EE 255 - Chapter 6b

EE 255 - Chapter 6b - 160 2 50 36.87 A A = ∠ ° I The...

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Unformatted text preview: 160 2 50 36.87 A A = ∠ ° I The internal generated voltage required to produce this current would be 2 2 2 A A A S A R jX φ =-- E V I I ( )( ) 2 120 0 V 0.8 50 36.87 A A j = ∠ °- Ω ∠ ° E 2 147.5 12.5 V A = ∠ - ° E The internal generated voltage A E is directly proportional to the field flux, and we have assumed in this problem that the flux is directly proportional to the field current. Therefore, the required field current is ( ) 2 2 1 1 147 V 2.7 A 3.20 A 124 V A F F A E I I E = = = (c) The new torque angle δ of this machine is –12.5 ° . 6-8. A synchronous machine has a synchronous reactance of 2.0 Ω per phase and an armature resistance of 0.4 Ω per phase. If E A =460 ∠-8 ° V and φ V = 480 ∠ ° V, is this machine a motor or a generator? How much power P is this machine consuming from or supplying to the electrical system? How much reactive power Q is this machine consuming from or supplying to the electrical system? S OLUTION This machine is a motor, consuming power from the power system, because A E is lagging φ V . It is also consuming reactive power, because cos A E V φ δ < . The current flowing in this machine is 480 0 V 460 8 V 33.6 9.6 A 0.4 2.0 A A A S R jX j φ- ∠ °- ∠ - ° = = = ∠ - ° + + Ω V E I Therefore the real power consumed by this motor is ( )( ) ( ) 3 cos 3 480 V 33.6 A cos 9.6 47.7 kW A P V I φ θ = = ° = and the reactive power consumed by this motor is ( )( ) ( ) 3 sin 3 480 V 33.6 A sin 9.6 8.07 kVAR A Q V I φ θ = = ° = 6-9. Figure P6-2 shows a synchronous motor phasor diagram for a motor operating at a leading power factor with no R A . For this motor, the torque angle is given by cos tan = sin S A S A X I V X I φ θ δ θ +-1 cos =tan sin S A S A X I V X I φ θ δ θ + Derive an equation for the torque angle of the synchronous motor if the armature resistance is included . 161 S OLUTION The phasor diagram with the armature resistance considered is shown below. jX S I A V φ I A E A I A R A I A R A cos θ θ θ θ X S I A sin θ } X S I A cos θ } } δ Therefore, cos sin tan sin cos S A A A S A A A X I R I V X I R I φ θ θ δ θ θ + = +- 1 cos sin tan sin cos S A A A S A A A X I R I V X I R I φ θ θ δ θ θ- + = +- 6-10. A 480-V 375-kVA 0.8-PF-lagging Y-connected synchronous generator has a synchronous reactance of 0.4 Ω and a negligible armature resistance. This generator is supplying power to a 480-V 80-kW 0.8-PF- leading Y-connected synchronous motor with a synchronous reactance of 1.1 Ω and a negligible armature resistance. The synchronous generator is adjusted to have a terminal voltage of 480 V when the motor is drawing the rated power at unity power factor. (a) Calculate the magnitudes and angles of A E for both machines....
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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.

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EE 255 - Chapter 6b - 160 2 50 36.87 A A = ∠ ° I The...

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