Electric Machinery Fundamentals (Solutions Manual) Part 7 pdf

Ngày tải lên: 05/08/2014, 11:20

... I φ =+ + ()( )( )( ) 79 67 0 1.5 418 36. 87 A 12.0 418 36. 87 A A jE =∠°+Ω∠− °+ Ω∠− ° 12,040 17. 6 V A E =∠° 123 () ()( ) 11nl1sys 0.1 MW/Hz 60.5 Hz 59.8 Hz 70 kW P Ps f f=−= − = ... () 6250 kVA 8 67 A 3 3 4160 V AL T S II V == = = The power factor is 0.85 lagging, so 8 67 31.8 A A I =∠− °. Therefore, AAASA RjXEV I I φ =+ + ()()()() 2402 0 0.04 8 67 31.8 A 0...

Ngày tải lên: 05/08/2014, 20:22

... Fractional and Subfractional Horsepower Electric Motors. New York: McGraw- Hill, 1 970 . 4. Werninck, E. H. (ed.). Electric Motor Handbook. London: McGraw-Hill, 1 978 . cha65239_ch10.qxd 11/14/03 12:41 ... stepper motors. REFERENCES 1. Fitzgerald, A. E., and C. Kingsley, Jr. Electric Machinery. New York: McGraw-Hill, 1952. 2. National Electrical Manufacturers Association. Motors and Ge...

Ngày tải lên: 05/08/2014, 20:22

... 249 .74 0 .73 3 255.08 0 .76 7 259.2 0.8 263 .74 0.833 2 67. 6 0.8 67 270 .8 0.9 273 .6 0.933 276 .14 0.966 278 1 279 .74 1.033 281.48 1.0 67 282.94 1.1 284.28 1.133 285.48 1.1 67 286.54 ... 0. 07 m 0. 07 m r l A µµ π − == =⋅ × R () ()()() 4 4 7 04 0.0005 m 77 .3 kA t/Wb 4 10 H/m 0. 07 m 0. 07 m 1.05 l A µ π − == = ⋅ × R () () ()() 5 5 7 05 0. 37...

Ngày tải lên: 05/08/2014, 20:22

... 8000/230 = 34 .78 . Thus the load impedance referred to the primary side is ()( ) 2 34 .78 3.2 1.5 3 871 1815 L Zjj ′ =+Ω=+Ω The referred secondary current is ()( ) 79 67 0 V 79 67 0 V 1 .78 28.2 ... source are ( ) ( ) ( ) cos 120 V 47. 59 A cos 37. 5 4531 WPVI θ == −°= ()( )( ) cos 120 V 47. 59 A sin 37. 5 3 477 varQVI θ == −°=− ( ) ( ) 120 V 47. 59 A 571 1 VASVI== =...

Ngày tải lên: 05/08/2014, 20:22

... to the primary side): 35 8 573 1.2 V 246.5 1.2 V 34 .78 S S a ′ ∠− ° == = ∠−° V V The voltage regulation is 79 67 8 573 VR 100% 7. 07% 8 573 − =×=− 2 -7. A 5000-kVA 230/13.8-kV single-phase ... M YGjB j=− = ∠− °= − 1/ 75 7 CC RG==Ω 1/ 3 27 MM XB==Ω The excitation branch elements can be expressed in per-unit as 75 7 65 .7 pu 11.52 C R Ω == Ω 3 27 28....

Ngày tải lên: 05/08/2014, 20:22

... ( ) 2 2 1 base1 base1 3 3 277 V 0.238 1000 kVA V Z S φ == = Ω () 2 2 2 base2 base2 3 38314 V 2 07. 4 1000 kVA V Z S φ == =Ω ( ) 2 2 3 base3 base3 3 3 277 V 0.238 1000 kVA V Z S φ == ... the generator is ()( ) ,pu cos 1 0. 476 5 cos41.6 0.356 G PVI θ == °= ()( ) ,pu sin 1 0. 476 5 sin 41.6 0.316 G QVI θ == °= ()( ) ,pu 1 0. 476 5 0. 476 5 G SVI== = ()( ) ,pu base 0...

Ngày tải lên: 05/08/2014, 20:22

... 12/7T b c SCR 2 SCR 6 SCR 2 150° 12/7T 12/9T b a SCR 2 SCR 4 SCR 4 210 ° 12/9T 12/11T c a SCR 3 SCR 4 SCR 3 270 ° 12/11T 12/T c b SCR 3 SCR 5 SCR 5 330° 75 ... substitute the known values for R, C, ω, and V M , this equation becomes ( ) 83.3 35 .76 7. 91 sin 35 .76 cos t C vt e t t ωω − =+− This equation is plotted below: 83 Figure (b) (...

Ngày tải lên: 05/08/2014, 20:22

... = 1 377 ∠ 6.8° V A (e) The new impedance per phase will be half of the old value, so 3.333 30 Z = ∠ ° Ω . The magnitude of the phase current flowing in this generator is 1 377 V 1 377 V 335 ... 113 + - E A 0.15 Ω j1.1 Ω 6.6 67 30°Z + - V φ I A The magnitude of the phase current flowing in this generator is 1 377 V 1 377 V 186 A 0.15 1.1 6.6 67 30 1.829 A A AS E I RjXZ j...

Ngày tải lên: 05/08/2014, 20:22

... kVARQQQ=+= + = The overall power factor is 140 ()()()() 72 17 0 0.01 87 4619 36. 87 A 1 .71 6 4619 36. 87 A A j=∠°+ Ω ∠− °+ Ω ∠− °E 13,590 27. 6 V A =∠°E Therefore, the magnitude of the internal ... decrease, 2 12, 570 V A E = , and 11 1 22 2 13,230 V sin sin sin sin 27. 9 29.5 12, 570 V A A E E δδ −− == °=° Therefore, the new armature current is 2 12,...

Ngày tải lên: 05/08/2014, 20:22