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 =
I
SB
Z
VB
B
B
...(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 ) 2B
(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)
100
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
(3 f )
...(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) 2B
Sload
(3 f )
...(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) 2B Sload
(3f)
|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 ) 2B,old
(MVA) B, old (KV ) 2B, new
...(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,
Also
VS = ES – ZS IS
...(5.21)
Ep = Vp – ZpIp
...(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
ZS(pu) =
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.
108
Electrical Power Systems
Example 5.5: Three single phase transformers are given with their name plate ratings. Determine
the reactance diagram Y–Y and Y–, connections, picking the voltage and power bases for the
three-phase bank.
Transformer ratings (1 B) : 1000 kVA
12.66/66 kV
xl = 0.10 pu
xm = 50 pu
Solution: For single-phase transformer
ZB1 =
(KV) 2B1
(MVA) B
ZB2 =
(KV)2B2
(MVA) B
(MVA)B =
1000
= 1.0
1000
(KV)B1 = 12.66 kV
(KV)B2 = 66 kV
\
ZB1 =
(12.66) 2
W = 160.27 W
1
ZB2 =
(66) 2
W = 4356 W
1
Actual reactances (referred to the primary) are:
xl = 0.1 × 160.27 W = 16.027 W
xm = 50 × 160.27 = 8013.5 W.
Let us consider now the three-phase interconnections of these single-phase transformers. If
we connect the primaries in Y (secondaries can be Y or D) and assume (MVA)B, 3f and (KV)B, L L,
then
(MVA)B, 3f = 3 × 1 = 3.0
(KV)B, L–L =
Therefore
\
ZB1 =
3 × 12.66 KV
(KV) 2B, L - L
(MVA) B, 3f
=
3 ´ (12.66) 2
= 160.27 W
3
xl =
16.027
= 0.1 pu
160.27
xm =
8013.5
= 50 pu
160.27
Reactance diagram of Y–Y and Y–D connections
is shown in Fig. 5.16. Note that reactance diagram
for D–Y and D–D is also same.
Fig. 5.16: Reactance diagram of three-phase
transformers (Y–Y, Y–D, D–Y and D–D).
Power System Components and Per Unit System 109
Example 5.6: Draw the per-unit impedance diagram of the system shown in Fig. 5.17. Assumed
base values are 100 MVA and 100 kV.
Fig. 5.17: Sample power system.
G1 : 50 MVA, 12.2 kV, xg1 = 0.10 pu
G2 : 20 MVA, 13.8 kV, xg2 = 0.10 pu
T1 : 80 MVA, 12.2/132 kV, xT1 = 0.10 pu
T2 : 40 MVA, 13.8/132 kV, xT2 = 0.10 pu
Load : 50 MVA, 0.80 pf lagging operating at 124 kV.
Solution: Base kV in the transmission line = 100 kV
12.2
= 9.24 kV.
132
13.8
Base kV in the generator circuits G2 = 100 ´
= 10.45 kV
132
Now, For G1,
Base kV in the generator circuit G1 = 100 ×
xg1, new = xg1, old ×
(MVA) B, new
(MVA) B, old
´
(KV) 2B, old
(KV) 2B, new
(MVA)B, new = (MVA)B = 100
(MVA)B, old = Rated MVA of G1 = 50 MVA
(KV)B, old = 12.2 kV
(KV)B, new = 9.24 kV
xg1, old = xg1 = 0.10 pu
\
FGH IJK pu = 0.3486 pu.
2
xg1, new = 0.10 ´
100
12.2
´
50
9.24
xg2, new = 0.10 ´
100
13.8
´
20
10.45
Similarly for G2,
FGH
IJK
2
pu = 0.8719 pu
FGH IJK pu = 0.2179 pu.
100 F 13.8 I
= 0.1 ´
´G
J pu = 0.33 pu.
40 H 10.45K
For T1,
xT1, new = 0.1 ´
For T2,
xT2, new
100
12.2
´
80
9.24
2
2
110
Electrical Power Systems
Base impedance of the transmission-line circuit,
ZB, line =
Z12(pu) =
(100) 2
= 100 ohm
100
Z12 (ohm) (4 + j 16)
=
= (0.04 + j0.16) pu
ZB, line
100
Z13(pu) = Z23(pu) =
(2 + j 8)
= (0.02 + j0.08) pu
100
The load is specified as:
S = 50(0.8 + j0.6) = (40 + j30) MVA.
