ECE421/521 POWER SYSTEM ANALYSIS
Homework #2
599/421
2.1a
10’
2.1b
10’
2.1c
10’
2.2a
5’
2.2b
5’
2.2c
5’
2.2d
10’
2.5
10’
2.6a
10’
2.6b
10’
2.1
Run the following code,
%ECE421/599 Homework #2 2.1
Vm = input('Enter the value of Vm(v): ');
thetav = input('Enter the value of thetav(degree): ');
Z = input('Enter the value of Z(omega): ');
gama = input('Enter the value of gama(degree): ');
thetai = thetav - gama;
theta = (thetav - thetai)*pi/180;
Im = Vm/Z;
wt = 0:.05:2*pi;
v = Vm*cos(wt);
i = Im*cos(wt + thetai*pi/180);
p = v.*i;
V = Vm/sqrt(2);
I = Im/sqrt(2);
P = V*I*cos(theta);
Q = V*I*sin(theta);
S = P + 1i*Q;
pr = P*(1 + cos(2*(wt +thetav)));
px = Q*sin(2*(wt + thetav));
PP = P*ones(1,length(wt));
xline = zeros(1, length(wt));
wt = 180/pi*wt;
subplot(2,2,1),plot(wt,v,wt,i,wt,xline),grid
title(['v(t)=Vm coswt, i(t)=Im cos(wt+',num2str(theta),')'])
xlabel('wt degree')
subplot(2,2,2),plot(wt, p, wt, xline),grid
title('p(t)=v(t) i(t)'), xlabel('wt,degree')
subplot(2,2,3),plot(wt, pr, wt, PP, wt, xline),grid
title('pr(t)'),xlabel('wt, degree')
subplot(2,2,4),plot(wt, px, wt, xline),grid
title('px(t)'),xlabel('wt, degree'),subplot(111)
For an inductive load, Z=1.25, gama=60 degree, input the following quantities,
2.7a
5’
2.7b
5’
2.7c
5’
Total
100’
We have the plots of 𝑖(𝑡), 𝑣(𝑡), 𝑝(𝑡), 𝑝𝑟 (𝑡), 𝑝𝑥 (𝑡),
(1’)
Also, in the same way, we can have plots of a capacitive load, Z = 2.0, gama = -30 degree,
(1’)
Finally we have figures of a resistive load Z = 2.5, gama = 0 degree,
(1’)
(a)
From the 𝑝𝑟 (𝑡) and 𝑝𝑥 (𝑡) plots, the real and reactive power for each load can be estimated,
Inductive load: 𝑃 = 2000𝑊, 𝑄 = 3400𝑣𝑎𝑟
(2’)
Capacitive load: 𝑃 = 2100𝑊, 𝑄 = −1200𝑣𝑎𝑟
Resistive load: 𝑃 = 2000𝑊, 𝑄 = 0𝑣𝑎𝑟
(2’)
(2’)
The conclusion regarding the sign of reactive power for inductive and capacitive loads is,
Inductive load: 𝑄 > 0
Capacitive load: 𝑄 < 0
(0.5’)
(0.5’)
(b)
Inductive load:
𝐼=
𝑉
𝑍
=
100
∠0°
√2
1.25∠60°
= 56.7∠ − 60°
(1’)
𝑆1 = 𝑉𝐼 ∗ = 4000∠60° = 2000𝑊 + 𝑗3464𝑣𝑎𝑟
(2’)
Capacitive load:
𝐼=
𝑉
𝑍
=
100
∠0°
√2
2∠−30°
= 35.4∠30°
(1’)
𝑆2 = 𝑉𝐼 ∗ = 2500∠ − 30° = 2165𝑊 − 𝑗1250𝑣𝑎𝑟
(2’)
Resistive load:
𝐼=
𝑉
𝑍
=
100
∠0°
√2
2.5∠0°
= 28.3∠0°
(1’)
𝑆3 = 𝑉𝐼 ∗ = 2000∠0° = 2000𝑊 + 𝑗0𝑣𝑎𝑟
(2’)
The results are close to the ones obtained from curves.
