Chapter 5. Power System Components and Per Unit System

Power System Components and Per Unit System 97



For the reference phase a,

Ea = (Zg + ZL)Ia



...(5.1)



Figure 5.2 gives the single-phase

equivalent of a balanced three-phase network

of Fig. 5.1.

As the system is balanced, the voltage

and currents in the other phases have the

same magnitude but are shifted in phase by

120°.

Now consider the case where a threephase transformer forms part of a three-phase

system. Three identical single-phase two

winding transformers may be connected to

form a three-phase transformer. The windings

can be connected in four ways, Y–Y, Y–D,

D–Y and D–D. The phase of Y–Y and D–D

transformer, there is no phase shift between

corresponding quantities on the low and high

voltage windings. However, for D–Y or Y–D

transformers, there is always a phase shift.

Figure 5.3 shows a three-phase Y–Y bank.

Schematic representation of this

transformer is shown in Fig. 5.4(a) and Fig.

5.4(b) shows the single-phase equivalent of

three-phase Y–Y transformer and Fig. 5.4(c)

shows the single-line diagram.



Fig. 5.3: Three-phase two-winding Y–Y

transformer bank (a) Arrangements of core

and coil (b) Single-line diagram.



Fig. 5.4: (a) Schematic representation of three-phase g-g transformer, (b) Single-phase equivalent

of Y–Y transformer, (c) Single-line diagram of transformer.



98 Electrical Power Systems



If the transformer is Y–D connected as shown in Fig. 5.5 and for obtaining the single-phase

equivalent circuit, the delta side has to be replaced by an equivalent star connection as shown

in Fig. 5.6. From Fig. 5.6, it is also seen that VAN leads Van by 30°.



Fig. 5.5: Three-phase two-winding Y–D transformer bank.



Fig. 5.6: Y–D transformer with equivalent star connection.



Figure 5.7(a) shows the single-phase equivalent of g-D transformer and Fig. 5.7(b) shows the

single-line diagram.



Fig. 5.7: (a) Single-phase equivalent of Y-D transformer, (b Single-line diagram.



Power System Components and Per Unit System 99



5.3 THE PER-UNIT (pu) SYSTEM

Power system quantities such as current, voltage, impedance and power are often expressed in

per-unit values. For example, if base voltage if 220 KV is specified, then the voltage 210 KV is

210/220 = 0.954 pu. One major advantage of the per-unit is that by properly specifying base

quantities, the equivalent circuit of transformer can be simplified. When expressed in per-unit

values, the equivalent impedance of a transformer whether referred to primary or secondary,

is the same. Another advantage of the per-unit system is that the comparison of the characteristics

of the various electrical apparatus of different types and ratings is facilitated by expressing the

impedances in per-unit based on their ratings. When all the quantities are converted in per-unit

values, the different voltage levels disappear and power network involving synchronous

generators, transformers and lines reduces to a system of simple impedances.

Per-unit quantities are calculated as follows:

per-unit quantity =



actual quantity

base value of quantity



...(5.2)



I

S

V

Z

, Vpu =

, Ipu =

and Zpu =

IB

SB

ZB

VB



...(5.3)



Let us define,

Spu =



where, S (apparent power), V(voltage), I(current) and Z(impedance) are phasor or complex

quantities and denominators (i.e., SB, VB, IB and ZB) are always real numbers. To completely

define a per-unit system, minimum four base quantities are required.

Two independent base values can be arbitrarily selected at one point in a power system.

Usually, the three-phase base volt-ampere SB or (MVA)B and the line-to-line base voltage VB or

(kV)B are selected. The base value has the same units as the actual quantity and hence making

the per-unit quantity dimensionless. Then, in order for electrical laws to be valid in the per-unit

system, following relations must be used for other base values:

IB =



and



ZB =



(MVA) B



...(5.4)



3 (KV ) B



(KV ) B

IB



3



...(5.5)



Now substituting for IB from eqn. (5.4), the base impedance becomes

ZB =



(KV ) 2

B

(MVA) B



...(5.6)



Note that phase and line quantities expressed in per-unit values are the same, and the

circuit laws are valid, i.e.,

*

Spu = Vpu · I pu



Here



Spu = per-unit complex power = Ppu + j Qpu

Vpu = per-unit voltage

*

Ipu = complex conjugate of per-unit current Ipu



...(5.7)



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Electrical Power Systems



and also



Vpu = Zpu Ipu



...(5.8)



The power consumed by the load at its rated voltage can also be expressed by per-unit

impedance. The three-phase complex load power can be given as:

*

Sload(3f) = 3 Vphase I L



Here



...(5.9)



Sload(3f) = three-phase complex load power

Vphase = phase voltage

*

I L = complex conjugate of per-phase load current IL.