(a) Series combination of resistance and inductance: Using eqn. (5.11),
*
ZLOAD
(ohm) =
*
ZLOAD
(pu) =
\
\
(124) 2
= 307.52 -36.87° ohm
(40 + j 30)
ZLOAD (ohm) 307.52 -36.87°
pu
=
ZB, line
100
ZLoad (pu) = (2.46 + j 1.845)pu
Rseries = 2.46 pu ; Xseries = 1.845 pu.
(b) Parallel combination of resistance and reactance
Rparallel =
(124) 2
= 384.4 ohm; \ Rparallel(pu) = 3.844 pu
40
(124) 2
= 512.5 ohm; Xparallel(pu) = 5.125 pu.
30
The reactance diagram is shown in Fig. 5.18. The load is represented as series combination
of R and L.
Xparallel =
Fig. 5.18: Reactance diagram of example 5.6.
Power System Components and Per Unit System 111
Example 5.7: Figure 5.19 shows a sample power system networks. Find the current supplied by
the generator, the transmission line current, the load current, the load voltage and the power
consumed by the load.
5 MVA
11.2/132 kV
NT1 = 0.10 pu
10 MVA
138/69 kV
NT2 = 0.10 pu
Fig. 5.19: Sample power system network for Ex-5.7.
Solution: Choose
(MVA)B = 100
(KV)B = 138
\
VB1 =
FGH 138IJK × 11.2 = 11.71 kV
132
VB2 = 138 kV
VB3 = 69 kV
\
\
ZB2 =
(138) 2
= 190.44 W
100
ZB3 =
(69) 2
= 47.61 W
100
Z (pu)
Line =
10 + j 10 10 + j 10
=
= 0.0525 (1 + j1) pu
ZB2
190.44
Z (pu)
Load =
30
30
= 0.63 pu
=
ZB3 47.61
Now using eqn. (5.16)
xT1, new =
Here
xT1,old ´ (KV)2B, old
(MVA) B, old
´
(MVA) B, new
(KV)2B, new
xT1, old = 0.10 pu, (KV)B, old = 11.2 kV,
(KV)B, new = 11.71 kV, (MVA)B, old = 5, (MVA)B,
\
xT1, new = 0.10 ´
xT2 = 0.10 ´
= 100
(112
. )2
100
´
pu = 1.83 pu.
5
(1171
. )2
100
= 1 pu
10
Note that old and new base voltage are name here.
Similarly
new
112
Electrical Power Systems
Finally the source voltage in per-unit,
11.2
= 0.956 pu.
|ES| =
1171
.
Figure 5.20 shows the impedance diagram of example 5.7.
Fig. 5.20: Impedance diagram of example 5.7.
0.956 0°
\
I(pu) =
\
I(pu) =
\
I(pu) = 0.3227 -76.68° pu
j 183
. + 0.0525 + j 0.0525 + j 1 + 0.63
pu
0.956 0°
0.956
=
pu
(0.6825 + j 2.8825) 2.962 76.68°
Load voltage V Load ( pu ) = 0.63 × 0.3227 -76.68° = 0.203 -76.68° pu.
\
PLoad ( pu ) = Z Load ( pu )|I ( pu )|2 = 0.63 × (0.3227)2 = 0.0656 pu
= 0.0656 × 100 = 6.56 MW load
Now
IB1 =
100 ´ 106
3
= 4934.6 Amp
´
3
117
. ´ 10 3
IB2 =
11.2
´ 4934.6 = 418.7 Amp
132
IB3 =
138
´ 418.7 = 837.4 Amp.
69
Generator current
\
Ig = |I(pu)| × IB1 = 0.3227 × 4934.6 = 1592.4 Amp
Transmission line current
|I2| = 0.3227 × 418.7 Amp = 135.11 Amp
Load current
|I3| = 0.3227 × 837.4 = 270.23 Amp
Load voltage,
VL(pu) = I(pu) × ZL(pu) = 0.3227 -76.68° × 0.63 pu
Power System Components and Per Unit System 113
VL(pu) = 0.2033 -76.68° pu
\
\
|VL| = 0.2033 × 69 KV = 14.02 KV (Line-to-line)
Example 5.8: The single line diagram of a three-phase power system is shown in Fig. 5.21.
Select a common base of 100 MVA and 13.8 kV on the generator side. Draw per-unit impedance
diagram.
Fig. 5.21: Single line diagram of example 5.8.
