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ELG3311: Solutions for Assignment 1
Problem 2-6:
A 15-kVA 8000/230-V distribution transformer has an impedance referred to the primary of 80 + j 300 Ω. The components of the excitation branch referred to the primary side are RC = 350 kΩ and XM = 70 kΩ. (a) If the primary voltage is 7967 V and the load impedance is ZL = 3.0 + j 1.5 Ω, what is the secondary voltage of the transformer? What is the voltage regulation of the transformer? (b) If the load is disconnected and a capacitor of –j 4.0 Ω is connected in its place, what is the secondary voltage of the transformer? What is its voltage regulation under these conditions?
Solution:
(a) The turns ratio is
a = 8000 / 230 = 34. 78
The load impedance referred to the primary side is
Z (^) LP = a^2 ZL =( 34. 78 )^2 ( 3. 0 + j 1. 5 )= 3629 + j 1815 Ω
The referred secondary current is
1. 87 29. 7 A
Z Z j j
V
I P
L
P i
P P S
The referred secondary voltage is
V (^) SP = ISPZLP =( 1. 87 ∠− 29. 7 °)( 3629 + j 1815 )= 7588 ∠− 3. 1 °V
The actual secondary voltage is
218. 2 3. 1 V
a
V
V
P S S
The voltage regulation is
× =
= P
S
P P S V
V V
VR
(b) The turns ratio is
a = 8000 / 230 = 34. 78
The load impedance referred to the primary side is
Z (^) LP = a^2 ZL =( 34. 78 )^2 (− j 4. 0 )=− j 4839 Ω
The referred secondary current is
1. 75 89. 0 A
Z Z j j
V
I P
L
P i
P P S
The referred secondary voltage is
V (^) SP = ISPZLP =( 1. 75 ∠ 89. 0 °)(− j 4839 )= 8468 ∠− 1. 0 °V
The actual secondary voltage is
243. 2 1. 0 V
a
V
V
P S S
The voltage regulation is
× =−
= P
S
P P S V
V V
VR
,
, C S
EQS R
R
,
, M S
EQS X
X
(b) The secondary current is
362. 3 36. 87 A
cos cos ( 0. 8 ) 36. 87
362. 3 A
( 13. 8 kV)( 0. 8 )
4000 kW
1 1
= ∠− °
− −
S
S
out S
I
PF
V PF
P
I
θ
The voltage on the primary side of the transformer (referred to the secondary side) is
V (^) PS = VS + ISZEQ , S = 13800 ∠ 0 °+( 362. 3 ∠− 36. 87 °)( 0. 38 + j 1. 9 )= 14330 ∠ 1. 9 °V
The voltage regulation of the transformer is
100 % × =
× =
S
S
S P V
V V
VR
The transformer copper losses and core losses are
- 4 kW 4240
( 362. 3 ) ( 0. 38 ) 49. 9 kW (^22)
2 ,
2
C
S P core
copper S EQS
R
V
P
P I R
Therefore the efficiency of this transformer at these conditions is
4000 kW 49. 9 kW 48. 4 kW
4000 kW 100 % × =
× =
out copper core
out P P P
P
Problem 2-8:
A 200-MVA, 15/200-kV single-phase power transformer has a per-unit resistance of 1. percent and a per-unit reactance of 5 percent (data taken from the transformer’s nameplate). The magnetizing impedance is j80 per unit. (a) Find the equivalent circuit referred to the low-voltage side of this transformer. (b) Calculate the voltage regulation of this transformer for a full-load current at power factor of 0.8 lagging. (c) Assume that the primary voltage of this transformer is a constant 15 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 0.8 lagging, 1.0, and 0. leading.
Solution:
(a) The base impedance of this transformer referred to the low-voltage side (primary) is
M
k S
V
Z
base
base base
So that
EQ base
EQ base X Z
R Z
And
XM = ( 80 ) Zbase =( 80 )( 1. 125 )= 90 Ω
The equivalent circuit referred to the low-voltage side of this transformer is shown below
R not spcified
R
CP
EQP = −
,
, 0.^0135
,
, M P
EQP X
X
I(3,:) = amps .* ( 0.8 + j*0.6); % 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:
Problem 2-14:
A 13.2-kV single-phase generator supplies power to a load through a transmission line. The load’s impedance is Zload = 500 ∠ 36. 87 °Ω, and the transmission line’s impedance is Zline = 60 ∠ 53. 1 °Ω.
