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CHAPTER 6 PROBLEMS

6.1. A power system network is shown in Figure 47. The generators at buses 1
and 2 are represented by their equivalent current sources with their reactances in
per unit on a 100-MVA base. The lines are represented by π model where series
reactances and shunt reactances are also expressed in per unit on a 100 MVA base.
The loads at buses 3 and 4 are expressed in MW and Mvar.
(a) Assuming a voltage magnitude of 1.0 per unit at buses 3 and 4, convert the loads
to per unit impedances. Convert network impedances to admittances and obtain the
bus admittance matrix by inspection.
(b) Use the function Y = ybus(zdata) to obtain the bus admittance matrix. The
function argument zdata is a matrix containing the line bus numbers, resistance
and reactance. (See Example 6.1.)
º·
1 2 º·
..................... .....................

¹¸ ¹¸
.................................... j0.25 ....................................
j0.25 ....................
......
......
......
j0.2
......
−j4 ......
......
......
......
−j4
......
......
......
j0.4
......
......
......
......
......
......
......
j0.1 ......
......
......
......
j0.16
......
......
......
......
......
......
j0.2 .....
...
...
...
..... ...
...
−j4 3 ...
S3...
.. −j4 4 ...
...
. S4
......... .........
.. ..

100 MW +j25 Mvar 200 MW +j50 Mvar

FIGURE 47
One-line diagram for Problem 6.1.

107
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108 CONTENTS

The load impedance in per unit is found from

|VL−L |2 |VB |2 |Vpu |2
Z= Ω & ZB = ∗ Ω or Z= pu
SL∗ SB ∗
Spu

(1.0)2
Z3 = = 0.9412 + j0.2353 pu
1 − j0.25
(1.0)2
Z4 = = 0.4706 + j0.11765 pu
2 − j0.5
Converting all impedances to admittances results in the admittance diagram shown
in Figure 48
¶³
1 2 ¶³
................... . ..................
µ´ µ´
................................. −j4 .................................
−j4 ...................
......
......
−j5
......
j0.25 ......
......
......
......
j0.25
......
−j2.5
......
......
......
......
......
......
......
−j10 ......
......
......
......
−j6.25
......
......
......
......
......
−j5 ......
...
...
... .. ...
... ...
j0.25 3 y3 ........
...
... j0.25 y4 4
..........
...
.
.. ..
1 − j0.25 pu 2 − j0.50 pu

FIGURE 48
The admittance diagram for problem 6.1.

The self admittances are

Y11 = −j4 + j0.25 − j4 − j10 − j2.5 = −j20.25
Y22 = −j5 + j0.25 − j4 − j6.25 = −j15
Y33 = (1 − j0.25) + j0.25 − j10 − j5 = 1 − j15
Y44 = (2 − j0.5) + j0.25 − j2.5 − j6.25 − j5 = 2 − j14
Therefore, the bus admittance matrix is
 
−j20.25 j4 j10 j2.5
 j4 −j15 0 j6.25 
Ybus =



j10 0 1 − j15 j5
j2.5 j6.25 j5 2 − j14
From the impedance diagram the following data is constructed for use with the
function Y = ybus(Z)
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CONTENTS 109

z = [0 1 0 0.25
0 1 0 -4.0
0 2 0 0.2
0 2 0 -4.0
0 3 0 -4.0
0 3 0.9412 0.2353
0 4 0 -4.0
0 4 0.4706 0.1176
1 2 0 0.25
1 3 0 0.10
1 4 0 0.40
2 4 0 0.16
3 4 0 0.20];
Y=ybus(z)

The result is

Y =
0 -20.25i 0 + 4.00i 0 +10.00i 0 + 2.50i
0 + 4.00i 0 -15.00i 0 0 + 6.25i
0 +10.00i 0 1 -15.00i 0 + 5.00i
0 + 2.50i 0 + 6.25i 0 + 5.00i 2 -14.00i

6.2. A power system network is shown in Figure 49.

0
... ... ... ...
.. .. ... ..
º·
.... ......
...
..
..... º·
....
.... ....... ........ ...
I1 ..........
..
.......
........
j0.5 j1.0 .......
........
..........
.. I2
¹¸
..... ....
.... 0.02 + j0.04 ....
.... ¹¸
.....
.... .. ..................................................................................................................................................... ... ....
. .. ... ... .. .
. .
....... ...
1 .....
..... .
.
..
.
.
......
. 2
..... ..
..... .....
..... ....
..... .....
..... .....
..... ..
.....
0.01 + j0.03 .....
.....
.....
...
.....
.....
0.0125 + j0.025
..... .....
.....
..... ..
......
..... .....
..... .....
....
...
... 3
................
.....
.
...
.
..........
.
............
..
..........
..
.....
0.4 + j0.2
..
..
.....
.
.......
....
...

0
FIGURE 49
One-line diagram for Problem 6.2.
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110 CONTENTS

The values marked are impedances in per unit on a base of 100 MVA. The currents
entering buses 1 and 2 are
I1 = 1.38 − j2.72 pu
I2 = 0.69 − j1.36 pu
(a) Determine the bus admittance matrix by inspection.
(b) Use the function Y = ybus(zdata) to obtain the bus admittance matrix. The
function argument zdata is a matrix containing the line bus numbers, resistance and
reactance. (See Example 6.1.) Write the necessary MATLAB commands to obtain
the bus voltages.
Converting all impedances to admittances results in the admittance diagram shown
in Figure 50

0
.... .... .... ....
. . ... .
º·
.... ......
...
...
.... º·
....
.... ....... ........ ...
I1 ..........
..
.......
........
.
−j2 −j1 .......
........
.
..........
.. I2
¹¸
..... ....
...
. 10 − j20 ....
.... ¹¸
.....
... ... .................................................................................................................................................
... ... ... ...
.. .. ... ... .. ..
.
....... ..
1 .....
.....
.
.
....
.
...
..
.
. 2
..... ..
..... ...
..... .....
..... .....
..... ....
..... ..
.....
.
10 − j30 .....
.....
.....
.....
....
.....
16 − j32
..... .....
.....
..... ..
.....
..... .....
..... .....
....
...
... 3
................
.....
.
...
.
..........
..
............
..
..........
.......
. 2 − j1
.
. ..
....
...
.....
....
...

0
FIGURE 50
The admittance diagram for problem 6.2.

(a) The self admittances are

Y11 = −j2 + (10 − j20) + (10 − j30) = 20 − j52
Y22 = −j1 + (10 − j20) + (16 − j32) = 26 − j53
Y33 = (2 − j1) + (10 − j30) + (16 − j32) = 28 − j63
Therefore, the bus admittance matrix is
 
20 − j52 −10 + j20 −10 + j30
Ybus =  −10 + j20 26 − j53 −16 + j32 
−10 + j30 −16 + j32 28 − j63
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CONTENTS 111

(b) To obtain the bus admittance matrix using Y = ybus(Z), and the bus voltages,
we use the following commands
z = [0 1 0.0 0.5
0 2 0.0 1.0
0 3 0.4 0.2
1 2 0.02 0.04
1 3 0.01 0.03
2 3 0.0125 0.025];
Y=ybus(z)
I=[1.38-j*2.72; 0.69-j*1.36; 0];
V=Y\I;
Vm=abs(V)
phase = 180/pi*angle(V)

The result is
Y =
20.0000-52.000i -10.0000+20.000i -10.0000+30.000i
-10.0000+20.000i 26.0000-53.000i -16.0000+32.000i
-10.0000+30.000i -16.0000+32.000i 28.0000-63.000i
Vm =
1.0293
1.0217
1.0001
phase =
1.4596
0.9905
-0.0150
6.3. Use Gauss-Seidel method to find the solution of the following equations
x1 + x1 x2 = 10
x1 + x2 = 6
with the following initial estimates
(0) (0)
(a) x1 = 1 and x2 = 1
(0) (0)
(b) x1 = 1 and x2 = 2
(k) (k)
Continue the iterations until |∆x1 | and |∆x2 | are less than 0.001.

