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

... 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

... ()() EQ 11 5 0 V 0 .14 0 0.5 32 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 =ì= (2) 1. 0 PF: ()() EQ 11 5 0 V 0 .14 0 0.5 32 8.7 0 A PS ... (1. 2 T)(0.04 m)(0.04 m) 0.0 019 2 WbBA φ == = (c) The total reluctance of the core is: 28 11 6.3 -11 5 VR 10 0% 1. 1% 11 5 =ì= (3) 0.8 PF Leading:...

Ngày tải lên: 05/08/2014, 20:22

... Y-Y 19 .9 kV 7.97 kV 200 kVA 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 34 6 kVA 2.50 :1 ... ,SC SC 16 00 VVV φ == , ,SC SC / 3 1. 1547 AII φ == , and ,SC SC / 3 3 83 WPP φ == . , , 16 00 V 13 85 1. 155 A SC EQ SC V Z I φ φ == = Ω ()( ) , 1...

Ngày tải lên: 05/08/2014, 20:22

... ()() ()() EQ 1. 513 1. 1 34 3.36 0. 010 0. 040 0.00723 0. 048 2 0. 040 0 .17 0 1. 513 1. 1 34 3.36 jj Zj j j jj +− =+ + + ++ + ++− EQ 0. 010 0. 040 0.00788 0.0525 0. 040 0 .17 0 (2.358 0 .10 9)Zj j j j=+++++++ EQ 2. 41 5 ... Load 0 .47 65 41 . 6 1. 513 1. 1 34 0.9 01 4. 7Zj== ∠−°+ =∠−°VI ()( ) Load Load,pu base3 0.9 01 48 0 V 43 2 VVVV== = The power supplied to...

Ngày tải lên: 05/08/2014, 20:22

... 12 /T− 12 /T c b SCR 3 SCR 5 SCR 5 -30 ° 12 /T 12 /3T a b SCR 1 SCR 5 SCR 1 30° 12 /3T 12 /5T a c SCR 1 SCR 6 SCR 6 90 ° 12 /5T 12 /7T b c SCR 2 SCR 6 SCR 2 15 0° 12 /7T ... SCR 6 SCR 2 15 0° 12 /7T 12 /9T b a SCR 2 SCR 4 SCR 4 210 ° 12 /9T 12 /11 T c a SCR 3 SCR 4 SCR 3 270 ° 12 /11 T 12 /T c b SCR 3 SCR 5 SCR 5...

Ngày tải lên: 05/08/2014, 20:22

... this generator is 13 77 V 13 77 V 18 6 A 0 .15 1. 1 6. 667 30 1. 829 A A AS E I RjXZ j == == ++ ++ ∠° Ω Therefore, the magnitude of the phase voltage is ()( ) 18 6 A 6. 667 12 40 V A VIZ φ == Ω= ... ( ) ( ) ( ) ( ) 12 40 0 0 .15 18 6 30 A 1. 1 18 6 30 A A j=∠°+ Ω∠−°+ Ω∠−°E 13 77 6. 8 V A =∠°E The resulting phasor diagram is shown below (not to scale): I...

Ngày tải lên: 05/08/2014, 20:22

... ( ) () ( ) () ( ) () sys sys sys 7 MW 1. 5 61. 0 1. 676 61. 5 1. 9 61 60.5fff=−+ −+ − sys sys sys 7 MW 91. 5 1. 5 10 3. 07 1. 676 11 8.64 1. 961fff=−+− +− sys 5 .13 7 306.2f = sys 59. 61 Hzf = The power supplied ... AAASA RjX φ =+ +EV I I ()( )()( ) 13 28 0 0 .15 2 51 36. 87 A 1. 1 2 51 36. 87 A A jE =∠°+ Ω∠°+ Ω∠ ° 12 17 11 .5 V A E =∠° The vo...

Ngày tải lên: 05/08/2014, 20:22

... generator is 15 1 () 11 1 21 2 384 V sin sin sin sin 36.4 31. 1 4 41. 6 V A A E E δδ −− == −°=−° The new armature current is 2 2 480 0 V 4 41. 6 31. 1 V 227 24 .1 A 1. 1 A A S jX ... kW 419 kWPPP=+= + = TOT 1 2 14 5 kVAR 11 5 kVAR 260 kVARQQQ=+= + = The overall power factor is 14 0 ()()()() 7 217 0 0. 0 18 7 4 619 36 .87 A 1. 716 4 619 36 .87...

