Question 11.13: Consider again the induction motor of Example 11.12. Assumin......
Consider again the induction motor of Example 11.12. Assuming the motor speed and electromagnetic power remain constant (at 1680 r/min and 9.7 kW), use MATLAB to plot the per-unit armature current I_a and terminal voltage V_a as a function of i_D as (λ_{DR})_{ref} is varied between 0.8 and 1.2 per unit, where 1.0 per unit corresponds to the rated peak value.
The desired plot is given in Fig. 11.22. Note that the armature current decreases and the terminal voltage increases as λ_{DR} is increased. This clearly shows how i_D, which controls λ_{DR}, can be chosen to optimize the tradeoff between such quantifies as armature current, armature flux linkages, and terminal voltage.
Here is the MATLAB script:
Script File
clc
clear
%Motor rating and characteristics
Prated = 12e3;
Vrated = 230;
Varated = 230/sqrt(3) ;
ferated = 60;
omegaerated = 2*pi*ferated;
Lambdarated = sqrt(2)*Varated/omegaerated;
Irated = Prated/(sqrt(3)*Vrated) ;
Ipeakbase = sqrt(2)*Irated;
poles : 4 ;
%Here are the 60-Hz motor parameters
V10 = Vrated/sqrt(3) ;
X10 = 0.680;
X20 = 0.672;
Xm0 = 18.7;
R1 = 0.095;
R2 = 0.2;
%Calculate required dq0 parameters
Lm = Xm0/omegaerated;
LS = Lm + X10/omegaerated;
LR = Lm + X20/omegaerated;
Ra = R1;
RaR = R2 ;
% Operating point
n = 1680 ;
omegam = n*pi/30;
omegame = (poles/2)*omegam;
Pmech = 9.7e3;
Tmech = Pmech/omegam;
% Loop to plot over lambdaDR
for n = 1:41
lambdaDR = (0.8 + (n-1)*0.4/40)*Lambdarated;
lambdaDRpu (n) = lambdaDR/Lambdarated;
iQ = (2/3) * (2/poles) * (LR/Lm) * (Tmech/lambdaDR) ;
iD = (lambdaDR/Lm) ;
iDpu(n) = iD/Ipeakbase;
iQR = - (Lm/LR)*iQ;
Ia = sqrt((iD^2 + iQ^2)/2) ;
Iapu(n) = Ia/Irated;
omegae = omegame - (RaR/LR)*(iQ/iD) ;
fe(n) = omegae/ (2*pi) ;
Varms = sqrt( ((Ra*iD-omegae*(LS-Lm^2/LR)*iQ) ^2 +(Ra*iQ+ omegae*LS*iD)^2) /2) ;
Vapu(n) = Varms/Varated;
end
%Now plot
plot (iDpu, Iapu)
hold
plot(iDpu,Vapu, ' :')
hold
xlabel('i_D [per unit] ')
ylabel('per unit')
text(.21,1.06, 'Ia')
text (.21, .83, 'Va')
