Post on 21-Jun-2015
description
transcript
1 | P a g e
EXPERIMENT-1[Computation & Modeling of Transmission Line Parameters]
PROBLEM-1: A three-phase transposed line composed of one ACSR, 1,43,000 cmil, 47/7 Bobolink conductor per phase with flat horizontal spacing of 11m between phases a and b and between phases b and c. The conductors have a diameter of 3.625 cm and a GMR of 1.439cm. The line is to be replaced by a three-conductor bundle of ACSR 477,000-cmil, 26/7Hawk conductors having the same cross sectional area of aluminum as the single-conductor line. The conductors have a diameter of 2.1793 cm and a GMR of 0.8839 cm. The new line will also have a flat horizontal configuration, but it is to be operated at a higher voltage and therefore the phase spacing is increased to 14m as measured from the center of the bundles. Spacing between the conductors in the bundle is 45 cm.
(a) Determine the inductance and capacitance per phase per kilometer of the above two lines.
(b) Verify the results using the MATLAB program. [Practice Problem-4.9]
MATLAB CODE:
GMD=input('Enter GMD Value: '); GMRL= input('Enter GMRL Value: '); GMRC= input('Enter GMRC Value: '); disp('Value of L in mF/KM:’); L = 0.2*log(GMD/GMRL)
disp('Value of C in µF/KM:’); C = 0.0556/log(GMD/GMRC)
MANUEL SOLUTION:
2 | P a g e
3 | P a g e
PROBLEM-2: A three phase overhead line 200km long R = 0.16 ohm/km and Conductor diameter of
2cm with spacing 4, 5, 6 m transposed. Find A, B, C, D constants, sending end voltage, current,
power factor and power when the line is delivering full load of 50MW at 132kV, 0.8 pf lagging,
transmission efficiency, receiving end voltage and regulation.
MATLAB CODE:
clc clear all z=[]; z=input('Enter the length of line(km),Receiving
voltage(v),Power(W),Powerfactor,L,R&C(per phase per km)'); l=z(1);vr=z(2);p=z(3);pf=z(4);L=z(5);R=z(6);c=z(7); vrph=(vr*1000)/1.732; ir=(p*1000000)/(3*vrph); z=(R+(2*50*3.14*L*i))*l; y=i*2*3.14*50*c*l; disp('Medium Transmission line by nominal T method:') A=1+(z*y)/2;C=y;B=z*(1+(z*y)/4);D=1+(z*y)/2; disp('Generalized ABCD Constants are:') disp(A); disp(B); disp(C); disp(D); vs=(A*vrph)+(ir*B); fprintf('Sending end voltage Vs is %f\n',vs); is=(vrph*C)+(D*ir); fprintf('Sending end current Is is %f\n',is); reg=(3*conj(vs)* conj(is))*100/A; fprintf('Voltage Regulation is %f\n',reg); pows=3* conj(vs)* conj(is)*0.808; eff=p*100/pows; fprintf('Efficiency is %f\n',eff);
4 | P a g e
MANUEL SOLUTION:
5 | P a g e
EXPERIMENT-2 [ YBUS,ZBUS matrix Formation ]
PROBLEM-1: 1) (i) Determine the Y bus matrix for the power system network shown in fig.
(ii) Check the results obtained in using MATLAB.
MATLAB CODE:
clear all; clc; n=input('enter the number of buses:'); for i=1:n for j=i+1:n y(i,j)=input(['enter the line admittance
y',num2str(i),num2str(j),':']); y(j,i)=y(i,j); end end for i=1:n y1(i)=input( ['enter the admittance to ground y',num2str(i),':']); end for i=1:n for j=1:n if i==j ybus(i,j)=0; for k=1:n ybus(i,j)=ybus(i,j)+y(i,k); end else ybus(i,j)=-1*y(i,j); end end end for i=1:n ybus(i,i)=ybus(i,i)+y1(i);
6 | P a g e
end disp('y bus matrix is:'); disp(ybus); disp('z bus matrix is:'); z=inv(ybus); disp(z);
MANUEL SOLUTION:
Y11=y11+y12+y13+y14 ; Y22=y22+y21+y23+y24 ; Y33=y33+y32+y31+y34
Y44=y41+y42+y43+y44
7 | P a g e
PROBLEM-2:
2. (i) Determine Z bus matrix for the power system network shown in fig.
(ii) Check the results obtained using MATLAB.
