ECE6320_HW01SolMatlab (1)
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Dec 6, 2023
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1
ECE6320 POWER SYSTEM OPERATION AND CONTROL
FALL 2023
Homework 1:
SOLUTIONS
(3 points)
Due Date: Wednesday, September 6
Problem 1
a) Develop a computer program to build the Ybus matrix for a system with the following
data. Present the source code and the output matrix.
Transmission Line Data (values in p.u.)
From Bus
To Bus
R
X
Tot. Charging
1
2
0
0.2
0
1
3
0
0.5
0
1
4
0
0.1
0
2
3
0
0.25
0
2
4
0
0.4
0
3
5
0
0.6
0
b) How would Ybus change if line 1-2 is upgraded to a line with the following
parameters?
From Bus
To Bus
R (p.u.)
X (p.u.)
Tot. Charging
1
2
0.10
0.250
0.100
Solution:
clear
all
j = sqrt(-1);
% 1.a
% INPUT = [
% 1 2
0
0.2 0;
% 1 3
0
0.5 0;
% 1 4
0
0.1 0;
% 2 3
0
0.25
0;
% 2 4
0
0.4 0;
% 3 5
0
0.6 0];
% 1.b
INPUT = [
1
2
0.1 0.25
0.1;
1
3
0
0.5 0;
1
4
0
0.1 0;
2
3
0
0.25
0;
2
4
0
0.4 0;
3
5
0
0.6 0];
% build Ybus
nlines = size(INPUT,1);
Ybus = zeros(5,5);
fr_node = INPUT(:,1);
to_node = INPUT(:,2);
r = INPUT(:,3);
x = INPUT(:,4);
2
bc = INPUT(:,5);
for
i = 1:nlines
row = fr_node(i);
col = to_node(i);
Ybus(row,col) = -1/(r(i)+j*x(i));
% negative of admittance
= 1/jX = -j(1/X)
end
% Add transpose to create symmetric matrix
Ybus = Ybus + transpose(Ybus)
d = -sum(Ybus,2);
Ybus = Ybus + diag(d)
% add line charging
for
i = 1:nlines
row = fr_node(i);
col = to_node(i);
Ybus(row,row) = Ybus(row,row) + j*0.5*bc(i);
Ybus(col,col) = Ybus(col,col) + j*0.5*bc(i);
end
Output:
a)
Ybus =
0.0000 -17.0000i
0.0000 + 5.0000i
0.0000 + 2.0000i
0.0000 +10.0000i
0.0000 + 0.0000i
0.0000 + 5.0000i
0.0000 -11.5000i
0.0000 + 4.0000i
0.0000 + 2.5000i
0.0000 + 0.0000i
0.0000 + 2.0000i
0.0000 + 4.0000i
0.0000 - 7.6667i
0.0000 + 0.0000i
0.0000 + 1.6667i
0.0000 +10.0000i
0.0000 + 2.5000i
0.0000 + 0.0000i
0.0000 -12.5000i
0.0000 + 0.0000i
0.0000 + 0.0000i
0.0000 + 0.0000i
0.0000 + 1.6667i
0.0000 + 0.0000i
0.0000 - 1.6667i
b)
Ybus =
1.3793 -15.4483i
-1.3793 + 3.4483i
0.0000 + 2.0000i
0.0000 +10.0000i
0.0000 + 0.0000i
-1.3793 + 3.4483i
1.3793 - 9.9483i
0.0000 + 4.0000i
0.0000 + 2.5000i
0.0000 + 0.0000i
0.0000 + 2.0000i
0.0000 + 4.0000i
0.0000 - 7.6667i
0.0000 + 0.0000i
0.0000 + 1.6667i
0.0000 +10.0000i
0.0000 + 2.5000i
0.0000 + 0.0000i
0.0000 -12.5000i
0.0000 + 0.0000i
0.0000 + 0.0000i
0.0000 + 0.0000i
0.0000 + 1.6667i
0.0000 + 0.0000i
0.0000 - 1.6667i
3
Problem 2
Write a computer program to reproduce the values shown on slide L02_30 (solution of a
power flow problem using Gauss). Report the program code and the program output.
Solution:
Matlab M-file:
clear all;
i
= sqrt(-1);
E2 = 1 + i*0;
% Flat start
v
=0;
err=999;
while err > 0.00001,
Res = [v E2 err]
v = v+1;
E2new = ((-1+i*0.5)/conj(E2)-(-5+i*15)*(1.0+i*0.0))/(5-i*14.70);
err=abs(E2new-E2);
E2=E2new;
end;
Matlab Result
>> HW01_2
Res =
0
1
999
Res =
1.0000
0.9671 - 0.0568i
0.0657
Res =
2.0000
0.9624 - 0.0553i
0.0049
Res =
3.0000
0.9622 - 0.0556i
0.0004
Res =
4.0000
0.9622 - 0.0556i
0.0000
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4
Problem 3
Reproduce the values shown on slide L02_39 (solution of a power flow problem
including a PV bus). Report the program code and the program output.
