Electric machinery fundamentals

... Manual to accompany Chapman Electric Machinery Fundamentals Fourth Edition Stephen J Chapman BAE SYSTEMS Australia i Instructor’s Manual to accompany Electric Machinery Fundamentals, Fourth Edition ... APPENDIX D: ERRATA FOR ELECTRIC MACHINERY FUNDAMENTALS 4/E 301 iii PREFACE TO THE INSTRUCTOR This Instructor’s Manual is intended to accompany the fourth edition of Electric...

Ngày tải lên: 24/12/2013, 08:16

... The overall efficiency of the power system will be the ratio of the output power to the input power The output power supplied to the load is POUT = 90 kW The input power supplied by the source ... questions about this power system (a) Assume that the switch shown in the figure is open, and calculate the current I, the power factor, and the real, reactive, and apparent power be...

Ngày tải lên: 21/06/2014, 21:20

... equivalent circuit of this power system 59 (b) With the switch opened, find the real power P, reactive power Q, and apparent power S supplied by the generator What is the power factor of the generator? ... generator? (c) With the switch closed, find the real power P, reactive power Q, and apparent power S supplied by the generator What is the power factor of the generator? (...

Ngày tải lên: 21/06/2014, 21:20

... line losses This problem is a good example of the advantages of power factor correction in power systems 62 Chapter 3: Introduction to Power Electronics 3-1 Calculate the ripple factor of a three-phase ... Pline,pu S base = ( 0.00121)(1000 kVA ) = 1.21 kW Load improved the power factor of the system, increasing the load voltage and the total power supplied to the loads, while simult...

Ngày tải lên: 21/06/2014, 21:20

... loop is equal to the amount of electrical power created by the loop This machine is acting as a generator, converting mechanical power into electrical power 4-2 Develop a table showing the speed ... power being consumed by the loop is: ( ) P = τ indω = 0.0515 sin ω t V (103 rad/s ) = 5.30 sin 2ω t W Note that the amount of mechanical power consumed by the loop is equal to the amount...

Ngày tải lên: 21/06/2014, 21:20

... The input power to this generator is equal to the output power plus losses The rated output power is POUT = (1000 kVA )( 0.8) = 800 kW PCU = I A2 RA = ( 251 A ) ( 0.15 Ω ) = 28.4 kW PF&W = 24 kW ... The input power to this generator is equal to the output power plus losses The rated output power is POUT = (1000 kVA )( 0.8) = 800 kW PCU = I A2 RA = ( 251 A ) ( 0.15 Ω ) = 28.4 kW PF...

Ngày tải lên: 21/06/2014, 21:20

... real power from or supplying real power to the power system? Is it consuming reactive power from or supplying reactive power to the power system? (b) Calculate the real power P and reactive power ... real power from or supplying real power to the power system? Is it consuming reactive power from or supplying reactive power to the power system? (d) Calculate the rea...

Ngày tải lên: 21/06/2014, 21:20

... find (a) The line current I L (b) The stator copper losses PSCL (c) The air-gap power PAG (d) The power converted from electrical to mechanical form Pconv (e) The induced torque τind (f) The load ... the same as the power consumed by the original circuit.) (Note that I A2 RF is equal to I 2 PAG = I 2 (d) R2 = 3I A RF = ( 44.8 A ) ( 2.220 Ω ) = 13.4 kW s The power converted from electr...

Ngày tải lên: 21/06/2014, 21:20

... Ao 134 V The power converted from electrical to mechanical form is Pconv = E A I A = ( 236.2 V )( 25.3 A ) = 5976 W Since the rotational losses are ignored, this is also the output power of the ... r/min The power converted from electrical to mechanical form is Pconv = E A I A = ( 232.4 V )(50.7 A ) = 11, 780 W Since the rotational losses are ignored, this is also the output power of...

Ngày tải lên: 21/06/2014, 21:20

... 0.15 Ω ) + ( 230 V )(1.35 A ) = 851 W The power converted from electrical to mechanical form is Pconv = E A I A = ( 221 V )( 60 A ) = 13,260 W The output power is POUT = (15 hp )( 746 W/hp ) = 11,190 ... the actual speed will be n= EA 221 V no = (1800 r/min ) = 1800 r/min E Ao 221 V (b) The output power is 7.5 hp and the output speed is 1800 r/min at rated conditions, therefore, the tor...

Ngày tải lên: 21/06/2014, 21:20

... SOLUTION (a) The power supplied by the ac machine to the ac power system is PAC = S cos θ = (50 kVA )( 0.8) = 40 kW 266 and the reactive power supplied by the ac machine to the ac power system is ... = 222.4 V Therefore, the power into the dc machine is VT I A = 41.38 kW , while the power converted from electrical to mechanical form (which is equal to the output power) is E A I A...

Ngày tải lên: 21/06/2014, 21:20

... the power- versus-torque-angle curve for this generator At what angle δ is the power of the generator maximum? (d) How does the maximum power out of this generator compare to the maximum power ... cos θ (e) The real and reactive power supplied by the generator when the switch is closed is just the sum of the real and reactive powers of Loads 1, 2, and The powers of Loads and have alr...

Ngày tải lên: 21/06/2014, 21:20

... 0.933 0.966 1. 033 1. 067 1. 1 1. 133 1. 167 1. 2 1. 233 1. 267 1. 3 1. 333 1. 367 1. 4 1. 433 1. 466 1. 5 10 4.4 11 8.86 13 2.86 14 6.46 15 9.78 17 2 .18 18 3.98 19 5.04 205 .18 214 .52 223.06 2 31. 2 238 244 .14 249.74 255.08 ... R3 + R4 ) R1 + R2 + R3 + R4 ( R1 + R2 ) φ right = φTOT = R1 + R2 + R3 + R4 (90 .1 + 77.3) 90 .1 + 10 8.3 + 90 .1 + 77.3 φTOT = (90 .1 +...

Ngày tải lên: 05/08/2014, 20:22

... 11 6.3 2. 28° V 27 VR = (3) 11 6.3 -11 5 × 10 0% = 1. 1% 11 5 0.8 PF Leading: VP ′ = VS + Z EQ IS = 11 5∠0° V + ( 0 .14 0 + j 0.5 32 Ω )(8.7∠36.87° A ) VP ′ = 11 3.3 2. 24° V 11 3.3 -11 5 VR = × 10 0% = 1. 5% 11 5 ... 11 5∠0° V + ( 0 .14 0 + j 0.5 32 Ω )(8.7∠ − 36.87° A ) VP ′ = 11 8.8 1. 4° V 11 8.8 -11 5 VR = × 10 0% = 3.3% 11 5 (2) 1. 0 PF: VP ′ = VS + Z EQ I S = 11 5∠0...

Ngày tải lên: 05/08/2014, 20:22

... 19 .9 kV 7.97 kV 200 kVA 2.50 :1 19.9 kV 13 .8 kV 200 kVA 1. 44 :1 Y-∆ 34 .5 kV 7.97 kV 200 kVA 4 .33 :1 ∆-Y 34 .5 kV 13 .8 kV 200 kVA 2.50 :1 ∆-∆ 34 .5 kV 13 .8 kV 34 6 kVA 2.50 :1 open-∆ 19 .9 kV 13 .8 kV 34 6 ... (low-voltage) side is Vbase (15 kV ) = 1. 125 Ω = S base 200 MVA Z base = so REQ = ( 0. 012 ) (1. 125 Ω ) = 0. 0 13 5 Ω X EQ = ( 0.05) (1. 125 Ω ) = 0.05 63 Ω...

Ngày tải lên: 05/08/2014, 20:22