Electric Machinery Fundamentals (Solutions Manual) Part 11 ppsx

... decreased by 10%, the new value if E A = (0.9)(603 V) = 543 V. To simplify this part of the problem, we will ignore R A . Then the quantity E A sin ™ will be constant ... power P and reactive power Q supplied or consumed by the machine under the conditions in part (a) . Is the machine operating within its ratings under these circumstance...

Ngày tải lên: 06/08/2014, 11:20

... field current in amps, and column 2 contains short-circuit termina current in amps. 5 -11. (a) What is the saturated synchronous reactance of this generator at the rated conditions? ... to read these files and calculate and plot X S is shown below. % M-file: prob5_11c.m % M-file to calculate and plot the saturated % synchronous reactance of...

Ngày tải lên: 06/08/2014, 11:20

... A ) E A j = ° + &  ° E A 11, 120 19.9 V = ° The resulting voltage regulation is 11, 120  7621 RV 100% 45.9% = ⋅ = 7621 ... power rating) of the generator will be reduced by a factor of 5/6. 5 ( ) 13.2 kV 11. 0 kV = = T V ,rated 6 (b) If armature resistance is ignored, the pow...

Ngày tải lên: 06/08/2014, 11:20

... 2 ( = = ) 2 ( ) & = 111 CU 3 A A P I 3 R 251 A 0.15 28.4 kW F& P = W 24 kW 110 114 ⎜ OUT P 100% = ⋅ 554 ... ) 335 A = = ⎞ A V I Z and the terminal voltage is ( 3.333 ) 11 & 17 = V 3 3 ( ) 111 7 V 1934 V T V V ⎞ = = = (f) To restore the te...

Ngày tải lên: 06/08/2014, 11:20

... 119 4 40.6 A = = = ° A 1.5 12.0 + + & A S R jX j The reactive power produced by the generator at this point is 3 sin 3 ( ) 7967 V ( 119 = ... plot is shown below: 117 system? What is the generator’s rate of shaft rotation after paralleling occurs? 118 5-5. Assume that the field current...

Ngày tải lên: 06/08/2014, 11:20

... will be supplying its rated real and reactive power. (d) Under the conditions of part (c) , which are the rated conditions of the generators, the internal generated ... 1. = 5.137 f sys = 306.2 f sys = 59.61 Hz 5 sys 103.07 1.676 sys 118 .64 1.961 sys f f f The power supplied by each generator will be...

Ngày tải lên: 06/08/2014, 11:20

... 5247 = A 3 3 ( ) 11. 0 kV A L T V The power factor is 0.8 lagging, so I A 5247 36.8 = 7  A ° . The phase voltage is 11. 0 kV / 3 = 6351 V. ... = jX S j 8.18 ° =  ° (e) Repeating part (d) : With a 10% decrease, E A2 = 11, 907 V , and sin 1 1 E 1 A sin sin 13,...

Ngày tải lên: 06/08/2014, 11:20

... = ° j  &  ° =  ° (b) This motor has 6 poles and an electrical frequency of 60 Hz, so its rotation speed is m n = 1200 r/min. The induced

Ngày tải lên: 06/08/2014, 11:20

... maximum power that the motor can produce at rated speed with the value of E A from part (b) is 158 ⎤ m S X ( ) 3600 r/min  ( ) 2.8 & 60 s ... when calculating the V-curves. d_nl = -1.27 * pi/180; % delta at no-load d_half = -11. 2 * pi/180; % delta at half-load d_full = -22.7 * pi/180; % delta at full...

Ngày tải lên: 06/08/2014, 11:20

... consuming from or supplying to the electrical system? How much reactive power Q is this machine consuming from or supplying to the electrical system? S OLUTION This ... This is consistent with what we learned about reactive power sharing in Chapter 5. 6 -11. A 480-V, 100-kW, 50-Hz, four-pole, Y-connected synchronous motor has a rated power ......

Ngày tải lên: 06/08/2014, 11:20