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ECE320, Spring 2006, Exam 2a
24th February, 2006
Name
PID Number
Problem
1
2
3
Total
Max Points
40
40
20
100
Points
Instructions:

Closed book, but you are allowed to have one-page single-sided handwritten notes.

You can have 3 pages of blank papers for calculation purpose.

Write your solutions and steps neatly and orderly. No credits for random side
writings.

Box your answers.
Problem 1. (40 points) In the magnetic circuit below, the air-gap is g = 5 mm, i = 40 A, N =
250 turns, cross section of the iron everywhere A = 40 mm*50 mm.
a) Draw an analog electric schematic for the magnetic circuit for both cases of iron core’s µ =
∞ and µ ≠ ∞. (10 points)
Assuming that the iron core has an infinite permeability, solve the following problems:
b) Calculate the flux density in the air-gaps. (10 points)
100 mm
c) What would be flux density in the air gaps
if the current is i = 40 sin (377 t) A. (10 points)
d) For the above (c), what would be the voltage induced
in the coil? (10 points)
110 mm
SOLUTION:
g
a) The re-labeled magnetic circuit is:
iron
Core 2
f
i
g
iron
gap 1
gap 2
Core 1
i
The electric schematic for the magnetic circuit for both cases of iron core’s µ = ∞ and µ ≠ ∞.
Rcore1
Rgap1
Rgap1
Φ
Ni
Ni
Rgap2
Figure 1. µ ≠ ∞
Rgap2
Figure 2. µ = ∞
b) Assuming that the iron core has an infinite permeability;
A  (0.04)(0.05)m2
g  0.005m
g
0.005
Rgap1  Rgap 2  R 
R
 1.989 MH 1
7
0 A
(4 10 )(0.04  0.05)
Ni
Ni
250  40
f


 0.00251 wb
Rgap1  Rgap 2 2R 2 1.989 M
f
0.00251
B 
 1.2566 T
A 0.05  0.04
Given:
c)
B  1.2566sin(377t ) T
d)
v(t ) 
d  d ( NBA) d

  250 1.2566sin(377t )  0.04  0.05   236.869 cos(377t ) V
dt
dt
dt
Problem 2. (40 points) A 50 kVA, 10kV/250 V, 60Hz, single-phase transformer gave the
following test result: Open-circuit test (open the high voltage side): 250V, 5A, and 250 watts
measured on low voltage side and Short-circuit test (short the low voltage side): 300V, rated
current, and 500 watts measured on high voltage side.
a) Draw the simplified high voltage side referred equivalent circuit in actual values. (8 points)
b) Calculate the winding resistance in actual values. (8 points)
c) Draw the simplified equivalent circuit in pu values. (8 points)
d) Calculate the leakage reactance in pu values. (8 points)
e) A load is connected to the low voltage side, drawing 50 kVA with pf =.866 lagging at 250 V.
Using per-unit values to calculate the high-voltage side voltage. (8 points)
SOLUTION:
(a) Draw the simplified high voltage side referred equivalent circuit in actual values.
X12
I1
R12
+
I2
+
+
V1
Xm
Rc
V2
_
_
(b)
_
Calculate the winding resistance in actual values.
50 kVA
5 A
10 kV
P
500
R12  2sc  2  20 
I1sc
5
I1sc 
(c)
Draw the simplified equivalent circuit in pu values
I1, pu
X12 , pu
+
_
I2, pu
+
Rc , pu
V1, pu
R12 , pu
Xm , pu
_
(d)
Calculate the leakage reactance in pu values
Z1base 
(V1,base )2
Sbase
 2k 
2
V 
 300 
2
X12   1sc   R122  
  20  56.57 
 5 
 I1sc 
X
56.57
X 12 pu  12 
 0.0283 pu
Z1base
2k
(e)
2
A load is connected to the low voltage side, drawing 50 kVA with pf =0.866
lagging at 250 V. Using per-unit values to calculate the high-voltage side
voltage.
S2 pu 
S2
50kVA

 1 pu
Sbase 50kVA
I 2 pu 
S2 pu
V2 pu
Z1base 
V2 pu 
V2
250

 1 pu
V2base 250
1
  cos 1 pf    30  1  30 pu
1
(V1base )
 2 k
Sbase
20
 0.01 pu
2k
56.57

 0.0283 pu
2k
R12, pu 
2
Therefore,
X 12, pu
And finally,
V1 pu  ( R12 pu  jX12 pu )  I 2 pu  V2 pu  (0.01  j 0.0283) 1   30  10  1.0231.092 pu
Problem 3. (20 points)
A three-phase transformer is made from three single-phase ones, each-rated at 10 kVA,
20kV/200V; rated winding losses 100 W, core losses 50 W. The three-phase transformer is
connected in delta on the high voltage side and in star on the low voltage side.
(a) What are the voltage ratings, current ratings, core loss (under rated voltage), winding loss
(under rated current) for the 3-phase transformer? (10 points)
(b) Draw a phasor diagram to express the phase relationship between the high-voltage side and
low-voltage side line-to-line voltages assuming the transformer is ideal. (10 points)
SOLUTION:
(a) Each transformer is 10 kVA 20kV / 200V .
The high-voltage side is , thus the high-voltage side voltage rating is 20 kV and the
high-voltage side current rating is 1.732*(10 kVA)/(20 kV) = 0.866 A.
The low-voltage side is Y, thus the low-voltage side voltage rating is 1.732*200 V =
346 V and the low-voltage side current rating is (10 kVA)/(200 V) = 50 A.
The 3-phase transformer’s core loss = 3 * 50 W = 150 W.
The 3-phase transformer’s winding loss = 3 * 100 W = 300 W.
(b)
__
V CA
__
A
V
__
a
ab
VC
__
V ca
__
V
30°
B
Va
__
C
AB
__
b
Vb
c
__
VBC
__
V bc