All Chapters Covered
SOLUTION MANUAL
, CONTENTS
1 THE POWER SYSTEM: AN OVERVIEW 1
2 BASIC PRINCIPLES 5
3 GENERATOR AND TRANSFORMER MODELS;
THE PER-UNIT SYSTEM 25
4 TRANSMISSION LINE PARAMETERS 52
5 LINE MODEL AND PERFORMANCE 68
6 POWER FLOW ANALYSIS 107
7 OPTIMAL DISPATCH OF GENERATION 147
8 SYNCHRONOUS MACHINE TRANSIENT ANALYSIS 170
9 BALANCED FAULT 181
10 SYMMETRICAL COMPONENTS AND UNBALANCED FAULT 208
11 STABILITY 244
12 POWER SYSTEM CONTROL 263
i
, CHAPTER 1 PROBLEMS
1.1 The demand estimation is the starting point for planning the future electric
power supply. The consistency of demand growth over the years has led to numer-
ous attempts to fit mathematical curves to this trend. One of the simplest curves
is
P = P0 ea(t−t0 )
where a is the average per unit growth rate, P is the demand in year t, and P0 is
the given demand at year t0 .
Assume the peak power demand in the United States in 1984 is 480 GW with
an average growth rate of 3.4 percent. Using MATLAB, plot the predicated peak
demand in GW from 1984 to 1999. Estimate the peak power demand for the year
1999.
We use the following commands to plot the demand growth
t0 = 84; P0 = 480;
a =.034;
t =(84:1:99)’;
P =P0*exp(a*(t-t0));
disp(’Predicted Peak Demand - GW’)
disp([t, P])
plot(t, P), grid
xlabel(’Year’), ylabel(’Peak power demand GW’)
P99 =P0*exp(a*(99 - t0))
The result is
1
,2 CONTENTS
Predicted Peak Demand - GW
84.0000 480.0000
85.0000 496.6006
86.0000 513.7753
87.0000 531.5441
88.0000 549.9273
89.0000 568.9463
90.0000 588.6231
91.0000 608.9804
92.0000 630.0418
93.0000 651.8315
94.0000 674.3740
95.0000 697.6978
96.0000 721.8274
97.0000 746.7916
98.0000 772.6190
99.0000 799.3398
P99 =
799.3398
The plot of the predicated demand is shown n Figure 1.
800 . . . . . . . ... . . . . . . ... . . . . . . ... . . . . . . ... . . . . . . ... . . . . . . ... . . . . . . ... . . . . . . ..
. . . . . . ..... .
. . . . . . ...... . .
......
750 . . . . . . . .. . . . . . . ... . . . . . . ... . . . . . . ... . . . . . . ... . . . . . . ... ............................ . . ... . . . . . . ..
. . . . . ....... . .
......
. . . . . ...... . . .
......
700 . . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . .............. . . .. .
.
.
.
.
.
.
.
.
. ......
..............
. .
.
. . . . . ..
.
.
. . . . . ..
.
.
Peak
.
. . . . ........ . . .
....... .
650
. . . . .
...
. . . .
. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . ............... . . .. . . . . . . .. . . . . . . .. . . . . . ..
Power . . .
......
.
..........
...
...
. . . .
. . . ....... . . . . .
Demand 600 . . .
. . . . . . . . . . . . . . .. . . . . . . .. ........ . . . . .. . . . . . . .. . . . . . . .. .
...
.
.
...
..
.......
....... . . . .
. . . . . ..
.
. . . . . ..
GW
. .
..
...
...... . . . . .
..... .
. . ........ . . . . .
. . .............. . . . . . .
550
.
...
...
. . . . . . . .. . . . . . .............. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . ..
.
...
...
..... . . . . . . .
......
. ........ . . . . . . .
. ....... . . . . . . .
.
...........
.
500
...
.. . . . . . . .
. . . .................. . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . ..
.
...
...
.....
........ . . . . . . . .
. . . . . . . .
. . . . . . . .
450 . . . . . . . . . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . ..
84 86 88 90 92 94 96 98 100
Year
FIGURE 1
Peak Power Demand for Problem 1.1
1.2 In a certain country, the energy consumption is expected to double in 10 years.
, CONTENTS 3
Assuming a simple exponential growth given by
P = P0 eat
calculate the growth rate a.
2P0 = P0 e10a
ln 2 = 10a
Solving for a, we have
0.693
a = = 0.0693 = 6.93%
10
1.3. The annual load of a substation is given in the following table. During each
month, the power is assumed constant at an average value. Using MATLAB and
the barcycle function, obtain a plot of the annual load curve. Write the necessary
statements to find the average load and the annual load factor.
