Electric Machinery Fundamentals (Solutions Manual) Part 5 ppsx

... () ( ) ( ) 2 22 BO 0.00 05 A 150 0 ln 0.00 15 ln 5. 5 ms 30 V H IR tRC V Ω =− =− = Therefore, the period of the relaxation oscillator is T = 178 ms + 5. 5 ms = 183 .5 ms, and the frequency of ... 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 2 150 °...

Ngày tải lên: 05/08/2014, 20:22

... ()() 9 15. 00.962 0. 951 === dpw kkk 276 1.4 Ω j1.9 Ω + - V = 220∠0° V I 1 R 1 jX 1 s R 2 5. 0 j0.5X 2 j0.5X M j1.90 Ω j30 Ω s R −2 5. 0 2 jX 2 j1.90 Ω j0.5X M j100 Ω { { { { Forward Reverse 0.5Z B 0.5Z F ... motors, °= 45 e θ . Therefore, Number of poles Mechanical Step Size 3-phase ( 60 e θ =°) 4-phase ( 45 e θ =°) 2 60 ° 45 4 30 ° 22 .5 ° 6 20 ° 15 ° 8 15...

Ngày tải lên: 05/08/2014, 20:22

... () 2 2 base base base 15 kV 1.1 25 200 MVA V Z S == =Ω so ( ) ( ) EQ 0.012 1.1 25 0.01 35 R =Ω=Ω ()( ) EQ 0. 05 1.1 25 0. 056 3 X =Ω=Ω 49 6 250 VA 52 .1 A 120 V S S S I V == = 2-16. A 50 00-VA 480/120-V ... 15 kV 0.8 S S P I V ′ == = 16,667 36.87 A S ′ =∠−° I The voltage on the primary side of the transformer is EQ,PSSP Z ′′ =+ VV I ()() 15, 000 0 V 16,6...

Ngày tải lên: 05/08/2014, 20:22

... 146.46 0.367 159 .78 0.4 172.18 0.433 183.98 0.467 1 95. 04 0 .5 2 05. 18 0 .53 3 214 .52 0 .56 7 223.06 0.6 231.2 0.633 238 0.667 244.14 0.7 249.74 0.733 255 .08 0.767 259 .2 0.8 263.74 ... lengths are 1 l = 2(27 .5 cm) = 55 cm, 2 l = 30 cm, and 3 l = 30 cm. The reluctances of these regions are: () () ()() 1 7 0 .55 m 58 .36 kA t/Wb 1000 4 10 H/m 0....

Ngày tải lên: 05/08/2014, 20:22

... () 0.010 Wb 50 0 t 2 s 2 .50 V 2 < t < 5 s () 0.020 Wb 50 0 t 3 s − -3.33 V 5 < t < 7 s () 0.010 Wb 50 0 t 2 s 2 .50 V 7 < t < 8 s () 0.010 Wb 50 0 t 1 s 5. 00 V The ... velocity for 50 Hz freq = 50 ; % Freq (Hz) w = 2 * pi * freq; % Calculate flux versus time time = 0:1/ 250 0:1/ 25; % 0 to 1/ 25 sec 22 ( ) ( ) ind 80 V - 66.7 A 0 .50...

Ngày tải lên: 05/08/2014, 20:22

... ()() ()() EQ 1 .51 3 1.134 3.36 0.010 0.040 0.00723 0.0482 0.040 0.170 1 .51 3 1.134 3.36 jj Zj j j jj +− =+ + + ++ + ++− EQ 0.010 0.040 0.00788 0. 052 5 0.040 0.170 (2. 358 0.109)Zj j j j=+++++++ EQ 2.4 15 ... 0. 356 G PVI θ == °= ()( ) ,pu sin 1 0.47 65 sin 41.6 0.316 G QVI θ == °= ()( ) ,pu 1 0.47 65 0.47 65 G SVI== = ()( ) ,pu base 0. 356 1000 kVA 356 kW GG PPS==...

Ngày tải lên: 05/08/2014, 20:22

... poles and electrical frequency is 120 e m f n P = The resulting table is Number of Poles e f = 25 Hz 2 150 0 r/min 4 750 r/min 6 50 0 r/min 8 3 75 r/min 10 300 r/min 12 250 r/min ... rad/s 0 .5 T 0 .5 m sin103et t= () ind 5. 15 sin103 Vet t= (b) If a 5 Ω resistor is connected as a load across the terminals of the loop, the current flow would be: () ind 5....

Ngày tải lên: 05/08/2014, 20:22

... MW/Hz 60 .5 Hz 0. 15 MW/Hz 60.0 Hzff=−+ − () () sys sys 100 kW 6 050 kW 0.10 MW/Hz 9000 kW 0. 15 MW/Hzff=− +− () sys 0. 25 MW/Hz 6 050 kW 9000 kW 100 kWf =+− sys 14, 950 kW 59 .8 Hz 0. 25 MW/Hz f ... MW/Hz 60 .5 Hz 59 .5 Hz P P s ff == = −− The power curve’s slope for generator 1 is 2 nl fl 0.0 75 MW 0. 150 MW/Hz 60.00 Hz 59 .50 Hz P P s ff == = −− The no-load f...

Ngày tải lên: 05/08/2014, 20:22

... new armature current is 152 The resulting plot is shown below 350 400 450 50 0 55 0 600 650 700 200 210 220 230 240 250 260 E A (V) I A (A) Synchronous Motor V-Curve 6-3. A 2300-V 1000-hp ... 4 75 A is 15. 2 kV, and the short-circuit current is 155 0 A. Since this generator is Y-connected, the corresponding phase voltage is 15. 2 kV/ 3 8776 V V φ == and the armature curr...

Ngày tải lên: 05/08/2014, 20:22

... V 129 V 50 Hz A E == (e) If the applied voltage φ V is derated by the same amount as A E , then V φ = ( 15/ 50)(277) = 83.1 V. Also, note that S X = ( 15/ 50)(1 .5 Ω) = 0. 45 Ω. The ... equation, () ()( ) ()( ) ,, 11 ,, 1 .5 80 kW sin sin 25. 9 3 3 280V 327 V Sg Sm Ag Am XXP EE γ −− + Ω == =° Therefore, ,, ,, 280 0 V 327 25. 9 V 95. 7 5. 7 A () 1 .5 Ag Am A Sg...

Ngày tải lên: 05/08/2014, 20:22