(a) If the generator is directly connected to the load, what is the ratio of the load voltage to the generated voltage? What are the transmission losses of the system? (b) If a 1:10 step-up transformer is placed at the output of the generator and a 10: transformer is placed at the load end of the transmission line, what is the new ratio of the load voltage to the generated voltage? What are the transmission losses of the system now? (Note: The transformers may be assumed to be ideal.)
Solution:
(a) In the case of the directly-connected load, the line current is
23. 66 38. 6 A
- 2 0 kV = ∠− ° ∠ °+ ∠ °
line load
G line load Z Z
V
I I
The load voltage is
Ploss = Iline^2 Rline =( 2. 637 )^2 ( 36 )= 250 W
Problem 2-15:
A 5000-VA, 480/120-V conventional transformer is to be used to supply power from a 600-V source to a 120-V load. Consider the transformer to be ideal, and assume that all insulation can handle 600 V. (a) Sketch the transformer connection that will do the required job. (b) Find the kilovoltampere rating of the transformer in the configuration. (c) Find the maximum primary and secondary currents under these conditions.
Solution:
(a) The common winding must be the smaller of the two windings, and NSE = 4 NC. The transformer connection is shown below:
(b) The kVA rating of the transformer is
VA
N
N N
S
N
N N
S
C
C C W C
SE C IO (^5000 )^6250
(c) The maximum primary current for this configuration is
A
V
S
I
P
IO P (^) 600 10.^4
The maximum secondary current is
A
V
S
I
S
IO S^52.^1 120
Problem 2-19:
A 20-kVA, 20,000/480-V, 60-Hz distribution transformer is tested with the following results: Open-circuit test (measured from secondary side)
Short-circuit test (measured from primary side) VOC = 480 V VSC = 1130 V I (^) OC = 1.60 A I (^) SC = 1.00 A POC = 305 W PSC = 260 W (a) Find the per-unit equivalent circuit for this transformer at 60 Hz. (b) What would be the rating of this transformer be if it were operated on a 50-Hz power system? (c) Sketch the per-unit 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
20 kΩ 20
, =^ S = k =
V
Z (^) baseP P
The base impedance of this transformer referred to the secondary side is
, (^) S k
V
Z (^) baseS S
The open-circuit test yields the values for the excitation branch (referred to the secondary side)
− −
cos cos
1 1
M
M
C
C
EX EX C M
OC OC
OC
OC
OC EX
B
X
G
R
Y Y G jB j
V I
P
V
I
Y
(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
V V V
V V V
S S k
S S
P P ( 50 / 60 ) ( 50 / 60 )( 480 ) 400
( 50 / 60 ) ( 50 / 60 )( 20 ) 16. 67 kVA
, 50 , 60
, 50 , 60
50 60
(c) The transformer parameters referred to the primary side at 60 Hz are
X Z X k k
R Z R k M
X Z X k
R Z R k
M base Mpu
C base Cpu
EQ base EQpu
EQ base EQpu
,
,
,
,
At 50 Hz, the resistances will be unaffected but the reactances are reduced in direct proportion to the decrease in frequency.
X X k
R R M
X X
R R
M M
C C
EQ EQ
EQ EQ
, 50 , 60
, 50 , 60
, 50 , 60
, 50 , 60
The base impedance of the transformer operating at 50 Hz referred to the primary side is
= = = k Ω S k
V
Z (^) base P P 16. 67
- 67
,, 50
The resulting per-unit equivalent circuit referred to the primary at 50 Hz is shown below
pu k
M
Z
R
R
pu Z k
R
R
baseP
C Cpu
baseP
EQ EQpu
,, 50
, 50 , , 50
,, 50
, 50 , , 50
pu k
k Z
X
X
pu Z k
X
X
baseP
M Mpu
baseP
EQ EQpu
,, 50
, 50 , , 50
,, 50
, 50 , , 50
