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Course
421
Subject
Electrical Engineering
Date
Dec 6, 2023
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1
ECE 421/521 FALL 2013 - MIDTERM EXAM 1
Problem 1 (25 points):
A load
Z
L
=50+
j
40
Ω
is supported by a 300V rms, 60-Hz source
through a line with impedance
Z
T
=0+
j
10
Ω
.
a.
Find the total current, the total real and reactive powers absorbed by the load, and
the voltage and overall power factor on the load side
b.
Find the capacitance of the shunt capacitor to improve the overall power factor on
the load side to 0.9 lagging. Then, what is the new voltage on the load side?
a.
Z=Z
T
+Z
L
=50+j50
Ω
2
I=V/Z=300/(50+j50)=3-j3A
2
V
L
=I
×
Z
L
=(3-j3)(50+j40)=270-j30=271.66
∠
-6.34
o
V
2
S
L
=P
L
+jQ
L
=V
L
×
I*=(270-j30)(3+j3)=900+j720 VA
3
|S
L
|=1152.6W
PF=P
L
/|S
L
|=0.7809
2
Another method for b: If the load and shunt capacitor has a
terminal voltage =V
L
Q
L
+Q
C
=P
L
×
tan(25.84
o
)=436MVar
Q
C
=436-270=-284Mvar
X
C
=|V
L
|
2
/Q
C
=-259.7
Ω
C=-1/(
ω
X
C
)= -1/(377
×
X
C
)=10.21
µ
F
b.
Z
Lnew
=Z
L
//X
C
= Z
L
×
jX
C
/(Z
L
+jX
C
)
2
=(50+j40)
×
jX
C
/(50+j40+jX
C
)
=(50+j40)
×
(40/X
C
+1-j50/X
C
)
Angle(Z
Lnew
)=cos
-1
(0.9)=0.451 rad = 25.84
o
2
= Angle(Z
L
) – Angle(40/X
C
+1-j50/X
C
)
2
Angle(Z
L
)=0.6747 rad
Angle(40/X
C
+1-j50/X
C
)=0.6747-0.451=0.2237 rad
(-50/X
C
)/(40/X
C
+1)=-50/(40+X
C
)=tan(0.2237)=0.2275
Solve Xc:
X
C
=-259.7
Ω
2
C=-1/(
ω
X
C
)= -1/(377
×
X
C
)=10.21
µ
F
2
Z
Lnew
=66.42+j32.17
Ω
2
V
Lnew
=Z
Lnew
/(Z
T
+Z
Lnew
)
×
V
=279.56-j32.19=281.41
∠
-6.57
o
V
2
2
Problem 2 (25 points):
A three-phase line has an impedance of
Z
T
=j2
Ω
and feeds two balanced
three-phase loads connected in parallel. The first load is
∆
-connected and has an impedance of
Z
1
=24
Ω
. The second load is Y-connected and has an impedance of Z
2
=12
Ω
. The line is energized
at the sending end from a three-phase balanced supply of line-to-line voltage |
V
L
|=415.69V. Take
the line voltage
V
AB
on the supply side as the reference, i.e.
V
AB
=|
V
L
|
∠
0
o
.
a.
Determine phase current
I
A
, three-phase real power and reactive power drawn from the
supply, and the power factor on the supply side.
b.
Determine line voltages
V
ab
and phase voltage
V
an
at the combined load side. Draw a
phasor diagram having
V
an
,
V
ab
,
I
A
and
V
AB
.
c.
Determine the total real and reactive powers absorbed by each three-phase load and the three-phase line
a.
Z
1Y
=24/3=8
Ω
2
Z
L
=Z
1Y
//Z
2
=12
×
8/(12+8)=4.8
Ω
2
V
AB
=415.69V
V
An
lags V
AB
by 30 degree
V
An
=415.69/
√
3
∠
-30
o
=240
∠
-30
o
V
2
Z=Z
T
+Z
L
= 4.8+j2
Ω
2
I
a
=V
An
/Z=28.02-j36.67=46.15
∠
-52.62
o
2
S
3
φ
=P
3
φ
+jQ
3
φ
=3V
An
I
a
*
=30675+j12781VA=33230
∠
22.62
o
2
|S
3
φ
|=33230W
PF=P
3
φ
/|S
3
φ
|=0.9231
2
b.
V
an
=I
a
Z
L
=134.5-j176.0=221.54
∠
-52.62
o
V
2
V
ab
= V
an
×√
3
∠
30
o
=204.5-j85.21=383.72
∠
-22.62
o
V
2
Phasor diagram
1
c.
S
line
=3|I
a
|
2
Z
T
=j12781
1
P
line
=0
Q
line
=12781 var
1
S
3
φ
1
= P
3
φ
1
+jQ
3
φ
1
=3|V
an
|
2
/Z
1
*=18405 VA
1
P
3
φ
1
=18405W
Q
3
φ
1
=0
1
S
3
φ
2
= P
3
φ
2
+jQ
3
φ
2
=3|V
an
|
2
/Z
2
*=12270 VA
1
P
3
φ
2
=12270W
Q
3
φ
2
=0
1
3
Problem 3 (30 points):
Short answers
a.
Why did AC win over DC in the “AC vs. DC battle” in the 1890s? List at least two reasons.
Give any two of the three:
1.
Voltage levels can easily be transformed in AC
2.
AC generators are simpler
3.
HVAC is easier to implement in order to reduce transmission loss
b.
Which of these generation resources utilize steam turbines in generating electric power?
Coal-fired power plant
Combined-cycle power plant
Parabolic Trough
Solar Tower
Pressurized water reactor
Check all
c.
A load supplied by a voltage source has 0.7 power factor lagging. A shunt capacitor connected across the load will
necessarily improve the power factor of the load side. True or false? Briefly explain why.
False. If the capacitance is too large, it may become <1 power factor leading.
d.
A two-winding transformer is first operated as a conventional two-winding transformer and then as an autotransformer to
supply the same load. When it is operated as the autotransformer, it has a higher rating, a higher efficiency and a lower
loss than it is operated as the conventional two-winding transformer. True or false? Briefly explain why.
True. For the autotransformer,
•
rating (not the actual loading) will be increased,
•
loss will decrease since the currents in windings will decrease to generate the same amount of total current as the
two-winding transformer does
•
then the efficiency will increase considering the loading is the same.
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