MATLAB CODE:
clc clear all close all g=[1.2 0.2 0.15 1.5 0.3]; z1=[g(1)]; disp(z1); disp('TYPE I MODIFICATION') z2=[g(1) g(1) g(1) g(1)+g(2)]; disp(z2); disp('TYPE II MODIFICATION') z3=[g(1) g(1) g(1) g(1) g(1)+g(2) g(1)+g(2) g(1) g(1)+g(2) g(1)+g(2)+g(3)]; disp(z3); disp('TYPE III MODIFICATION') z4=[g(1) g(1) g(1) g(1) g(1) g(1)+g(2) g(1)+g(2) g(1)+g(2) g(1) g(1)+g(2) g(1)+g(2)+g(3) g(1)+g(2)+g(3) g(1) g(1)+g(2) g(1)+g(2)+g(3) g(1)+g(2)+g(3)+g(4)] z4=[z4];
8 | P a g e
disp(z4); disp('Actual Zbus matrix is:') n=4; for i=1:1:n for j=1:1:n z4(i,j)=z4(i,j)-(z4(i,n)*z4(n,j)/z4(n,n)); end end z4(:,4)=[]; z4(4,:)=[]; disp(z4); disp('TYPE IV MODIFICATION MATRIX') l=4; p=2;q=4; for i=1:l-1 z4(l,i)=z4(p-1,i)-z4(q-1,i); z4(i,l)=z4(l,i); end z4(l,l)=g(5)+z4(1,1)+z4(3,3)-2*z4(1,3); disp(z4); n=4; for i=1:1:n for j=1:1:n z4(i,j)=z4(i,j)-(z4(i,n)*z4(n,j)/z4(n,n)); end end z4(:,4)=[]; z4(4,:)=[]; disp('THE REQUIRED ZBUS MATRIX IS:'); ZBUS=z4*1i; disp(ZBUS);
9 | P a g e
MANUEL SOLUTION:
10 | P a g e
EXPERIMENT-3[Gauss-Seidal]
Problem:
The magnitude at bus 1 is adjusted to 1.05pu. The scheduled loads at buses 2 and 3 are marked on the diagram. Line impedances are marked in p.u. The base value is 100kVA. The line charging susceptances are neglected. Determine the phasor values of the voltage at the Load bus 2 and 3. (i) Find the slack bus real and reactive power. (ii) Verify the result using MATLAB.
MANUEL CALCULATION:
11 | P a g e
12 | P a g e
MATLAB CODING:
y12=10-j*20;
y13=10-j*30;
y23=16-j*32;
V1=1.05+j*0;
V2=1+j*0;
V3=1+j*0;
S2=-2.566-j*1.102;
S3=-1.386-j*.452;
iter =0;
for I=1:10;
iter=iter+1;
V2 = (conj(S2)/conj(V2)+y12*V1+y23*V3)/(y12+y23);
V3 = (conj(S3)/conj(V3)+y13*V1+y23*V2)/(y13+y23);
disp([iter, V2, V3])
end
13 | P a g e
V2= .98-j*.06;
V3= 1-j*.05;
I12=y12*(V1-V2); I21=-I12;
I13=y13*(V1-V3); I31=-I13;
I23=y23*(V2-V3); I32=-I23;
S12=V1*conj(I12); S21=V2*conj(I21);
S13=V1*conj(I13); S31=V3*conj(I31);
S23=V2*conj(I23); S32=V3*conj(I32);
I1221=[I12,I21]
I1331=[I13,I31]
I2332=[I23,I32]
S1221=[S12, S21 (S12+S13) S12+S21]
S1331=[S13, S31 (S31+S32) S13+S31]
S2332=[S23, S32 (S23+S21) S23+S32]
14 | P a g e
EXPERIMENT-4[Newton-Raphson Method]
Problem:
Consider the 3 bus system each of the 3 line bus a series impedance of 0.02 + j0.08 p.u and a
total shunt admittance of j0.02 p.u. The specified quantities at the bus are given below.
Verify the result using MATLAB
MANUEL CALCULATIONS:
15 | P a g e
16 | P a g e
MATLAB CODING: V = [1.04; 1; 1.04]; d = [0; 0; 0]; Ps=[0.5; -1.5]; Qs= 1.0; YB = [ 5.882-j*23.528 -2.941-j*11.76 -2.941-j*11.76 -2.941-j*11.76 5.882-j*23.528 -2.941-j*11.76 -2.941-j*11.76 -2.941-j*11.76 5.882-j*23.528]; Y= abs(YB); t = angle(YB); iter=0; pwracur = 0.00025; % Power accuracy DC = 10; % Set the maximum power residual to a high value while max(abs(DC)) > pwracur iter = iter +1 P=[V(2)*V(1)*Y(2,1)*cos(t(2,1)-d(2)+d(1))+V(2)^2*Y(2,2)*cos(t(2,2))+ ...
