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

... ()() 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 =×= (2) 1. 0 PF: ()() EQ 11 5 0 V 0 .14 0 0.5 32 8.7 0 A PS ... be (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 =×= (3) 0.8 PF Leadin...

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 22 7 24 .1 A 1. 1 A A S jX ... a 10 % decrease, 2 11 , 907 V A E = , and 11 1 22 2 13 ,23 0 V sin sin sin sin 27 .9 31. 3 11 ,907 V A A E E δδ −− == °=° Therefore, the new arm...

Ngày tải lên: 05/08/2014, 20:22

... 0.7 24 9.74 0.733 25 5.08 0.767 25 9 .2 0.8 26 3.74 0.833 26 7.6 0.867 27 0.8 0.9 27 3.6 0.933 27 6 .14 0.966 27 8 1 27 9.74 1. 033 2 81. 48 1. 067 28 2.94 1. 1 28 4 .28 1. 133 28 5.48 1. 167 ... 1. 133 28 5.48 1. 167 28 6.54 1. 2 287.3 1. 23 3 28 7.86 1. 26 7 28 8.36 1. 3 28 8. 82 1. 333 28 9 .2 1. 367 28...

Ngày tải lên: 05/08/2014, 20:22

... () 2 2 base base base 15 kV 1. 125 20 0 MVA V Z S == =Ω so ( ) ( ) EQ 0. 0 12 1. 125 0. 013 5 R =Ω=Ω ()( ) EQ 0.05 1. 125 0.0563 X =Ω=Ω 49 625 0 VA 52 .1 A 12 0 V S S S I V == = 2 -16 . A ... ( ) ( ) -1 11 tan 12 0 kW tan cos 0.8 90 kvarQP θ == = 2 50 kWP = ( ) ( ) -1 22 tan 50 kW tan cos 0.9 24 .2 kvarQP θ == = TOT 17 0 kW P =...

Ngày tải lên: 05/08/2014, 20:22

... 2 rms 315 15 sin sin 22 664 3 3 M V V ππ π π π =−−− 22 rms 315 313 3 sin sin 23 4 3 3 23 422 MM VV V πππ π ππ =−−=−−− 22 rms 313 33 3 0.8407 23 422 23 4 MM M VV VV ππ ππ ... ()() ()() EQ 1. 513 1. 134 3.36 0. 010 0.040 0.00 723 0.04 82 0.040 0 .17 0 1. 513 1. 134 3.36 jj Zj j j jj +− =+ + + ++ + ++− EQ 0. 010 0.040 0.0078...

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 ... voltage will be: 86 1 22 2 1 M V B RC ω = + and 2 222 1 M RC V B RC ω ω − = + Therefore, the forced solution to the equation is () , 22 2 22 2 s...

Ngày tải lên: 05/08/2014, 20:22

... ()()()() 13 28 0 0 .15 2 51 36.87 A 1. 1 2 51 36.87 A A j=∠°+ Ω∠− °+ Ω ∠− °E 15 37 7.4 V A =∠°E (c) The equivalent open-circuit terminal voltage corresponding to an A E of 15 37 volts is () ,oc 3 15 27 ... ()() OUT 10 00 kVA 0.8 800 kWP == ( ) ( ) 2 2 CU 3 3 2 51 A 0 .15 28 .4 kW AA PIR== Ω= F&W 24 kWP = 11 4 OUT IN 554 kW 10 0% 10 0% 90.5%...

Ngày tải lên: 05/08/2014, 20:22

... 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 voltage regulation would be: 12 17 13 28 VR 10 0% 8.4% 13 28 − =×=− (d) If the ... MW/Hz 2 Hz PA S == 3 MW 1. 676 MW/Hz 1. 79 Hz PB S == 3 MW 1. 9 61 MW/Hz 1. 53 Hz PC S == 11 6 ()()()() 13 28 0 0 .15 2 51 0 A...

Ngày tải lên: 05/08/2014, 20:22

... ( ) ( ) ,nl 13 28 0 V 2. 8 10 .5 0 A 13 28 .3 1. 27 V A j=∠°− Ω ∠°= ∠−°E ()( ) ,half 13 28 0 V 1. 5 93.6 0 A 13 54 11 .2 V A j=∠°− Ω ∠°=∠−°E ( ) ( ) ,full 13 28 0 V 2. 8 19 8 0 A 14 39 22 .7 V A j=∠°− ... ) () OUT load 21 0 00 hp 746 W/hp 12 4,700 N m 1 min 2 rad 12 00 r/min 60 s 1 r m P τ π ω == = ⋅ () () load 525 2 21 0 00 hp 525 2 91, 91...

Ngày tải lên: 05/08/2014, 20:22

... RjXX +−+ = ++ −+ () 22 2 2 11 11 1 1 1 1 TH 2 2 11 MMM MMM M RX X RX X RX j R X X X X X Z RXX −+++ ++ = ++ () () 22 22 1 111 TH TH TH 22 22 11 11 MMMM MM RX R X X X X X ZRjX ... is () 2 1 TH 2 2 11 M M RX R RXX = ++ . If 1M XR>> , then ()() 22 2 11 1 MM RXX XX++ ≈+ , so 2 TH 1 1 M M X RR XX ≈ + The Thevenin re...

Ngày tải lên: 05/08/2014, 20:22