ECE421/521 POWER SYSTEM ANALYSIS
Homework #3
2.7a / 2.7b / 2.7c / 2.8 / 2.9 / 2.10a / 2.10b / 2.14a / 2.14b / 2.15a / 2.15b / 2.16a / 2.16b / Total521 / 5’ / 5’ / 5’ / 10’ / 10’ / 5’ / 5’ / 10’ / 10’ / 10’ / 10’ / 10’ / 5’ / 100’
421 / 10’ / 10’ / 10’ / 10’ / 10’ / 10’ / 10’ / 5’ / 5’ / 10’ / 10’ / 10’ / 10’ / 100’+20’
2.7(a)
(1’/2’)
(1’/2’)
(1’/2’)
(1’/2’)
(1’/2’)
(b)
(1’/2’)
(2’/4’)
lagging (2’/4’)
(c)
(1’/2’)
(1’/2’)
(1’/2’)
(1’/2’)
(1’/2’)
2.8
(1’/1’)
(1’/1’)
(2’/2’)
(2’/2’)
From the above results, is positive and is negative, machine 1 is delivering 28kW real power and machine 2 is receiving 25kW real power.is negative and is positive, machine 1 is receiving 21kvar reactive power and machine 2 is delivering 33kvar reactive power. (2’/2’)
(2’/2’)
2.9
Run the following codes, (2’/2’)
%HW3 2.9
E1 = 120;
a1 = -5;
E2 = 100;
a2 = 0;
R = 1;
X = 7;
Z = R +1i*X;
E1 = (0.75*E1:1:E1)';
k = length(E1);
a1 = ones(k,1)*a1;
a1r = a1*pi/180;
a2 = ones(k,1)*a2;
a2r = a2*pi/180;
V1 = E1.*cos(a1r) + 1i*E1.*sin(a1r);
V2 = E2.*cos(a2r) + 1i*E2.*sin(a1r);
I12 = (V1 - V2)./Z;
I21 = -I12;
S1 = V1.*conj(I12); P1 = real(S1); Q1 = imag(S1);
S2 = V2.*conj(I21); P2 = real(S2); Q2 = imag(S2);
SL = S1+S2; PL = real(SL); QL = imag(SL);
Result1 = [E1, Q1, Q2, QL];
disp(' |V| 1 Q-1 Q-2 Q-L ')
disp(Result1)
plot(E1, Q1, E1, Q2, E1, QL)
xlabel('Source #1 Voltage Magnitude, V')
ylabel('Q, var')
text(115, 300, 'Q1'), text(115, -250, 'Q2'), text(115, 0, 'QL')
The result is, (2’/2’)
ECE521HW3_2_9
|V| 1 Q-1 Q-2 Q-L
90.0000 -130.7167 145.7984 15.0817
91.0000 -119.4291 131.7447 12.3156
92.0000 -107.8616 117.6910 9.8294
93.0000 -96.0140 103.6372 7.6233
94.0000 -83.8864 89.5835 5.6971
95.0000 -71.4788 75.5298 4.0510
96.0000 -58.7912 61.4760 2.6848
97.0000 -45.8236 47.4223 1.5987
98.0000 -32.5760 33.3686 0.7926
99.0000 -19.0484 19.3148 0.2664
100.0000 -5.2408 5.2611 0.0203
101.0000 8.8468 -8.7926 0.0541
102.0000 23.2144 -22.8464 0.3680
103.0000 37.8620 -36.9001 0.9618
104.0000 52.7895 -50.9538 1.8357
105.0000 67.9971 -65.0076 2.9896
106.0000 83.4847 -79.0613 4.4234
107.0000 99.2523 -93.1151 6.1373
108.0000 115.2999 -107.1688 8.1311
109.0000 131.6275 -121.2225 10.4050
110.0000 148.2351 -135.2763 12.9588
111.0000 165.1227 -149.3300 15.7927
112.0000 182.2903 -163.3837 18.9066
113.0000 199.7379 -177.4375 22.3004
114.0000 217.4655 -191.4912 25.9743
115.0000 235.4731 -205.5449 29.9281
116.0000 253.7606 -219.5987 34.1620
117.0000 272.3282 -233.6524 38.6758
118.0000 291.1758 -247.7061 43.4697
119.0000 310.3034 -261.7599 48.5436
120.0000 329.7110 -275.8136 53.8974
(3’/3’)
As the voltage magnitude of increase, the reactive power delivered from machine 1 increases and the reactive power delivered from machine 2 decreases. As, . So, the reactive power flows from higher terminal voltage to lower terminal voltage. (3’/3’)
2.10(a)
Run the following code, (1’/2’)
%HW3 2.10
wt = 0:0.05:2*pi;
van = 2500*cos(wt);
vbn = 2500*cos(wt-120*pi/180);
vcn = 2500*cos(wt-240*pi/180);
Z = 250;
delta = 36.87*pi/180;
ian = 2500*cos(wt - delta)/Z;
ibn = 2500*cos(wt-120*pi/180-delta)/Z;
icn = 2500*cos(wt-240*pi/180-delta)/Z;
pa = van.*ian/1000;
pb = vbn.*ibn/1000;
pc = vcn.*icn/1000;
p = pa+pb+pc;
plot(wt,pa,'*',wt,pb,'-',wt,pc,'+',wt,p,'o')
xlabel('Time, s')
ylabel('instantaneous power, kW')
text(0.3, 25, 'pa'), text(2.4, 25, 'pb')
text(4.5, 25, 'pc'), text(6, 28, 'pa+pb+pc')
(2’/4’)
The plot shows that pc lags pb 120 degree and pb lags pa 120 degree. The sum of them is a constant value 30kVA. (2’/4’)
(b)
(1’/2’)
(2’/4’)
The result agrees to the total power obtained in (a). (2’/4’)
2.14(a)
(2’/1’)
(2’/1’)
(2’/1’)
(1’/0.5’)
Total system kW:
Total system kvar:
System PF: (1’/0.5’)
Supply current per phase: (2’/1’)
(b)
If the capacitor bank is switched off,
(4’/2’)
New system PF: lagging (2’/1’)
New supply current per phase: (4’/2’)
2.15(a)
(2’/2’)
(2’/2’)
(2’/2’)
(1’/1’)
System PF: lagging (1’/1’)
Supply current per phase: (2’/2’)
(b)
To improve the overall PF from 0.6 to 0.8,
The new total kvar is: (1’/1’)
The kvar been compensated per phase: (1’/1’)
(2’/2’)
(2’/2’)
New line current: (4’/4’)
2.16(a)
The Δ-connected resistive load is transformed into equivalent Y-connection,
(1’/1’)
The phase voltage is,
(1’/1’)
The total impedance in single phase is,
(2’/2’)
The current per phase is,
(2’/2’)
The three phase power supplied is,
(2’/2’)
(b)
The line-to-neutral voltage of phase a at the combined load terminals is,
(3’/6’)
The line-to-line voltage of phase a at the combined load terminals is,
(2’/4’)