V(2)*V(3)*Y(2,3)*cos(t(2,3)-d(2)+d(3));
V(3)*V(1)*Y(3,1)*cos(t(3,1)-d(3)+d(1))+V(3)^2*Y(3,3)*cos(t(3,3))+ ...
V(3)*V(2)*Y(3,2)*cos(t(3,2)-d(3)+d(2))];
Q= -V(2)*V(1)*Y(2,1)*sin(t(2,1)-d(2)+d(1))-V(2)^2*Y(2,2)*sin(t(2,2))- ...
V(2)*V(3)*Y(2,3)*sin(t(2,3)-d(2)+d(3));
J(1,1)=V(2)*V(1)*Y(2,1)*sin(t(2,1)-d(2)+d(1))+...
V(2)*V(3)*Y(2,3)*sin(t(2,3)-d(2)+d(3));
J(1,2)=-V(2)*V(3)*Y(2,3)*sin(t(2,3)-d(2)+d(3));
J(1,3)=V(1)*Y(2,1)*cos(t(2,1)-d(2)+d(1))+2*V(2)*Y(2,2)*cos(t(2,2))+...
V(3)*Y(2,3)*cos(t(2,3)-d(2)+d(3));
J(2,1)=-V(3)*V(2)*Y(3,2)*sin(t(3,2)-d(3)+d(2));
J(2,2)=V(3)*V(1)*Y(3,1)*sin(t(3,1)-d(3)+d(1))+...
V(3)*V(2)*Y(3,2)*sin(t(3,2)-d(3)+d(2));
J(2,3)=V(3)*Y(2,3)*cos(t(3,2)-d(3)+d(2));
J(3,1)=V(2)*V(1)*Y(2,1)*cos(t(2,1)-d(2)+d(1))+...
V(2)*V(3)*Y(2,3)*cos(t(2,3)-d(2)+d(3));
17 | P a g e
J(3,2)=-V(2)*V(3)*Y(2,3)*cos(t(2,3)-d(2)+d(3));
J(3,3)=-V(1)*Y(2,1)*sin(t(2,1)-d(2)+d(1))-2*V(2)*Y(2,2)*sin(t(2,2))-...
V(3)*Y(2,3)*sin(t(2,3)-d(2)+d(3));
DP = Ps - P;
DQ = Qs - Q;
DC = [DP; DQ]
J
DX = J\DC
d(2) =d(2)+DX(1);
d(3)=d(3) +DX(2);
V(2)= V(2)+DX(3);
V, d, delta =180/pi*d;
end
P1= V(1)^2*Y(1,1)*cos(t(1,1))+V(1)*V(2)*Y(1,2)*cos(t(1,2)-d(1)+d(2))+...
V(1)*V(3)*Y(1,3)*cos(t(1,3)-d(1)+d(3))
Q1=-V(1)^2*Y(1,1)*sin(t(1,1))-V(1)*V(2)*Y(1,2)*sin(t(1,2)-d(1)+d(2))-...
V(1)*V(3)*Y(1,3)*sin(t(1,3)-d(1)+d(3))
Q3=-V(3)*V(1)*Y(3,1)*sin(t(3,1)-d(3)+d(1))-V(3)*V(2)*Y(3,2)*...
sin(t(3,2)-d(3)+d(2))-V(3)^2*Y(3,3)*sin(t(3,3))
18 | P a g e
EXPERIMENT-5[TRANSIENT AND SMALL SIGNAL STABILITY OF SINGLE MACHINE INFINITE BUS SYSTEM]
PROBLEM:
1. A 60Hz synchronous generator having inertia constant H = 5 MJ/MVA and a direct axis
transient reactance Xd1 = 0.3 per unit is connected to an infinite bus through a purely reactive
circuit as shown in figure. Reactances are marked on the diagram on a common system base.
The generator is delivering real power Pe= 0.8 per unit and Q = 0.074 per unit to the infinite bus
at a voltage of V = 1 per unit.
a) A temporary three-phase fault occurs at the sending end of the line at point F.When
the fault is cleared, both lines are intact. Determine the critical clearing angle and the
critical fault clearing time.
b) Verify the result using MATLAB program.
