Instructors Manual Electric Machinery Fundamentals 4th Edition Stephen J Chapman

... FOR ELECTRIC MACHINERY FUNDAMENTALS 4/E 301 iii PREFACE TO THE INSTRUCTOR This Instructor’s Manual is intended to accompany the fourth edition of Electric Machinery Fundamentals To make this manual ... Thank you Stephen J Chapman Melbourne, Australia January 4, 2004 Stephen J Chapman 278 Orrong Road Caulfield North, VIC 3161 Australia Phone +61-3-9527-9372 vi Cha...

Ngày tải lên: 21/12/2016, 10:34

... 0.933 0.966 1. 033 1. 067 1. 1 1. 133 1. 167 1. 2 1. 233 1. 267 1. 3 1. 333 1. 367 1. 4 1. 433 1. 466 1. 5 10 4.4 11 8.86 13 2.86 14 6.46 15 9.78 17 2 .18 18 3.98 19 5.04 205 .18 214 .52 223.06 2 31. 2 238 244 .14 249.74 255.08 ... R3 + R4 ) R1 + R2 + R3 + R4 ( R1 + R2 ) φ right = φTOT = R1 + R2 + R3 + R4 (90 .1 + 77.3) 90 .1 + 10 8.3 + 90 .1 + 77.3 φTOT = (90 .1 +...

Ngày tải lên: 05/08/2014, 20:22

... 11 6.3 2. 28° V 27 VR = (3) 11 6.3 -11 5 × 10 0% = 1. 1% 11 5 0.8 PF Leading: VP ′ = VS + Z EQ IS = 11 5∠0° V + ( 0 .14 0 + j 0.5 32 Ω )(8.7∠36.87° A ) VP ′ = 11 3.3 2. 24° V 11 3.3 -11 5 VR = × 10 0% = 1. 5% 11 5 ... 11 5∠0° V + ( 0 .14 0 + j 0.5 32 Ω )(8.7∠ − 36.87° A ) VP ′ = 11 8.8 1. 4° V 11 8.8 -11 5 VR = × 10 0% = 3.3% 11 5 (2) 1. 0 PF: VP ′ = VS + Z EQ I S = 11 5∠0...

Ngày tải lên: 05/08/2014, 20:22

... 19 .9 kV 7.97 kV 200 kVA 2.50 :1 19.9 kV 13 .8 kV 200 kVA 1. 44 :1 Y-∆ 34 .5 kV 7.97 kV 200 kVA 4 .33 :1 ∆-Y 34 .5 kV 13 .8 kV 200 kVA 2.50 :1 ∆-∆ 34 .5 kV 13 .8 kV 34 6 kVA 2.50 :1 open-∆ 19 .9 kV 13 .8 kV 34 6 ... (low-voltage) side is Vbase (15 kV ) = 1. 125 Ω = S base 200 MVA Z base = so REQ = ( 0. 012 ) (1. 125 Ω ) = 0. 0 13 5 Ω X EQ = ( 0.05) (1. 125 Ω ) = 0.05 63 Ω...

Ngày tải lên: 05/08/2014, 20:22

... Load = (0 .47 65∠ − 41 . 6°) (1. 513 + j1 .13 4 ) = 0.9 01 − 4. 7° VLoad = VLoad,puVbase3 = (0.9 01) (48 0 V ) = 43 2 V The power supplied to the load is PLoad,pu = I RLoad = ( 0 .47 65) (1. 513 ) = 0. 344 PLoad ... j0. 040 + 0.00723 + j0. 048 2 + 0. 040 + j 0 .17 0 + 1. 513 + j1 .13 4 Z EQ = 1. 5702 + j1.3922 = 2.099∠ 41 . 6° The resulting current is I= 1 0° = 0 .47 65∠ − 41 . 6° 2....

Ngày tải lên: 05/08/2014, 20:22

... Phase − T / 12 T / 12 3T / 12 5T / 12 7T / 12 9T / 12 11 T / 12 T / 12 3T / 12 5T / 12 7T / 12 9T / 12 11 T / 12 T / 12 c a a b b c c b b c c a a b 88 Conducting SCR (Positive) SCR3 SCR1 SCR1 SCR2 SCR2 ... from R1, the time at which iD(t) reaches IH is t2 = − R2C ln I H R2 (0.00 05 A ) ( 15 00 Ω ) = 5. 5 ms = − ( 0.0 0 15 ) ln 30 V VBO Therefore, the period of the relaxation...

Ngày tải lên: 05/08/2014, 20:22

... below: 11 2 0 .15 Ω j1 .1 Ω IA + EA + - Vφ Z 6. 667 ∠30° - The magnitude of the phase current flowing in this generator is IA = EA 13 77 V 13 77 V = = = 18 6 A RA + jX S + Z 0 .15 + j1 .1 + 6. 667 ∠30° 1. 829 ... A = 12 40∠0° + ( 0 .15 Ω ) (18 6 − 30° A ) + j (1. 1 Ω) (18 6 − 30° A ) E A = 13 77 6. 8° V The resulting phasor diagram is shown below (not to scale): E = 13 77 6. 8...

Ngày tải lên: 05/08/2014, 20:22

... MW = (1. 5) 61. 0 − f sys + (1. 676 ) 61. 5 − f sys + (1. 9 61) 60.5 − f sys ) MW = 91. 5 − 1. 5f sys + 10 3. 07 − 1. 676 f sys + 11 8.64 − 1. 961f sys 5 .13 7 fsys = 306.2 f sys = 59. 61 Hz The power supplied by ... SD B 3.0 +1 +1 100 10 0 f nl,C 60.5 Hz f fl,C = = = 58. 97 Hz SDC 2.6 +1 +1 100 10 0 and the slopes of the power-frequency curves are: MW S PA = = 1. 5 MW/Hz Hz...

