Electric Machinery Fundamentals (Solutions Manual) Part 9 ppt
... 96 .2 A = ,m ,mA L ( ) 3 PF 3 ( ) 480 V 1.0 T V so I A,m 96 .2 0 A = ° . This machine (d) E A 163 3 3 ( ) 277 V ( ) 4 29 ... ) 13.3 = 0 .9 ° 73 = leading , since a current angle of ) ( ) ° = motor 3 ⎞ A sin Q V I ⎝ 3 286 V 95 .7 A sin 13.3 18 .9 k...
Ngày tải lên: 06/08/2014, 11:20
... the armature current is A E V I ⎞ 12, 040 90 V ° - 796 7 0 V ° 1 194 40.6 A = = = ° A 1.5 12.0 + + & A S R ... & ° j + & ° E A 1 393 11.4 = V ° The voltage regulation would be: 1 393 1328 RV 100% 4 .9% = ⋅ = 1328 (e) For...
Ngày tải lên: 06/08/2014, 11:20
... j = ° + & ° E A 92 64 19. 9 = V ° The resulting voltage regulation is 92 64 6351 RV 100% 45 .9% = ⋅ = 6351 Because voltage, ... 290 MW 34 62.21 = 8.5 29 186.63 3 f sys f sys = 59. 37 Hz sys 34 62.21 sys 34 62.21 f f sys f Each generator will suppl...
Ngày tải lên: 06/08/2014, 11:20
Electric Machinery Fundamentals (Solutions Manual) Part 10 ppt
... 2 4. ° 9 0.663 20.2 pu = = = ° + A 0 .90 A S R jX j In actual amps, this current is A I ( ) 1404 A ( 0.663 20.2 ) 93 1 20 = ... armature resistance) is ( ) ( ) sin 1 1 S X P sin 0 .90 0. = = 623 24 .9 = ° ™ 3 ⎞ A V E ( ) 1.0 ( 1. ) 33 At rated volta...
Ngày tải lên: 06/08/2014, 11:20
Electric Machinery Fundamentals (Solutions Manual) Part 1 pdf
... Q = -4810; DE = 5567; S = 1000; % Get points for stator current limit theta = -95 :1 :95 ; % Angle in degrees rad = theta * pi / 180; % Angle in radians s_curve = S ... 1377 V 186 A = = A + + 0.15 1.1 + 6.667 + 30 1.82 ° A S R jX Z j 9 & Therefore, the magnitude of the phase voltage is ( ) 186 A = = ⎞ A V ....
Ngày tải lên: 06/08/2014, 11:20
Electric Machinery Fundamentals (Solutions Manual) Part 3 docx
... /Hz ( ) sys 90 00 kW + 0.15 MW /Hz sys f f 0.25 MW/Hz f sys 6050 kW 90 00 kW = + 100 kW f sys 14, 95 0 kW 59. 8 Hz = = 0.25 MW/Hz ... 61.5 sys 1 .96 1 60.5 f f sys f + + 7 MW 91 .5 1. = 5.137 f sys = 306.2 f sys = 59. 61 Hz 5 sys 103.07 1.676 sys 118.64...
Ngày tải lên: 06/08/2014, 11:20
Electric Machinery Fundamentals (Solutions Manual) Part 4 ppsx
... generator is 3 ⎞ V E P sin A 3 ( ) 277 V ( 5 09 V ) ( ) sin 471 kW sin ™ ™ ™ = = = S X G 0. 899 & The power supplied as a function of the torque ... ( 016 565.3 31.8 A ) ( ) 0. 899 ( 565.3 31.8 A ) E A = ° + & ° j + & ° E A 5 09 30.5 = V ° (b) Thi...
Ngày tải lên: 06/08/2014, 11:20
Electric Machinery Fundamentals (Solutions Manual) Part 5 ppsx
... I A A ⎞ A S A R jX 7621 0 ( ) 0 .9 ( 52 49 36.87 A ) E A j = ° + & ° E A 11,120 19. 9 V = ° The resulting voltage regulation ... flux_ratio = 0 .90 :0.01:1.00; % Flux ratio Ear = 695 ; % Ea at full flux dr = 38.4 * pi/180; % Torque ang at full flux Vp = 277; % Phase voltage...
Ngày tải lên: 06/08/2014, 11:20
Electric Machinery Fundamentals (Solutions Manual) Part 7 pdf
... 9, 92 3 V , and sin 1 1 E 1 A sin sin 13, 230 ™ ™ V sin = = 27 .9 38.6 ° = ° 2 2 E A 2 Therefore, the new armature current is 9, 923 ... 298 .4 kW = = 4 49 A = L 3 PF 3 ( ) 480 V ( 0.8 ) T V Because the motor is -connected, the corresponding phase current is 4 49 / 3 2 59...
Ngày tải lên: 06/08/2014, 11:20
Electric Machinery Fundamentals (Solutions Manual) Part 8 pps
... ( 141.5 31.8 A ) E A = ° j & ° E A 4 29 24 = .9 V ° So E A = 4 29 V at rated conditions. The resulting plot is shown below 260 ... before, but the phase angle has become positive. The new power factor is cos 3.5 ° = 0 .99 8 leading, and the reactive power supplied by the motor is 3 sin 3 ( 2300 V...
Ngày tải lên: 06/08/2014, 11:20