p.u. =P2
V × pf=
1
1× 1= 1
IL = IL1 + IL2 = 1.25 −36.87◦ + 1 0◦ = 2.14 −20.56◦ p.u.
Ib =Sb√
3× Vb=
5× 103√3× 400
= 7.22 A.
IL = 2.14 −20.56◦ × 7.22 = 15.44 −20.56◦ A

2. To make unity pf, QC = QT .
QTp.u. = PTp.u.× tanφ = 2× tan(20.56◦) = 0.75 p.u.
QC = QCp.u.× Sb = 0.75× 5 = 3.75 kVAr
Capacitors supply reactive power.
When the problems to be solved are more complex, andparticularly when transformers are involved, the advantages ofcalculations in per unit are more apparent .