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

(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 2 Part 1 pptx

... 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 00050 10 0 15 0200250300350400450nmτindInduction Motor Torque-Speed CharacteristicR2 = 0.0059 ohmsR2 = 0 .17 2 ohms ... Ω+-VφIA,startR 1 jX 1 RFjXF The equivalent impedance of the rotor circuit in parallel with MjX at starting conditions (s = 1. 0) is: ,start2 11 0 .16 7 0. 415 0.448 68 .1 11 1 130 0 .17 2 0.42FMZjjX ... )2ind223 262 V 0 .17 2 /1. 962 18 8.5 rad/s 0.3 21 0 .17 2 /1. 962 0. 418 0.420τΩ= +Ω + + ()( )()( )( )2ind223262 V 0.0877 18 8.5 rad/s 0.3 21 0.0877 0. 418 0.420τ= +++ ind 11 0 N m, opposite...

(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 2 Part 2 ppt

... = 11 5 V, and the internal generated voltage ,1AoE for a current of 58 A and a speed of on = 12 00 r/min is ,1AoE = 13 4 V. () ,1 2 21 1,2 10 7.4 V 13 4 V 10 50 r/min 13 26 r/min99 .1 V 11 5 ... ()()()240 V 54.3 A 0.44 216 .1 VATAA SEVIRR=− + = − Ω= 219 The resulting plot is shown below: 0 20 40 60 80 10 0 12 0800850900950 10 00 10 50 11 00 11 50 12 00 12 50τind (N-m)nm (r/min)Shunt ... (a) If AV = 12 0 V, then AE = 12 0 V, and () 12 0 V 12 00 r/min 5 31 r/min2 71 Vn == (a) If AV = 18 0 V, then AE = 18 0 V, and () 18 0 V 12 00 r/min 797 r/min2 71 Vn == ...

(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 2 Part 3 docx

... 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 OUTIN 11 ,19 0 W 10 0% 10 0% 79.4% 14 ,10 0 WPPη=× = × = (e) The no-load AE will be 230 V, so the no-load speed will be ()230 V 18 00 r/min 18 73 r/min2 21 VAoAoEnnE== ... motor with adjR = 12 0 Ω, which is 11 25 r/min. The speed regulation of this motor is thus nl flfl 11 25 r/min - 915 r/minSR 10 0% 10 0% 23.0% 915 r/minnnn−=×= ×= 9 -18 . The motor is reconnected...

(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 2 Part 4 pptx

... 60 1. 170 2.3 31 1. 282 2.40 60BjjZjjj+==+Ω++ (a) The input current is 1 11 I0.5 0.5FBRjX Z Z=++ +V ()( )( ) 1 120 0 VI4.0359.0 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 60BjjZjjj+==+Ω++ (a) The input current is 1 11 I0.5 0.5FBRjX Z Z=++ +V ()( )( ) 1 120 0 ... 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...

(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 2 Part 5 ppsx

... 19 0∠0° V, 1 11 I0.5 0.5FBRjX Z Z=++ +V ()( )( ) 1 190 0 VI 11 .2 27.0 A 1. 40 1. 90 0.5 26.59 9.69 0.5 0.7 41 1.870jjj∠°==∠−°++ ++ + ()()( )22AG, 1 0.5 11 .2 A 13 .29 16 67 WFFPIR== ... factor is ()() 915 .00.962 0.9 51 ===dpwkkk 276 1. 4 Ω j1.9 Ω+-V = 220∠0° VI 1 R 1 jX 1 sR25.0j0.5X2j0.5XMj1.90 Ωj30 ΩsR−25.02jX2j1.90 Ωj0.5XMj100 Ω{{{{ForwardReverse0.5ZB0.5ZF ... jX+=++ ()()30 1. 90 10 026.59 9.69 30 1. 90 10 0FjjZjjj+==+Ω++ ()()()2222/2/2MBMR s jX jXZR s jX jX =−+ + ()()0.769 1. 90 10 00.7 41 1.870 0.769 1. 90 10 0BjjZjjj+==+Ω++...

(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 2 Part 6 ppt

... ?FIGURE P1 13 The core of Problem 1 17 .t (ms) 12 3456780φ (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 2263420 0 .1 Open-circuit voltage, VField current, A 12 00 11 00 10 00900800700600500400300200 10 000.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1 .1 ... A 12 00 11 00 10 00900800700600500400300200 10 000.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1 .1 1.2 1. 3 1. 4 1. 5(a)0Armature current, AField current, A 16 00 14 00 12 00 10 0080060040020000.2 0.4 0.6 0.8 1 1.2 1. 4(b)Open Circuit CharacteristicShort...

(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 2 Part 7 ppt

... 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 680DC MOTORS ... Kingsley, Jr. Electric Machinery. New York: McGraw-Hill, 19 52.2. National Electrical Manufacturers Association. Motors and Generators, Publication No. MG1- 19 93. Washington, D.C.: NEMA, 19 93.3. ... magnetization curve for thismotor, taken at a speed of 18 00 r/min, is given in tabular form below:EA, V 8.5 15 0 18 0 215 226 242IF, A 0.00 0.80 1. 00 1. 28 1. 44 2.88(a) What is the speed of this motor...

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

... 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 0.567 223.06 0.6 2 31. 2 0.633 238 0.667 244 .14 ... ()()()34left TOT 12 3490 .1 77.30.0033 Wb 0.0 019 3 Wb90 .1 108.3 90 .1 77.3φφ++== =+++ + + +RRRR RR ()()() 12 right TOT 12 3490 .1 108.30.0033 Wb 0.00229 Wb90 .1 108.3 90 .1 77.3φφ++==...

(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 2 ppt

... Lagging: ()()EQ 11 5 0 V 0 .14 0 0.532 8.7 36 87 APS SZj.′=+ =∠°+ + Ω ∠− °VV I 11 8.8 1. 4 VP′=∠°V 11 8.8 -11 5VR 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 0VR 10 0% 10 0% 6.2% 10 0PSSVaVaV−−=×= ×= The input power to this transformer is ()() ()IN cos 10 6.2 V 11 .1 A cos 2.6 41. 0PPPVIθ== ()()IN 10 6.2 V 11 .1 A cos ... would be (1. 2 T)(0.04 m)(0.04 m) 0.0 019 2 WbBAφ== = (c) The total reluctance of the core is: 28 11 6.3 -11 5VR 10 0% 1. 1% 11 5=×= (3) 0.8 PF Leading: ()()EQ 11 5 0 V 0 .14 0 0.532...

(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 3 ppsx

... its ratings be? 54 11 30 V 11 30 1. 00 ASCEQSCVZI== = Ω ()() 11 260 Wcos cos 76.7 11 30 V 1. 00 ASCSC SCPVIθ−− == =° 11 30 76.7 260 11 00 EQ EQ EQZRjX j=+ = ... ()22basebasebase 15 kV 1. 125 200 MVAVZS== =Ω so ()()EQ0. 012 1. 125 0. 013 5 R =Ω=Ω ()( )EQ0.05 1. 125 0.0563 X =Ω=Ω 49 6250 VA52 .1 A 12 0 VSSSIV== = 2 -16 . A 5000-VA 480 /12 0-V ... 2.50 :1 Y-∆ 19 .9 kV 13 .8 kV 200 kVA 1. 44 :1 ∆-Y 34.5 kV 7.97 kV 200 kVA 4.33 :1 ∆-∆ 34.5 kV 13 .8 kV 200 kVA 2.50 :1 open-∆ 34.5 kV 13 .8 kV 346 kVA 2.50 :1 open-Y—open-∆ 19 .9 kV 13 .8...

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