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Unformatted text preview: 295 Appendix C : Salient Pole Theory of Synchronous Machines C1. A 480V 200kVA 0.8PFlagging 60Hz fourpole Yconnected synchronous generator has a directaxis reactance of 0.25 Ω , a quadratureaxis reactance of 0.18 Ω , and an armature resistance of 0.03 Ω . Friction, windage, and stray losses may be assumed negligible. The generator’s opencircuit characteristic is given by Figure P51. (a) How much field current is required to make V T equal to 480 V when the generator is running at no load? (b) What is the internal generated voltage of this machine when it is operating at rated conditions? How does this value of E A compare to that of Problem 52 b ? (c) What fraction of this generator’s fullload power is due to the reluctance torque of the rotor? S OLUTION (a) If the noload terminal voltage is 480 V, the required field current can be read directly from the open circuit characteristic. It is 4.55 A. (b) At rated conditions, the line and phase current in this generator is ( ) A 6 . 240 V 480 3 kVA 200 3 = = = = L L A V P I I at an angle of –36.87 ° 296 ′′ = + + E V I I A A A q A R jX φ ( )( ) ( )( ) A 87 . 36 6 . 240 18 . A 87 . 36 6 . 240 03 . 277 ° ∠ Ω + ° ∠ Ω + ° ∠ = ′ ′ j A E V 61 . 5 310 ° ∠ = ′ ′ A E Therefore, the torque angle δ is 5.61 ° . The directaxis current is ( ) ° ∠ + = 90 sin δ δ θ A d I I ( ) ( ) ° ∠ ° = 4 . 84 48 . 42 sin A 6 . 240 d I A 4 . 84 5 . 162 ° ∠ = d I The quadratureaxis current is ( ) δ δ θ ∠ + = cos A q I I ( ) ( ) ° ∠ ° = 61 . 5 48 . 42 cos A 6 . 240 q I A 61 . 5 4 . 177 ° ∠ = q I Therefore, the internal generated voltage of the machine is q q d d A A A jX jX R I I I V E + + + = φ ( )( ) ( )( ) ( )( ) ° ∠ + ° ∠ + ° ∠ + ° ∠ = 61 . 5 4 . 177 18 . 4 . 84 5 . 162 25 . 87 . 36 6 . 240 03 . 277 j j A E V 61 . 5 322 ° ∠ = A E A E is approximately the same magnitude here as in Problem 52 b , but the angle is about 2.2 ° different. (c) The power supplied by this machine is given by the equation P V E X V X X X X A d d q d q = + g167 g169 g168 g168 g183 g185 g184 g184 3 3 2 2 φ φ δ δ sin sin 2 ( )( ) ( ) ( )( ) ° g184 g184 g185 g183 g168 g168 g169 g167 + ° 5 = 1.22 1 sin 18 . 25 ....
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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
 Shaban

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