ECE 476 – Power System Analysis Fall 2013 Homework 2
Due Date: Thursday September 12, 2013 Problem 1. In PowerWorld Simulator Problem 2.32 (see Figure 1 below) a 8 MW/4 Mvar load is supplied at 13.8
kV through a feeder with an impedance impedance of (1+ j2) Ω. The load is compensated compensated with a capacitor whose output, output, Q cap , can be varied in 0.5 Mvar steps between 0 and 10.0 Mvars. What value of Q cap minimizes the real power line losses? What value of Q cap minimizes the MVA power flow into the feeder? The complex power drawn by the load is 8 + j 4 MVA. The capacitor can provide [0,10] MVar of compensation. So the resulting complex power drawn by the load is S load Qcap) MVA. Then, the current into the feeder load = 8 + j (4 is ∗ ∗ ¯ S load 8 + j (4 Qcap ) 8 j (4 Qcap ) load ¯ I = = = kA. ¯ 13.8 13.8 V The complex power loss due to the line impedance is
¯loss ¯ 2 ¯ = (8 + j (4 S loss = I Z =
||
The real power lost is
−
cap ))(8
−Q
13.8
2
− −
− j (4 − Q
cap ))
(1 + j 2) =
−
64 + (4 Qcap )2 (1 + j 2). 13.82
−
64 + (4 Qcap )2 P loss MW, loss = 13.82
−
which which is minimized minimized when Q cap = 4 MVar. For the second part of the question, the complex power into the feeder is the sum of the complex power lost and absorbed by the combined load as follows: 64 + (4 Qcap )2 2 ¯ ¯ ¯ ¯ ¯ ¯ + S load (1 + j 2) + 8 + j (4 Qcap ). S feeder feeder = S loss loss + S load load = I Z + load = 13.82 ¯feeder We would like to minimize S complicated ed expression, expression, we resort to feeder with respect to Qcap . Since this is a very complicat ¯ a graphical solution via MATLAB. The plot below shows that S feeder MVar. ar. The feeder is minimized when Qcap = 4.5 MV code is also provided.
−
||
|
|
−
|
|
8.6
8.55
> > > > > > > >
Qcap Qcap = 0:0.5: 0:0.5:10; 10; Sload Sload = 8+1i*( 8+1i*(4-Q 4-Qcap cap); ); I = conj(S conj(Sloa load/1 d/13.8 3.8); ); Sloss = conj(I).*I. conj(I).*I.*(1+1 *(1+1i*2) i*2); ; Sfee Sfeede der r = Sloa Sload d + Slos Sloss; s;
8.5
|
r e d e e8.45 f
S |
8.4
8.35
2
3
4
5
Q cap
1
6
7
Assume balanced operation.
Figure 1: Screen for Problem 1. Problem 2. 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:
√
¯2 = Z
60
− j 45 = 20 − j 15 Ω 3
1. The current, real power, and reactive power delivered by the sending-end source. The per-phase total impedances of the line and loads is ¯ = Z ¯line + Z ¯1 Z ¯2 = 2 + j 4 + Z
||
Then, the source current is
1 1 + (30 + j 40) (20 j 15)
−
1
−
= 24∠0◦ Ω .
◦ ¯ ¯s = V = 120∠0 = 5 ∠0◦ A. I ¯ 24◦ Z
The complex power delivered by the source is ¯s = 3V ¯ ¯ S I ∗ = 3(120∠0◦)(5∠0◦ ) = 1800 ∠0◦ VA, with P s = 1800 W and Q s = 0 MVar. 2. The line-to-line voltage at the load. The phase voltage at the load is ¯L = V ¯s V
− Z ¯
¯∗
line I s =
120∠0◦
− (2 + j 4)(5∠0◦) = 110 − j 20 V = 111.80∠ − 10.3◦ V.
Therefore the line-to-line voltage at the load is ¯L,L−L = V
√ ¯
3V L ∠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 ◦ ¯ ¯1 = V L = 111.80∠ 10.3 = 2.236∠ I ¯1 30 + j 40 Z
−
− 63.4◦ A.
The per-phase current through the Y -connected equivalent of the ∆-connected load is ◦ ¯ ¯2,φ = V L = 111.80∠ 10.3 = 4.472∠26.57◦ A . I ¯2 20 j 15 Z
−
−
So the per-phase current of the ∆-connected load is ¯2,φ I
.472 ◦ ◦ ◦ √ 3 ∠30◦ = 4√ ∠26.57 + 30 = 2.582∠56.57 A . 3
¯2,∆ = I
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 ¯1 = 3V ¯L I ¯1∗ = 3(111.80∠ S
− 10.3◦)(2.236∠ − 63.4◦)∗ = 450.3 + j 599.7 VA.
The 3φ complex power absorbed by the ∆-connected load is ¯2 = 3V ¯L,L−L I ¯2∗,∆ = 3(193.65∠19.7◦)(2.582∠56.57◦)∗ = 1200 S
− j 900 VA.
The complex power absorbed by the line impedance is ¯line = 3 ¯ S Z line I 2 = 3(2 + j 4)52 = 150 + j 300 VA. The sum of the three quantities above is 1800 + j 0 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 3. 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 G 1 is ¯1,3φ = S
15 −1 0.8 = 18.75∠36.87◦ kVA. ∠ cos 0.8
The current supplied by G 1 is ¯G I
1
¯ ∗ 18 75 36 87◦ ∗ 0◦ = 23 53 = = 3¯ 3 √ S 1,3φ V G
.
∠
460
1
3
.
.
∠
∠
− 36.87◦ A.
So the phase voltage at the load is ¯L = V ¯G V
1
−
¯l I ¯G = Z 1
1
460 √ 3
∠0
◦
− (1.4 + j 1.6)(23.53∠ − 36.87◦) = 216.9∠ − 2.74◦ V,
and the line-to-line voltage at the load is ¯L,L−L = V
√
3(216.9)∠27.26◦ V,
2. The voltage at the terminals of generator G2. The complex power absorbed by the load is ¯L,3φ = S
30 −1 0.8 = 37.5∠36.87◦ kVA. ∠ cos 0.8
The current into the load is ∗ ∗ ¯ 37.5∠36.87◦ S L,3φ ¯ I L = = = 57.63∠ ¯L 3(216.9∠ 2.74◦) 3V
−
− 39.61◦ A.
The current into the load is the sum of the currents supplied by the two generators, so ¯G = I ¯L I 2
− ¯I
G1
= 57 .63∠
− 39.61◦ − 23.53∠ − 36.87◦ = 34.15∠ − 41.5◦ A.
¯l : And finally the voltage at the terminal of G 2 is the sum of the voltage drop across the load and Z 2
¯G = V ¯L + Z ¯l I ¯G = 216.9∠ V 2
2
2
− 2.74◦ + (0.8 + j 1)(34.15∠ − 41.5◦) = 259.8∠ − 0.638◦ V.
And the line-to-line voltage at the terminal of G 2 is ¯G V
2
,L−L =
√
3(259.8)∠29.36◦ V .
3. The real and reactive power supplied by generator G2. The complex power supplied by G 2 is ¯G S
2
,3φ =
∗ = 3(259.8∠ ¯G I ¯G 3V 2
2
Hence, the real power supplied is P G
2
,3φ =
− 0.638◦)(34.15∠41.5◦) = 20.1 + j 17.4 kVA.
20.1 kW and the reactive power supplied is Q G
2
,3φ =
17.4 kVar.
