Question

Q1: (30 points) A shunt DC generator, running at a constant speed of 1400 rpm, has a field resistance of 200 22 and armature resistance of 0.5 12. Ignore armature reaction and assume a 400 W rotational loss. a) [15] While supplying no load, the induced armature voltage is 280 V. Calculate the input power. b) [15] When the generator is loaded, the developed power is found to be 4.1 kW at the same magnetic flux of the field found in part a). Calculate the generator's efficiency. Consider the generator given in Q1. a) [6] Explain how one can find the load resistance so that the generator is operated at its maximum efficiency. b) [7] What is your opinion about the generator, i.e. would you recommend a customer to buy it or not? c) [7] Explain what happens to the speed if the armature reaction is considered.

Answer 1

Que Generator data RA 2oor Ra=0.5 Sheed N =1400 Part@ :- at No wad E =0 G Cgiven) = 280V RA = 0o 5 20o I ao No load asent & Eg = 940 Tao 59 Rf + Ra 280 Tao 1-3965 and 200 +0.5 so Input Power Pin = Ploss & Protation was 400 Eg Taot = 280 x 1.3965 +400 Pin 791.02 wath 1 Input Powe wer.

Post: developed 4.1kW 280 Eg amature arrent so = Pdey Big H - 4100 = 14.623 Amp. IL y IA Load V RA 200 0.5 Rq 5 = 280 so terminal Voltage VL = Eg - Fara = 280 14.623 X 0.5 ve - 27 2.678 Volt and field Current ТР 272.6785 = V2 Rf = goo a Anup If 1.3633 = I1 = Iq - IL so load Grosent 13.28 Amp so efficiency Oll 272,6785 *13.28 IIP 410ot 400 80.47% efficiency (%) Ang