MANUEL CALCULATION:
19 | P a g e
PROGRAM
%(a) Fault at the sending end. Both lines intact when fault is cleared
Pm = 0.8; E= 1.17; V = 1.0; X1 = 0.65; X2 = inf; X3 = 0.65;
eacfault(Pm, E, V, X1, X2, X3)
%(b) Fault at the mid-point of one line. Faulted line is isolated
X2 = 1.8; X3 = 0.8;
eacfault(Pm, E, V, X1, X2, X3)
20 | P a g e
EXPERIMENT-6[TRANSIENT AND SMALL SIGNAL STABILITY OF MULTI MACHINE INFINITE BUS SYSTEM]
PROBLEM:
MATLAB CODE:
basemva = 100; accuracy = 0.0001; maxiter = 10;
busdata=[1 1 1.06 0.0 00.00 00.00 0.00 00.00 0 0 0
2 2 1.04 0.0 00.00 00.00 150.00 00.00 0 140 0
3 2 1.03 0.0 00.00 00.00 100.00 00.00 0 90 0
4 0 1.0 0.0 100.00 70.00 00.00 00.00 0 0 0
5 0 1.0 0.0 90.00 30.00 00.00 00.00 0 0 0
6 0 1.0 0.0 160.00 110.00 00.00 00.00 0 0 0];
21 | P a g e
linedata=[1 4 0.035 0.225 0.0065 1.0
1 5 0.025 0.105 0.0045 1.0
1 6 0.040 0.215 0.0055 1.0
2 4 0.000 0.035 0.0000 1.0
3 5 0.000 0.042 0.0000 1.0
4 6 0.028 0.125 0.0035 1.0
5 6 0.026 0.175 0.0300 1.0];
lfybus % form the bus admittance matrix
lfnewton % Power flow solution by Newton-Raphson method
busout % Prints the power flow solution on the screen
% Gen. Ra Xd' H
gendata=[ 1 0 0.20 20
2 0 0.15 4
3 0 0.25 5];
trstab %clearing time of the fault
OUTPUT:
NOTE: Fault Bus No: 6 Bus to Bus Line to be removed:[5,6]
22 | P a g e
EXPERIMENT-7 [ECONOMIC DISPATCH IN POWER SYSTEMS]
PROGRAM-1:
MANUEL CALCULATION:
23 | P a g e
MATLAB CODE:
alpha =[500; 400; 200]; beta = [5.3; 5.5; 5.8]; gama=[.004; .006; .009];
PD=800;
DelP = 10; % Error in DelP is set to a high value
lambda = input('Enter estimated value of Lambda = ');
fprintf('\n ')
disp([' Lambda P1 P2 P3 DP'... ' grad Delambda'])
iter = 0; % Iteration counter
while abs(DelP) >= 0.001 % Test for convergence
iter = iter + 1; % No. of iterations
P = (lambda - beta)./(2*gama);
24 | P a g e
DelP =PD - sum(P); % Residual
J = sum( ones(length(gama), 1)./(2*gama)); % Gradient sum
Delambda = DelP/J; % Change in variable
disp([lambda, P(1), P(2), P(3), DelP, J, Delambda])
lambda = lambda + Delambda; % Successive solution
end
totalcost = sum(alpha + beta.*P + gama.*P.^2)
OUTPUT:
PROGRAM-2:
25 | P a g e
MANUEL CALCULATION:
26 | P a g e
27 | P a g e
MATLAB CODE:
cost = [200 7.0 0.008
180 6.3 0.009
140 6.8 0.007];
mwlimits =[10 85
10 80
10 70];
28 | P a g e
Pdt = 150;
B = [0.0218 0 0
0 0.0228 0
0 0 0.0179];
basemva = 100;
dispatch
gencost
OUTPUT:
29 | P a g e
Experiment 8: compute the average load and the daily load factor. Using
MATLAB
MANUEL CALCULATION:
30 | P a g e
MATLAB CODE:
data=[0 2 6; 2 6 5; 6 9 10; 9 12 15; 12 14 12; 14 16 14; 16 18 16;
18 20 18; 20 22 16; 22 23 12; 23 24 6];
p=data(:,3);
Dt=data(:,2)-data(:,1);
w=p'*Dt;
pavg=w/sum(Dt)
peak=max(p)
LF=pavg/peak*100
OUTPUT:
pavg = 11.5417 peak = 18 LF = 64.1204
31 | P a g e
EXPERIMENT-9 [FAULT ANALYSIS] PROBLEM:
The one line diagram of a simple power system is shown in figure. The neutral of each generator is grounded through a current limiting reactor of 0.25/3 per unit on a 100MVA base. The system data expressed in per unit on a common 100 MVA base is tabulated below. The generators are running on no load at their rated voltage and rated frequency with their emf in phase. Determine the fault current for the following faults. (a) A balanced three phase fault at bus 3 through a fault impedance, Zf = j0.1 per unit. (b) A single line to ground fault at bus3 through a fault impedance, Zf = j0.1 per unit. (c) A line to line fault at bus3 through a fault impedance, Zf = j0.1 per unit. (d) A double line to ground fault at bus3 through a fault impedance, Zf = j0.1 per unit.