The phase load current can be given as:

IL =



Vphase



...(5.10)



ZL



where ZL is load impedance per phase.

Substituting IL from eqn. (5.10) in eqn. (5.9), we get,

Sload(3f) = 3. Vphase



\



Sload(3f) =



\



ZL =



FG V IJ

HZ K

phase



*



L



3 Vphase|2

|

*

ZL



3|Vphase|2

*

Sload (3f)



...(5.11)



Also, load impedance in per-unit can be given as

Zpu =



ZL

ZB



...(5.12)



Substituting ZL from eqn. (5.11) and ZB from eqn. (5.6) into eqn. (5.12), we obtain

Zpu =

Now

\



|VL–L| =



3|Vphase|2 (MVA) B

×

*

(KV) 2

Sload (3f)

B



...(5.13)



3 |Vphase|



2



3|Vphase| = |VLL|2



...(5.14)



Using eqns. (5.13) and (5.14), we get

Zpu =



\



Zpu =



|VL - L|2 (MVA) B

× *

(KV) 2 Sload(3f)

B

|Vpu|2

*

Sload (pu)



...(5.15)



Power System Components and Per Unit System 101



The impedance of generators, transformers and motors supplied by the manufacturer are

generally given in per-unit values on their own ratings. For power system analysis, all impedances

must be expressed in per-unit values on a common base.

When base quantities are changed from (MVA)B, old to (MVA)B, new and from (KV)B, old to

(KV)B, new, the new per-unit impedance can be given by

Zpu, new = Zpu, old



(MVA) B, new



×



(KV ) 2

B,old



(MVA) B, old (KV ) 2 new

B,



...(5.16)



5.4 PER-UNIT REPRESENTATION OF TRANSFORMER

It has been stated in section–5.2, that a threephase transformer can be represented by a

single-phase transformer for obtaining per phase

solution of the system. Figure 5.8 shows a singlephase transformer in terms of primary and

Fig. 5.8: Representation of single phase

secondary leakage reactances Zp and Zs and

transformer (magnetizing impedance neglected).

transformation ratio 1 : a.

Let us choose voltage base on the primary side VpB and on the secondary side VSB. Also

choose a common volt-ampere base of (VA)B.

Now



VpB

VSB



=



1

a



...(5.17)



As the (VA)B is common, we can also write



IpB

ISB



=a



ZpB =

ZSB =



VpB

IpB

VSB

ISB



...(5.18)

...(5.19)

...(5.20)



From Fig. 5.8, we can write,

VS = ES – ZS IS

Ep = Vp – ZpIp

Also



...(5.21)

...(5.22)



Es = a.Ep



...(5.23)



Substituting Es from eqn. (5.23) into eqn. (5.24), we obtain

VS = aEp – ZSIS



...(5.24)



Substituting Ep from eqn. (5.22) into eqn. (5.24), we get,

VS = a(Vp – ZpIp) – ZSIS

Eqn. (5.25) can be converted in per-unit form, i.e.,



...(5.25)



102



Electrical Power Systems



Vs(pu) VSB = a Vp (pu )VpB - Zp (pu ) ZpB Ip (pu ) IpB

– ZS(pu).ZSB IS(pu) ISB



...(5.28)



or

Dividing eqn. (5.26) by VSB and using the base relationships of eqns. (5.17), (5.18), (5.19) and

(5.20), we get.

VS(pu) = Vp(pu) – Ip(pu) Zp(pu) – IS(pu) ZS(pu)



...(5.27)



Now we can write



Ip

Is

\

\



Ip

IpB



=



=



IpB

ISB



=a



Is

ISB



Ip(pu) = Is(pu) = I(pu)



...(5.28)



Using eqns. (5.27) and (5.28) we get,

Vs(pu) = Vp(pu) – I(pu) Z(pu)

where



Z(pu) = Zp(pu) + Zs(pu)



...(5.29)

...(5.30)



Figure 5.9 shows the per-unit equivalent circuit of the Fig. 5.9: Per-unit equivalent circuit

transformer.

of single-phase transformer.

Z(pu) can be determined from the equivalent impedance

on primary or secondary side of a transformer.

