Electric Machinery Fundamentals (Solutions Manual) Part 3 docx

... 91.5 1.5 sys 1 03. 07 1.6 P f 76 sys 118.64 1.96 f =  5. 137 sys 31 3.2 f P LOAD 1 sys f f sys = 31 3.2  L P OAD 5. 137 A MATLAB ... kVA)(0.85) = 5 .31 MW. Generator 3 will be the first machine to reach that limit. Generator 3 will supply this power at a frequency of ( ) ( ) =  5 .3...

Ngày tải lên: 06/08/2014, 11:20

... 3. 333 3  0 ° Ω . The magnitude of the phase current flowing in this generator is E A 137 7 V 137 7 V 33 5 A A I = = = = 0.15 1.1 3. 333 30 ... the phase voltage is ( ) 33 5 A = = ⎞ A V I Z and the terminal voltage is ( 3. 333 ) 11 & 17 = V 3 3 ( ) 1117 V 1 934 V T V V ⎞ =...

Ngày tải lên: 06/08/2014, 11:20

... ° j + & ° E A 139 3 11.4 = V ° The voltage regulation would be: 139 3  132 8 RV 100% 4.9% = ⋅ = 132 8 (e) For this problem, ... angle of 36 .87 ° A L ( 3 3 230 0 ) V L V = = . The internal generated voltage of the machineThe phase voltage of this machine is ⎞ T / V V 3...

Ngày tải lên: 06/08/2014, 11:20

... 565 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 = 277 V. ... 0 ( ) 0. ( 016 565 .3 31 .8 A ) ( ) 0.899 ( 565 .3 31 .8 A ) E A = ° + &  ° j + &  ° E A 509 30 .5 = V °...

Ngày tải lên: 06/08/2014, 11:20

... 5249 = A 3 3 ( ) 13. 2 kV A L T V The power factor is 0.8 lagging, so I A 5249 36 .8 = 7  A ° . The phase voltage is 13. 2 kV / 3 = 7621 V. ... 134 7217 0 ( 0.0187 )( ) 4619 36 .87 A ( 1.716 )( ) 4619 36 .87 A E A = ° + &  ° j + &  ° E A 13, 590 27.6...

Ngày tải lên: 06/08/2014, 11:20

... f 3 n s l3 sysP f f ( ) ( ) ( ) ( ) =  + ( ) ( )  +  290 MW 34 62.21 =  8.529 186. 63 3 f sys f sys = 59 .37 Hz sys 34 ... test Field current, A 32 0 36 5 38 0 475 570 Line voltage, kV 13. 0 13. 8 14.1 15.2 16.0 Extrapolated air-gap voltage, kV 15.4 17.5 18 .3 22.8 27.4...

Ngày tải lên: 06/08/2014, 11:20

... sin sin 13, 230 ™ ™ V sin = = 27.9   38 .6 ° = ° 2 2 E A 2 Therefore, the new armature current is 9,9 23 V  A2 ⎞ 9, 9 23 38 .6 70 ... 0 V ( 1.1 )( 259 36 .87 A ) 38 4 36 .4 V E A = ° j  &  ° =  ° (b) This motor has 6 poles and an electrical frequ...

Ngày tải lên: 06/08/2014, 11:20

... OUT P 100% = ⋅ 746 k = W 100% ⋅ 89 .3% = 3 ( ) 137 6 = 238 3 V . This voltage IN P 835 kW (c) To solve this problem, we will ... new power factor is cos 3. 5 ° = 0.998 leading, and the reactive power supplied by the motor is 3 sin 3 ( 230 0 V )( 21 = = 4.5 A ) sin ( 3. 5 ) 52.2 kVAR °...

Ngày tải lên: 06/08/2014, 11:20

... 33 .6 9.6 A = = =  ° A 0.4 2.0 + + & A S R jX j Therefore the real power consumed by this motor is 3 cos 3 ( ) 480 V ( 3 = = 3. 6 ... V ⎞ 286 0 V = ° E , A m 32 7 18. = 3  V ° I A 95.7 13. 3 = A ° The new terminal voltage is ( ) 3 286 V 495 V T V = =...

Ngày tải lên: 06/08/2014, 11:20

... j  ° E A 1 .33 42 = .6  pu ° The base phase voltage of this motor is 6600 / 3 = 38 10 V, so E A is A E ( ) 1 .33 42.6 =  ( 38 1 ° 0 V ) 5067 = ... 1  .33 2 4.  ° 9 0.6 63 20.2 pu = = = ° + A 0.90 A S R jX j In actual amps, this current is A I ( ) 1404 A ( 0.6 63...

Ngày tải lên: 06/08/2014, 11:20