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 volt-ampere (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

But|ST | 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 single-phase inductive load draws to 1 kW at 0.6power-factor 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 R-L load

The average/real power in a 3-phase 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 3-phase balanced system isconstant, it does not mean that there is no reactive power. Stillthe instantaneous power of individual phases is pulsating.

The 3-phase 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 line-line 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 (L-L 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 line-line voltage is 1p.u. and the line-neutral 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 .