On the primary side,

Z 1 = Zp +

\



ZS

a2



Zp

Z1

ZS

=

+

Z pB

ZpB ZpB a 2

ZS

ZSB



\



Z1(pu) = Zp(pu) +



\



Z1(pu) = Zp(pu) + ZS(pu) = Z(pu)



...(5.31)



Similarly on the secondary side,

Z2(pu) = ZS(pu) + Zp(pu) = Z(pu)



...(5.32)



Therefore, per-unit impedance of a transformer is the same whether computed from primary

or secondary side.

Example 5.1: A single phase two-winding transformer is rated 25 kVA, 1100/440 volts, 50 Hz.

The equivalent leakage impedance of the transformer referred to the low voltage side is 0.06 78°

9. Using transformer rating as base values, determine the per-unit leakage impedance referred

to low voltage winding and referred to high voltage winding.



Power System Components and Per Unit System 103



Solution: Let us assume high voltage side is primary and low voltage side is secondary

windings.

Transformer rating = 25 kVA = 0.025 MVA

Vp = 1100 volt = 1.1 kV; VS = 440 volt = 0.44 kV

(MVA)B = 0.025, VpB = 1.1 kV, VSB = 0.44 kV.

Base impedance on the 440 volt side of the transformer is

ZSB =



2

VSB

(0.44) 2

=

= 7.744 ohm

(MVA) B (0.025)



Per-unit leakage impedance referred to the low voltage side is

(pu)

ZS =



Zs, eq

ZSB



=



0.06 78°

7.744



= 7.74 ´ 10 -3 78° pu .



If Zp, eq referred to primary winding (HV side),



FG N IJ × Z

HN K

2



Zp, eq = a2.Zs, eq =

\



1

2



S, eq



=



.

FG 11 IJ

H 0.44 K



2



´ 0.06 78°



Zp, eq = 0.375 78° ohm.



Base impedance on the 1.1 KV side is

ZpB =

Zp(pu) =



2

VpB



(MBA) B



Zp, eq

ZpB



=



=



(1.1) 2

= 48.4 W

0.025



0.375 78°

= 7.74 × 10–3 78° pu

48.4



Therefore, per-unit leakage impedance remains unchanged and this has been achieved by

specifying



VpB

VSB



=



Vp, rated

Vs, rated



=



1.1

= 2.5.

0.44



Example 5.2: Figure 5.10 shows single line diagram of a single- phase circuit. Using the base

values of 3 kVA and 230 volts, draw the per-unit circuit diagram and determine the per-unit

impedances and the per-unit source voltage. Also calculate the load current both in per unit and

in Amperes.



T1 : 3 kVA, 230/433 volts, Neq = 0.10 pu

T2: 2 kVA, 440/120 volts, Neq = 0.10 pu



Fig. 5.10: Single-phase circuit.



104



Electrical Power Systems



Solution: First base values in each section have to be obtained.

Base MVA =

\



3

= 0.003 and this base value will remain same for the entire network.

1000

(MVA)B = 0.003



Also,



VB1 = 230 volts = 0.23 kV, as specified in Section-1.



When moving across a transformer, the voltage base is changed in proportion to the

transformer voltage ratings. Therefore,



FGH 433IJK ´ 230 = 433 volt = 0.433 kV

230

FH 120IJK ´ 433 volts = 118.09 volts = 0.11809 kV.

=G



VB2 =

and



VB3



440



ZB1 =



(VB1 )2

(0.23)2

= 17.63 ohm

=

( MVA ) B

0.003



ZB2 =



(VB2 )2

(0.433)2

=

= 62.5 ohm

( MVA )B

0.003



ZB3 =



(V B3 )2

(0.11809 )2

= 4.64 W

=

( MVA )B

0.003



Base current in Section-3 is

IB3 =

Given that

\



(MVA) B

0.003

=

kA = 25.4 Amp

(V B3 )

0.11809



x1, old = xeq = 0.10 pu

x1, new = 0.10 pu



Therefore, for transformer-T1, no change in per-unit value of leakage reactance.

For transformer, T2,

ZBT =

2



(0.44) 2

= 96.8 ohm

2

1000



FGH IJK



x2(ohm) = 0.1 × 96.8 = 9.68 ohm

ZB2 = 62.5 W

\



x2, new =

xline (pu) =



9.68

= 0.1548 pu

62.5

xline (ohm)

3

=

= 0.048 pu

ZB2

62.5



Power System Components and Per Unit System 105



ZL(pu) =



Z L (ohm) (0.8 + j 0.3)

=

ZB3

4.64



= (0.1724 + j 0.0646) pu.

