Electric Machinery Fundamentals (Solutions Manual) Part 1 pdf
... 10 00 kVA 0.8 800 kW = = 2 ( = = ) 2 ( ) & = 11 1 CU 3 A A P I 3 R 2 51 A 0 .15 28.4 kW F& P = W 24 kW 11 0 11 4 ... A S A R jX 12 40 0 ( 0 .1 )( 5 18 6 30 A ) ( 1. 1 )( 18 6 30 A ) E A = ° + & ° j + & ° E A 13 77 6.8 V...
Ngày tải lên: 06/08/2014, 11:20
... 2 1 1 .15 1. 15 E E 38 A A 4 V 4 41. 6 V E A is increased by 15 %, the new torque angle can 15 0 14 7 With a 20% decrease, E A2 = 10 ,584 ... 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 V...
Ngày tải lên: 06/08/2014, 11:20
... to a frequency of 13 28 0 ( 0 .15 )( ) 2 51 0 A ( 1. 1 )( ) 2 51 0 A E A = ° + & ° j + & ° E A 13 93 11 .4 = V ° The ... A ⎞ A S A R jX 13 28 0 ( 0 .15 )( ) 2 51 0 A ( 1. 1 )( ) 2 51 0 A E A = ° + & ° j + & ° E A 13 93 11 .4 =...
Ngày tải lên: 06/08/2014, 11:20
Electric Machinery Fundamentals (Solutions Manual) Part 3 docx
... LOAD 1. 5 61. 0 sys 1. 676 61. 5 P f sys 1. 9 61 60.5 f sys f = + + LOAD 91. 5 1. 5 sys 10 3.07 1. 6 P f 76 sys 11 8.64 1. 96 f = 5 .13 7 ... MW 1. 5 61. 0 sys 1. 676 61. 5 sys 1. 9 61 60.5 f f sys f + + 7 MW 91. 5 1. = 5 .13 7 f sys = 306.2 f sys = 59....
Ngày tải lên: 06/08/2014, 11:20
Electric Machinery Fundamentals (Solutions Manual) Part 4 ppsx
... interp1(p52_occ(: ,1) ,p52_occ(:,2),0.534) vt = 880.400 12 9 generators would have more margin than this. 13 1 E A 2825 10 .9 = V ° Problems 5 -11 to 5- 21 refer ... if_scc = p52(: ,1) ; ia_scc = p52(:,2); % Calculate Xs if = 0.0 01: 0.02 :1; % Current steps vt = interp1(if_occ,vt_occ,If); % Terminal voltage ia = interp...
Ngày tải lên: 06/08/2014, 11:20
Electric Machinery Fundamentals (Solutions Manual) Part 5 ppsx
... + & ° E A 11 ,12 0 19 .9 V = ° The resulting voltage regulation is 11 ,12 0 76 21 RV 10 0% 45.9% = ⋅ = 76 21 (b) If the generator ... 13 4 7 217 0 ( 0. 018 7 )( ) 4 619 36.87 A ( 1. 716 )( ) 4 619 36.87 A E A = ° + & ° j + & ° E A 13 , 590...
Ngày tải lên: 06/08/2014, 11:20
Electric Machinery Fundamentals (Solutions Manual) Part 6 ppt
... = TOT 1 2 P P 29 P 9 kW 12 0 kW 419 kW = + = + = TOT 1 2 Q Q 14 Q 5 kVAR 11 5 kVAR 260 kVAR The overall power factor is 14 1 14 5 f sys ... 320 365 380 475 570 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 The short-circu...
Ngày tải lên: 06/08/2014, 11:20
Electric Machinery Fundamentals (Solutions Manual) Part 8 pps
... will be 2450 V / 3 = 14 15 V. sin E 1 A sin sin 13 76 V ™ ™ sin ( ) 1 1 = = 25. 3 24.6 ° = ° 2 1 E A2 14 15 V Therefore, the new armature ... ) 6 e f m n P = = 12 0 12 0 15 Hz = A speed of 10 00 r/min corresponds to a frequency of ( ) 10 00 r/min e f m n P = = 12 0 12 0...
Ngày tải lên: 06/08/2014, 11:20
Electric Machinery Fundamentals (Solutions Manual) Part 9 ppt
... current is E A 2 14 7 V ( ) 2 = = .7 A 3.20 A = 2 1 F F I I E 1 A 12 4 V (c) The new torque angle ™ of this machine is 12 .5 ° . 6-8. A synchronous ... note that be P X S = (15 /50) (1. 5 & ) = 0.45 & . The maximum power that the motor could supply would 3 3 ( ) 83 .1 V ( 12 9 V )...
Ngày tải lên: 06/08/2014, 11:20
Electric Machinery Fundamentals (Solutions Manual) Part 10 ppt
... ) 12 00 r/min 1 min 2 rad 60 s 1 r ⎮ load 5252 P 5252 ( ) 210 00 hp ( ) 12 00 r/min 91, 910 lb ft = = = ⊕ m n 6 -13 . A 440-V ... rated PF S 200 kVA 0.8 16 0 kW ⎮ OUT P = = 16 0 kW 19 1 N = m ⊕ load ⎤ m ( ) 8000 r/min 1 min 2 rad 60 s 1 r (c)...
Ngày tải lên: 06/08/2014, 11:20