Page 138 - Solutions Manual to accompany Electric Machinery Fundamentals

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Calculate the armature currents associated with
% each value of Ea for the half-load case.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% First, calculate delta.
delta = asin ( Ea_half ./ Ea .* sin(d_half) );

% Calculate the phasor Ea
Ea2 = Ea .* (cos(delta) + j .* sin(delta));

% Now calculate Ia
Ia_half = ( Vp - Ea2 ) / (j * Xs);

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Calculate the armature currents associated with
% each value of Ea for the full-load case.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% First, calculate delta.
delta = asin ( Ea_full ./ Ea .* sin(d_full) );

% Calculate the phasor Ea
Ea2 = Ea .* (cos(delta) + j .* sin(delta));

% Now calculate Ia
Ia_full = ( Vp - Ea2 ) / (j * Xs);

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Plot the v-curves
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
hold off;
plot(if1,abs(Ia_nl),'k-','Linewidth',2.0);
hold on;
plot(if1,abs(Ia_half),'b--','Linewidth',2.0);
plot(if1,abs(Ia_full),'r:','Linewidth',2.0);
xlabel('\bfField Current (A)');
ylabel('\bfArmature Current (A)');
title ('\bfSynchronous Motor V-Curve');
legend('No load','Half load','Full load');
grid on;

The resulting plot is shown below. The flattening visible to the right of the V-curves is due to magnetic
saturation in the machine.

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