Electric Machinery Fundamentals (Solutions Manual) Part 8 pps
... W/hp 24 kW 18 kW 78 = + 8 kW + = Therefore, the line and phase current at unity power factor is P I I = = = 788 kW 1 98 = A A L ( ... 3 PF 3 ( ) 480 V 0 .85 T V 5 A so I A 141.5 31 .8 = A ° . This machine is Y-connected, so the phase voltage is V ⎞ = 480 / 3 = 2...
Ngày tải lên: 06/08/2014, 11:20
... circuit voltage at 0.532 A is 88 0 V 4 , so the open-circuit phase voltage (= 5 08 V. The approximate saturated synchronous reactance X S is E A ) is 88 0/ 3 = 3 ... A = 3 3 ( ) 480 V A L T V The power factor is 0 .85 lagging, so I A 565.3 31 .8 = A ° . The rated phase voltage is V ⎞ = 480 V / 3 =...
Ngày tải lên: 06/08/2014, 11:20
... % XS (ohms) % Calculate impedance angle theta in degrees theta = -31 .8: 0.3 18: 31 .8; th = theta * pi/ 180 ; % In radians % Calculate the phase voltage and terminal voltage ... OUT P 0 .85 100 MVA 85 MW = = 120 120 ( ) 50 Hz n f e = = = sync P 3000 r/min 2 the applied torque would be ⎮ ⎮ = = 85 ,...
Ngày tải lên: 06/08/2014, 11:20
Electric Machinery Fundamentals (Solutions Manual) Part 11 ppsx
... IN C P P U 83 P .8 kW 4.16 kW 79.6 kW (g) If E A is decreased by 10%, the new value if E A = (0.9)(603 V) = 543 V. To simplify this part of the problem, ... = 19.5 21 .8 ° = ° 2 1 Therefore, I E A2 ⎞ A V E 440 0 V ° 543 V 5 43 2 1 .8 °...
Ngày tải lên: 06/08/2014, 11:20
Electric Machinery Fundamentals (Solutions Manual) Part 1 pdf
... 1240 V 186 A 0 .8 554 kW 2 ( ) & = CU 3 A A P I 3 18 R 6 A 0.15 15.6 kW F& P = W 24 kW co P re = 18 kW co P re = 18 kW ... 1240 0 ( 0.1 )( 5 186 30 A ) ( 1.1 )( 186 30 A ) E A = ° + & ° j + & ° E A 1377 6 .8 V = ° The...
Ngày tải lên: 06/08/2014, 11:20
Electric Machinery Fundamentals (Solutions Manual) Part 2 ppt
... + A A E V ⎞ A I S A R jX I 7967 0 ( ) 1.5 ( 4 18 36 .87 A ) ( 12.0 )( 4 18 36 .87 A ) E A = ° + & ° j + & ... 1000 kVA 0 .8 80 0 kW = = 2 ( ) = = 2 ( ) & = CU 3 A A P I 3 R 251 A 0.15 28. 4 kW F& P = W 24 kW co P re = 18 kW...
Ngày tải lên: 06/08/2014, 11:20
Electric Machinery Fundamentals (Solutions Manual) Part 3 docx
... 1. ⋅ 68% = n fl 3570 r/min The speed droop of generator 2 is given by n n DS 2 nl fl 100% 18 = ⋅ 00 r/m = in 1 785 r/min 100% 0. ⋅ 84 % = ... I = = 62 = 50 kVA 86 7 = A 3 3 ( ) 4160 V A L T V The power factor is 0 .85 lagging, so I A 86 7 31. = 8 A ° . Therefore...
Ngày tải lên: 06/08/2014, 11:20
Electric Machinery Fundamentals (Solutions Manual) Part 7 pdf
... 1 48 A2 ⎞ 10, 584 35 .8 70 ° 44 0 780 14.0 A E V A I = = jX S 8. 18 j ° = ° With ... P 41.6 kW = = 62.5 A = L 3 PF 3 ( ) 480 V ( 0 .8 ) T V 6-2. A 480 -V, 60 Hz, 400-hp 0 .8- PF-leading six-pole -connected synchronous motor...
Ngày tải lên: 06/08/2014, 11:20
Electric Machinery Fundamentals (Solutions Manual) Part 9 ppt
... & 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 & ... g A 280 0 V j 0.4 95.7 5.7 A 286 7.6 V If we make ⎞ V the reference (as we usually do), these voltages and currents become: E , A g...
Ngày tải lên: 06/08/2014, 11:20