This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 121 The magnitude of A E is 12,040 V. (b) The torque angle of the generator at rated conditions is = 17.6 . (c) Ignoring A R , the maximum output power of the generator is given by ( )( ) MAX 3 3 7967 V 12,040 V 24.0 MW 12 A S V E P X = = = The power at maximum load is 8 MW, so the maximum output power is three times the full load output power. (d) The phasor diagram at these conditions is shown below: jX S I A V I A E A I A R A Under these conditions, the armature current is 12,040 90 V  7967 0 V 1194 40.6 A 1.5 12.0 A A A S R jX j E V I  = = = + + The reactive power produced by the generator at this point is ( )( ) ( ) 3 sin 3 7967 V 1194 A sin 0 40.6 18.6 MVAR A Q V I = =  =  The generator is actually consuming reactive power at this time. 58. A 480V, 100kW, twopole, threephase, 60Hz synchronous generators prime mover has a noload speed of 3630 r/min and a fullload speed of 3570 r/min. It is operating in parallel with a 480V, 75kW, four pole, 60Hz synchronous generator whose prime mover has a noload speed of 1800 r/min and a fullload speed of 1785 r/min. The loads supplied by the two generators consist of 100 kW at 0.85 PF lagging. (a) Calculate the speed droops of generator 1 and generator 2. (b) Find the operating frequency of the power system. (c) Find the power being supplied by each of the generators in this system. (d) If V T is 460 V, what must the generators operators do to correct for the low terminal voltage? S OLUTION The noload frequency of generator 1 corresponds to a frequency of ( )( ) nl1 3630 r/min 2 60.5 Hz 120 120 m n P f = = = The fullload frequency of generator 1 corresponds to a frequency of 122 ( )( ) fl1 3570 r/min 2 59.5 Hz 120 120 m n P f = = = The noload frequency of generator 2 corresponds to a frequency of ( )( ) nl2 1800 r/min 4 60.00 Hz 120 120 m n P f = = = The fullload frequency of generator 2 corresponds to a frequency of ( )( ) fl2 1785 r/min 4 59.50 Hz 120 120 m n P f = = = (a) The speed droop of generator 1 is given by nl fl 1 fl 3630 r/min 3570 r/min SD 100% 100% 1.68% 3570 r/min n n n = = = The speed droop of generator 2 is given by nl fl 2 fl 1800 r/min 1785 r/min SD 100% 100% 0.84% 1785 r/min n n n = = = (b) The power supplied by generator 1 is given by ( ) 1 1 nl1 sys P P s f f = and the power supplied by generator 1 is given by ( ) 2 2 nl2 sys P P s f f = The power curves slope for generator 1 is 1 nl fl 0.1 MW 0.1 MW/Hz 60.5 Hz 59.5 Hz P P s f f = = = The power curves slope for generator 1 is 2 nl fl 0.075 MW 0.150 MW/Hz 60.00 Hz 59.50 Hz P P s f f = = = The noload frequency of generator 1 is 60.5 Hz and the noload frequency of generator 2 is 60 Hz. The total power that they must supply is 100 kW, so the system frequency can be found from the equations LOAD 1 2 P P P = + ( ) ( ) LOAD 1 nl1 sys 2 nl2 sys P P P s f f s f f = + ( ) ( ) ( ) ( ) sys sys 100 kW 0.1 MW/Hz 60.5 Hz 0.15 MW/Hz 60.0 Hz0....
View
Full
Document
 Spring '05
 Shaban

Click to edit the document details