📄 ex8_4.m
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% Example 8.4
clc
clear
%(a) First plot the lambda-i characteristics
for m = 1:10
theta(m) = 10*(m-1);
for n=1:101
i(n) = 30*(n-1)/100;
Lambda(n) = i(n)*(0.005 + 0.09*((90-theta(m))/90)*(8/(i(n)+8)));
end
plot(i,Lambda)
if m==1
hold
end
end
hold
xlabel('Current [A]')
ylabel('Lambda [Wb]')
title('Family of lambda-i curves as theta_m varies from 0 to 90 degrees')
text(17,.7,'theta_m = 0 degrees')
text(20,.06,'theta_m = 90 degrees')
%(b) Now integrate to find the areas.
%Peak lambda at 0 degrees, 25 Amps
lambdamax = 25*(0.005+0.09*(8/(25+8)));
AreaWnet = 0;
AreaWrec = 0;
% 100 integration step
deli = 25/100;
for n=1:101
i(n) = 25*(n-1)/100;
AreaWnet = AreaWnet + deli*i(n)*(0.09)*(8/(i(n)+8));
AreaWrec = AreaWrec + deli*(lambdamax - i(n)*(0.005+0.09*(8/(i(n)+8))));
end
Ratio = (AreaWrec + AreaWnet)/AreaWnet;
fprintf('\nPart(b) Ratio = %g',Ratio)
%(c) Calculate the power
rpm = 2500;
rps = 2500/60;
T = 1/rps;
Pphase = 2*AreaWnet/T;
Ptot = 2*Pphase;
fprintf('\n\nPart(c) AreaWnet = %g [Joules]',AreaWnet)
fprintf('\n Pphase = %g [W] and Ptot = %g [W]\n',Pphase,Ptot)
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