Solving for x1 , and x2 from the first and second equation respectively, results in
10
x1 =
1 + x2
x2 = 6 − x1
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112 CONTENTS

(0) (0)
(a) With initial estimates x1 = 1 and x2 = 1, the iterative sequence becomes
(1) 10
x1 = =5
1+1
(1)
x2 = 6 − 5 = 1
(2) 10
x1 = =5
1+1
(2)
x2 = 6 − 5 = 5

(0) (0)
(b) With initial estimates x1 = 1 and x2 = 2, the iterative sequence becomes
(1) 10
x1 = = 3.3333
1+2
(1)
x2 = 6 − 3.3333 = 2.6666
(2) 10
x1 = = 2.7272
1 + 2.6666
(2)
x2 = 6 − 2.7272 = 3.2727
(3) 10
x1 = = 2.3404
1 + 3.2727
(3)
x2 = 6 − 2.3404 = 3.6596
(4) 10
x1 = = 2.1461
1 + 3.6596
(4)
x2 = 6 − 2.1461 = 3.8539
(5) 10
x1 = = 2.0602
1 + 3.8539
(5)
x2 = 6 − 2.0602 = 3.9398
(6) 10
x1 = = 2.0244
1 + 3.9398
(6)
x2 = 6 − 2.0244 = 3.9756
(7) 10
x1 = = 2.0098
1 + 3.9756
(7)
x2 = 6 − 2.00098 = 3.9902
(8) 10
x1 = = 2.0039
1 + 3.9902
(8)
x2 = 6 − 2.0039 = 3.9961
(9) 10
x1 = = 2.0016
1 + 3.9961
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CONTENTS 113

(9)
x2 = 6 − 2.0244 = 3.9984
(10) 10
x1 = = 2.0006
1 + 3.9984
(10)
x2 = 6 − 2.0006 = 3.9994

The above procedure may be placed in a MATLAB file as follows

x=input(’Enter initial estimates [x1; x2] -> ’);

dx=[1;1] % Errors in x1, x2 are set to a high value
xc=x;
while max(abs(dx)) >=.001
xc(1)=10/(1+x(2));
xc(2)=6-xc(1);
dx=x-xc
x=xc
end
disp(’Graphical display of of x1 and x2’)
x2=0:.1:10;
x1=10*ones(1,length(x2))./(1+x2);
x3 =-x2+6;
plot(x2,x1,x2,x3),grid

The result is

Enter initial estimates [x1; x2] -> [1; 1]

x =
5
1
x =
5
1
Graphical display of of x1 and x2

Running ch6p3 for the next initial estimates results in

Enter initial estimates [x1; x2] -> [1; 2]

x =
3.3333
2.6667
x =
2.7273
3.2727
x =
2.3404
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114 CONTENTS

3.6596
x =
2.1461
3.8539
x =
2.0602
3.9398
x =
2.0244
3.9756
x =
2.0098
3.9902
x =
2.0039
3.9961
x =
2.0016
3.9984
x =
2.0006
3.9994
Graphical display of of x1 and x2

6.4. A fourth-order polynomial equation is given by

x4 − 21x3 + 147x2 − 379x + 252 = 0
(a) Use Newton-Raphson method and hand calculations to find one of the roots of
the polynomial equation. Start with the initial estimate of x(0) = 0 and continue
until |∆x(k) | < 0.001.
(b) Write a MATLAB program to find the roots of the above polynomial by Newton-
Raphson method. The program should prompt the user to input the initial estimate.
Run using the initial estimates of 0, 3, 6, 10.
(c) Check your answers using the MATLAB function r = roots(A), where A is a
row vector containing the polynomial coefficients in descending powers.

df (x)
= 4x3 − 63x2 + 294x − 379
dx
(a) for x(0) = 0, we have
∆c(0) = c − f (x(0) ) = 0 − (252) = −252
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CONTENTS 115

µ ¶(0)
df
= −379
dx

∆c(0) −252
∆x(0) = ³ ´(0) = = 0.6649
df −379
dx

Therefore, the result at the end of the first iteration is

x(1) = x(0) + ∆x(0) = 0 + 0.6649 = 0.6649

The subsequent iterations result in

∆c(1) = c − f (x(1) ) = 0 − [(0.6649)4 − 21(0.6649)3 + 147(0.6649)2

−379(0.6649) + 252] = −59.0114

µ ¶(1)
df
= 4(0.6649)3 − 63(0.6649)2 + 294(0.6649) − 379 = −210.194
dx

∆c(1) −59.0114
∆x(1) = ³ ´(1) = = 0.28075
df −210.194
dx

x(2) = x(1) + ∆x(1) = 0.6649 + 0.28075 = 0.9457

∆c(2) = c − f (x(2) ) = 0 − [(0.9457)4 − 21(0.9457)3 + 147(0.9457)2

−379(0.9457) + 252] = −8.0942

µ ¶(2)
df
= 4(0.9457)3 − 63(0.9457)2 + 294(0.9457) − 379 = −153.9333
dx

∆c(2) −8.0942
∆x(2) = ³ ´(2) = = 0.05258
df −153.9333
dx

x(3) = x(2) + ∆x(2) = 0.9457 + 0.05258 = 0.9982

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116 CONTENTS

∆c(3) = c − f (x(3) ) = 0 − [(0.9982)4 − 21(0.9982)3 + 147(0.9982)2

−379(0.9982) + 252] = −0.2541

µ ¶(3)
df
= 4(0.9982)3 − 63(0.9982)2 + 294(0.9982) − 379 = −144.3174
dx

∆c(3) −0.2541
∆x(3) = ³ ´(3) = = 0.0018
df −144.3174
dx

x(4) = x(3) + ∆x(3) = 0.9982 + 0.0018 = 1.0000

∆c(4) = c − f (x(4) ) = 0 − [(1.0000)4 − 21(1.0000)3 + 147(1.0000)2

−379(1.0000) + 252] = −0.0003

µ ¶(4)
df
= 4(1.0000)3 − 63(1.0000)2 + 294(1.0000) − 379 = −144.0003
dx

∆c(4) −0.0003
∆x(4) = ³ ´(4) = = 0.0000
df −144.0003
dx

x(5) = x(4) + ∆x(4) = 1.0000 + 0.0000 = 1.0000

The following commands show the procedure for the solution of the given equation
by the Newton-Raphson method.

dx=1; % Change in variable is set to a high value

x=input(’Enter the initial estimate -> ’); % Initial estimate
iter = 0; % Iteration counter
disp(’iter Dc J dx x’)%Heading for result
while abs(dx) >= 0.001 & iter < 100 % Test for convergence
iter = iter + 1; % No. of iterations
Dc=0 - (x^4-21*x^3+147*x^2-379*x+252); % Residual
J = 4*x^3-63*x^2+ 294*x-379; % Derivative
dx= Dc/J; %Change in variable
x=x+dx; % Successive solution
fprintf(’%g’, iter), disp([Dc, J, dx, x])
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CONTENTS 117

The result is

Enter the initial estimate -> 0

iter Dc J dx x
1 -252.0000 -379.0000 0.6649 0.6649

2 -59.0114 -210.1938 0.2807 0.9457

3 -8.0942 -153.9333 0.0526 0.9982

4 -0.2541 -144.3174 0.0018 1.0000

5 -0.0003 -144.0003 0.0000 1.0000

Repeating for x(0) = 3, we have

Enter the initial estimate -> 3

iter Dc J dx x
1 48.0000 44.0000 1.0909 4.0909

2 -4.0128 43.2427 -0.0928 3.9981

3 0.0850 45.0339 0.0019 4.0000

4 0.0000 45.0000 0.0000 4.0000

Repeating for x(0) = 6, we have

Enter the initial estimate -> 6

iter Dc J dx x
1 -30.0000 -19.0000 1.5789 7.5789

2 19.3714 -28.1850 -0.6873 6.8917

3 -3.8917 -35.7586 0.1088 7.0005

4 0.0175 -36.0000 -0.0005 7.0000

Repeating for x(0) = 10, we have

Enter the initial estimate -> 10

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118 CONTENTS

iter Dc J dx x
1 -162.0000 261.0000 -0.6207 9.3793

2 -40.6800 136.7617 -0.2975 9.0819

3 -6.9992 91.1091 -0.0768 9.0050

4 -0.4046 80.6659 -0.0050 9.0000

5 -0.0017 80.0027 0.0000 9.0000

The above initial estimates have converged to the values of 1, 4, 7, and 9. As we

can see from the MATLAB roots function, these are the roots of the above fourth
order polynomial.
A = [1 -21 147 -379 252];
r = roots(A)

r =
9.0000
7.0000
4.0000
1.0000
6.5. Use Newton-Raphson method and hand calculation to find the solution of the
following equations:
x21 − 2x1 − x2 = 3
x21 + x22 = 41
(0) (0)
(a) Start with the initial estimates of x1 = 2, x2 = 3. Perform three iterations.
(b) Write a MATLAB program to find one of the solutions of the above equations by
Newton-Raphson method. The program should prompt the user to input the initial
estimates. Run the program with the above initial estimates.

