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Unformatted text preview: 149 Chapter 6 : Synchronous Motors 6-1. A 480-V, 60 Hz, four-pole synchronous motor draws 50 A from the line at unity power factor and full load. Assuming that the motor is lossless, answer the following questions: (a) What is the output torque of this motor? Express the answer both in newton-meters and in pound-feet. (b) What must be done to change the power factor to 0.8 leading? Explain your answer, using phasor diagrams. (c) What will the magnitude of the line current be if the power factor is adjusted to 0.8 leading? S OLUTION (a) If this motor is assumed lossless, then the input power is equal to the output power. The input power to this motor is ( )( )( ) IN 3 cos 3 480 V 50 A 1.0 41.6 kW T L P V I θ = = = The output torque would be ( ) OUT LOAD 41.6 kW 221 N m 1 min 2 rad 1800 r/min 60 s 1 r m P τ π ω = = = ⋅ In English units, ( )( ) ( ) OUT LOAD 7.04 41.6 kW 7.04 163 lb ft 1800 r/min m P n τ = = = ⋅ (b) To change the motor’s power factor to 0.8 leading, its field current must be increased. Since the power supplied to the load is independent of the field current level, an increase in field current increases A E while keeping the distance δ sin A E constant. This increase in A E changes the angle of the current A I , eventually causing it to reach a power factor of 0.8 leading. V φ E A 1 jX S I A E A 2 I A 2 I A 1 Q ∝ I sin θ A } ∝ P } ∝ P (c) The magnitude of the line current will be ( )( ) 41.6 kW 62.5 A 3 PF 3 480 V 0.8 L T P I V = = = 6-2. A 480-V, 60 Hz, 400-hp 0.8-PF-leading six-pole Δ-connected synchronous motor has a synchronous reactance of 1.1 Ω and negligible armature resistance. Ignore its friction, windage, and core losses for the purposes of this problem. 150 (a) If this motor is initially supplying 400 hp at 0.8 PF lagging, what are the magnitudes and angles of E A and I A ? (b) How much torque is this motor producing? What is the torque angle δ ? How near is this value to the maximum possible induced torque of the motor for this field current setting? (c) If E A is increased by 15 percent, what is the new magnitude of the armature current? What is the motor’s new power factor? (d) Calculate and plot the motor’s V-curve for this load condition. S OLUTION (a) If losses are being ignored, the output power is equal to the input power, so the input power will be ( )( ) IN 400 hp 746 W/hp 298.4 kW P = = This situation is shown in the phasor diagram below: V φ E A jX S I A I A The line current flow under these circumstances is ( )( ) 298.4 kW 449 A 3 PF 3 480 V 0.8 L T P I V = = = Because the motor is Δ-connected, the corresponding phase current is 449 / 3 259 A A I = = . The angle of the current is ( ) 1 cos 0.80 36.87-- = - ° , so 259 36.87 A A = ∠ - ° I . The internal generated voltage A E is A S A jX φ =- E V I ( ) ( )( ) 480 0 V 1.1 259 36.87 A 384 36.4 V A j = ∠ °- Ω ∠ - ° = ∠ - ° E (b) This motor has 6 poles and an electrical frequency of 60 Hz, so its rotation speed is m n = 1200 r/min. = 1200 r/min....
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This note was uploaded on 10/15/2009 for the course EE 255 taught by Professor Shaban during the Spring '05 term at Cal Poly.
- Spring '05