Annual System Load
Interval – Month Load – MW
January 8
February 6
March 4
April 2
May 6
June 12
July 16
August 14
September 10
October 4
November 6
December 8
The following commands
data = [ 0 1 8
1 2 6
2 3 4
3 4 2
4 5 6
5 6 12
,4 CONTENTS
6 7 16
7 8 14
8 9 10
9 10 4
10 11 6
11 12 8];
P = data(:,3); % Column array of load
Dt = data(:, 2) - data(:,1); % Column array of demand interval
W = P’*Dt; % Total energy, area under the curve
Pavg = W/sum(Dt) % Average load
Peak = max(P) % Peak load
LF = Pavg/Peak*100 % Percent load factor
barcycle(data) % Plots the load cycle
xlabel(’time, month’), ylabel(’P, MW’), grid
result in
Pavg =
8
Peak =
16
LF =
50
16 .
.
.
.
....................................
... ...
.
.
.
.
. . ... ...
. .
.
. ...
14
. . .. . . .
. . . . . . . . . . .. . . . . . . . . . .. . . . . . . . . . .... . . . . .................................... . . . . . . . . . .. . . . . . . . . .
.
. ...
. . . .
... .
..
.
. . ... .
....
12
. . .. .
. . . . . . . . . . .. . . . . . . . . . .. . . . . .................................... . . . . . . . . . ..... . . . . . . . . . .. . . . . . . . . .
.. .
. . ... . ..... .
. . .
. . .
.. .
.... ...
10 . . . . . . . . . . . . . . . . . . . . . . . . . ... . . . . . . . . . . . . . ................................... . . . . .. . . . . . . . . .
. . .
. . .
. . .. . . ... .
. . ... . . ... .
P ...
. . . . ... .
. ...
8
.
.
.................................. . . . . .. . . . . . . . . . .. . . . . .... . . . . . . . . . . . . . . .. . . . . ..... . . . . .. . . . . .................................
MW
... . . ... . . ... . ...
... . . ... . . ... . ...
... ... ... ....
... . . . . . ... .
6
. .
. . . . . ..................................... . . . . . . . . . .................................... . . . . . . . . . . . . . . .. . . . . ..... . . . . ..................................... . . . .
... ... . . ... ..
.... ... . . ... ...
... ... . .
...
... ....
4
. .. ..
. . . . . . . . . . .................................... . . . . .... . . . . . . . . . . . . . . . . . . . .. . . . . .................................... . . . . . . . . .
. ... ..
. . . .
. .... ..
. . . .
.
...
.. .... . . .
2
.
. . . . . . . . . . .. . . . . .................................... . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . .. . . . . . . . . .
. . . . .
. . . . .
. . . . .
0 . . . . .
0 2 4 6 8 10 12
time, month
FIGURE 2
Monthly load cycle for Problem 1.3
, CHAPTER 2 PROBLEMS
2.1. Modify the program in Example 2.1 such that the following quantities can be
entered by the user:
The peak amplitude Vm , and the phase angle θv of the sinusoidal supply v(t) =
Vm cos(ωt + θv ). The impedance magnitude Z, and its phase angle γ of the load.
The program should produce plots for i(t), v(t), p(t), pr (t) and px (t), similar to
Example 2.1. Run the program for Vm = 100 V, θv = 0 and the following loads:
An inductive load, Z = 1.256 60◦ Ω
A capacitive load, Z = 2.06 −30◦ Ω
A resistive load, Z = 2.56 0◦ Ω
(a) From pr (t) and px (t) plots, estimate the real and reactive power for each load.
Draw a conclusion regarding the sign of reactive power for inductive and capaci-
tive loads.
(b) Using phasor values of current and voltage, calculate the real and reactive power
for each load and compare with the results obtained from the curves.
(c) If the above loads are all connected across the same power supply, determine
the total real and reactive power taken from the supply.
The following statements are used to plot the instantaneous voltage, current, and
the instantaneous terms given by(2-6) and (2-8).
