Chapter 5 Synchronous Motors - KFUPM

Chapter 5: Synchronous Motors5-1.A 480-V, 60 Hz, 400-hp 0.8-PF-leading eight-pole -connected synchronous motor has a synchronousreactance of 0.6 and negligible armature resistance. Ignore its friction, windage, and core losses for thepurposes of this problem. Assume that E A is directly proportional to the field current I F (in otherwords, assume that the motor operates in the linear part of the magnetization curve), and that E A 480V when I F 4 A.(a) What is the speed of this motor?(b) If this motor is initially supplying 400 hp at 0.8 PF lagging, what are the magnitudes and anglesof E A and I A ?(c) How much torque is this motor producing? What is the torque angle ? How near is this valueto the maximum possible induced torque of the motor for this field current setting?(d) If E A is increased by 30 percent, what is the new magnitude of the armature current? What isthe motor’s new power factor?(e) Calculate and plot the motor’s V-curve for this load condition.SOLUTION(a)The speed of this motor is given bynm (b)120 f se 120 60 Hz 900 r/min8PIf losses are being ignored, the output power is equal to the input power, so the input power will bePIN 400 hp 746 W/hp 298.4 kWThis situation is shown in the phasor diagram below:V IAjX IS AEAThe line current flow under these circumstances isIL P298.4 kW 449 A3 VT PF3 480 V 0.8 Because the motor is -connected, the corresponding phase current is I A 449 / 3 259 A . The angleof the current is cos 1 0.80 36.87 , so I A 259 36.87 A . The internal generated voltage E AisE A V jX S I AE A 480 0 V j 0.6 259 36.87 A 406 17.8 V123

(c) This motor has 6 poles and an electrical frequency of 60 Hz, so its rotation speed is nm 1200r/min. The induced torque is ind POUT m 298.4 kW 3166 N m1 min 2 rad 900 r/min 60 s 1 r The maximum possible induced torque for the motor at this field setting is the maximum possible powerdivided by m ind,max 3V E A m X S 3 480 V 406 V 10,340 N m 1 min 2 rad 900 r/min 0.6 60 s 1 r The current operating torque is about 1/3 of the maximum possible torque.(d)If the magnitude of the internal generated voltage E A is increased by 30%, the new torque anglecan be found from the fact that E A sin P constant .E A 2 1.30 E A1 1.30 406 V 487.2 V E A1 406 V sin 1 sin 1 sin 17.8 14.8 487.2 V E A2 2 sin 1 The new armature current isI A2 V E A 2jX S 480 0 V 487.2 14.8 V 208 4.1 Aj 0.6 The magnitude of the armature current is 208 A, and the power factor is cos (-24.1 ) 0.913 lagging.(e)A MATLAB program to calculate and plot the motor’s V-curve is shown below:% M-file: prob5 1e.m% M-file create a plot of armature current versus Ea%for the synchronous motor of Problem 5-1.% Initialize valuesEa (0.90:0.01:1.70)*406;% Magnitude of Ea voltsEar 406;% Reference Eadeltar -17.8 * pi/180;% Reference torque angleXs 0.6;% Synchronous reactance (ohms)Vp 480;% Phase voltage at 0 degreesEar Ear * (cos(deltar) j * sin(deltar));% Calculate delta2delta2 asin ( abs(Ear) ./ abs(Ea) .* sin(deltar) );% Calculate the phasor EaEa Ea .* (cos(delta2) j .* sin(delta2));% Calculate IaIa ( Vp - Ea ) / ( j * Xs);% Plot the v-curve124

Chapter 5: Synchronous Motors 5-1. A 480-V, 60 Hz, 400-hp 0.8-PF-leading eight-pole -connected synchronous motor has a synchronous reactance of 0.6 and negligible armature resistance. Ignore its friction, w