(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 1 pot

... will be 19 6 16 00 16 20 16 40 16 60 16 80 17 00 17 20 17 40 17 60 17 80 18 00 0 50 10 0 15 0 200 250 300 350 400 450 n m τ ind Induction Motor Torque-Speed Characteristic R2 = 0.0059 ohms R2 = 0 .17 2 ohms ... Ω + - V φ I A, start R 1 jX 1 R F jX F The equivalent impedance of the rotor circuit in parallel with M jX at starting conditions (s = 1. 0) is: ,start 2...

Ngày tải lên: 05/08/2014, 20:22

... = 11 5 V, and the internal generated voltage ,1Ao E for a current of 58 A and a speed of o n = 12 00 r/min is ,1Ao E = 13 4 V. () ,1 2 21 1,2 10 7.4 V 13 4 V 10 50 r/min 13 26 r/min 99 .1 V 11 5 ... () ( ) ( ) 240 V 54.3 A 0.44 216 .1 V ATAA S EVIRR=− + = − Ω= 219 The resulting plot is shown below: 0 20 40 60 80 10 0 12 0 800 850 900 950 10 00 10 50 11 00 1...

Ngày tải lên: 05/08/2014, 20:22

... current'); axis ([0 5 0 15 0]); set(gca,'YTick',[0 10 20 30 40 50 60 70 80 90 10 0 11 0 12 0 13 0 14 0 15 0]') set(gca,'XTick',[0 0.5 1. 0 1. 5 2.0 2.5 3.0 3.5 4.0 4.5 ... efficiency is OUT IN 11 ,19 0 W 10 0% 10 0% 79.4% 14 ,10 0 W P P η =× = × = (e) The no-load A E will be 230 V, so the no-load speed will be () 230 V 18 00 r/min 18...

Ngày tải lên: 05/08/2014, 20:22

... 60 1. 170 2.3 31 1. 282 2.40 60 B jj Zj jj + ==+Ω ++ (a) The input current is 1 11 I 0.5 0.5 FB RjX Z Z = ++ + V ()( )( ) 1 120 0 V I4.0359.0 A 1. 80 2.40 0.5 25. 91 43.83 0.5 1. 170 2.331jj ... voltage of 11 6 V, as shown below. 2 71 ()() 1. 282 2.40 60 1. 185 2.332 1. 282 2.40 60 B jj Zj jj + ==+Ω ++ (a) The input current is 1 11 I 0.5 0.5 FB RjX Z Z...

Ngày tải lên: 05/08/2014, 20:22

... 19 0∠0° V, 1 11 I 0.5 0.5 FB RjX Z Z = ++ + V ()( )( ) 1 190 0 V I 11 .2 27.0 A 1. 40 1. 90 0.5 26.59 9.69 0.5 0.7 41 1.870jjj ∠° ==∠−° ++ ++ + () ()( ) 2 2 AG, 1 0.5 11 .2 A 13 .29 16 67 W FF PIR== ... factor is ()() 915 .00.962 0.9 51 === dpw kkk 276 1. 4 Ω j1.9 Ω + - V = 220∠0° V I 1 R 1 jX 1 s R 2 5.0 j0.5X 2 j0.5X M j1.90 Ω j30 Ω s R −2 5.0 2 jX...

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... ? FIGURE P1 13 The core of Problem 1 17 . t (ms) 12 345678 0 φ (Wb) 0. 010 0.005 –0.005 – 0. 010 FIGURE P1 12 Plot of flux ␾ as a function of time for Problem 1 16 . cha65239_ch 01. qxd 10 /23/2003 ... 3 10 . cha65239_ch03.qxd 10 /30/2003 1: 15 PM Page 226 342 0 0 .1 Open-circuit voltage, V Field current, A 12 00 11 00 10 00 900 800 700 600 500 400 300 200 10 0 0 0.2 0.3 0.4...

Ngày tải lên: 05/08/2014, 20:22

... Subfractional Horsepower Electric Motors. New York: McGraw- Hill, 19 70. 4. Werninck, E. H. (ed.). Electric Motor Handbook. London: McGraw-Hill, 19 78. cha65239_ch10.qxd 11 /14 /03 12 : 41 PM Page 680 DC MOTORS ... Kingsley, Jr. Electric Machinery. New York: McGraw-Hill, 19 52. 2. National Electrical Manufacturers Association. Motors and Generators, Publication No. MG1- 19 93. W...

Ngày tải lên: 05/08/2014, 20:22

... 276 .14 0.966 278 1 279.74 1. 033 2 81. 48 1. 067 282.94 1. 1 284.28 1. 133 285.48 1. 167 286.54 1. 2 287.3 1. 233 287.86 1. 267 288.36 1. 3 288.82 1. 333 289.2 1. 367 289.375 1. 4 ... 0.233 10 4.4 0.267 11 8.86 0.3 13 2.86 0.333 14 6.46 0.367 15 9.78 0.4 17 2 .18 0.433 18 3.98 0.467 19 5.04 0.5 205 .18 0.533 214 .52 0.567...

Ngày tải lên: 05/08/2014, 20:22

... Lagging: ()() EQ 11 5 0 V 0 .14 0 0.532 8.7 36 87 A PS S Zj. ′ =+ =∠°+ + Ω ∠− °VV I 11 8.8 1. 4 V P ′ =∠° V 11 8.8 -11 5 VR 10 0% 3.3% 11 5 =×= (2) 1. 0 PF: ()() EQ 11 5 0 V 0 .14 0 0.532 8.7 0 ... is 10 6.2 10 0 VR 10 0% 10 0% 6.2% 10 0 PS S VaV aV −− =×= ×= The input power to this transformer is ()() () IN cos 10 6.2 V 11 .1 A cos 2.6 41. 0 PP PVI θ =...

Ngày tải lên: 05/08/2014, 20:22

... its ratings be? 54 11 30 V 11 30 1. 00 A SC EQ SC V Z I == = Ω ()() 11 260 W cos cos 76.7 11 30 V 1. 00 A SC SC SC P VI θ −− == =° 11 30 76.7 260 11 00 EQ EQ EQ ZRjX j=+ = ... () 2 2 base base base 15 kV 1. 125 200 MVA V Z S == =Ω so ( ) ( ) EQ 0. 012 1. 125 0. 013 5 R =Ω=Ω ()( ) EQ 0.05 1. 125 0.0563 X =Ω=Ω 49 6250 VA 52 .1 A 12 0 V S S...

Ngày tải lên: 05/08/2014, 20:22