ECE 476 – Power System Analysis Fall 2017
Homework 2
In-class quiz: Thursday, September 14, 2017
Problem 1. A three-phase line, which has an impedance of (2 + j4) Ω per phase, feeds two balanced three-phase
loads that are connected in parallel. One of the loads is Y-connected with an impedance of (30 + j40) Ω per phase,
and the other is ∆-connected with an impedance of (60 – √
j45) Ω per phase. The line is energized at the sending end
from a 60-Hz, three-phase, balanced voltage source of 120 3 V (rms, line-to-line). Determine:
First, convert the ∆-connected load to its Y -connected equivalent:
60 − j45
= 20 − j15 Ω
3
1. The current, real power, and reactive power delivered by the sending-end source.
Z̄2 =
The per-phase total impedances of the line and loads is
−1
1
1
+
= 24∠0◦ Ω.
Z̄ = Z̄line + Z̄1 ||Z̄2 = 2 + j4 +
(30 + j40) (20 − j15)
Then, the source current is
120∠0◦
V̄
=
= 5∠0◦ A.
I¯s =
24◦
Z̄
The complex power delivered by the source is
S̄s = 3V̄ I¯∗ = 3(120∠0◦ )(5∠0◦ ) = 1800∠0◦ VA,
with Ps = 1800 W and Qs = 0 MVar.
2. The line-to-line voltage at the load.
The phase voltage at the load is
V̄L = V̄s − Z̄line I¯s∗ = 120∠0◦ − (2 + j4)(5∠0◦ ) = 110 − j20 V = 111.80∠ − 10.3◦ V.
Therefore the line-to-line voltage at the load is
√
√
V̄L,L−L = 3V̄L ∠30◦ = 3111.80∠ − 10.3◦ + 30◦ = 193.65∠19.7◦ V
3. The current per phase in each load.
The per-phase current through the Y -connected load is
V̄L
111.80∠ − 10.3◦
I¯1 =
=
= 2.236∠ − 63.4◦ A.
30 + j40
Z̄1
The per-phase current through the Y -connected equivalent of the ∆-connected load is
V̄L
111.80∠ − 10.3◦
I¯2,φ =
=
= 4.472∠26.57◦ A.
20 − j15
Z̄2
So the per-phase current of the ∆-connected load is
I¯2,φ
4.472
I¯2,∆ = √ ∠30◦ = √ ∠26.57◦ + 30◦ = 2.582∠56.57◦ A.
3
3
1
4. The total three-phase real and reactive powers absorbed by each load and by the line.
The 3φ complex power absorbed by the Y -connected load is
S̄1 = 3V̄L I¯1∗ = 3(111.80∠ − 10.3◦ )(2.236∠ − 63.4◦ )∗ = 450.3 + j599.7 VA.
The 3φ complex power absorbed by the ∆-connected load is
∗
S̄2 = 3V̄L,L−L I¯2,∆
= 3(193.65∠19.7◦ )(2.582∠56.57◦ )∗ = 1200 − j900 VA.
The complex power absorbed by the line impedance is
S̄line = 3Z̄line I 2 = 3(2 + j4)52 = 150 + j300 VA.
The sum of the three quantities above is 1800 + j0 VA, which matches the value obtained in Part 1.
Check that the total three-phase complex power delivered by the source equals the total three-phase
power absorbed by the line and loads.
Problem 2. Two three-phase generators supply a three-phase load through separate three-phase lines. The load
absorbs 30 kW at 0.8 power factor lagging. The line impedance is (1.4 + j1.6) Ω per phase between generator G1
and the load, and (0.8 + j1) Ω per phase between generator G2 and the load. If generator G1 supplies 15 kW at 0.8
power factor lagging, with a terminal voltage of 460 V line-to-line, determine:
1. The voltage at the load terminals.
The complex power supplied by G1 is
S̄1,3φ =
15
∠ cos−1 0.8 = 18.75∠36.87◦ kVA.
0.8
The current supplied by G1 is
S̄1,3φ
I¯G1 =
3V̄G1

∗
=
◦
∗
18.75∠36.87 
= 23.53∠ − 36.87◦ A.
460
◦
3 √3 ∠0
So the phase voltage at the load is
460
∠0◦ − (1.4 + j1.6)(23.53∠ − 36.87◦ ) = 216.9∠ − 2.74◦ V,
V̄L = V̄G1 − Z̄l1 I¯G1 = √
3
and the line-to-line voltage at the load is
V̄L,L−L =
√
3(216.9)∠27.26◦ V,
2. The voltage at the terminals of generator G2.
The complex power absorbed by the load is
S̄L,3φ =
30
∠ cos−1 0.8 = 37.5∠36.87◦ kVA.
0.8
The current into the load is
∗ ∗
37.5∠36.87◦
S̄L,3φ
¯
=
IL =
= 57.63∠ − 39.61◦ A.
3(216.9∠ − 2.74◦ )
3V̄L
The current into the load is the sum of the currents supplied by the two generators, so
I¯G2 = I¯L − I¯G1 = 57.63∠ − 39.61◦ − 23.53∠ − 36.87◦ = 34.15∠ − 41.5◦ A.
And finally the voltage at the terminal of G2 is the sum of the voltage drop across the load and Z̄l2 :
V̄G2 = V̄L + Z̄l2 I¯G2 = 216.9∠ − 2.74◦ + (0.8 + j1)(34.15∠ − 41.5◦ ) = 259.8∠ − 0.638◦ V.
And the line-to-line voltage at the terminal of G2 is
√
V̄G2 ,L−L = 3(259.8)∠29.36◦ V.
3. The real and reactive power supplied by generator G2. The complex power supplied by G2 is
∗
= 3(259.8∠ − 0.638◦ )(34.15∠41.5◦ ) = 20.1 + j17.4 kVA.
S̄G2 ,3φ = 3V̄G2 I¯G
2
Hence, the real power supplied is PG2 ,3φ = 20.1 kW and the reactive power supplied is QG2 ,3φ = 17.4 kVar.
Problem 3. An unbalanced three-phase, Y -connected power system is shown in the figure below. The three phases
have voltages V a = 100∠0◦ V, V b = 100∠ − 120◦ V, V c = 100∠120◦ V. The impedances of loads A, B, C are
Z a = 10 Ω, Z b = −j10 Ω, Z c = j10 Ω.
1. What are the currents of each phase I a , I b , I c ?
Ia =
Va
Za
=
100∠0◦
10
= 10∠0◦ A, I b =
Vb
Zb
=
100∠0◦
−j10
= 10∠ − 30◦ A, I c =
Vc
Zc
=
100∠0◦
j10
= 10∠30◦ A
2. What is the current on the neutral line I n ?
√
I n = I a + I b + I c = 10∠0◦ + 10∠ − 30◦ + 10∠30◦ = (10 + 10 3)∠0◦ = 27.3∠0◦ A
3. What are the line voltages V ab , V bc , V ca ?
√
√
√
V ab = 100 3∠30◦ = 173∠30◦ V, V bc = 100 3∠ − 90◦ = 173∠ − 90◦ V, V ca = 100 3∠150◦ = 173∠150◦ V
4. Provide the phasor diagram of the phasors including the phase voltages, line voltages, phase currents and the
current on the neutral line.
Figure 1: phasor diagram