Ngày tải lên: 05/08/2014, 20:22

... = 1. 15 E A1 = 1. 15 ( 384 V ) = 4 41. 6 V 15 0 δ = sin 1 E A1 384 V sin 1 = sin 1 sin ( −36.4° ) = − 31. 1° E A2 4 41. 6 V The new armature current is I A2 = Vφ − E A2 jX S = 480 ∠0° V − 4 41. 6∠ − 31. 1° ... = sin 1 E A1 13 ,230 V sin δ = sin 1 sin 27.9° = 31. 3° E A2 11 ,907 V Therefore, the new armature current is IA = E A2 − Vφ jX S = 11 ,907∠ 31. 3° − 7044∠0° = 84 8∠ − 26 .8...

Ngày tải lên: 05/08/2014, 20:22

... PIN = (c) Pout POUT 10 0 kW = = 11 0 kW 0. 91 η The mechanical speed is nm = 15 00 r/min (d) The armature current is IA = IL = P 11 0 kW = = 15 6 A VT PF ( 480 V )(0.85) I A = 15 6∠ 31. 8° A Therefore, ... English units is τ load = POUT ωm = ( 210 00 hp)(746 W/hp) (12 00 r/min ) 2π rad 60 s τ load = 6 -13 = 12 4,700 N ⋅ m 1r 5252 P 5252 ( 210 00 hp ) = = 91 , 91 0 lb ⋅ ft nm (12 00 r/...

Ngày tải lên: 05/08/2014, 20:22

... = 0. 211 Ω and X = 0. 317 Ω Therefore, X M = 5.455 Ω − 0. 211 Ω = 5.244 Ω The resulting equivalent circuit is shown below: 19 2 IA R1 + Vφ jX1 0 .10 5 Ω jX2 j0. 211 Ω j0. 317 Ω j5.244 Ω R2 0.0 71 Ω jXM ... M ( R1 + jX ) ( j15 Ω )( 0.20 Ω + j 0. 41 Ω ) = = 0 .18 95 + j0.4 016 Ω = 0.444∠64.7° Ω R1 + j ( X + X M ) 0.20 Ω + j ( 0. 41 Ω + 15 Ω ) VTH = jX M ( j15 Ω ) Vφ = (12 0∠0° V ) = 11 6...

Ngày tải lên: 05/08/2014, 20:22

... (18 8.5 rad/s) (0. 3 21 + 0 .1 72 Ω /1. 9 62) 2 + (0. 418 + 0. 420 )2 ( 26 2 V ) (0.0877 ) τ ind = (18 8.5 rad/s) (0. 3 21 + 0.0877 )2 + ( 0. 418 + 0. 420 )2 τ ind = 11 0 N ⋅ m, opposite the direction of motion 20 3 ... Motor Torque-Speed Characteristic 450 R2 = 0.0059 ohms R2 = 0 .1 72 ohms 400 350 300 τ ind 25 0 20 0 15 0 10 0 50 16 00 1 620 16 40 16 60 16 80 17 00 n 1...

Ngày tải lên: 05/08/2014, 20:22

... know that % % Ea2 K' phi2 n2 Eao2 n2 % - = = -% Ea1 K' phi1 n1 Eao1 n1 % % Ea2 Eao1 % ==> n2 = - n1 % Ea1 Eao2 % % where Ea0 is the internal generated voltage at 120 0 r/min % for ... V A = 24 0 V, then E A = 24 0 V, and n= 9-9 120 V 27 1 V If V A = 180 V, then E A = 180 V, and n= (a) no If V A = 120 V, then E A = 120 V, and n= (a) EA E Ao 24 0 V 27 1 V ( 120 0 r/min ) = 1063 r/min ... g...

Ngày tải lên: 05/08/2014, 20:22

... Rstart,1 = R1 + R2 + R3 = 1. 628 Ω Rstart ,2 = R2 + R3 = 0.561 Ω Rstart ,3 = R3 = 0. 134 Ω Therefore, the starting resistances are R1 = 1.067 Ω R2 = 0. 427 Ω R3 = 0. 134 Ω 24 2 9 -21 A 15-hp 120 -V 1800 r/min ... rises to E A = 24 0 V − ( 0 .28 4 Ω )(54 A ) = 22 4.7 V If the resistance is cut out when E A reaches 22 8,6 V, the resulting current is IA = 24 0 V − 22 4.7 V = 1 02 A < 13...

Ngày tải lên: 05/08/2014, 20:22

... I1 = 120 ∠ 0° V = 5 .23 ∠ − 44 .2 A 1.80 + j 2. 40 ) + 0.5 ( 28 .15 + j 24 .87 ) + 0.5 (1.185 + j 2. 3 32 ) ( PIN = VI cos θ = ( 120 V )(5 .23 A ) cos 44 .2 = 45 0 W (b) The air-gap power is PAG,F = I 12 ( ... ) = 28 .15 + j 24 .87 Ω 50 + j 2. 40 + j 60 R2 / ( s ) jX ( jX M ) R2 / ( − s ) + jX + jX M 27 0 { I1 { 10-1 Reverse ZB = (a) (1 .28 2 + j 2. 40 )( j 60) = 1.185 + j 2...

Ngày tải lên: 05/08/2014, 20:22