Vm = input(’Enter voltage peak amplitude Vm = ’);
thetav =input(’Enter voltage phase angle in degree thetav = ’);
Vm = 100; thetav = 0; % Voltage amplitude and phase angle
Z = input(’Enter magnitude of the load impedance Z = ’);
gama = input(’Enter load phase angle in degree gama = ’);
thetai = thetav - gama; % Current phase angle in degree
5
,6 CONTENTS
theta = (thetav - thetai)*pi/180; % Degree to radian
Im = Vm/Z; % Current amplitude
wt=0:.05:2*pi; % wt from 0 to 2*pi
v=Vm*cos(wt); % Instantaneous voltage
i=Im*cos(wt + thetai*pi/180); % Instantaneous current
p=v.*i; % Instantaneous power
V=Vm/sqrt(2); I=Im/sqrt(2); % RMS voltage and current
pr = V*I*cos(theta)*(1 + cos(2*wt)); % Eq. (2.6)
px = V*I*sin(theta)*sin(2*wt); % Eq. (2.8)
disp(’(a) Estimate from the plots’)
P = max(pr)/2, Q = V*I*sin(theta)*sin(2*pi/4)
P = P*ones(1, length(wt)); % Average power for plot
xline = zeros(1, length(wt)); % generates a zero vector
wt=180/pi*wt; % converting radian to degree
subplot(221), plot(wt, v, wt, i,wt, xline), grid
title([’v(t)=Vm coswt, i(t)=Im cos(wt +’,num2str(thetai),’)’])
xlabel(’wt, degrees’)
subplot(222), plot(wt, p, wt, xline), grid
title(’p(t)=v(t) i(t)’), xlabel(’wt, degrees’)
subplot(223), plot(wt, pr, wt, P, wt,xline), grid
title(’pr(t) Eq. 2.6’), xlabel(’wt, degrees’)
subplot(224), plot(wt, px, wt, xline), grid
title(’px(t) Eq. 2.8’), xlabel(’wt, degrees’)
subplot(111)
disp(’(b) From P and Q formulas using phasor values ’)
P=V*I*cos(theta) % Average power
Q = V*I*sin(theta) % Reactive power
The result for the inductive load Z = 1.256 60◦ Ω is
Enter voltage peak amplitude Vm = 100
Enter voltage phase angle in degree thatav = 0
Enter magnitude of the load impedance Z = 1.25
Enter load phase angle in degree gama = 60
(a) Estimate from the plots
P =
2000
Q =
3464
(b) For the inductive load Z = 1.256 60◦ Ω, the rms values of voltage and current
are
1006 0◦
V = = 70.716 0◦ V
1.414
, CONTENTS 7
v(t) = Vm cos ωt, i(t) = Im cos(ωt − 60) p(t) = v(t)i(t)
100 ................ ......
....
....
6000 ...
........
... .....
...
.........
... ....
... ...
... ........ ..
......... .......... ... .... ... .... ...
..... .. .... ... .
..
... .
..
...
.... ..... .... .. . ... . ...
4000
.
50
. . . .
...
. ...
...
...
... ... ... ...
... ... ...
...
... ... ... ... ... .
.
.
. ...
... ... ... . ... .
. ... ..
... ... ..
. ... ...
.
.
. ... ...
... ...
... ..
..
..
.. ... .... ... ...
... . . ... ... ..
0 2000
. . . ... .
....................................................................................................................................................................... ... .. ... ...
... ... .. .. ... ... ...
... ...
...
...
...
... ... ... ... ... ... ..
..
... .. ... .. .
.
... ... .. ... . ...
...
... .... ..
. ..
. ... ... ...
... ... ... ...
−50 0
... ... . ..
... ... ..
.. ... ....................................................................................................................................................................
... ...
.... .... ... ... . ... .
... . . ....... ... .
... ... .
...
... ...... .... ... .. ...
... .................. ... .. ... ...
.... ...
.... ... .... ... .....
−100
.....
−2000
................ ... ..
......
... ..
........
0 100 200 300 400 0 100 200 300 400
ωt, degrees ωt, degrees
pr (t), Eq. 2.6 px (t), Eq. 2.8
4000 .....
...
...
.........
... ....
..
... 4000
... ... ... ... ........... ...........
... .... ... ..
. ... ..... ... ....
... .. ... ... ... ... ... ...
... ... ... .. .. ...
3000 2000
... ... . ... .
... ... ... ..
.. ... ... ... ...
...
... ..
. ... .
. .... ... ..
. ...
...
... ... ... ... ... ...
...
..
. ...
... ... ... ... .
. ...
.
.
. ...
.. ... .. .... .
. ...
... .. .. ... .
. . . ....
2000 0
. . . . ... .. ..
...................................................................................................................................................................... ...............................................................................................................................................................
... .. ... .. ... .
. ...
