Download ECE320-Exam1

25th January, 2006
ECE320, Spring 2006, Exam 1
Duration 1 hour
Name
PID Number
Problem
1
2
3
Total
SOLUTIONS
Max Points
30
40
30
100
Points
Instructions


Read all the problems first.
Attempt to solve first the problems you can do-don’t spend too much time on one
problem.
 Read carefully the statement of the problems.
 Use the previous page’s back if more space is needed.
 Write neatly and show all the steps, no marks otherwise.
Problem 1. For the circuit below, calculate the load I L using Thevenin’s theorem.
 j100
VS  12000 [V]
A
Ζ L  10  j10
Load
j10
B
IL
 j 200
j 20
Solution:
The voltages at nodes A and B can be found by voltage division.
Open Circuit Voltage ( VTH )
C
j 10
-j 100
120 V
A
+
-j 200
Vth
-
B
j 20
D
 j 200
VAD 
1200
 j 200  j100
 j 20
VBD 
1200
 j 20  j10
VTH  VAB  VAD  VBD = 0.
Thevenin Impedance ( ZTH )
1
1
ZTH 

1
1
1
1


 j100  j 200 j10 j 20
ZTH  j (66.66)  j (6.66)  j 60
Since VTH  0 , no voltage drop appears across terminals A and B; therefore, no current flows
through Z L .
IL 
VTH
0
Z TH  Z L
.
Problem 2. For a balanced 3-phase circuit below, the source voltage is 208 V. Calculate (1)
the real (active) and reactive power of each load, (2) the total real and reactive power of both
loads, and (3) the real and reactive power provided by the source. Explain why the source
reactive power differs from the total load reactive power. What about the source real power
and the total load real power (should they be equal)?
j0.2 
ISa
VSa
IL2a
IL1a
VSb
VSc
6 j6 
2+ j2 
Load 1
Load 2
Solution:
Converting delta connected load to star and then considering only a single phase, the circuit
will be as shown in figure.
_____
j0.2
I sa
_____
I La 2
_____
I La1
_____
AC
Vsa  1200
(2  j 2)
(2  j 2)
(1)
Total Impedance
j 0.2  [(2  j 2) (2  j 2)]  2  j 0.2
__
I sa 
We have
1200   59.7  5.71
2  j 0.2
A
__


2  j2
I La1  I sa 
  42.2139.2 A
 2  j2  2  j2 
__
Using Current Division
__


2  j2
I La 2  I sa 
  42.21  50.7 A
 2  j2  2  j2 
__
Load Voltages
__
__
__
__
V La1  I La1  2  j 2  119.4  5.72 V
V La 2  I La 2  2  j 2  119.4  5.72 V
Transmission Line Voltage Drop
__
I sa  j 0.2  11.9484.2
A
Load 1
Real Power
P L1  3V La1 I La1 cos  5.72  39.2   10.692 kW
Reactive Power
Q L1  3V La1 I La1 sin  5.72  39.2  10.692 kVAR
Real Power
P L 2  3V La 2 I La 2 cos  5.72  50.7   10.692 kW
Reactive Power
Q L 2  3V La 2 I La 2 sin  5.72  50.7   10.692 kVAR
Load 2
(2)
Total Real and Reactive Power of Both Loads:
PTL  P L1  P L 2  (10.69  10.69)kW  21.38 kW
QTL  Q L1  Q L 2  (10.69  10.69)kVAR  0
(3)
The Real and Reactive Power Provided by the Source:
P S  3V sa I sa cos  0  5.72  21.38 kW
QS  3V sa I sa sin  0  5.72   2.138 kVAR
Source Reactive Power differs from the total load reactive power (which is zero in this case)
because it is the power associated with the transmission line.
Transmission Line Reactive Power = 3 (11.94) (59.7) sin (84.2+5.72) = 2.138 kVAR
Or = 3 (59.7 A)2 (0.2 ) = 2.138 kVAR
Since, Transmission Line Real Power = 0
Therefore, Source Real Power = Total Load Real Power.
Problem 3
In a 3-phase system below (Fig. a), the source is 230 V and the load (Load_1) is drawing 50
kW at 0.866 pf lagging.
a) A second load (Load_2) is added and its real power is equal to 30 kW. What should the
pf of the second load be in order for the total pf of the both loads together to be unity?
b) Draw a power triangle diagram to represent the above-obtained real, reactive, and
apparent power of the source, Load_1, and Load_2.
Load_1
3-f
Source
Fig. a
Load_1
3-f
Source
Fig. b
Solution:
(a)
P L1  50 kW
P f  L1  0.866(lagging )
50kW
S L1 
 57.73 kVA
0.866
Q L1  57.73 sin(cos 1 0.866)   28.87 kVAR
P L 2  30 kW [given]
PTL  P L1  P L 2  (50  30)kW  80 kW
P f TL  unity
 QTL  Q L1  Q L 2  0
 Q L1   Q L 2
 Q L 2  28.87 kVA




 28.87k  
1  Q L 2  

P f  L 2  cos tan
 cos  tan 1 




 30k  

P
L
2





Therefore, P f  L 2  0.72(leading ) , since  Q L 2  0 


Load_2
(b)
Power Triangles
30 kW
57.73 kVA
j28.8kVAR
41.6 kVA
50 kW
Load 1
Load 2
-j28.8kVAR