% the leading
I(1,:) = amps
I(2,:) = amps
I(3,:) = amps
currents.
.* ( 0.8 - j*0.6);
.* ( 1.0
);
.* ( 0.8 + j*0.6);
% Lagging
% Unity
% Leading
% Calculate VS referred to the primary side
% for each current and power factor.
aVS = VP - (Req.*I + j.*Xeq.*I);
% Refer the secondary voltages back to the
% secondary side using the turns ratio.
VS = aVS * (200/15);
% Plot the secondary voltage (in kV!) versus load
plot(amps,abs(VS(1,:)/1000),'b-','LineWidth',2.0);
hold on;
plot(amps,abs(VS(2,:)/1000),'k--','LineWidth',2.0);
plot(amps,abs(VS(3,:)/1000),'r-.','LineWidth',2.0);
title ('\bfSecondary Voltage Versus Load');
xlabel ('\bfLoad (A)');
ylabel ('\bfSecondary Voltage (kV)');
legend('0.8 PF lagging','1.0 PF','0.8 PF leading');
grid on;
hold off;
The resulting plot of secondary voltage versus load is shown below:
1.
A three-phase transformer bank is to handle 600 kVA and have a 34.5/13.8-kV voltage ratio. Find the
rating of each individual transformer in the bank (high voltage, low voltage, turns ratio, and apparent
power) if the transformer bank is connected to (a) Y-Y, (b) Y-∆, (c) ∆-Y, (d) ∆-∆, (e) open-∆, (f) openY—open-∆.
SOLUTION For the first four connections, the apparent power rating of each transformer is 1/3 of the total
apparent power rating of the three-phase transformer. For the open-∆ and open-Y—open-∆ connections,
the apparent power rating is a bit more complicated. The 600 kVA must be 86.6% of the total apparent
power rating of the two transformers, implying that the apparent power rating of each transformer must be
231 kVA.
The ratings for each transformer in the bank for each connection are given below:
Connection
Primary Voltage Secondary Voltage Apparent Power Turns Ratio
Y-Y
19.9 kV
7.97 kV
200 kVA
2.50:1
19.9 kV
13.8 kV
200 kVA
1.44:1
Y-∆
34.5 kV
7.97 kV
200 kVA
4.33:1
∆-Y
34.5 kV
13.8 kV
200 kVA
2.50:1
∆-∆
34.5 kV
13.8 kV
346 kVA
2.50:1
open-∆
19.9 kV
13.8 kV
346 kVA
1.44:1
open-Y—open-∆
Note: The open-Y—open-∆ answer assumes that the Y is on the high-voltage side; if the Y is on the lowvoltage side, the turns ratio would be 4.33:1, and the apparent power rating would be unchanged.
2.
A 13,800/480 V three-phase Y-∆-connected transformer bank consists of three identical 100-kVA
7967/480-V transformers. It is supplied with power directly from a large constant-voltage bus. In the
short-circuit test, the recorded values on the high-voltage side for one of these transformers are
VSC = 560 V
I SC = 12.6 A
PSC = 3300 W
(a) If this bank delivers a rated load at 0.85 PF lagging and rated voltage, what is the line-to-line voltage on
the primary of the transformer bank?
(b) What is the voltage regulation under these conditions?
(c) Assume that the primary voltage of this transformer bank is a constant 13.8 kV, and plot the secondary
voltage as a function of load current for currents from no-load to full-load. Repeat this process for
power factors of 0.85 lagging, 1.0, and 0.85 leading.
(d) Plot the voltage regulation of this transformer as a function of load current for currents from no-load to
full-load. Repeat this process for power factors of 0.85 lagging, 1.0, and 0.85 leading.