MATLAB PROGRAM:
Z133 = j*0.22; Z033 = j*0.35; Zf = j*0.1; disp('(a) Balanced three-phase fault at bus 3') Ia3F = 1.0/(Z133+Zf) disp('(b) Single line-to-ground fault at bus 3') I03 = 1.0/(Z033 + 3*Zf + Z133 + Z133); I012=[I03; I03; I03] %sctm; global sctm a =cos(2*pi/3)+j*sin(2*pi/3); sctm = [1 1 1; 1 a^2 a; 1 a a^2]; Iabc3 = sctm*I012 disp('(c) Line-to-line fault at bus 3') I13 = 1.0/(Z133 + Z133 + Zf); I012 = [0; I13; -I13] Iabc3 = sctm*I012 disp('(d) Double line-to-ground fault at bus 3') I13 = 1/(Z133 + Z133*(Z033+3*Zf)/(Z133+Z033+3*Zf)); I23 = -(1.0 - Z133*I13)/Z133; I03 = -(1.0 - Z133*I13)/(Z033+3*Zf); I012 = [I03; I13; I23] Iabc3 = sctm*I012
32 | P a g e
MANUEL CALCULATION
33 | P a g e
ZERO SEQUENCE NETWORK:
34 | P a g e
35 | P a g e
36 | P a g e
EXPERIMENT-10 [LOAD – FREQUENCY DYNAMICS OF SINGLE AREA POWER SYSTEMS]
Problem: An isolated power station has the following parameters
Turbine time constant, τ T= 0.5sec
Governor time constant, τ g= 0.2sec
Generator inertia constant, H = 5sec Governor speed regulation = R per unit The load varies by 0.8 percent for a 1 percent change in frequency, i.e, D = 0.8 (a) Use the Routh – Hurwitz array to find the range of R for control system
stability. (b) Use MATLAB to obtain the root locus plot. (c) The governor speed regulation is set to R = 0.05 per unit.The turbine rated output is 250MW at nominal frequency of 60Hz. A sudden load change of 50 MW (∆PL= 0.2 per unit) occurs.
(i) Find the steady state frequency deviation in Hz. (ii) Use MATLAB to obtain the time domain performance specifications and the frequency deviation step response.
SOLUTION MANUEL
K=1/R; CHARACTERISTIC EQUATION IS
37 | P a g e
MATLAB CODE: disp('Root-locus') num = 1; den = [1 7.08 10.56 .8]; figure (1), rlocus(num, den) disp('Frequency deviation step response') PL = 0.2; numc = [0.1 0.7 1]; denc = [1 7.08 10.56 20.8]; t = 0:.02:10; c = -PL*step(numc, denc, t); figure(2), plot(t, c), grid xlabel('t, sec'), ylabel('pu') title('Frequency deviation step response') timespec(numc, denc)
38 | P a g e
SIMULINK DIAGRAM:
39 | P a g e
EXPERIMENT-11 [LOAD – FREQUENCY DYNAMICS OF TWO AREA POWER SYSTEMS]
Problem: A two area system connected by a tie line has the following parameters on a 1000MVA common base
The units are operating in parallel at the nominal frequency of 60Hz.The synchronizing co-efficient is computed from the initial operating condition and is gain to be PS=20p.u. A load change of 187.5 MW in area 1.
a) Determine the new steady state frequency and change in tie line flow. b) Construct the SIMULINK block diagram and obtain the frequency deviation response for the condition specified.
SOLUTION:
40 | P a g e
MATLAB CODE:
R1 = 0.05; R2 = 0.0625; D1 = 0.6; D2 = 0.9; DPL1 = 187.5/1000; Dw = -DPL1/(1/R1 + D1+ 1/R2+D2) Df = Dw*60, f = 60+Df DPm1 = -Dw/R1 DPm2 = -Dw/R2 DP12 = Dw*(1/R2 + D2)
SIMULINK DIAGRAM:
41 | P a g e
EXPERIMENT-12 [AVR SYSTEM OF GENERATOR]
MATLAB CODE:
num=500; den=[1 33.5 307.5 775 500]; figure(1), rlocus(num, den); kA= 10; numc=kA*[25 500]; denc=[1 33.5 307.5 775 500+ 500*kA]; t=0:.05:20; c=step(numc, denc, t); figure(2), plot(t, c), xlabel('t, sec'), grid timespec(numc, denc)
SIMULINK MODEL
42 | P a g e
MANUEL SOLUTION: