Polytechnic Institute of NYU, Dept. of ECE
EL5613
Electric Power Systems Midterm Exam.
10/21/09
a. Answer both problems.
b. Written material is allowed.
c. Time: 1 1/2 hours
Good luck!
1. A three-phas
EL5613
Final Exam Solution
12/17/08
1. For a PF = 0.85!lag!cfw_! = 31.8 at the load, the 3 reactive power would be:
QL = PL tan (! ) = 300 tan 31.8 = 186!kVAr
For a PF = 0.92!lag!cfw_!1 = 23.1 at the
EL5613 ePoly
Final solution
12/16/2010
1. From delta to wye: divide the value by
3 , and deduct 30 from the phase angle.
13.8
Therefore, the primary phase-voltage is: V A =
! 150 " 30 = 7.98!120 kV .
EL5613
Final Solution
12/23/09
345 2
345 2 345 2
1. (a) and (b) From the power equations: 240 =
sin , and 21 =
cos
X
X
X
We get: = 10 , and X = 86.1
345
345
10
3 = 0.4035 kA
(c) The current is: I a
NYU Tandon School of Engineering - Department of Electrical Engineering
EL5613
a.
b.
c.
d.
Elect. Power Systems Midterm Exam.
10/31/2016
Answer the two problems.
Written material is allowed.
Calculato
EL5613
Polytechnic Institute Electrical Engineering Department
Electric Power Systems - Final Exam
1. Answer the three problems
2. Written material is allowed.
3. Time: 3 hours
12/23/09
Good luck!
1.
Polytechnic Institute, Dept. of ECE
EL5613
Electric Power Systems Midterm Exam.
a. Answer both problems.
b. Written material is allowed.
c. Time: 1 1/2 hours
10/29/08
Good luck!
1. A 2400 V (line-to-l
Polytechnic University, Dept. of ECE
EL5613
Electric Power Systems Midterm Exam.
a. Answer both problems.
b. Written material is allowed.
c. Time: 1 1/2 hours
11/1/07
Good luck!
1. Three loads are con
Polytechnic University, Dept. of ECE
EL561
Electric Power Systems Midterm Exam.
a. Answer both problems.
b. Written material is allowed.
c. Time: 1 1/2 hours
10/24/06
Good luck!
1. A 3 line has an imp
EL5613
Midterm Solution
10/21/09
1
a. The equivalent single phase circuit is
Ia
0.6+j4.8
13,800
=
VLan
Van
3
0
400 kVA
200 kVA
The P, Q powers of each load are:
P1 =S1 *PF1 = (400 kVA)(0.6) = 240 kW
EL561
Midterm Solution
10/24/06
156
117
+ j
= 52 + j 39kVA ph
1. The apparent power in load 1 is: S1 ph =
3
3
3 2600
= 30 A
The phase current in the -connected load 2 is: I 2 =
144 2 + 42 2
The appare
EL5613
Midterm Solution
10/29/08
1. The total complex power per-phase is:
300
240
S ph1 =
36.87 +
53.13 = 128.1 1.8kVA
3
3
V
2400
The phase voltage is: Van = ab 30 =
( 30 30 ) = 1385.60V
3
3
S ph1 1
Polytechnic Institute of NYU ECE Department
EE5613 ePoly
Electric Power Systems Final Exam
a. Answer the three problems
b. Written material is allowed.
c. Time: 3 hours
12/16/10
Good luck!
1. A 3, del