📄 ex6_6.m
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% Example 6.6
clc
clear
% Here are the two sets of parameters
R1(1) = 0.262; R1(2) = 0.262;
R2(1) = 0.447; R2(2) = 0.444;
X1(1) = 0.633; X1(2) = 0.603;
X2(1) = 1.47; X2(2) = 1.41;
Xm(1) = 21.2; Xm(2) = 21.2;
Nph = 3;
poles = 4;
Prot = 354;
%Here is the operating condition
V1 = 220/sqrt(3);
fe = 60;
rpm = 1746;
%Calculate the synchronous speed
ns = 120*fe/poles;
omegas = 4*pi*fe/poles;
slip = (ns-rpm)/ns;
omegam = omegas*(1-slip);
%Calculate stator Thevenin equivalent
%Loop over the two motors
for m = 1:2
Zgap = j*Xm(m)*(j*X2(m)+R2(m)/slip)/(R2(m)/slip+j*(Xm(m)+X2(m)));
Zin = R1(m) + j*X1(m) + Zgap;
I1 = V1/Zin;
I2 = I1*(j*Xm(m))/(R2(m)/slip+j*(Xm(m)+X2(m)));
Tmech = Nph*abs(I2)^2*R2(m)/(slip*omegas); %Internal torque
Pmech = omegam*Tmech; %Internal power
Pshaft = Pmech - Prot;
if (m == 1)
fprintf('\nFor parameters of Example 7-5:')
else
fprintf('\nFor parameters of Example 7-6(a):')
end
fprintf('\n Pmech = %.1f [W], Pshaft = %.1f [W]',Pmech,Pshaft)
fprintf('\n I1 = %.1f [A]\n',abs(I1));
end % end of "for m = 1:2" loop
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