Taking partial derivatives of the above equations results in the Jacobian matrix
· ¸
2x1 −2 −1
J=
2x1 2x2
(0) (0)
(a) Starting with initial estimates x1 = 2, and x2 = 3, the analytical solution
given by the Newton-Raphson method is
· ¸ · ¸
(0) 3 − [(2)2 − 2(2) − 3] 6
∆C = =
41 − [(2)2 + (3)2 ] 28
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CONTENTS 119

· ¸ · ¸
2(2) − 2 −1 2 −1
J= =
2(2) 2(3) 4 6

· ¸−1 · ¸ · ¸
(0) 2 −1 6 4
∆X = =
4 6 28 2

· ¸ · ¸ · ¸
2 4 6
X (0) = + =
3 2 5
For the second iteration, we have
· ¸ · ¸
(1) 3 − [(6)2 − 2(6) − 5] −16
∆C = =
41 − [(6)2 + (5)2 ] −20

· ¸ · ¸
2(6) − 2 −1 10 −1
J (1) = =
2(6) 2(5) 12 10

· ¸−1 · ¸ · ¸
10 −1 −16 −1.6071
∆X (1) = =
12 10 −20 −0.0714

· ¸ · ¸ · ¸
(2) 6 −1.6071 4.3929
X = + =
5 −0.0714 4.9286
and for the third iteration, we have
· ¸ · ¸
(2) 3 − [(4.3929)2 − 2(4.3929) − 4.9286] −2.5829
∆C = =
41 − [(4.3929)2 + (4.9286)2 ] −2.5880

· ¸ · ¸
2(4.3929) − 2 −1 6.7857 −1.0000
J (2) = =
2(4.3929) 2(4.9286) 8.7857 10.000

· ¸−1 · ¸ · ¸
(1) 6.7857 −1.0000 −2.5829 −0.3706
∆X = =
8.7857 10.000 −2.5880 0.0678

· ¸ · ¸ · ¸
4.3929 −0.3706 4.0222
X (3) = + =
4.9286 0.0678 4.9964
The following MATLAB program is used to find one of the solutions of the above
equations.
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120 CONTENTS

iter = 0; % Iteration counter

x=input(’Enter initial estimates, col. vector [x1; x2]-> ’);
Dx = [1; 1];
C=[3; 41];
disp(’Iter DC Jacobin matrix Dx x’);
% Heading for results
while max(abs(Dx)) >= .0001 & iter < 10 % Convergence test
iter=iter+1; % Iteration counter
f = [x(1)^2-2*x(1)-x(2); x(1)^2+x(2)^2]; % Functions
DC = C - f; % Residuals
J = [2*x(1)-2 -1 % Jacobian matrix
2*x(1) 2*x(2)];
Dx=J\DC; % Change in variables
x=x+Dx; % Successive solutions
fprintf(’%g’, iter), disp([DC, J, Dx, x]) % Results
end

The result is
Enter initial estimates, col. vector [x1; x2] -> [2; 3]
Iter DC Jacobin matrix Dx x
1 6 2 -1 4 6
28 4 6 2 5

2 -16.0000 10.0000 -1.0000 -1.6071 4.3929

-20.0000 12.0000 10.0000 -0.0714 4.9286

3 -2.5829 6.7857 -1.0000 -0.3706 4.0222

-2.5880 8.7857 9.8571 0.0678 4.9964

4 -0.1374 6.0444 -1.0000 -0.0221 4.0001

-0.1420 8.0444 9.9928 0.0036 5.0000

5 -0.0005 6.0002 -1.0000 -0.0001 4.0000

-0.0005 8.0002 10.0000 0.0000 5.0000

6.6. In the power system network shown in Figure 51, bus 1 is a slack bus with
V1 = 1.06 0◦ per unit and bus 2 is a load bus with S2 = 280 MW + j60 Mvar. The
line impedance on a base of 100 MVA is Z = 0.02 + j0.04 per unit.
(0)
(a) Using Gauss-Seidel method, determine V2 . Use an initial estimate of V2 =
1.0 + j0.0 and perform four iterations.
(b) If after several iterations voltage at bus 2 converges to V2 = 0.90 − j0.10,
determine S1 and the real and reactive power loss in the line.
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CONTENTS 121

¾»

½¼
...
...
.. Z12 = 0.02 + j0.04
S1............. ...
...
..... ...
.. ...
...
2 ...
...
1 ...
... S2 = 280 MW +j60 Mvar
...
.........
..

FIGURE 51
One-line diagram for Problem 6.6.

1
y12 = = 10 − j29
0.02 + j0.04
The per unit load at bus 2 is
280 + j60
S2 = − = −2.8 − j0.60
100
(0)
Starting with an initial estimate of V2 = 1.0+j0.0, the voltage at bus 2 computed
from (6.28) for three iterations are
−2.8+j0.60
(1) 1.00000−j0.00000 + (10 − j20)(1)
V2 = = 0.92000 − j0.10000
10 − j20
−2.8+j0.60
(2) 0.92000+j0.10000 + (10 − j20)(1)
V2 = = 0.90238 − j0.09808
10 − j20
−2.8+j0.60
(3) 0.90238−j0.09808 + (10 − j20)(1)
V2 = = 0.90050 − j0.10000
10 − j20
(b) Assuming voltage at bus 2 converges to V2 = 0.9 − j0.1, the line flows are
computed as follows

I12 = y12 (V1 − V2 ) = (10 − j20)[(1 + j0) − (0.9 − j0.10] = 3.0 − j1.0

I21 = −I12 = −3.0 + j1.0

∗
S12 = V1 I12 = (1.0 + j0.0)(3.0 + j1.0) = 3 + j1 pu
= 300 MW + j100 Mvar
∗
S21 = V2 I21 = (0.9 − j0.1)(−3.0 − j1.0) = −2.8 − j0.6 pu
= −280 MW − j60 Mvar
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122 CONTENTS

The line loss is

SL 12 = S12 + S21 = (300 + j100) + (−280 − j60) = 20 MW + j40 Mvar

The slack bus real and reactive power are P1 = 300 MW, and Q1 = 100 Mvar. The
above calculations is placed in a MATLAB file, and solution is obtained by typing
ch6p6 at the MATLAB prompt.

6.7. Figure 6.6 shows the one-line diagram of a simple three-bus power system with
generation at bus 1. The voltage at bus 1 is V1 = 1.06 0◦ per unit. The scheduled
loads on buses 2 and 3 are marked on the diagram. Line impedances are marked in
per unit on a 100 MVA base. For the purpose of hand calculations, line resistances
and line charging susceptances are neglected.
(0) (0)
(a) Using Gauss-Seidel method and initial estimates of V2 = 1.0+j0 and V3 =
1.0 + j0, determine V2 and V3 . Perform two iterations.
(b) If after several iterations the bus voltages converge to

V2 = 0.90 − j0.10 pu
V3 = 0.95 − j0.05 pu

determine the line flows and line losses and the slack bus real and reactive power.
Construct a power flow diagram and show the direction of the line flows.
(c) Check the power flow solution using the lfgauss and other required programs.
(Refer to Example 6.9.) Use a power accuracy of 0.00001 and an acceleration factor
of 1.0.
1
V = 1 6 0◦ 1
1
j 30 2
¾» .......................................
400 MW

½¼........................ ..
.......................................................
..... 320 Mvar
..... ....
Slack .....
..... ..
.....
...
.... ....
.... ....
....
..... .....
..... .
......
j0.0125 .....
....
.... j0.05 ...
.....
.....
.... ....
..... .
.
.....
..... .
..... .....
..... .....
.... ....
.... ..
.....
.... .
... .....
... ...

....
...
... 3 ..
....
...
... ...
... ...
... ...
.. ...
.......... ...........
.. ..

300 MW 270 Mvar

FIGURE 52
One-line diagram for Problem 6.7.
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CONTENTS 123

(a) Line impedances are converted to admittances

y12 = −j30
1
y13 = = −j80
j0.0125
1
y23 = = −j20
j0.05
At the P-Q buses, the complex loads expressed in per units are
(400 + j320)
S2sch = − = −4.0 − j3.2 pu
100
(300 + j270)
S3sch = − = −3.0 − j2.7 pu
100
For hand calculation, we use (6.28). Bus 1 is taken as reference bus (slack bus).
(0) (0)
Starting from an initial estimate of V2 = 1.0 + j0.0 and V3 = 1.0 + j0.0, V2
and V3 are computed from (6.28) as follows
∗
S2sch (0)
(0) ∗ + y12 V1 + y23 V3
(1) V2
V2 =
y12 + y23
−4.0+j3.2
1.0−j0 + (−j30)(1.0 + j0) + (−j20)(1.0 + j0)
=
−j50
= 0.936 − j0.08
and
∗
S3sch (1)
(0) ∗ + y13 V1 + y23 V2
(1) V3
V3 =
y13 + y23
−3.0+j2.7
1−j0 + (−j80)(1.0 + j0) + (−j20)(0.936 − j0.08)
=
−j100
= 0.9602 − j0.046
For the second iteration we have
−4.0+j3.2
(2) 0.936+j0.08 + (−j30)(1.0 + j0) + (−j20)(0.9602 − j0.046)
V2 =
−j50
= 0.9089 − j0.0974
and
−3.0+j2.7
(2) 0.9602+j0.046 + (−j80)(1.0 + j0) + (−j20)(0.9089 − j0.0974)
V3 =
(−j100)
= 0.9522 − j0.0493
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124 CONTENTS

The process is continued and a solution is converged with an accuracy of 5 × 10−5

per unit in seven iterations as given below.
(3) (3)
V2 = 0.9020 − j0.0993 V3 = 0.9505 − j0.0498
(4) (4)
V2 = 0.9004 − j0.0998 V3 = 0.9501 − j0.0500
(5) (5)
V2 = 0.9001 − j0.1000 V3 = 0.9500 − j0.0500
(6) (6)
V2 = 0.9000 − j0.1000 V3 = 0.9500 − j0.0500
(7) (7)
V2 = 0.9000 − j0.1000 V3 = 0.9500 − j0.0500

The final solution is

V2 = 0.90 − j0.10 = 0.9055546 −6.34◦ pu

V3 = 0.95 − j0.05 = 0.95136 −3.0128◦ pu

(b) With the knowledge of all bus voltages, the slack bus power is obtained from
(6.27)

P1 − jQ1 = V1∗ [V1 (y12 + y13 ) − (y12 V2 + y13 V3 )]

= 1.0[1.0(−j30 − j80) − (−j30)(0.9 − j0.1) −
(−j80)(0.95 − j0.05)]
= 7.0 − j7.0

or the slack bus real and reactive powers are P1 = 7.0 pu = 700 MW and Q1 = 7.0
pu = 700 Mvar.

