Electrical Power Systems

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
V p = 1100 volt = 1.1 kV; V S = 440 volt = 0.44 kV
(MVA) B = 0.025, V pB = 1.1 kV, V SB = 0.44 kV.
Base impedance on the 440 volt side of the transformer is
Z SB =
2
VSB (MVA) B
2
(. 044)
= = 7.744 ohm
(. 0025)
Per-unit leakage impedance referred to the low voltage side is
Z
(pu) s, eq
ZS =
ZSB
If Z p, eq referred to primary winding (HV side),
Zp, eq = a 2 .Zs, eq = N1
N 2
\ Zp, eq = 0. 375 78°
ohm.
Base impedance on the 1.1 KV side is
Z pB =
V pB 2
(MBA) B
Zp (pu) = Zp,
eq
Z
pB
006 . 78°
-3
= = 774 . ´ 10 78°
pu.
7. 744
HG
I
2 2
11
× Z = 006 78
KJ 044
. I S, eq
´ . ° HG . KJ
2
(.) 11
= = 48.4 W
0025 .
0. 375 78°
=
48. 4
= 7.74 × 10 –3 78° pu
Therefore, per-unit leakage impedance remains unchanged and this has been achieved by
specifying
VpB
=
V
Vp,
rated
=
V
11 .
= 2.5.
044 .
SB
s, rated
Example 5.2: igure 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.
T 1 : 3 kVA, 230/433 volts, x eq = 0.10 pu
T 2: 2 kVA, 440/120 volts, x eq = 0.10 pu
ig. 5.10: Single-phase circuit.