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.

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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:

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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')