Chapter 11
Q. 11.13
Q. 11.13
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.
Step-by-Step
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')