To find the line flows, first the line currents are computed. With line charging ca-
pacitors neglected, the line currents are

I12 = y12 (V1 − V2 ) = (−j30)[(1.0 + j0) − (0.90 − j0.10)] = 3.0 − j3.0

I21 = −I12 = −3.0 + j3.0
I13 = y13 (V1 − V3 ) = (−j80)[(1.0 + j0) − (0.95 − j.05)] = 4.0 − j4.0
I31 = −I13 = −4.0 + j4.0
I23 = y23 (V2 − V3 ) = (−j20)[(0.90 − j0.10) − (0.95 − j.05)] = −1.0 + j1.0
I32 = −I23 = 1.0 − j1.0

The line flows are

∗
S12 = V1 I12 = (1.0 + j0.0)(3.0 + j3) = 3.0 + j3.0 pu
= 300 MW + j300 Mvar
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CONTENTS 125

∗
S21 = V2 I21 = (0.90 − j0.10)(−3 − j3) = −3.0 − j2.4 pu
= −300 MW − j240 Mvar
∗
S13 = V1 I13 = (1.0 + j0.0)(4.0 + j4.0) = 4.0 + j4.0 pu
= 400 MW + j400 Mvar
∗
S31 = V3 I31 = (0.95 − j0.05)(−4.0 − j4.0) = −4.0 − j3.6 pu
= −400 MW − j360 Mvar
∗
S23 = V2 I23 = (0.90 − j0.10)(−1.0 − j1.0) = −1.0 − j0.80 pu
= −100 MW − j80 Mvar
∗
S32 = V3 I32 = (0.95 − j0.05)(1 + j1) = 1.0 + j0.9 pu
= 100 MW + j90 Mvar

and the line losses are

SL 12 = S12 + S21 = 0.0 MW + j60 Mvar

SL 13 = S13 + S31 = 0.0 MW + j40 Mvar
SL 23 = S23 + S32 = 0.0 MW + j10 Mvar

The power flow diagram is shown in Figure 6.7, where real power direction is
indicated by → and the reactive power direction is indicated by 7→. The values
within parentheses are the real and reactive losses in the line.

1 ..300 300 2
....................................... (0) .........................................

700
.........................................
¾» 300
......................................... (60) ......................................... .........................................

240 400
½¼ 400 400 100 100
......................................... ......................................... (0) ................................ ................................ (0) ......................................... .........................................

700 ......................................... (40) ................................. ................................. (10) ......................................... 320

400 360 90 80
3 ...
...
...
...
...
...
... ...
... ...
.. ..
.......... ..........
.. ..

300 270
FIGURE 53
Power flow diagram of Problem 6.7 (powers in MW and Mvar).

(c) The power flow program lfgauss is used to obtain the solution, with the follow-
ing statements:

clear
basemva = 100; accuracy = 0.000001; accel = 1.1; maxiter = 100;
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126 CONTENTS

% Problem 6.7(c)
% Bus Bus Voltage Angle -Load--- -Generator-- Injected
% No code Mag. Degree MW MVAR MW MVAR Qmin Qmax Mvar
busdata=[1 1 1.0 0.0 0.0 0.0 0.0 0.0 0 0 0
2 0 1.0 0.0 400 320 0.0 0.0 0 0 0
3 0 1.0 0.0 300 270 0.0 0.0 0 0 0];

% Line code
% Bus bus R X 1/2 B = 1 for lines
% nl nr pu pu pu >1 or <1 tr. tap at bus nl
linedata=[1 2 0.0 1/30 0.0 1
1 3 0.0 0.0125 0.0 1
2 3 0.0 0.050 0.0 1];
disp(’Problem 6.7(c)’)
lfybus % form the bus admittance matrix
lfgauss % Load flow solution by Gauss-Seidel method
busout % Prints the power flow solution on the screen
lineflow % Computes and displays the line flow and losses

The above statements are saved in the file ch6p7c.m. Run the program to obtain
the solution.

6.8. Figure 54 shows the one-line diagram of a simple three-bus power system with
generation at buses 1 and 3. The voltage at bus 1 is V1 = 1.0256 0◦ per unit. Volt-
age magnitude at bus 3 is fixed at 1.03 pu with a real power generation of 300 MW.
A load consisting of 400 MW and 200 Mvar is taken from bus 2. Line impedances
are marked in per unit on a 100 MVA base. For the purpose of hand calculations,
line resistances and line charging susceptances are neglected.
(0) (0)
(a) Using Gauss-Seidel method and initial estimates of V2 = 1.0+j0 and V3 =
1.03 + j0 and keeping |V3 | = 1.03 pu, determine the phasor values of V2 and V3 .
Perform two iterations.
(b) If after several iterations the bus voltages converge to

V2 = 1.0012436 −2.1◦ = 1.000571 − j0.0366898 pu

V3 = 1.036 1.36851◦ = 1.029706 + j.0246 pu

determine the line flows and line losses and the slack bus real and reactive power.
Construct a power flow diagram and show the direction of the line flows.
(c) Check the power flow solution using the lfgauss and other required programs.
(Refer to Example 6.9.)
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CONTENTS 127

V1 = 1.0256 0◦ P..3................= 300 MW

¾» j0.05 ¾»
.....................

1 3
½¼........................ .................
....
....
½¼
....
Slack ....
....
.... .
..
.
.
....
.... |V3 | = 1.03
..... ....
.... ....
.... .....
....
.... .
......
...
j0.025 .....
....
....
....
j0.025
....
....
....
..... ..
.....
.
.... ....
.... ....
.... ....
.... ....
.....
..... .
.....
.
.
.. ....
...
... .....

....
...
.
2 ...
..
...
... ...
... ...
... ...
.. ...
.......... ..........
.. ..

400 MW 200 Mvar

FIGURE 54
One-line diagram of Problem 6.8.

(a) Line impedances converted to admittances are y12 = −j40, y13 = −j20 and
y23 = −j40. The load and generation expressed in per units are
(400 + j200)
S2sch = − = −4.0 − j2.0 pu
100
300
P3sch = = 3.0 pu
100
Bus 1 is taken as the reference bus (slack bus). Starting from an initial estimate of
(0) (0)
V2 = 1.0 + j0.0 and V3 = 1.03 + j0.0, V2 and V3 are computed from (6.28).
∗
S sch (0)
2
(0) ∗ + y12 V1 + y23 V3
(1) V2
V2 =
y12 + y23
−4.0+j2.0
1.0−j0 + (−j40)(1.025 + j0) + (−j40)(1.03 + j0)
=
(−j80)
= 1.0025 − j0.05

Bus 3 is a regulated bus where voltage magnitude and real power are specified. For
the voltage-controlled bus, first the reactive power is computed from (6.30)
(1) (0) ∗ (0) (1)
Q3 = −={V3 [V3 (y13 + y23 ) − y13 V1 − y23 V2 ]}
= −={(1.03 − j0)[(1.03 + j0)(−j60) − (−j20)(1.025 + j0)
−(−j40)(1.0025 − j0.05)]}
= 1.236
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128 CONTENTS

(1)
The value of Q3 is used as Qsch 3 for the computation of voltage at bus 3. The
(1)
complex voltage at bus 3, denoted by Vc3 , is calculated
S sch
3 ∗ (1)
(0) + y13 V1 + y23 V2
(1) V3 ∗
Vc3 =
y13 + y23
3.0−j1.236
1.03−j0 + (−j20)(1.025 + j0) + (−j40)(1.0025 − j0.05)
=
(−j60)
= 1.0300 + j0.0152
(1)
Since |V3 | is held constant at 1.03 pu, only the imaginary part of Vc3 is retained,
(1)
i.e, f3 = 0.0152, and its real part is obtained from
q
(1)
e3 = (1.03)2 − (0.0152)2 = 1.0299

Thus
(1)
V3 = 1.0299 + j0.0152

For the second iteration, we have

S sch
2 ∗ (1)
(1) ∗ + y12 V1 + y23 V3
(2) V2
V2 =
y12 + y23
−4.0+j2.0
1.0025+j.05 + (−j40)(1.025) + (−j40)(1.0299 + j0.0152)
=
(−j80)
= 1.0001 − j0.0409

(2) (1) ∗ (1) (2)

Q3 = −={V3 [V3 (y13 + y23 ) − y13 V1 − y23 V2 ]}
= −={(1.0299 − j0.0152)[(1.0299 + j0.0152)(−j60)
−(−j20)(1.025 + j0) − (−j40)(1.0001 − j0.0409)]}
= 1.3671

S sch
3 ∗ (2)
(1) ∗ + y13 V1 + y23 V2
(2) V3
Vc3 =
y13 + y23
3.0−j1.3671
1.0299−j0.0152 + (−j20)(1.025) + (−j40)(1.0001 − j.0409)
=
(−j60)
= 1.0298 + j0.0216
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CONTENTS 129