... ... ...
... ... ... ..
. ...
... ..
...
... ..
.. ... ..
..
...
... ..
.. ... ...
. . . ..
... .
...
... .
...
... .
.. ... ..
−2000
.
1000
... ... ... . .
... .
... ... ... ... ... ... ... ...
... ... ... ... ... ... ... ...
... .... ... .... ... ... ... ...
... .. ... . ... .... ... ...
... .. ... .. .... .... ..... .....
... .... ... ..... ... ..
0 −4000
.......... .... ...
..
0 100 200 300 400 0 100 200 300 400
ωt - degrees ωt, degrees
FIGURE 3
Instantaneous current, voltage, power, Eqs. 2.6 and 2.8.
70.716 0◦
I= = 56.576 −60◦ A
1.256 60◦
Using (2.7) and (2.9), we have
P = (70.71)(56.57) cos(60) = 2000 W
Q = (70.71)(56.57) sin(60) = 3464 Var
Running the above program for the capacitive load Z = 2.06 −30◦ Ω will result in
(a) Estimate from the plots
P =
2165
Q =
-1250
SOLUTION MANUAL
, CONTENTS
1 THE POWER SYSTEM: AN OVERVIEW 1
2 BASIC PRINCIPLES 5
3 GENERATOR AND TRANSFORMER MODELS;
THE PER-UNIT SYSTEM 25
4 TRANSMISSION LINE PARAMETERS 52
5 LINE MODEL AND PERFORMANCE 68
6 POWER FLOW ANALYSIS 107
7 OPTIMAL DISPATCH OF GENERATION 147
8 SYNCHRONOUS MACHINE TRANSIENT ANALYSIS 170
9 BALANCED FAULT 181
10 SYMMETRICAL COMPONENTS AND UNBALANCED FAULT 208
11 STABILITY 244
12 POWER SYSTEM CONTROL 263
i
, CHAPTER 1 PROBLEMS
1.1 The demand estimation is the starting point for planning the future electric
power supply. The consistency of demand growth over the years has led to numer-
ous attempts to fit mathematical curves to this trend. One of the simplest curves
is
P = P0 ea(t−t0 )
where a is the average per unit growth rate, P is the demand in year t, and P0 is
the given demand at year t0 .
Assume the peak power demand in the United States in 1984 is 480 GW with
an average growth rate of 3.4 percent. Using MATLAB, plot the predicated peak
demand in GW from 1984 to 1999. Estimate the peak power demand for the year
1999.
We use the following commands to plot the demand growth
t0 = 84; P0 = 480;
a =.034;
t =(84:1:99)’;
P =P0*exp(a*(t-t0));
disp(’Predicted Peak Demand - GW’)
disp([t, P])
plot(t, P), grid
xlabel(’Year’), ylabel(’Peak power demand GW’)
P99 =P0*exp(a*(99 - t0))
The result is
1
,2 CONTENTS
Predicted Peak Demand - GW
84.0000 480.0000
85.0000 496.6006
86.0000 513.7753
87.0000 531.5441
88.0000 549.9273
89.0000 568.9463
90.0000 588.6231
91.0000 608.9804
92.0000 630.0418
93.0000 651.8315
94.0000 674.3740
95.0000 697.6978
96.0000 721.8274
97.0000 746.7916
98.0000 772.6190
99.0000 799.3398
P99 =
799.3398
The plot of the predicated demand is shown n Figure 1.
800 . . . . . . . ... . . . . . . ... . . . . . . ... . . . . . . ... . . . . . . ... . . . . . . ... . . . . . . ... . . . . . . ..
. . . . . . ..... .
. . . . . . ...... . .
......
750 . . . . . . . .. . . . . . . ... . . . . . . ... . . . . . . ... . . . . . . ... . . . . . . ... ............................ . . ... . . . . . . ..
. . . . . ....... . .
......
. . . . . ...... . . .
......
700 . . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . .............. . . .. .
.
.
.
.
.
.
.
.
. ......
..............
. .
.
. . . . . ..
.
.
. . . . . ..
.
.
Peak
.
. . . . ........ . . .
....... .
650
. . . . .
...
. . . .
. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . ............... . . .. . . . . . . .. . . . . . . .. . . . . . ..
Power . . .
......
.
..........
...
...
. . . .
. . . ....... . . . . .
Demand 600 . . .
. . . . . . . . . . . . . . .. . . . . . . .. ........ . . . . .. . . . . . . .. . . . . . . .. .
...
.
.
...