Solution:
Matlab M-file
clear all;
v = -1;
Y = [5-1i*14.95 , -5+1i*15 ; -5+1i*15 , 5-1i*14.7];
E1 = 1;
P2 = 0;
E2 = 1.05;
E2_temp = 0;
out = [];
for k = 0:2
Q2 = -imag(conj(E2)*(Y(2,1)*E1+Y(2,2)*E2));
E2_temp = 1/Y(2,2)*((P2-1i*Q2)/conj(E2)-Y(2,1)*E1);
E2 = abs(E2)*exp(1i*angle(E2_temp));
v=v+1;
out = [out;v,P2,Q2,abs(E2_temp),180/pi*angle(E2_temp),
abs(E2),180/pi*angle(E2)];
end;
Matlab Result;
v P2 Q2 V2_t Theta2_t V2 Theta2
0 0 0.457 1.045 -0.836 1.050 -0.836
1 0 0.535 1.049 -0.941 1.050 -0.941
2 0 0.545 1.050 -0.955 1.050 -0.955
5
Problem 4
Determine the solution of the following simultaneous system of equations using
Newton’s method.
2
2
1
1
2
3
2
2
2
2
1
2
1
2
3
2
2
2
3
1
2
2
3
3
( )
2
4
0
( )
2
5
0
( )
3
2
7
0
f
x
x
x
f
x
x
x x
x
f
x
x
x x
x
=
+
+
−
=
=
−
+
+
−
=
=
−
+
+
−
=
x
x
x
Calculate the Jacobian explicitly
Report the code and the program output for the functions f and variables x at each
iteration. Assume the initial condition is
1
2
3
1;
2;
1.
x
x
x
=
=
Solution:
Matlab M-file
clear
all
clc
%format short e % exponential
v=0;
x=[1; 1; 1];
% column vector
err = 999;
iter = 0
while
err > 0.001,
iter = iter + 1;
f = [x(1)^2+2*x(2)^2+x(3)-4;
2*x(1)^2-x(2)^2+x(1)*x(2)+x(3)^2-5;
x(1)^2-3*x(2)^2+x(2)*x(3)+2*x(3)^2-7];
J = [
2*x(1)
4*x(2)
1;
4*x(1)+x(2)
x(1)-2*x(2)
2*x(3);
2*x(1)
x(3)-6*x(2)
4*x(3)-x(2)];
dx
= -inv(J)*f;
I(iter,:) = iter;
X(iter,:) = x';
F(iter,:) = f';
E(iter,:) = err;
err = max(abs(f));
x
= x + dx;
v
= v + 1;
end
;
[I X F E]
Matlab Result
ans =
1.0000
1.0000
1.0000
1.0000
0
-2.0000
-6.0000
999.0000
2.0000
0.6226
0.7358
2.8113
0.2820
3.5956
9.6390
6.0000
3.0000
1.2326
0.7190
1.8193
0.3726
1.7176
0.8961
9.6390
4.0000
0.9106
0.8531
1.8547
0.1396
0.1474
0.1079
1.7176
5.0000
0.8931
0.8285
1.8309
0.0015
0.0010
-0.0404
0.1474
6.0000
0.8859
0.8294
1.8395
0.0001
0.0002
0.0145
0.0404
7.0000
0.8883
0.8290
1.8364
0.0000
0.0000
-0.0051
0.0145
8.0000
0.8874
0.8291
1.8375
0.0000
0.0000
0.0018
0.0051
9.0000
0.8878
0.8291
1.8371
0.0000
0.0000
-0.0007
0.0018
6
Problem 5
Reproduce the results of the two-bus system Newton power flow shown on slides L03_22 and
L03_23. Report the code and the program output. Show the Jacobian matrix at each iteration.
Solution:
Matlab M-file
clear all;
format short
v=0;
x=[0; 1];
% [theta2; V2] column vector
err = 999;
while err > 0.0001,
f
= [x(2)*10*sin(x(1))+2; x(2)*(-10*cos(x(1)))+x(2)^2*10+1];
J
= [10*x(2)*cos(x(1))
10*sin(x(1)); 10*x(2)*sin(x(1)) -
10*cos(x(1))+20*x(2)]
dx
= -inv(J)*f;
Res
= [v x' f' err]
err
= max(abs(f));
x
= x + dx;
v
= v + 1;
end;
Matlab Result:
>> HW01_5
J =
10
0
0
10
Res =
0
0
1
2
1
999
J =
8.8206
-1.9867
-1.7880
8.1993
Res =
1.0000
-0.2000
0.9000
0.2120
0.2794
2.0000
J =
8.3538
-2.3123
-1.9855
7.4441
Res =
2.0000
-0.2333
0.8587
0.0145
0.0190
0.2794
J =
8.3169
-2.3380
-1.9999
7.3850
Res =
3.0000
-0.2360
0.8554
0.0001
0.0001
0.0190
J =
8.3166
-2.3382
-2.0000
7.3846
Res =
4.0000
-0.2360
0.8554
0.0000
0.0000
0.0001
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