(2)
Since |V3 | is held constant at 1.03 pu, only the imaginary part of Vc3 is retained,
(2)
i.e, f3 = 0.0216, and its real part is obtained from
q
(2)
e3 = (1.03)2 − (0.0216)2 = 1.0298

or
(2)
V3 = 1.0298 + j0.0216

The process is continued and a solution is converged with an accuracy of 5 × 10−5

pu to

V2 = 1.0012436 −2.1◦ = 1.000571 − j.0366898 pu

S3 = 3.0 + j1.3694 pu = 300 MW + j136.94 Mvar
V3 = 1.036 1.36851◦ pu = 1.029706 + j0.0246

(b) Line flows and line losses are computed as in Problem 6.7, and the results
expressed in MW and Mvar are

S12 = 150.43 + j100.16 S21 = −150.43 − j92.39 SL 12 = 0.0 + j7.77

S13 = −50.43 − j9.65 S31 = 50.43 + j10.90 SL 13 = 0.0 + j1.25
S23 = −249.58 − j107.61 S32 = 249.58 + j126.03 SL 23 = 0.0 + j18.42

The slack bus real and reactive powers are

S1 = S12 + S13 = (150.43 + j100.16) + (−50.43 − j9.65)

= 100 MW + j90.51 Mvar

50.43 50.43
1 ................................... (0.0) ..........................................
3 300
100
.................................................
º·9.65
................................... (1.25) .......................................... ...
........................
º·
10.9
¹¸150.43 150.43 249.58 249.58 ¹¸
.................................................
....
(0.0) ...................................
................................ (0.0) ....................................... ............................
...............................
90.51 (7.77) ...................................
................................... (18.42)...................................... 136.93
...............................
.

100.16 92.39 107.61 126.03

2 ........ ...
...
...
... ...
.. .
.......... ..........
.. ..

400 200
FIGURE 55
Power flow diagram of Problem 6.8 (powers in MW and Mvar).
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130 CONTENTS

The power flow diagram is shown in Figure 55, where real power direction is indi-
cated by → and the reactive power direction is indicated by 7→. The values within
parenthesis are the real and reactive losses in the line.

(c) The power flow program lfgauss is used to obtain the solution, with the follow-
ing statements:

clear
basemva = 100; accuracy = 0.000001; accel = 1.1; maxiter = 100;

% Problem 6.8(c)
% Bus Bus Voltage Angle -Load--- -Generator-- Injected
% No code Mag. Degree MW MVAR MW MVAR Qmin Qmax Mvar
busdata=[1 1 1.025 0.0 0.0 0.0 0.0 0.0 0 0 0
2 0 1.0 0.0 400 200 0.0 0.0 0 0 0
3 2 1.03 0.0 0.0 0.0 300 0.0 0 160 0];

% Line code
% Bus bus R X 1/2 B = 1 for lines
% nl nr pu pu pu >1 or <1 tr. tap at bus nl
linedata=[1 2 0.0 0.025 0.0 1
1 3 0.0 0.05 0.0 1
2 3 0.0 0.025 0.0 1];
disp(’Problem 6.8(c)’)
lfybus % form the bus admittance matrix
lfgauss % Load flow solution by Gauss-Seidel method
busout % Prints the power flow solution on the screen
lineflow % Computes and displays the line flow and losses

The above statements are saved in the file ch6p8c.m. Run the program to obtain
the solution.

6.9. The one-line diagram of a four-bus power system is as shown in Figure 56.
Reactances are given in per unit on a common MVA base. Transformers T1 and
T2 have tap settings of 0.8:1, and 1.25:1 respectively. Obtain the bus admittance
matrix.
From Figure 6.15 in the text, the sending end transformer π model is
yt −j80
y12 = = = −j100
µa 0.8
¶
1−a 1 − 0.8
y10 = 2
yt = (−j80) = −j25
a 0.64
µ ¶
a−1 0.8 − 1
y30 = yt = (−j80) = j20
a 0.8
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CONTENTS 131

j0.2
Xt = j0.0125 Xt = j0.16
........ .. ........ ..
... ....... ... .......
. .. . ..
.. ...... ....... .. ...... .......
.. ..
............................................. .
..... .. ..
............................................. .
.....
... ...... ... ......
. .. . ..
.......
... ........
.. j0.25 .......
... ........
..

1 0.8 : 1 1.25 : 1 2
3 4

FIGURE 56
One-line diagram for Problem 6.9.

and the receiving end transformer π model is

yt −j6.25
y42 = = = −j5
µa 1.25
¶
1−a 1 − 1.25
y40 = 2
yt = (−j6.25) = j1
a 1.5625
µ ¶
a−1 1.25 − 1
y20 = yt = (−j6.25) = −j1.25
a 1.25
All line impedances are converted to admittances, and the admittance diagram is
constructed as shown in Figure 57.

3 −j5 4
1 2
−j100
. . . .
−j5
. . . .
.. .. .. .. .. .. .. ..
.... ........ ........ ........ ..... .... ........ ........ ........ .....

−j4
.............. ............... .............. ...............
... ... ... ...
... ... ... ...
... .. ... ..
. .

−j25 j20 j1 −j1.25

FIGURE 57
Admittance diagram for the system of Problem 6.9.

The bus admittance matrix for the network of Figure 57 obtained by inspection is
 
−j125 0 j100 0
 0 −j6.25 0 j5 
Ybus =
 j100

0 −j89 j9 
0 j5 j9 −j13

The script file lfybus can be used to compute the bus admittance matrix. We use
the following statements.
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132 CONTENTS

clear
% Line code
% Bus bus R X 1/2 B = 1 for lines
% nl nr pu pu pu >1 or <1 tr. tap at bus nl
linedata=[1 3 0.0 0.0125 0.0 0.8
3 4 0.0 0.20 0.0 1
3 4 0.0 0.25 0.0 1
4 2 0.0 0.16 0.0 1.25];
lfybus % form the bus admittance matrix
Ybus % displays Ybus on the screen

Run ch6p9 to obtain the bus admittance matrix.

6.10. In the two-bus system shown in Figure 58, bus 1 is a slack bus with V1 =
1.06 0◦ pu. A load of 150 MW and 50 Mvar is taken from bus 2. The line admittance
is y12 = 106 −73.74◦ pu on a base of 100 MVA. The expression for real and
reactive power at bus 2 is given by

P2 = 10|V2 ||V1 | cos(106.26◦ − δ2 + δ1 ) + 10|V2 |2 cos(−73.74◦ )

Q2 = −10|V2 ||V1 | sin(106.26◦ − δ2 + δ1 ) − 10|V2 |2 sin(−73.74◦ )

Using Newton-Raphson method, obtain the voltage magnitude and phase angle of
bus 2. Start with an initial estimate of |V2 |(0) = 1.0 pu and δ2 (0) = 0◦ . Perform
two iterations. Partial derivatives of P2 , and Q2 with respect to |V2 |, and δ2 are

2
¾»1 ........................................
150 MW
y12 = 2.8 − j9.6
½¼ ........................................
50 Mvar
V1 = 1.06 0◦

FIGURE 58
One-line diagram for Problem 6.10.

∂P2
= 10|V2 ||V1 | sin(106.26◦ − δ2 + δ1 )
∂δ2
∂P2
= 10|V1 | cos(106.26◦ − δ2 + δ1 ) + 20|V2 | cos(−73.74◦ )
∂|V2 |
∂Q2
= 10|V2 ||V1 | cos(106.26◦ − δ2 + δ1 )
∂δ2
∂Q2
= −10|V1 | sin(106.26◦ − δ2 + δ1 ) − 20|V2 | sin(−73.74◦ )
∂|V2 |
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CONTENTS 133

The load expressed in per unit is

(150 + j50)
S2sch = − = −1.5 − j0.5 pu
100
The slack bus voltage is V1 = 1.06 0 pu. Starting with an initial estimate of
(0) (0)
|V2 | = 1.0, δ2 = 0.0, the power residuals are computed from (6.63) and
(6.64)
(0) (0)
∆P2 = P2sch − P2 = −1.5 − [10 cos(106.26◦ ) + 10 cos(−73.74◦ )]
= −1.5 pu
(0) (0)
∆Q2 = Qsch ◦ ◦
2 − Q2 = −0.5 − [−10 sin(106.26 ) − 10 sin(−73.74 )]
= −0.5 pu

The elements of the Jacobian matrix at the initial estimate are

(0)
J1 = 10(1)(1) sin(106.26◦ ) = 9.6
(0)
J2 = 10(1) cos(106.26◦ ) + 20(1) cos(−73.74◦ ) = 2.8
(0)
J3 = 10(1)(1) cos(106.26◦ ) = −2.8
(0)
J4 = −10(1) sin(106.26◦ ) − 20(1) sin(−73.74◦ ) = 9.6