G
: 90 MVA, 13.8 kV, xg = 18%
T1 : 50 MVA, 13.8/220 kV, xT1 = 10%
T2
: 50 MVA, 220/11 kV, xT2 = 10%
T3 : 50 MVA, 13.8/132 kV, xT3 = 10%
T4 : 50 MVA, 132/11 kV, xT4 = 10%
M
: 80 MVA, 10.45 kV, xm = 20%
Load : 57 MVA, 0.8 pf (lagging) at 10.45 kV.
xline1 = 50 W
xline2 = 70 W
Solution: The generator rated voltage is given as the base voltage at bus 1. This fixes the
voltage bases for the other buses in accordance to the transformer turns ratios.
\
VB1 = 13.8 kV
VB2 = 13.8
FGH 220 IJK = 220 kV
13.8
Base voltage on the high voltage side of T2 is
VB3 = 220 kV
and on its Low voltage side,
VB4 = 220
Similarly,
Now Base MVA = 100
\
VB5 = VB6
FGH 11 IJK = 11 kV
220
FH 132 IJK = 132 kV.
= 13.8 G
13.8
xg1 = 0.18 ´
100
= 0.20 pu
90
114
Electrical Power Systems
FGH 100IJK = 0.20 pu
50
FH 100IJK = 0.20 pu
= 0.10 G
50
FH 100IJK = 0.20 pu
= 0.10 G
50
FH 100IJK = 0.20 pu
= 0.10 G
xT1 = 0.10
xT2
xT3
xT4
50
For motor, using eqn. (5.16)
xm, new (pu) = xm, old (pu) ×
Here
(MVA) B, new
(MVA) B, old
´
(KV) 2B, old
(KV) 2B, new
xm, old (pu) = 0.20, (MVA)B, old = 80, (KV)B, old = 10.45 kV
(MVA)B, new = 100, (KV)B, new = 11 kV
\
xm, new (pu) = 0.2 ´
FGH
100
10.45
´
80
11
IJK
2
= 0.2256 pu.
Base impedance for lines
\
ZB, 2–3 =
(VB2 ) 2
(220 ) 2
=
= 484 W
100
(MVA) B
ZB, 5–6 =
(VB5 ) 2
(132) 2
=
= 174.24 W .
(MVA) B
100
xline-1(pu) =
50
pu = 0.1033 pu
484
xline-2(pu) =
70
pu = 0.4017 pu.
174.24
The load is at 0.8 pf lagging is given by
SL(3f) = 57 36.87°
Load impedance is given by
ZL =
\
(V L - L )2
SL* ( 3f)
=
(10.45)2
57 -36.87°
ZL = (1.532 + j1.1495)W.
Power System Components and Per Unit System 115
Base impedance for the load is
\
ZB, load =
(11) 2
W = 1.21 W
100
ZL(pu) =
(1.532 + j 1.1495)
pu = (1.266 + j 0.95) pu
1.21
The per-unit equivalent circuit diagram is shown in Fig. 5.22.
Fig. 5.22: Per-unit impedance diagram of example 5.8.
5.5 METHODS OF VOLTAGE CONTROL
The methods for voltage control are the use of (i) Tap changing transformer, (ii) Regulating
transformers or Boosters, (iii) Shunt capacitors, (iv) Series capacitors, (v) FACTS devices.
In this book, first four category will be described.
5.5.1 Tap Changing Transformer
The main purpose of all power transformer and distribution transformers is to transform
electric energy from one voltage level to another. Practically all power and many distribution
transformers have taps for changing the turns ratio. Voltage magnitude is altered by changing
the tap setting and affects the distribution of VARS and may be used to control the flow of
reactive power. There are two types of tap changing transformers.
(i) Off-load tap changing transformers.
(ii) Tap changing under load (TCUL) transformers.
The off-load tap changing transformer requires the disconnection of the transformer when
the tap setting is to be changed. Figure 5.23 gives the connection of off-load tap changing
transformer. A typical off-load tap changing transformer might have four taps in addition to the
nominal setting.
Tap changing under load is used when changes in turn ratio may be frequent. Basically, a
TCUL transformer is a transformer with the ability to change taps while power is connected.
Figure 5.24 gives the diagram of on-load tap changing transformer. In the position shown in
Fig. 5.24, the voltage is maximum. To change the voltage, following operations are required: (i)
open A1, (ii) move selector switch P1 to the next contact, (iii) Close A1, (iv) open A2, (v) move
selector switch P2 to the next contact, (vi) close A2. These operations are required for one
change in tap position. Step-down units usually have TCUL in the low voltage winding and
de-energized taps in the high voltage winding.