Ngày tải lên: 05/08/2014, 20:22

... () 2 2 2 205,9 32 / 38,774 0. 3 21 / 0.7 02 Rs Rs = ++ () 2 22 0. 3 21 / 0.7 02 5. 311 /Rs Rs ++= () 2 22 2 0 .10 3 0.6 42 / / 0.7 02 5. 311 /Rs Rs Rs +++= 2 22 4.669 0.7 02 0 RR ss ... ()() 2 TH 2 ind 22 sync TH 2 TH 2 3/ / VRs RRs X X τ ω = +++ ()( ) ()( )( ) 2 ind 22 3 26 2 V 0 .1 72 /1. 9 62 18 8.5 rad/s 0. 3 21 0 ....

Ngày tải lên: 05/08/2014, 20:22

... () ()( ) 1 120 V 58 A 0 .20 0.16 99.1 V ATAAS EVIRR=− + = − Ω+ Ω= The final speed is given by the equation ,2 2 22 2 122 ,11 Ao A AAo En EK EK En φω φω == since the ratio ,2 ,1 / Ao ... 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...

Ngày tải lên: 05/08/2014, 20:22

... start,1 1 2 3 1. 628 RRRR=++= Ω start ,2 2 3 0.561 RRR=+= Ω start ,3 3 0. 134 RR== Ω Therefore, the starting resistances are 1 1.067 R =Ω 2 0. 427 R =Ω 3 0. 134 R =Ω 25 0 % Calculate ... Ao E of about 24 2 V at a speed of o n = 1800 r/min. The actual A E is 23 0 V, so the actual speed will be 24 5 () 23 0 V 1800 r/min 1711 r/min 24 2 V A o Ao E...

Ngày tải lên: 05/08/2014, 20:22

... jX + = ++ ( ) ( ) 50 2. 40 60 28 .15 24 .87 50 2. 40 60 F jj Zj jj + ==+Ω ++ () () () 22 22 /2 /2 M B M R s jX jX Z R s jX jX = −+ + 27 2 () () () 22 22 /2 /2 M B M R s jX jX Z R ... 27 1 ()() 1 .28 2 2. 40 60 1.185 2. 3 32 1 .28 2 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 12...

Ngày tải lên: 05/08/2014, 20:22

... ,load 27 7 0 V 25 3 .2 -7.3 41.3 52 V φφ ∆=−=∠°−∠°=∠°VV V (c) The real and reactive power of each load is () 2 2 1 25 3 .2 V 3 cos 3 cos 36.87 61.6 kW 2. 5 V P Z φ θ == °= Ω () 2 2 1 25 3 .2 V 3 ... V 3 sin 3 sin 36.87 46 .2 kvar 2. 5 V Q Z φ θ == °= Ω () () 2 2 2 253 .2 V 3 cos 3 cos -20 108.4 kW 1.67 V P Z φ θ == °= Ω () () 2 2 2 253 .2 V 3...

Ngày tải lên: 05/08/2014, 20:22

... holding the infinite bus is 460 -V. 29 8 ()() () ()() δδ 2 sin 40.0 62 . 0 40.0 62 . 0 2 7 62 1 3 sin 62 . 0 98907 62 1 3 2         − +=P MW2 sin 3.77 sin 7. 364 δ δ +=P A plot of power ... X XX A d dq dq =+ −         33 2 2 φφ δδ sin sin 2 ()() () ()() °         − +°5= 1 .22 1 sin 18. 025 .0 18. 025 .0 2 2773 .61 sin 25 .0 322 27...

Ngày tải lên: 05/08/2014, 20:22

... 180 21 5 22 6 24 2 I F , A 0.00 0.80 1.00 1 .28 1.44 2. 88 (a) What is the speed of this motor when it is running at the rated conditions spec- ified above? (b) The output power from the motor is 7. 5 ... Subfractional Horsepower Electric Motors. New York: McGraw- Hill, 1 970 . 4. Werninck, E. H. (ed.). Electric Motor Handbook. London: McGraw-Hill, 1 978 . cha6 523 9_ch10.qxd 11/14/0...

Ngày tải lên: 05/08/2014, 20:22