The set of linear equations in the first iteration becomes

· ¸ · ¸" (0)
#
−1.5 9.6 2.8 ∆δ2
= (0)
−0.5 −2.8 9.6 ∆|V2 |
Obtaining the solution of the above matrix equation, voltage at bus 2 in the first
iteration is
(0) (1)
∆δ2 = −0.13 δ2 = 0 + (−0.13) = −0.13 radian
(0) (1)
∆|V2 | = −0.09 |V2 | = 1 + (−0.09) = 0.91 pu
For the second iteration, we have
(1) (1)
∆P2 = P2sch − P2 = −1.5 − (−1.3403) = −0.1597 pu
(1) (1)
∆Q2 = Qsch
2 − Q2 = −0.5 − (−0.3822) = −0.1178 pu

Also, computing the elements of the Jacobian matrix, the set of linear equations in
the second iteration becomes
· ¸ · ¸" (1)
#
−0.1597 8.332 1.0751 ∆δ2
= (1)
−0.1178 −3.659 8.3160 ∆|V2 |
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134 CONTENTS

Obtaining the solution of the above matrix equation, voltage at bus 2 in the second
iteration is
(1) (2)
∆δ2 = −0.0164 δ2 = −0.13 + (−0.0164) = −0.1464 radian
(1) (2)
∆|V2 | = −0.0214 |V2 | = 0.91 + (−0.0214) = 0.8886 pu

6.11. In the two-bus system shown in Figure 59, bus 1 is a slack bus with V1 =
1.06 0◦ pu. A load of 100 MW and 50 Mvar is taken from bus 2. The line impedance
is z12 = 0.12 + j0.16 pu on a base of 100 MVA. Using Newton-Raphson method,
obtain the voltage magnitude and phase angle of bus 2. Start with an initial estimate
of |V2 |(0) = 1.0 pu and δ2 (0) = 0◦ . Perform two iterations.

2
¾»1 ........................................
100 MW
z12 = 0.12 + j0.16
½¼ ........................................
50 Mvar
V1 = 1.06 0◦

FIGURE 59
One-line diagram for Problem 6.11.

The power flow equation with voltages and admittances expressed in polar form is
n
X
Pi = |Vi ||Vj ||Yij | cos (θij − δi + δj )
j=1
n
X
Qi = − |Vi ||Vj ||Yij | sin (θij − δi + δj )
j=1

The bus admittance matrix is

· ¸
56 −53.13◦ 56 126.87◦
Ybus =
56 126.87◦ 56 −53.13◦

Substituting for admittances, the expression for real and reactive power at bus 2
becomes

P2 = 5|V2 ||V1 | cos(126.87◦ − δ2 + δ1 ) + 5|V2 |2 cos(−53.13◦ )

Q2 = −5|V2 ||V1 | sin(126.87◦ − δ2 + δ1 ) − 5|V2 |2 sin(−53.13◦ )

Partial derivatives of P2 , and Q2 with respect to |V2 |, and δ2 are

∂P2
= 5|V2 ||V1 | sin(126.87◦ − δ2 + δ1 )
∂δ2
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CONTENTS 135

∂P2
= 5|V1 | cos(126.87◦ − δ2 + δ1 ) + 10|V2 | cos(−53.13◦ )
∂|V2 |
∂Q2
= 5|V2 ||V1 | cos(126.87◦ − δ2 + δ1 )
∂δ2
∂Q2
= −5|V1 | sin(126.87◦ − δ2 + δ1 ) − 10|V2 | sin(−73.74◦ )
∂|V2 |
The load expressed in per units is
(100 + j50)
S2sch = − = −1.0 − j0.5 pu
100
The slack bus voltage is V1 = 1.06 0 pu. Starting with an initial estimate of
(0) (0)
|V2 | = 1.0, δ2 = 0.0, the power residuals are computed from (6.63) and
(6.64)
(0) (0)
∆P2 = P2sch − P2 = −1.0 − [5 cos(126.87◦ ) + 5 cos(−53.13◦ )]
= −1.0 pu
(0) (0)
∆Q2 = Qsch ◦ ◦
2 − Q2 = −0.5 − [−5 sin(126.87 ) − 5 sin(−53.13 )]
= −0.5 pu
The elements of the Jacobian matrix at the initial estimate are
(0)
J1 = 5(1)(1) sin(126.87◦ ) = 4
(0)
J2 = 5(1) cos(126.87◦ ) + 10(1) cos(−53.13◦ ) = 3
(0)
J3 = 5(1)(1) cos(126.87◦ ) = −3
(0)
J4 = −5(1) sin(126.87◦ ) − 10(1) sin(−53.13◦ ) = 4
The set of linear equations in the first iteration becomes
· ¸ · ¸" (0)
#
−1.0 4 3 ∆δ2
= (0)
−0.5 −3 4 ∆|V2 |
Obtaining the solution of the above matrix equation, voltage at bus 2 in the first
iteration is
(0) (1)
∆δ2 = −0.10 δ2 = 0 + (−0.10) = −0.10 radian
(0) (1)
∆|V2 | = −0.2 |V2 | = 1 + (−0.2) = 0.8 pu
For the second iteration, we have
(1) (1)
∆P2 = P2sch − P2 = −1.0 − (−0.7875) = −0.2125 pu
(1) (1)
∆Q2 = Qsch
2 − Q2 = −0.5 − (−0.3844) = −0.1156 pu
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136 CONTENTS

Also, computing the elements of the Jacobian matrix, the set of linear equations in
the second iteration becomes
· ¸ · ¸" (1)
#
−0.2125 2.9444 1.4157 ∆δ2
= (1)
−0.1156 −2.7075 2.7195 ∆|V2 |

Obtaining the solution of the above matrix equation, voltage at bus 2 in the second
iteration is
(1) (2)
∆δ2 = −0.0350 δ2 = −0.1 + (−0.0350) = −0.135 radian
(1) (2)
∆|V2 | = −0.0773 |V2 | = 0.8 + (−0.0773) = 0.7227 pu

6.12. Figure 60 shows the one-line diagram of a simple three-bus power system
with generation at buses 1 and 2. The voltage at bus 1 is V = 1.06 0◦ per unit.
Voltage magnitude at bus 2 is fixed at 1.05 pu with a real power generation of
400 MW. A load consisting of 500 MW and 400 Mvar is taken from bus 3. Line
admittances are marked in per unit on a 100 MVA base. For the purpose of hand
calculations, line resistances and line charging susceptances are neglected.

1 y12 = −j40 2

¾» P2 =
¾»400 MW
............................

½¼ ½¼
y13 = −j20 y23 = −j20

Slack Bus 3 ...

...
...
...
...
...
| V2 |= 1.05
V1 = 1.06 0◦ ...
...
.
...
...
.
.......... ..........
.. ..

500 400
MW Mvar
FIGURE 60
One-line diagram for problem 6.12.

(a) Show that the expression for the real power at bus 2 and real and reactive power
at bus 3 are

P2 = 40|V2 ||V1 | cos(90◦ − δ2 + δ1 ) + 20|V2 ||V3 | cos(90◦ − δ2 + δ3 )

P3 = 20|V3 ||V1 | cos(90◦ − δ3 + δ1 ) + 20|V3 ||V2 | cos(90◦ − δ3 + δ2 )
Q3 = −20|V3 ||V1 | sin(90◦ −δ3 +δ1 )−20|V3 ||V2 | sin(90◦ −δ3 +δ2 )+40|V3 |2

(b) Using Newton-Raphson method, start with the initial estimates of V2 (0) =
1.05 + j0 and V3 (0) = 1.0 + j0, and keeping |V2 | = 1.05 pu, determine the
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CONTENTS 137

phasor values of V2 and V3 . Perform two iterations.

(c) Check the power flow solution for Problem 6.12 using the lfnewton and other
required programs. Assume the regulated bus (bus # 2) reactive power limits are
between 0 and 600 Mvar.

By inspection, the bus admittance matrix in polar form is

 
606 − π2 406 π2 206 π2
Ybus =  406 π2 606 − π2 206 π2 
206 π2 206 π2 406 − π2
(a) The power flow equation with voltages and admittances expressed in polar form
is
n
X
Pi = |Vi ||Vj ||Yij | cos (θij − δi + δj )
j=1
n
X
Qi = − |Vi ||Vj ||Yij | sin (θij − δi + δj )
j=1

Substituting the elements of the bus admittance matrix in the above equations for
P2 , P3 , and Q3 will result in the given equations.

