Electric Machinery Fundamentals (Solutions Manual) Part 2 ppt

... r/min ( 2 ) f m n P = = = nl1 120 120 60.5 Hz The full-load frequency of generator 1 corresponds to a frequency of 1 328 0 ( 0.15 )( ) 25 1 0 ... 2 ( ) = + cos + sin ( ) 2 + co ⎝ ⎝ s  sin ⎝ 2 ⎝ A A E V ⎞ R A I S A X I S A X I A R S I 2 ( ) =  cos s  in...

Ngày tải lên: 06/08/2014, 11:20

... 2 sin 2A A E E With a 5% decrease, E A2 = 12, 570 V , and sin 1 1 E 1 A sin sin 13, 23 0 ™ ™ V sin = = 27 .9   29 .5 ° = ° 2 2 E A2 ... ) =  + ( ) ( )  +  29 0 MW 34 62. 21 =  8. 529 186.63 3 f sys f sys = 59.37 Hz sys 34 62. 21 sys 34 62. 21 f f sys f...

Ngày tải lên: 06/08/2014, 11:20

... current is E A 2 147 V ( ) 2 = = .7 A 3 .20 A = 2 1 F F I I E 1 A 124 V (c) The new torque angle ™ of this machine is – 12. 5 ° . 6-8. A synchronous ...  E V  I I 2 2A A ⎞ 2A S A R jX ( )( ) E A 2 120 0 V = ° j 0.8  50 & 36.8 7 A ° E A2 147.5 12. 5 = V...

Ngày tải lên: 06/08/2014, 11:20

... 2 = = 1000 hp 746 W/hp 124 , 700 N = m ⊕ load ⎤ m ( ) 120 0 r/min 1 min 2  rad 60 s 1 r ⎮ load 525 2 P 525 2 ( ) 21 000 hp ... .33 2 4.  ° 9 0.663 20 .2 pu = = = ° + A 0.90 A S R jX j In actual amps, this current is A I ( ) 1404 A ( 0.663 20 .2...

Ngày tải lên: 06/08/2014, 11:20

... 1000 kVA 0.8 800 kW = = 2 ( = = ) 2 ( ) & = 111 CU 3 A A P I 3 R 25 1 A 0.15 28 .4 kW F& P = W 24 kW 110 114 ⎜ ... ) 60 Hz = = s n ync P P 1 2 2 P 6 12 = = P 1 5 10 Therefore, a 10-pole synchronous motor must be coupled to a 12- pole synchronous genera...

Ngày tải lên: 06/08/2014, 11:20

... 121 (c) The power supplied by generator 1 is 122 ( ) 3570 r/min ( 2 ) f m n P = = = fl1 120 120 59.5 Hz The no-load frequency of generator 2 ... = nl2 120 120 60.00 Hz The full-load frequency of generator 2 corresponds to a frequency of ( ) 1785 r/min ( 4 ) f m n P = = = fl2 1...

Ngày tải lên: 06/08/2014, 11:20

... voltage. load p 52_ occ.dat vt = interp1(p 52_ occ(:,1),p 52_ occ(: ,2) ,0.534) vt = 880.400 129 generators would have more margin than this. 131 E A 28 25 10.9 = V ° ... I S A R I AA By the Pythagorean Theorem, ( 2 2 = + ) 2 A S E V ⎞ 2 A X I ( ) 2 V ⎞ = A E  X S I A At leading...

Ngày tải lên: 06/08/2014, 11:20

... 12 = 0 MVA 524 9 = A 3 3 ( ) 13 .2 kV A L T V The power factor is 0.8 lagging, so I A 524 9 36.8 = 7  A ° . The phase voltage is 13 .2 kV ... r/min 2 the applied torque would be ⎮ ⎮ = = 85, 000, 000 W 27 0, = 000 N m ⊕ app ind ( ) 3000 r/min ( 2  rad/r )( 1 min/60 s ) 5 -...

Ngày tải lên: 06/08/2014, 11:20

... sin sin 13, 23 0 ™ ™ V sin = = 27 .9   38.6 ° = ° 2 2 E A 2 Therefore, the new armature current is 9, 923 V  A2 ⎞ 9, 923 38.6 70 ... 10,584 V , and sin 1 1 E 1 A sin sin 13, 23 0 ™ ™ V sin = = 27 .9   35.8 ° = ° 2 2 E A 2 Therefore, the new armature current is...

Ngày tải lên: 06/08/2014, 11:20

... E A 429 24 = .9  V ° So E A = 429 V at rated conditions. The resulting plot is shown below 26 0 Synchronous Motor V-Curve 25 0 24 0 23 0 ... Motor V-Curve 25 0 24 0 23 0 22 0 21 0 20 0 350 400 450 500 550 600 650 700 E (V) A 6-3. A 23 00-V 1000-hp 0.8-PF leading 60-Hz two-pol...

Ngày tải lên: 06/08/2014, 11:20