Per-unit circuit is shown in Fig. 5.11.

IL(pu) =



VS (pu)

ZT (pu)



ZT(pu) = j 0.10 + j0.048 + j0.1548

+ 0.1724 + j0.0646



Fig. 5.11: Per-unit circuit.



= 0.4058 64.86°

\ IL(pu) =



0.956 0°

0.4058 64.86°



= 2.355 - 64.86° pu



IL(Amp) = IL(pu) × IB3 = 2.355 -64.86° ´ 25.4

= 59.83 -64.86° Amp.

Example 5.3: Figure 5.12 shows single-line diagram of a power system. The ratings of the

generators and transformers are given below:

G1 : 25 MVA, 6.6 kV, xg1 = 0.20 pu

G2 : 15 MVA, 6.6 kV, xg2 = 0.15 pu

G3 : 30 MVA, 13.2 kV, xg3 = 0.15 pu

T1 : 30 MVA, 6.6 D – 115 Y kV, xT1 = 0.10 pu

T2 : 15 MVA, 6.6 D – 115 Y kV, xT2 = 0.10 pu

T3 : Single-phase unit each rated 10 MVA, 6.9/69 kV, xT3 = 0.10 pu.

Draw per-unit circuit diagram using base values of 30 MVA and 6.6 kV in the circuit of

generator-1.



Fig. 5.12: Single-line diagram.



Solution: The chosen base values are 30 MVA and 6.6 kV in the generator 1 circuit.

Consequently, the transmission line base voltage of Line-1 is 115 kV. For generator-2 base

voltage is also 6.6 kV.



106



Electrical Power Systems



As the transformer T3 is rated 6.9 kV and 69 kV per phase, the line voltage ratio is

12

´ 115 =

6.9 3 69 3 = 12/120 kV. Therefore, base line voltage for generator-3 circuit is

120

11.5 kV.

Therefore, line kV base on H.V. side of transformer T3 is the same as that of transmission

line, i.e., 115 kV.

(MVA)B = 30



FGH IJK



xg1 = 0.2 ´



30

= 0.24 pu

25



xg2 = 0.15 ´

xg3 = 0.15 ´



30

= 0.30 pu

15



FGH 13.2IJK

115

.



2



= 0.20 pu



xT1 = 0.10 pu



FG 30IJ = 0.20 pu

H 15K

FH 120IJK = 0.11 pu

= 0.10 G

115



xT2 = 0.10



xT3



2



ZB, line =



(115) 2

= 440 W

30



xLine-1 =



120

= 0.27 pu

440



xLine-2 =



90

= 0.205 pu.

440



Figure 5.13 shows the per-unit circuit

diagram.



Fig. 5.13: Per-unit circuit diagram.



Example 5.4: A 100 MVA, 33 kV, three phase generator has a reactance of 15%. The generator

is connected to the motors through a transmission line and transformers as shown in Fig. 5.14.

Motors have rated inputs of 40 MVA, 30 MVA and 20 MVA at 30 kV with 20% reactance-each.

Draw the per-unit circuit diagram.

Solution:



Fig. 5.14: Single-line diagram.



Power System Components and Per Unit System 107



Solution: Assuming,

(MVA)B = 100 and (KV)B = 33 in the generator circuit.

\



xg = 0.15 pu

(KV)B, line = 33 ´



110

= 113.43 kV

32



In the motor circuit

(KV)B, motor = 113.43 ´

Now



ZB =



(33) 2

= 10.89 W

100



ZB, T1 = ZB,

\



32

= 33 kV.

110



T2



=



(32) 2

W = 9.309 W

110



xT1(W) = 0.08 × 9.309 W = 0.744 W



0.744

= 0.0683 pu

10.89



\



xT1, new (pu) =



\



xT2, new (pu) = 0.0683 pu

ZB, line =



\



xline(pu) =



(113.43) 2

= 128.66 W

100



60

= 0.466 pu

128.66



FG IJ

H K

100 F 30I

(pu) = 0.20 ´

´G J

30 H 33K

100 F 30I

(pu) = 0.20 ´

´G J

H K



xmotor 1 (pu) = 0.20 ´

xmotor-2

xmotor-3



100

30

´

40

33



2



= 0.413 pu

2



= 0.551 pu



2



= 0.826 pu.

20

33

Figure 5.15 shows the per-unit reactance diagram.



Fig. 5.15: Per-unit reactance diagram.