(b) Elements of the Jacobian matrix are obtained by taking partial derivatives of
the given equations with respect to δ2 , δ3 and |V3 |.
∂P2 π π
= 40|V2 ||V1 | sin( − δ2 + δ1 ) + 20|V2 ||V3 | sin( − δ2 + δ3 )
∂δ2 2 2
∂P2 π
= −20|V2 ||V3 | sin( − δ2 + δ3 )
∂δ3 2
∂P2 π
= 20|V2 | cos( − δ2 + δ3 )
∂|V3 | 2
∂P3 π
= −20|V3 ||V2 | sin( − δ3 + δ2 )
∂δ2 2
∂P3 π π
= 20|V3 ||V1 | sin( − δ3 + δ1 ) + 20|V3 ||V2 | sin( − δ3 + δ2 )
∂δ3 2 2
∂P3 π π
= 20|V1 | cos( − δ3 + δ1 ) + 20|V2 | cos( − δ3 + δ2 )
∂|V3 | 2 2
∂Q3 π
= −20|V3 ||V2 | cos( − δ3 + δ2 )
∂δ2 2
∂Q3 π π
= 20|V3 ||V1 | cos( − δ3 + δ1 ) + 20|V3 ||V2 | cos( − δ3 + δ2 )
∂δ3 2 2
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138 CONTENTS

∂Q3 π π
= −20|V1 | sin( − δ3 + δ1 ) − 20|V2 | sin( − δ3 + δ2 ) + 80|V3 |
∂|V3 | 2 2
The load and generation expressed in per units are
400
P2sch = = 4.0 pu
100
(500 + j400)
S3sch =− = −5.0 − j4.0 pu
100
The slack bus voltage is V1 = 1.06 0 pu, and the bus 2 voltage magnitude is |V2 | =
(0) (0) (0)
1.05 pu. Starting with an initial estimate of |V3 | = 1.0, δ2 = 0.0, and δ3 =
0.0, the power residuals are
(0) (0)
∆P2 = P2sch − P2 = 4.0 − (0) = 4.0
(0) (0)
∆P3 = P3sch − P3 = −5.0 − (0) = −5.0
(0) (0)
∆Q3 = Qsch
3 − Q3 = −4.0 − (−1.0) = −3.0
Evaluating the elements of the Jacobian matrix with the initial estimate, the set of
linear equations in the first iteration becomes
 
    (0)
−2.8600 63 −21 0  ∆δ2 
(0) 
 1.4384  =  −21 41 0  
 ∆δ3 
−0.2200 0 0 39 (0)
∆|V3 |
Obtaining the solution of the above matrix equation, the new bus voltages in the
first iteration are
(0) (1)
∆δ2 = 0.0275 δ2 = 0 + 0.0275 = 0.0275 radian = 1.5782◦
(0) (1)
∆δ3 = −0.1078 δ3 = 0 + (−0.1078) = −0.1078 radian = −6.1790◦
(0) (1)
∆|V3 | = −0.0769 |V3 | = 1 + (−0.0769) = 0.9231 pu
For the second iteration, we have
 
    (1)
0.2269 61.1913 −19.2072 2.8345  ∆δ2 
(1) 
 −0.3965  =  −19.2072 37.5615 −4.9871  
 ∆δ3 
−0.5213 2.6164 −4.6035 33.1545 (1)
∆|V3 |
and
(1) (2)
∆δ2 = 0.0006 δ2 = 0.0275 + 0.0006 = 0.0281 radian = 1.61◦
(1) (2)
∆δ3 = −0.0126 δ3 = −0.1078 + (−0.0126) = −0.1204 radian = −6.898◦
(1) (2)
∆|V3 | = −0.0175 |V3 | = 0.9231 + (−0.0175) = 0.9056 pu
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CONTENTS 139

(c) The power flow program lfnewton is used to obtain the solution, with the fol-
lowing statements:

clear
basemva = 100; accuracy = 0.000001; maxiter = 10;

% Problem 6.12(c)
% Bus Bus Voltage Angle -Load--- -Generator-- Injected
% No code Mag. Degree MW MVAR MW MVAR Qmin Qmax Mvar
busdata=[1 1 1.0 0.0 0.0 0.0 0.0 0.0 0 0 0
2 2 1.05 0.0 0 0 400 0.0 600 0 0
3 0 1.0 0.0 500 400 0.0 0.0 0 0 0];

% Line code
% Bus bus R X 1/2 B = 1 for lines
% nl nr pu pu pu >1 or <1 tr. tap at bus nl
linedata=[1 2 0.0 0.025 0.0 1
1 3 0.0 0.05 0.0 1
2 3 0.0 0.05 0.0 1];
disp(’Problem 6.12(c)’)
lfybus % form the bus admittance matrix
lfnewton % Power flow solution by Gauss-Seidel method
busout % Prints the power flow solution on the screen
lineflow % Computes and displays the line flow and losses

The above statements are saved in the file ch6p12c.m. Run the program to obtain
the solution.

6.13. For Problem 6.12:

(a) Obtain the power flow solution using the fast decoupled algorithm. Perform two
iterations.
(b)Check the power flow solution for Problem 6.12 using the decouple and other
required programs. Assume the regulated bus (bus # 2) reactive power limits are
between 0 and 600 Mvar.

(a) In this system, bus 1 is the slack bus and the corresponding bus susceptance
matrix for evaluation of phase angles ∆δ2 and ∆δ3 form the bus admittance matrix
in Problem 6.12 is
· ¸
0 −60 20
B =
20 −40
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140 CONTENTS

The inverse of the above matrix is

· ¸
−0.02 −0.01
[B 0 ]−1 =
−0.01 −0.03

The expressions for real power at bus 2 and 3 and the reactive power at bus 3 are
given in Problem 6.12. The slack bus voltage is V1 = 1.06 0 pu, and the bus 2
(0)
voltage magnitude is |V3 | = 1.05 pu. Starting with an initial estimate of |V3 | =
(0) (0)
1.0, δ2 = 0.0, and δ3 = 0.0, the power residuals are computed from (6.63) and
(6.64)
(0) (0)
∆P2 = P2sch − P2 = 4 − (0) = 4
(0) (0)
∆P3 = P3sch − P3 = −5 − (0) = −5
(0) (0)
∆Q3 = Qsch
3 − Q3 = −4 − (−1) = −3

The fast decoupled power flow algorithm given by (6.77) becomes

  " #" # " #
(0) 4
∆δ2 −0.02 −0.0 1.05 .0262
 =− =
(0) −0.01 −0.03 −5 −0.1119
∆δ3 1.0

Since bus 2 is a regulated bus, the corresponding row and column of B 0 are elimi-
nated and we get

B 00 = [−40]

From (6.78), we have

· ¸· ¸
−1 −3
∆|V3 | = − = −0.075
40 1.0
The new bus voltages in the first iteration are
(0) (1)
∆δ2 = 0.0262 δ2 = 0 + (0.0262) = 0.0262 radian = 1.5006◦
(0) (1)
∆δ3 = −0.1119 δ3 = 0 + (−0.1119) = −0.1119 radian = −6.4117◦
(0) (1)
∆|V3 | = −0.075 |V2 | = 1 + (−0.075) = 0.925 pu

For the second iteration, the power residuals are

(1) (1)
∆P2 = P2sch − P2 = 4 − (3.7739) = 0.2261
(1) (1)
∆P3 = P3sch − P3 = −5 − (−4.7399) = −.2601
(1) (1)
∆Q3 = Qsch
3 − Q3 = −4 − (−3.3994) = −0.6006
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CONTENTS 141

" (1)
# · ¸" # · ¸
0.2261
∆δ2 −0.02 −0.0 1.05 0.0015
(1) =− −0.2601 =
∆δ3 −0.01 −0.03 0.925
−0.0063

From (6.78), we have

· ¸· ¸
−1 −0.6006
∆|V3 | = − = −0.0162
40 0.925
The new bus voltages in the second iteration are
(1) (2)
∆δ2 = 0.0015 δ2 = 0.0262 + (0.0015) = 0.0277 radian = 1.5863◦
(1) (2)
∆δ3 = −0.0063 δ3 = −0.1119+(−0.0063) = −0.1182 radian = −6.7716◦
(1) (2)
∆|V3 | = −0.0162 |V3 | = 0.925 + (−0.0162) = 0.9088 pu

(c) The power flow program decouple is used to obtain the solution. The statements
are the same as in Problem 6.12, except the lfnewton is replaced by decouple. The
statements are saved in the file ch6p13b. Run this program to obtain the power
flow solution.

6.14. The 26-bus power system network of an electric utility company is shown in
Figure 61. Obtain the power flow solution by the following methods:
(a) Gauss-Seidel power flow (see Example 6.9).
(b) Newton-Raphson power flow (see Example 6.11).
(c) Fast decoupled power flow (see Example 6.13).

The load data is as follows.