(1’)
(c)
If the above loads are all connected across the same power supply,
𝑆 = 𝑆1 + 𝑆2 + 𝑆3 = 6165𝑊 + 𝑗2213𝑣𝑎𝑟
(10’)
2.2(a)
We have,
𝑣(𝑡) = 200cos(377𝑡)
Rearrange instantaneous power p(t),
𝑝(𝑡) = 800 + 1000 cos(754𝑡 − 36.87°) = 800 + 1000 cos(754𝑡) cos(36.87°) +
1000 sin(754𝑡) sin(36.87°) =
sin(36.87°)sin(2 ∗ 377𝑡)
200 10
∗ 2
√2
√
∗ cos(36.87°) [1 + cos(2 ∗ 377𝑡)] +
(1’)
Compare it with the standard form of p(t),
𝑝(𝑡) = 𝑃[1 + cos(2𝜔𝑡)] + 𝑄𝑠𝑖𝑛(2𝜔𝑡)
We have,
P = 800W
Q = 600var
(1’)
(1’)
Thus the complex power supplied to the load is
200 10
∗
√2 √2
∗
S = P+ jQ = 800W + j600var
(2’)
(b)
According to the rearranged p(t) equation, the instantaneous current is,
𝑖(𝑡) = 10 cos(337𝑡 − 36.87°) 𝐴
(3’)
The rms value of it is
𝐼 = 5√2 𝐴
(2’)
(c)
The load impedance is,
𝑍=
𝑉
𝐼
200
= 10 √2
√2
∠0°
∠−36.87°
= 20∠36.87° 𝛺
(5’)
(d)
Run the following code,
(2’)
%ECE421/521 Homework#2 2.2
t = 0:.1*10^-3:16.67*10^-3;
v = 200*cos(377*t);
p = 800 + 1000*cos(754*t - 36.87*pi/180);
i = p./v;
subplot(3,1,1),plot(t,v),grid
title('v(t)=200cos(377t)')
xlabel('t/s')
ylabel('v/V')
subplot(3,1,2),plot(t,p),grid
title('p(t)=800+1000cos(754t-36.87)')
xlabel('t/s')
ylabel('p/VA')
subplot(3,1,3),plot(t,i),grid
title('v(t)=10cos(337t-36.87)')
xlabel('t/s')
ylabel('i/A')
we have the plots of v(t), p(t) and i(t),
(2’)
From the current plot, we can estimate,
The peak amplitude is 10A
(2’)
The phase angle is approximately 𝜃 =
0.0015
∗
0.016
360° = 33.75°
1
1
(2’)
The angular frequency is approximately 𝜔 = 2𝜋 𝑇 = 2𝜋 0.016 = 390 𝑟𝑎𝑑/𝑠
(2’)
2.5
3
𝑄 = 𝑃𝑡𝑎𝑛𝜃 = 400 ∗ 4 = 300𝑘𝑣𝑎𝑟
𝑆 = 400𝑘𝑊 + 𝑗300𝑘𝑣𝑎𝑟
𝑡𝑎𝑛𝜃1 =
(2’)
(2’)
√1 − cos 2 𝜃1 √1 − 0.962
=
= 0.29
𝑐𝑜𝑠𝜃1
0.96
𝑄1 = 𝑃1 𝑡𝑎𝑛𝜃1 = 120 ∗ 0.29 = 35𝑘𝑣𝑎𝑟
𝑆1 = 120𝑘𝑊 − 𝑗35𝑘𝑣𝑎𝑟
(2’)
(2’)
𝑆2 = 𝑆 − 𝑆1 = (400 − 120)𝑘𝑊 + 𝑗(300 − (−35))𝑘𝑣𝑎𝑟 = 280𝑘𝑊 + 𝑗335𝑘𝑣𝑎𝑟
(2’)
2.6(a)
Given,
|V||I|=30kVA
Power factor 0.8 leading
𝑆 = |𝑉||𝐼|𝑐𝑜𝑠𝜃 + 𝑗|𝑉||𝐼|𝑠𝑖𝑛𝜃 = 30 ∗ 0.8 − 𝑗30 ∗ 0.6 = 24𝑘𝑊 − 𝑗18𝑘𝑣𝑎𝑟
𝑅=
𝑋=
|𝑉|2
𝑃
|𝑉|2
𝑄
12002
= 24∗103 = 60𝛺
=
12002
18∗103
= 80𝛺
(4’)
(3’)
(3’)
(b)
𝑉
𝑉
𝐼 = 𝑅 + −𝑗𝑋 =
1200
1200
+ −𝑗80
60
= 20 + 𝑗15 𝐴
(3’)
𝑉𝑙𝑖𝑛𝑒 = 𝑍𝐼 = (20 + 𝑗15)(8.4 + 𝑗11.2) = 0 + 𝑗350 𝑉
(3’)
𝑉𝑖𝑛 = 𝑉𝑙𝑖𝑛𝑒 + 𝑉 = (0 + 𝑗350)(1200 + 𝑗0) = 1200 + 𝑗350 𝑉
(4’)
2.7(a)
𝑉
200∠0°
𝐼1 = 𝑍 = 0.8+𝑗5.6 = 5 − 𝑗35 𝐴
1
𝑉
𝐼2 = 𝑍 =
2
200∠0°
8−𝑗16
= 5 + 𝑗10 𝐴
(1’)
(1’)
𝑆1 = 𝑉𝐼1∗ = 200 ∗ (5 + 𝑗35) = 1000𝑊 + 𝑗7000𝑣𝑎𝑟
(1’)
𝑆2 = 𝑉𝐼2∗ = 200 ∗ (5 − 𝑗10) = 1000𝑊 − 𝑗2000𝑣𝑎𝑟
(1’)
𝑆3 = 5000 ∗ 0.8 + 5000 ∗ 0.6 = 4000𝑊 + 𝑗3000𝑣𝑎𝑟
(1’)
(b)
𝑆 = 𝑆1 + 𝑆2 + 𝑆3 = 6000𝑊 + 𝑗8000𝑣𝑎𝑟
𝑆∗
𝐼3 = 𝑉3∗ =
3
4000−𝑗3000
200
(1’)
= 20 − 𝑗15 𝐴
𝐼 = 𝐼1 + 𝐼2 + 𝐼3 = 5 − 𝑗35 + 5 + 𝑗10 + 20 − 𝑗15 = 30 − 𝑗40 = 50∠ − 53.13° 𝐴
𝑃𝐹 = cos(−53.13°) = 0.6 lagging
(2’)
(2’)
(c)
𝜃 ′ = 𝑐𝑜𝑠 −1 (1) = 0°
𝑄 ′ = 𝑃𝑡𝑎𝑛𝜃 ′ = 0 𝑣𝑎𝑟
𝑄𝑐 = 8000 − 0 = 8000 𝑣𝑎𝑟 (1’)
𝑆𝑐 = −𝑗8 𝑘𝑣𝑎𝑟
𝑍=
|𝑉|2
𝑆𝑐∗
𝐶=
106
2𝜋60∗5
𝑆∗
=
𝐼 = 𝑉∗ =
2002
𝑗8000
(1’)
= −𝑗5 𝛺
= 531 𝜇𝐹
6000
200
(1’)
(1’)
= 30∠0° 𝐴
(1’)