..
.......
....... . . . .
. . . . . ..
.
. . . . . ..
GW
. .
..
...
...... . . . . .
..... .
. . ........ . . . . .
. . .............. . . . . . .
550
.
...
...
. . . . . . . .. . . . . . .............. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . ..
.
...
...
..... . . . . . . .
......
. ........ . . . . . . .
. ....... . . . . . . .
.
...........
.
500
...
.. . . . . . . .
. . . .................. . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . ..
.
...
...
.....
........ . . . . . . . .
. . . . . . . .
. . . . . . . .
450 . . . . . . . . . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . ..
84 86 88 90 92 94 96 98 100
Year
FIGURE 1
Peak Power Demand for Problem 1.1
1.2 In a certain country, the energy consumption is expected to double in 10 years.
, CONTENTS 3
Assuming a simple exponential growth given by
P = P0 eat
calculate the growth rate a.
2P0 = P0 e10a
ln 2 = 10a
Solving for a, we have
0.693
a = = 0.0693 = 6.93%
10
1.3. The annual load of a substation is given in the following table. During each
month, the power is assumed constant at an average value. Using MATLAB and
the barcycle function, obtain a plot of the annual load curve. Write the necessary
statements to find the average load and the annual load factor.
Annual System Load
Interval – Month Load – MW
January 8
February 6
March 4
April 2
May 6
June 12
July 16
August 14
September 10
October 4
November 6
December 8
The following commands
data = [ 0 1 8
1 2 6
2 3 4
3 4 2
4 5 6
5 6 12
,4 CONTENTS
6 7 16
7 8 14
8 9 10
9 10 4
10 11 6
11 12 8];
P = data(:,3); % Column array of load
Dt = data(:, 2) - data(:,1); % Column array of demand interval
W = P’*Dt; % Total energy, area under the curve
Pavg = W/sum(Dt) % Average load
Peak = max(P) % Peak load
LF = Pavg/Peak*100 % Percent load factor
barcycle(data) % Plots the load cycle
xlabel(’time, month’), ylabel(’P, MW’), grid
result in
Pavg =
8
Peak =
16
LF =
50
16 .
.
.
.
....................................
... ...
.
.
.
.
. . ... ...
. .
.
. ...
14
. . .. . . .
. . . . . . . . . . .. . . . . . . . . . .. . . . . . . . . . .... . . . . .................................... . . . . . . . . . .. . . . . . . . . .
.
. ...
. . . .
... .
..
.
. . ... .
....
12
. . .. .
. . . . . . . . . . .. . . . . . . . . . .. . . . . .................................... . . . . . . . . . ..... . . . . . . . . . .. . . . . . . . . .
.. .
. . ... . ..... .
. . .
. . .
.. .
.... ...
10 . . . . . . . . . . . . . . . . . . . . . . . . . ... . . . . . . . . . . . . . ................................... . . . . .. . . . . . . . . .
. . .
. . .
. . .. . . ... .
. . ... . . ... .
P ...
. . . . ... .
. ...
8
.
.
.................................. . . . . .. . . . . . . . . . .. . . . . .... . . . . . . . . . . . . . . .. . . . . ..... . . . . .. . . . . .................................
MW
... . . ... . . ... . ...
... . . ... . . ... . ...
... ... ... ....
... . . . . . ... .
6
. .
. . . . . ..................................... . . . . . . . . . .................................... . . . . . . . . . . . . . . .. . . . . ..... . . . . ..................................... . . . .
... ... . . ... ..
.... ... . . ... ...
... ... . .
...
... ....
4
. .. ..
. . . . . . . . . . .................................... . . . . .... . . . . . . . . . . . . . . . . . . . .. . . . . .................................... . . . . . . . . .
. ... ..
. . . .
. .... ..
. . . .
.
...
.. .... . . .
2
.
. . . . . . . . . . .. . . . . .................................... . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . .. . . . . . . . . .
. . . . .
. . . . .
. . . . .
0 . . . . .
0 2 4 6 8 10 12
time, month
FIGURE 2
Monthly load cycle for Problem 1.3
, CHAPTER 2 PROBLEMS
2.1. Modify the program in Example 2.1 such that the following quantities can be
entered by the user:
The peak amplitude Vm , and the phase angle θv of the sinusoidal supply v(t) =
Vm cos(ωt + θv ). The impedance magnitude Z, and its phase angle γ of the load.