LOAD DATA
Bus Load Bus Load
No. MW Mvar No. MW Mvar
1 51.0 41.0 14 24.0 12.0
2 22.0 15.0 15 70.0 31.0
3 64.0 50.0 16 55.0 27.0
4 25.0 10.0 17 78.0 38.0
5 50.0 30.0 18 153.0 67.0
6 76.0 29.0 19 75.0 15.0
7 0.0 0.0 20 48.0 27.0
8 0.0 0.0 21 46.0 23.0
9 89.0 50.0 22 45.0 22.0
10 0.0 0.0 23 25.0 12.0
11 25.0 15.0 24 54.0 27.0
12 89.0 48.0 25 28.0 13.0
13 31.0 15.0 26 40.0 20.0
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142 CONTENTS

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17

FIGURE 61
One-line diagram for Problem 6.14.
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CONTENTS 143

Voltage magnitude, generation schedule, and the reactive power limits for the regu-
lated buses are tabulated below. Bus 1, whose voltage is specified as V1 = 1.0256 0◦ ,
is taken as the slack bus.
GENERATION DATA
Bus Voltage Generation Mvar Limits
No. Mag. MW Min. Max.
1 1.025
2 1.020 79.0 40.0 250.0
3 1.025 20.0 40.0 150.0
4 1.050 100.0 40.0 80.0
5 1.045 300.0 40.0 160.0
26 1.015 60.0 15.0 50.0
The Mvar of the shunt capacitors installed at substations and the transformer tap
settings are given below.
SHUNT CAPACITORS
TRANSFORMER TAP
Bus No. Mvar
Designation Tap Setting
1 4.0
2− 3 0.960
4 2.0
2 − 13 0.960
5 5.0
3 − 13 1.017
6 2.0
4− 8 1.050
11 1.5
4 − 12 1.050
12 2.0
6 − 19 0.950
15 0.5
7− 9 0.950
19 5.0
The line and transformer data containing the series resistance and reactance in per
unit and one-half the total capacitance in per unit susceptance on a 100-MVA base
are tabulated below.
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144 CONTENTS

LINE AND TRANSFORMER DATA

1 1
Bus Bus R, X, 2 B, Bus Bus R, X, 2 B,
No. No. pu pu pu No. No. pu pu pu
1 2 0.0005 0.0048 0.0300 10 22 0.0069 0.0298 0.005
1 18 0.0013 0.0110 0.0600 11 25 0.0960 0.2700 0.010
2 3 0.0014 0.0513 0.0500 11 26 0.0165 0.0970 0.004
2 7 0.0103 0.0586 0.0180 12 14 0.0327 0.0802 0.000
2 8 0.0074 0.0321 0.0390 12 15 0.0180 0.0598 0.000
2 13 0.0035 0.0967 0.0250 13 14 0.0046 0.0271 0.001
2 26 0.0323 0.1967 0.0000 13 15 0.0116 0.0610 0.000
3 13 0.0007 0.0054 0.0005 13 16 0.0179 0.0888 0.001
4 8 0.0008 0.0240 0.0001 14 15 0.0069 0.0382 0.000
4 12 0.0016 0.0207 0.0150 15 16 0.0209 0.0512 0.000
5 6 0.0069 0.0300 0.0990 16 17 0.0990 0.0600 0.000
6 7 0.0053 0.0306 0.0010 16 20 0.0239 0.0585 0.000
6 11 0.0097 0.0570 0.0001 17 18 0.0032 0.0600 0.038
6 18 0.0037 0.0222 0.0012 17 21 0.2290 0.4450 0.000
6 19 0.0035 0.0660 0.0450 19 23 0.0300 0.1310 0.000
6 21 0.0050 0.0900 0.0226 19 24 0.0300 0.1250 0.002
7 8 0.0012 0.0069 0.0001 19 25 0.1190 0.2249 0.004
7 9 0.0009 0.0429 0.0250 20 21 0.0657 0.1570 0.000
8 12 0.0020 0.0180 0.0200 20 22 0.0150 0.0366 0.000
9 10 0.0010 0.0493 0.0010 21 24 0.0476 0.1510 0.000
10 12 0.0024 0.0132 0.0100 22 23 0.0290 0.0990 0.000
10 19 0.0547 0.2360 0.0000 22 24 0.0310 0.0880 0.000
10 20 0.0066 0.0160 0.0010 23 25 0.0987 0.1168 0.000

The power flow data and the command required are as follows.

clear
basemva = 100; accuracy = 0.0001; accel = 1.6; maxiter = 60;

% Bus Bus Voltage Angle --Load-- --Generator----Injected

% No code Mag. Degree MW Mvar MW Mvar Qmin Qmax Mvar
busdata=[1 1 1.025 0 51 41 0 0 0 0 4
2 2 1.020 0 22 15 79 0 40 250 0
3 2 1.025 0 64 50 20 0 40 150 0
4 2 1.050 0 25 10 100 0 40 80 2
5 2 1.045 0 50 30 300 0 40 160 5
6 0 1.00 0 76 29 0 0 0 0 2
7 0 1.00 0 0 0 0 0 0 0 0
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CONTENTS 145

8 0 1.00 0 0 0 0 0 0 0 0
9 0 1.00 0 89 50 0 0 0 0 3
10 0 1.00 0 0 0 0 0 0 0 0
11 0 1.00 0 25 15 0 0 0 0 1.5
12 0 1.00 0 89 48 0 0 0 0 2
13 0 1.00 0 31 15 0 0 0 0 0
14 0 1.00 0 24 12 0 0 0 0 0
15 0 1.00 0 70 31 0 0 0 0 0.5
16 0 1.00 0 55 27 0 0 0 0 0
17 0 1.00 0 78 38 0 0 0 0 0
18 0 1.00 0 153 67 0 0 0 0 0
19 0 1.00 0 75 15 0 0 0 0 5
20 0 1.00 0 48 27 0 0 0 0 0
21 0 1.00 0 46 23 0 0 0 0 0
22 0 1.00 0 45 22 0 0 0 0 0
23 0 1.00 0 25 12 0 0 0 0 0
24 0 1.00 0 54 27 0 0 0 0 0
25 0 1.00 0 28 13 0 0 0 0 0
26 2 1.015 0 40 20 60 0 15 50 0];
% Line code
% Bus bus R X 1/2 B 1 for line codes or
% nl nr pu pu pu tap setting value
linedata=[1 2 0.00055 0.00480 0.03000 1
1 18 0.00130 0.01150 0.06000 1
2 3 0.00146 0.05130 0.05000 0.96
2 7 0.01030 0.05860 0.01800 1
2 8 0.00740 0.03210 0.03900 1
2 13 0.00357 0.09670 0.02500 0.96
2 26 0.03230 0.19670 0.00000 1
3 13 0.00070 0.00548 0.00050 1.017
4 8 0.00080 0.02400 0.00010 1.050
4 12 0.00160 0.02070 0.01500 1.050
5 6 0.00690 0.03000 0.09900 1
6 7 0.00535 0.03060 0.00105 1
6 11 0.00970 0.05700 0.00010 1
6 18 0.00374 0.02220 0.00120 1
6 19 0.00350 0.06600 0.04500 0.95
6 21 0.00500 0.09000 0.02260 1
7 8 0.00120 0.00693 0.00010 1
7 9 0.00095 0.04290 0.02500 0.95
8 12 0.00200 0.01800 0.02000 1
9 10 0.00104 0.04930 0.00100 1
10 12 0.00247 0.01320 0.01000 1
10 19 0.05470 0.23600 0.00000 1
10 20 0.00660 0.01600 0.00100 1
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146 CONTENTS

10 22 0.00690
0.02980 0.00500 1
11 25 0.09600
0.27000 0.01000 1
11 26 0.01650
0.09700 0.00400 1
12 14 0.03270
0.08020 0.00000 1
12 15 0.01800
0.05980 0.00000 1
13 14 0.00460
0.02710 0.00100 1
13 15 0.01160
0.06100 0.00000 1
13 16 0.01793
0.08880 0.00100 1
14 15 0.00690
0.03820 0.00000 1
15 16 0.02090
0.05120 0.00000 1
16 17 0.09900
0.06000 0.00000 1
16 20 0.02390
0.05850 0.00000 1
17 18 0.00320
0.06000 0.03800 1
17 21 0.22900
0.44500 0.00000 1
19 23 0.03000
0.13100 0.00000 1
19 24 0.03000
0.12500 0.00200 1
19 25 0.11900
0.22490 0.00400 1
20 21 0.06570
0.15700 0.00000 1
20 22 0.01500
0.03660 0.00000 1
21 24 0.04760
0.15100 0.00000 1
22 23 0.02900
0.09900 0.00000 1
22 24 0.03100
0.08800 0.00000 1
23 25 0.09870
0.11680 0.00000 1];
lfybus % Forms the bus admittance matrix
lfgauss % Power flow solution by Newton-Raphson method
%lfnewton % Power flow solution by Newton-Raphson method
%decouple % Power flow solution by Newton-Raphson method
busout % Prints the power flow solution on the screen
lfybus % form the bus admittance matrix
lineflow % Computes and displays the line flow and losses
% (b) Comment lfgauss and uncomment lfnewton
% (c) Comment lfgauss and lfnewton and uncomment decouple

Uncomment lfgauss, lfnewton, or decouple in turn, and run to obtain the power
flow solution by these methods.
 www.mohandesyar.com

CHAPTER 7 PROBLEMS

7.1. Find a rectangle of maximum perimeter that can be inscribed in a circle of unit
radius given by

g(x, y) = x2 + y 2 − 1 = 0

Check the eigenvalues for sufficient conditions.

Consider point c with coordinates (x, y) as shown in Figure 62.

..
..
...
............ ........................
....
........... .... .......
..... ......
....
......
.
.... ...
... c .....
......

...
.
.
...
. .
.
..
.
..
x .....
.
.
.
..... .......
..... ...
...
. .
....
....
... ....
... ......... y .....
...
...
...
... ... ....... ...
.
. .......
. .
.................................................................................................................................................
...
...
...
0 .
.
.
.
.
. ....
..
... .... ...
... ... ...
... ... ...
... .
. .
....
....
..... .... ...
....
..... ... .....
......
... ......
.......
........... .
. .... .
.. .......
.
........................... .
...
..
...

FIGURE 62
Constraint function of Problem 7.1.

The problem is to find the minimum value of the function

f (x, y) = 4(x + y)

subject to the equality constraint

g(x, y) = x2 + y 2 − 1 = 0
147