EE 255 - Chapter 9j

# EE 255 - Chapter 9j - 295 Appendix C Salient Pole Theory of...

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Unformatted text preview: 295 Appendix C : Salient Pole Theory of Synchronous Machines C-1. A 480-V 200-kVA 0.8-PF-lagging 60-Hz four-pole Y-connected synchronous generator has a direct-axis reactance of 0.25 Ω , a quadrature-axis 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 open-circuit characteristic is given by Figure P5-1. (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 5-2 b ? (c) What fraction of this generator’s full-load power is due to the reluctance torque of the rotor? S OLUTION (a) If the no-load 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 direct-axis 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 quadrature-axis 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 5-2 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.

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EE 255 - Chapter 9j - 295 Appendix C Salient Pole Theory of...

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