Electric Machinery Fundamentals (Solutions Manual) Part 10 ppt

Electric Machinery Fundamentals (Solutions Manual) Part 10 ppt

Electric Machinery Fundamentals (Solutions Manual) Part 10 ppt

... efficiency of this motor at full load is ( )( ) ⎜ OUT P 100 % = ⋅ 2100 = 0 hp 746 W/hp 100 % ⋅ 97.6% = IN P 16.05 MW (f) The output torque in SI ... supplied to the motor is 10 MW / 16.05 MW = 0.623. The maximum E A is 5067 V or 1.33 pu, so with E A set to maximum and the motor consuming 10 MW, the to...

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Electric Machinery Fundamentals (Solutions Manual) Part 2 ppt

Electric Machinery Fundamentals (Solutions Manual) Part 2 ppt

... OUT P P CU P F&W P core P stray P 870.4 kW η OUT P 100 % 800 kW 100 % 91.9% = ⋅ = ⋅ = IN P 870.4 kW (b) If the generator is ...  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...

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Electric Machinery Fundamentals (Solutions Manual) Part 6 ppt

Electric Machinery Fundamentals (Solutions Manual) Part 6 ppt

... you were an engineer planning a new electric co-generation facility for a plant with excess process steam. You have a choice of either two 10 MW turbine-generators or a single ... and more efficient than two 10 MW generators, but if the 20 MW generator goes down all 20 MW of generation would be lost at once. If two 10 MW generators are chosen,...

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Electric Machinery Fundamentals (Solutions Manual) Part 9 ppt

Electric Machinery Fundamentals (Solutions Manual) Part 9 ppt

... 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 ... and angles of E A 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?...

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Electric Machinery Fundamentals (Solutions Manual) Part 1 pdf

Electric Machinery Fundamentals (Solutions Manual) Part 1 pdf

... P 1 5 10 Therefore, a 10- pole synchronous motor must be coupled to a 12-pole synchronous generator to accomplish this frequency conversion. 5-2. A 2300-V 100 0-kVA ... synchronous generator. % Calculate the waveforms for times from 0 to 1/30 s Q = -4 810; DE = 5567; S = 100 0; % Get points for stator current limit theta = -95:1:95;...

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Electric Machinery Fundamentals (Solutions Manual) Part 3 docx

Electric Machinery Fundamentals (Solutions Manual) Part 3 docx

... nl fl 100 % = ⋅ 3630 r/m = in 3570 r/min 100 % 1. ⋅ 68% = n fl 3570 r/min The speed droop of generator 2 is given by n n   DS 2 nl fl 100 % 18 = ... ) ( ) ( ) ( ) =  +  100 kW 0.1 MW/Hz 60.5 Hz sys 0.15 MW/Hz 60.0 Hz sys f f ( ) ( ) 100 kW 6050 kW 0 .10 MW =  /Hz ( ) sy...

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Electric Machinery Fundamentals (Solutions Manual) Part 4 ppsx

Electric Machinery Fundamentals (Solutions Manual) Part 4 ppsx

... at SCC is 532 A. I F = 0.50 A, the air-gap line voltage is 104 0 V, and the X Su ( ) 104 0 V / 3 1.13 = = & 532 A (c) This task can ... 880.400 129 generators would have more margin than this. 131 E A 2825 10. 9 = V ° Problems 5-11 to 5-21 refer to a two-pole Y-connected synchronous gene...

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Electric Machinery Fundamentals (Solutions Manual) Part 5 ppsx

Electric Machinery Fundamentals (Solutions Manual) Part 5 ppsx

... 75% to 100 % of the flux at rated conditions. (c) Plot the reactive power supplied by this generator as a function of the flux ⎞ as the flux is varied from 75% to 100 % of ... 19.9 V = ° The resulting voltage regulation is 11,120  7621 RV 100 % 45.9% = ⋅ = 7621 (b) If the generator is to be operated at 50 Hz with t...

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Electric Machinery Fundamentals (Solutions Manual) Part 7 pdf

Electric Machinery Fundamentals (Solutions Manual) Part 7 pdf

... 10, 584 V , and sin 1 1 E 1 A sin sin 13, 230 ™ ™ V sin = = 27.9   35.8 ° = ° 2 2 E A 2 Therefore, the new armature current is 10, 584 ... 148  A2 ⎞ 10, 584 35.8 70 ° 44  0 780 14.0 A E V A I = = jX S 8....

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Electric Machinery Fundamentals (Solutions Manual) Part 8 pps

Electric Machinery Fundamentals (Solutions Manual) Part 8 pps

... ( ) 6 e f m n P = = 120 120 15 Hz = A speed of 100 0 r/min corresponds to a frequency of ( ) 100 0 r/min e f m n P = = 120 120 ( ) 6 50 Hz = ... V-Curve 250 240 230 220 210 200 350 400 450 500 550 600 650 700 E (V) A 6-3. A 2300-V 100 0-hp 0.8-PF leading 60-Hz two-pole Y-connec...

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