Fall_2012_HW#7 Solution

76 1823 06 25903 1364 25939 301 the new power factor

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Unformatted text preview: 4 259.39∠ 3.01 The new power factor would be . e) cos 3.01 0.998 Calculate and plot the motor’s V curve for this load condition Synchronous Motor V-Curve 460 440 420 A I (A) 400 380 360 340 320 300 350 400 450 500 550 E (V) 600 650 700 750 A % M-file: prob5_1e.m % M-file create a plot of armature current versus Ea % for the synchronous motor of Problem 5-1. % Initialize values Ea = (0.90:0.01:1.70)*413.96; % Magnitude of Ea volts 2 ECE3602012FLHW#7 Ear = 496.76; % Reference Ea deltar = -22.05 * pi/180; % Reference torque angle Xs = 0.6; % Synchronous reactance (ohms) Vp = 480; % Phase voltage at 0 degrees Ear = Ear * (cos(deltar) + j * sin(deltar)); % Calculate delta2 delta2 = asin ( abs(Ear) ./ abs(Ea) .* sin(deltar) ); % Calculate the phasor Ea Ea = Ea .* (cos(delta2) + j .* sin(delta2)); % Calculate Ia Ia = ( Vp - Ea ) / ( j * Xs); % Plot the v-curve figure(1); plot(abs(Ea),abs(Ia),'b','Linewidth',2.0); xlabel('\bf\itE_{A}\rm\bf (V)'); ylabel('\b...
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