Electric Machinery Fundamentals (Solutions Manual) Part 4 ppsx

... 47 = 0 kVA 565 A = 3 3 ( ) 48 0 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 ⎞ = 48 0 ... short-circuit current may be read from the SCC. It is 0.5 34 A 3 . The open circuit voltage at 0.532 A is 880 V 4 , so the open-circuit phase voltage (= 508 V. The appr...

Ngày tải lên: 06/08/2014, 11:20

... 1 34 7217 0 ( 0.0187 )( ) 46 19 36.87 A ( 1.716 )( ) 46 19 36.87 A E A = ° + &  ° ... S I I = = 12 = 0 MVA 5 249 = A 3 3 ( ) 13.2 kV A L T V The power factor is 0.8 lagging, so I A 5 249 36.8 = 7  A ° . The phase ... 5 249 36.87 A ) E A j = °...

Ngày tải lên: 06/08/2014, 11:20

... V 43 0 13.5 44 0 0 V 34. 2 16.5 A = = = ° A I 0.22 3.0 + + A S R jX j The real power supplied by this machine is 3 cos 3 ( ) 44 0 V ... ( 3 = = 4. 2 A ) cos ( 16.5 ) 43 .  3 kW ° = ⎞ A P V I ⎝ The reactive power supplied by this machine is 3 sin 3 ( ) 44 0 V ( 3 = = 4. 2 A...

Ngày tải lên: 06/08/2014, 11:20

... 12 & 40 = V 3 3 ( ) 1 240 V 2 148 V T V V ⎞ = = = (b) Armature current is I A 186 30 =  A ° , and the phase voltage is ⎞ V 1 240 0 ... CU 3 A A P I 3 R 251 A 0.15 28 .4 kW F& P = W 24 kW 110 1 14 ⎜ OUT P 100% = ⋅ 5 54 kW = 100% ⋅ 90.5% = IN P...

Ngày tải lên: 06/08/2014, 11:20

... the armature current is A E V  I ⎞ 12, 040 90 V ° - 7967 0 V ° 11 94 40 .6 A = = = ° A 1.5 12.0 + + & A S R ... = ⎞ ⎝ A Q V I 4 A ) sin ( 0 40 .6 ) 18 ° .6  MVA ° R =  The generator is actually consuming reactive power at this time. 5-8. A 48 0-V, 100...

Ngày tải lên: 06/08/2014, 11:20

... nl1 f sys f 2.5 MW/Hz 61 Hz 59.23 Hz 4. 425 MW The total power supplied by the generators at this condition is 14. 16 MW. (c) To get each of the generators to supply ... generators is 41 60 V / 3 = 240 2 V, and since the generators are Y-connected, their rated current is S I I = = 62 = 50 kVA 867 = A 3 3 ( ) 4...

Ngày tải lên: 06/08/2014, 11:20

... generator is 12.2 kV / 3 = 7 044 V. The base impedance of this generator is 2 ( ) 2 Z base 3 V ⎞ ,base 3 7 044 V 7 .44 = = = & Therefore, ... sin = = 27.9   33 .4 ° = ° 2 2 E A 2 Therefore, the new armature current is 11, 246 V  A2 ⎞ 11, 246 33 .4 7 04 °...

Ngày tải lên: 06/08/2014, 11:20

... P 298 .4 kW = = 44 9 A = L 3 PF 3 ( ) 48 0 V ( 0.8 ) T V Because the motor is  -connected, the corresponding phase current is 44 9 / 3 259 ... 10,5 84 V 146 149 The resulting plot is shown below: 148  A2 ⎞ 10, 5 8...

Ngày tải lên: 06/08/2014, 11:20

... 3 19 R 8 A 0 .4 47 .0 kW Therefore, a better estimate of the input power at full load is ( )( ) IN P 1000 hp 746 W/hp 24 kW 18 kW 47 = + kW 83 + ... ° E A 42 9 24 = .9  V ° So E A = 42 9 V at rated conditions. The resulting plot is shown below 260 Synchronous Motor V-Curve 250...

Ngày tải lên: 06/08/2014, 11:20

... A 40 0 A = ° . =  E V  I I A A ⎞ A S A R jX 120 0 V ( ) 0.8 ( 40 0 A ) E A = ° j  & ° E A 1 24 14. 9 = ... reactance of 2.0 & per phase and an armature resistance of 0 .4 & per phase. If E A =46 0 -8 ° V and ⎞ V = 48 0 0 ° V, is this machine a motor or a gener...

Ngày tải lên: 06/08/2014, 11:20