The program should produce plots for i(t), v(t), p(t), pr (t) and px (t), similar to
Example 2.1. Run the program for Vm = 100 V, θv = 0 and the following loads:
An inductive load, Z = 1.256 60◦ Ω
A capacitive load, Z = 2.06 −30◦ Ω
A resistive load, Z = 2.56 0◦ Ω
(a) From pr (t) and px (t) plots, estimate the real and reactive power for each load.
Draw a conclusion regarding the sign of reactive power for inductive and capaci-
tive loads.
(b) Using phasor values of current and voltage, calculate the real and reactive power
for each load and compare with the results obtained from the curves.
(c) If the above loads are all connected across the same power supply, determine
the total real and reactive power taken from the supply.
The following statements are used to plot the instantaneous voltage, current, and
the instantaneous terms given by(2-6) and (2-8).
Vm = input(’Enter voltage peak amplitude Vm = ’);
thetav =input(’Enter voltage phase angle in degree thetav = ’);
Vm = 100; thetav = 0; % Voltage amplitude and phase angle
Z = input(’Enter magnitude of the load impedance Z = ’);
gama = input(’Enter load phase angle in degree gama = ’);
thetai = thetav - gama; % Current phase angle in degree
5
,6 CONTENTS
theta = (thetav - thetai)*pi/180; % Degree to radian
Im = Vm/Z; % Current amplitude
wt=0:.05:2*pi; % wt from 0 to 2*pi
v=Vm*cos(wt); % Instantaneous voltage
i=Im*cos(wt + thetai*pi/180); % Instantaneous current
p=v.*i; % Instantaneous power
V=Vm/sqrt(2); I=Im/sqrt(2); % RMS voltage and current
pr = V*I*cos(theta)*(1 + cos(2*wt)); % Eq. (2.6)
px = V*I*sin(theta)*sin(2*wt); % Eq. (2.8)
disp(’(a) Estimate from the plots’)
P = max(pr)/2, Q = V*I*sin(theta)*sin(2*pi/4)
P = P*ones(1, length(wt)); % Average power for plot
xline = zeros(1, length(wt)); % generates a zero vector
wt=180/pi*wt; % converting radian to degree
subplot(221), plot(wt, v, wt, i,wt, xline), grid
title([’v(t)=Vm coswt, i(t)=Im cos(wt +’,num2str(thetai),’)’])
xlabel(’wt, degrees’)
subplot(222), plot(wt, p, wt, xline), grid
title(’p(t)=v(t) i(t)’), xlabel(’wt, degrees’)
subplot(223), plot(wt, pr, wt, P, wt,xline), grid
title(’pr(t) Eq. 2.6’), xlabel(’wt, degrees’)
subplot(224), plot(wt, px, wt, xline), grid
title(’px(t) Eq. 2.8’), xlabel(’wt, degrees’)
subplot(111)
disp(’(b) From P and Q formulas using phasor values ’)
P=V*I*cos(theta) % Average power
Q = V*I*sin(theta) % Reactive power
The result for the inductive load Z = 1.256 60◦ Ω is
Enter voltage peak amplitude Vm = 100
Enter voltage phase angle in degree thatav = 0
Enter magnitude of the load impedance Z = 1.25
Enter load phase angle in degree gama = 60
(a) Estimate from the plots
P =
2000
Q =
3464
(b) For the inductive load Z = 1.256 60◦ Ω, the rms values of voltage and current
are
1006 0◦
V = = 70.716 0◦ V
1.414
, CONTENTS 7
v(t) = Vm cos ωt, i(t) = Im cos(ωt − 60) p(t) = v(t)i(t)
100 ................ ......
....
....
6000 ...
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4000
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−50 0
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−100
.....
−2000
................ ... ..
......
... ..
........
0 100 200 300 400 0 100 200 300 400
ωt, degrees ωt, degrees
pr (t), Eq. 2.6 px (t), Eq. 2.8
4000 .....
...
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..
... 4000
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3000 2000
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0 −4000
.......... .... ...
..
0 100 200 300 400 0 100 200 300 400
ωt - degrees ωt, degrees
FIGURE 3
Instantaneous current, voltage, power, Eqs. 2.6 and 2.8.
70.716 0◦
I= = 56.576 −60◦ A
1.256 60◦
Using (2.7) and (2.9), we have
P = (70.71)(56.57) cos(60) = 2000 W
Q = (70.71)(56.57) sin(60) = 3464 Var
Running the above program for the capacitive load Z = 2.06 −30◦ Ω will result in
(a) Estimate from the plots
P =
2165
Q =
-1250
