ex5_6.m

美国 A.E.Fitgerald著 刘新正,苏少平,高琳翻译,第六版的源代码

% Example 5.6

clc
clear

% Solution for part (b) 

%System parameters

Veq = 1.0;
Eaf = 1.0;
Xeq = .23;
Xs = 1.35;

% Solve for Va as delta varies from 0 to 90 degrees
for n = 1:101
	delta(n) = (pi/2.)*(n-1)/100;
	Ia(n) = (Eaf *exp(j*delta(n)) -  Veq)/(j*(Xs + Xeq));
	Va(n) = abs(Veq + j*Xeq*Ia(n));
	degrees(n) = 180*delta(n)/pi;
end

%Now plot the results
plot(degrees,Va)
xlabel('Power angle, delta [degrees]')
ylabel('Terminal voltage [per unit]')
title('Terminal voltage vs. power angle for part (b)')

fprintf('\n\nHit any key to continue\n')
pause

% Solution for part (c)

%Set terminal voltage to unity

Vterm = 1.0;

for n = 1:101
	P(n) = (n-1)/100;
	deltat(n) = asin(P(n)*Xeq/(Vterm*Veq));
	Ia(n) = (Vterm *exp(j*deltat(n)) -  Veq)/(j*Xeq);
	Eaf(n) = abs(Vterm + j*(Xs+Xeq)*Ia(n));
end

%Now plot the results
plot(P,Eaf)
xlabel('Power [per unit]')
ylabel('Eaf [per unit]')
title('Eaf vs. power for part (c)')