SOLUTION From the short-circuit information, it is possible to determine the per-phase impedance of the
transformer bank referred to the high-voltage side. The primary side of this transformer is Y-connected, so
the short-circuit phase voltage is
Vφ ,SC =
VSC 560 V
=
= 323.3 V
3
3
the short-circuit phase current is
I φ ,SC = I SC = 12.6 A
and the power per phase is
Pφ ,SC =
PSC
= 1100 W
3
Thus the per-phase impedance is
323.3 V
= 25.66 Ω
12.6 A
1100 W
= cos −1
= 74.3°
(323.3 V )(12.6 A )
Z EQ = REQ + jX EQ =
θ = cos−1
PSC
VSC I SC
Z EQ = REQ + jX EQ = 25.66∠74.3° Ω = 6.94 + j 24.7 Ω
2
REQ = 6.94 Ω
X EQ = j 24.7 Ω
(a) If this Y-∆ transformer bank delivers rated kVA (300 kVA) at 0.85 power factor lagging while the
secondary voltage is at rated value, then each transformer delivers 100 kVA at a voltage of 480 V and 0.85
PF lagging. Referred to the primary side of one of the transformers, the load on each transformer is
equivalent to 100 kVA at 7967 V and 0.85 PF lagging. The equivalent current flowing in the secondary of
one transformer referred to the primary side is
I φ ,S ′ =
100 kVA
= 12.55 A
7967 V
Iφ ,S ′ = 12.55∠ − 31.79° A
The voltage on the primary side of a single transformer is thus
′
′
Vφ ,P = Vφ ,S + I φ ,S Z EQ,P
Vφ ,P = 7967∠0° V + (12.55∠ − 31.79° A )( 6.94 + j 24.7 Ω ) = 8207∠1.52° V
The line-to-line voltage on the primary of the transformer is
VLL,P = 3 Vφ , P = 3 (8207 V ) = 14.22 kV
(b)
The voltage regulation of the transformer is
VR =
8207-7967
× 100% = 3.01%
7967
3
3. Three 25-kVA 24,000/277-V distribution transformers are connected in ∆-Y. The open-circuit test was
performed on the low-voltage side of this transformer bank, and the following data were recorded:
Vline,OC = 480 V
I line,OC = 4.10 A
P3φ ,OC = 945 W
The short-circuit test was performed on the high-voltage side of this transformer bank, and the following
data were recorded:
Vline,SC = 1600 V
I line,SC = 2.00 A
P3φ ,SC = 1150 W
(a) Find the per-unit equivalent circuit of this transformer bank.
(b) Find the voltage regulation of this transformer bank at the rated load and 0.90 PF lagging.
(c) What is the transformer bank’s efficiency under these conditions?
SOLUTION (a) The equivalent of this three-phase transformer bank can be found just like the equivalent
circuit of a single-phase transformer if we work on a per-phase bases. The open-circuit test data on the
low-voltage side can be used to find the excitation branch impedances referred to the secondary side of the
transformer bank. Since the low-voltage side of the transformer is Y-connected, the per-phase open-circuit
quantities are:
Vφ ,OC = 277 V
I φ ,OC = 4.10 A
Pφ ,OC = 315 W
The excitation admittance is given by
YEX =
I φ ,OC
=
Vφ ,OC
4.10 A
= 0.01480 S
277 V
The admittance angle is
θ = − cos −1
Pφ ,OC
Vφ ,OC I φ ,OC
= − cos −1
315 W
= −73.9°
( 277 V)( 4.10 A )
Therefore,
YEX = GC − jBM = 0.01483∠ − 73.9° = 0.00410 − j0.01422
RC = 1/ GC = 244 Ω
X M = 1/ BM = 70.3 Ω
The base impedance for a single transformer referred to the low-voltage side is
(V )
=
2
Z base,S
φ ,S
Sφ
=
( 277 V)2 = 3.069 Ω
25 kVA
so the excitation branch elements can be expressed in per-unit as
RC =
244 Ω
= 79.5 pu
3.069 Ω
XM =
4
70.3 Ω
= 22.9 pu
3.069 Ω
The short-circuit test data can be used to find the series impedances referred to the high-voltage side, since
the short-circuit test data was taken on the high-voltage side. Note that the high-voltage is ∆-connected, so
Vφ ,SC = VSC = 1600 V , I φ ,SC = I SC / 3 = 1.1547 A , and Pφ ,SC = PSC / 3 = 383 W .
Z EQ =
Vφ ,SC 1600 V
=
= 1385 Ω
I φ ,SC 1.155 A
θ = cos−1
Pφ ,SC
Vφ ,SC I φ ,SC
= cos −1
383 W
(1600 V )(1.155 A )
= 78.0°
Z EQ = REQ + jX EQ = 1385∠78.0° = 288 + j1355 Ω
The base impedance referred to the high-voltage side is
(V )
=
2
Z base,P
φ,P
Sφ
=
( 24,000 V )2 = 23, 040 Ω
25 kVA
The resulting per-unit impedances are
REQ =
288 Ω
= 0.0125 pu
23,040 Ω
X EQ =
1355 Ω
= 0.0588 pu
23,040 Ω
The per-unit, per-phase equivalent circuit of the transformer bank is shown below:
IP
+
VP
REQ
jXEQ
0.0125
j0.0588
RC
jXM
79.5
j22.9
-
IS
+
VS
-
(b) If this transformer is operating at rated load and 0.90 PF lagging, then current flow will be at an
angle of − cos −1 (0.9) , or –25.8°. The per-unit voltage at the primary side of the transformer will be
VP = VS + I S Z EQ = 1.0∠ 0° + (1.0∠ − 25.8°)(0.0125 + j 0.0588) = 1.038∠ 2.62°
The voltage regulation of this transformer bank is
VR =
(c)
1.038 − 1.0
× 100% = 3.8%
1.0
The output power of this transformer bank is
POUT = VS I S cos θ = (1.0 )(1.0)(0.9 ) = 0.9 pu
The copper losses are
PCU = I S 2 REQ = (1.0 ) (0.0125) = 0.0125 pu
2
5
The core losses are
V 2 (1.038)
= P =
= 0.0136 pu
79.5
RC
2
Pcore
Therefore, the total input power to the transformer bank is
PIN = POUT + PCU + Pcore = 0.9 + 0.0125 + 0.0136 = 0.926
and the efficiency of the transformer bank is
η=
4.
POUT
0.9
× 100% =
× 100% = 97.2%
PIN
0.926
A 20-kVA 20,000/480-V 60-Hz distribution transformer is tested with the following results:
Open-circuit test
Short-circuit test
(measured from secondary side)
(measured from primary side)
VOC = 480 V
VSC = 1130 V
IOC = 1.60 A
ISC = 1.00 A
PSC = 260 W
VOC = 305 W
(a) Find the per-unit equivalent circuit for this transformer at 60 Hz.
(b) What would the rating of this transformer be if it were operated on a 50-Hz power system?
(c) Sketch the equivalent circuit of this transformer referred to the primary side if it is operating at 50 Hz.
SOLUTION
(a)
The base impedance of this transformer referred to the primary side is
Z base,P =
(VP )2 = (20,000 V )2 = 20 kΩ
S
20 kVA
The base impedance of this transformer referred to the secondary side is
Z base,S
2
2
VS )
480 V )
(
(
=
=
S
20 kVA
= 11.52 Ω
The open circuit test yields the values for the excitation branch (referred to the secondary side):
YEX =
I φ ,OC
Vφ ,OC
θ = − cos−1
=
1.60 A
= 0.00333 S
480 V
POC
= − cos −1
VOC I OC
305 W
= −66.6°
(480 V )(1.60 A )
YEX = GC − jBM = 0.00333∠ − 66.6° = 0.00132 − j 0.00306
RC = 1/ GC = 757 Ω
X M = 1/ BM = 327 Ω
The excitation branch elements can be expressed in per-unit as
RC =
757 Ω
= 65.7 pu
11.52 Ω
XM =
327 Ω
= 28.4 pu
11.52 Ω
The short circuit test yields the values for the series impedances (referred to the primary side):
6
Z EQ =
VSC 1130 V
=
= 1130 Ω
I SC 1.00 A
θ = cos −1
PSC
= cos −1
VSC I SC
260 W
= 76.7°
(1130 V)(1.00 A )
Z EQ = REQ + jX EQ = 1130∠76.7° = 260 + j1100 Ω
The resulting per-unit impedances are
REQ =
260 Ω
= 0.013 pu
20,000 Ω
X EQ =
1100 Ω
= 0.055 pu
20,000 Ω
The per-unit equivalent circuit is
IP
+
VP
RC
jXM
65.7
j28.4
REQ
jXEQ
0.013
j0.055
-
IS
+
VS
-
(b) If this transformer were operated at 50 Hz, both the voltage and apparent power would have to be
derated by a factor of 50/60, so its ratings would be 16.67 kVA, 16,667/400 V, and 50 Hz.
(c)
The transformer parameters referred to the primary side at 60 Hz are:
RC = Z base RC ,pu = ( 20 kΩ )( 65.7 ) = 1.31 MΩ
X M = Z base X M ,pu = ( 20 kΩ )( 28.4 ) = 568 kΩ
REQ = Z base REQ ,pu = ( 20 kΩ )(0.013) = 260 Ω
X EQ = Z base X EQ ,pu = ( 20 kΩ )( 0.055) = 1100 Ω
At 50 Hz, the resistance will be unaffected but the reactances are reduced in direct proportion to the
decrease in frequency. At 50 Hz, the reactances are
50 Hz
60 Hz
50 Hz
=
60 Hz
XM =
(568 kΩ ) = 473 kΩ
X EQ
(1100 Ω) = 917 Ω
7
The resulting equivalent circuit referred to the primary at 50 Hz is shown below:
IP
+
VP
RC
1.31 MΩ
REQ
jXEQ
260 Ω
j917 Ω
jXM
IS
+
VS
j473 kΩ
-
-
8