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EE 255 - Chapter 5b

# EE 255 - Chapter 5b - The magnitude of E A is 12,040 V(b(c...

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121 The magnitude of A E is 12,040 V. (b) The torque angle of the generator at rated conditions is δ = 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 φ = = = Ω 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 φ 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 φ - ° ∠ ° = = = ° + + Ω 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 φ θ = = ° - ° = - The generator is actually consuming reactive power at this time. 5-8. A 480-V, 100-kW, two-pole, three-phase, 60-Hz synchronous generator’s prime mover has a no-load speed of 3630 r/min and a full-load speed of 3570 r/min. It is operating in parallel with a 480-V, 75-kW, four- pole, 60-Hz synchronous generator whose prime mover has a no-load speed of 1800 r/min and a full-load 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 generator’s operators do to correct for the low terminal voltage? S OLUTION The no-load frequency of generator 1 corresponds to a frequency of ( )( ) nl1 3630 r/min 2 60.5 Hz 120 120 m n P f = = = The full-load frequency of generator 1 corresponds to a frequency of

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122 ( )( ) fl1 3570 r/min 2 59.5 Hz 120 120 m n P f = = = The no-load frequency of generator 2 corresponds to a frequency of ( )( ) nl2 1800 r/min 4 60.00 Hz 120 120 m n P f = = = The full-load 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 curve’s 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 curve’s 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 no-load frequency of generator 1 is 60.5 Hz and the no-load 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 Hz f f = - + - ( ) ( ) sys sys 100 kW 6050 kW 0.10 MW/Hz 9000 kW 0.15 MW/Hz f f = - + - ( ) sys 0.25 MW/Hz 6050 kW 9000 kW 100 kW f = + - sys 14,950 kW 59.8 Hz 0.25 MW/Hz f = = (c) The power supplied by generator 1 is
123 ( ) ( )( ) 1 1 nl1 sys 0.1 MW/Hz 60.5 Hz 59.8 Hz 70 kW P P s f f = - = - = The power supplied by generator 2 is ( ) ( )( ) 2 2 nl2 sys 0.15 MW/Hz 60.0 Hz 59.8 Hz 30 kW P P s f f = - = - = (d) If the terminal voltage is 460 V, the operators of the generators must increase the field currents on both generators simultaneously.

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