EE 255 - Chapter 6b

# EE 255 - Chapter 6b - I A2 = 5036.87 A The internal...

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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 0 ° 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 .

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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. (b) If the flux of the motor is increased by 10 percent, what happens to the terminal voltage of the power system? What is its new value? (c) What is the power factor of the motor after the increase in motor flux?
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