A self-excited shunt DC generator is rated 10 kW. 125 V, and has fric

A self-excited shunt DC generator is rated 10 kW. 125 V, and has friction and windage losses of 1200 W and core losses of 750 W when operating at rated flux. A field current of 6.0A is required and the armature circuit resistance is 0.24 ohm. Assume the field current and machine speed arc constant. Calculate the power required to turn the generator at no load, with 125V at the generator terminals. Repeat the calculation in A) above for a separately excited generator. Calculate the power to turn the generator with no excitation. Calculate the efficiency of the generator when operating at rated conditions (self-excited). A 200 KW, 240V, separately excited DC generator is operated at rated load. The armature and field resistances are .02 ohm and 240 ohm respectively. The field current is adjusted so the generator produces 240 V with no load while at rated speed. The field current and speed arc kept constant while the generator is being loaded. Calculate the rated current produced by the generator. Calculate the output voltage of the generator at rated load. Which machine parameter could be adjusted to maintain the output voltage at 240 V? If the generator is 85% efficient, calculate the power in HP required by the prime mover. Solution In self excited generators, generated cucurrents drive the field components So current flowing the field windings comes from armature so aramature current = Ia=6A armature voltage V=125V as it is running on no load, power required to turn on the generatore is power deliverd from armature to field +losses =VIa+windage loss+coreloss= 125(6)+1200+750=2700W B) for a seperately excited generator, field and armature are seperately excited. As there is no load, means an open circuit at generator terminals(armature) so no current passes through aarmature only power required to start the generator is the power req to over come losses Power req to start it = 1200 +750 = 1950W c) with no excitation means field is aided by armature Power = Torque*speed torque = KIa Back emf = E=k*speed speed = E/kso power = Ia*E as said in the problem, when self excited field current is 6A hence armature current = 6A as generatore generates 125V E = IaR +V = 6*0.24 + 125 =126.44 Power = 6*126.44=758.64W D) efficeincy = (output power)/(input power) at rated conditions input power = 10KW output power = input - losses = 10000-(1200+750) = 8050W efficiency = (8050/10000) * 100 = 80.5% 2. A) rated power = 200KW rated voltage = 240V rated current = 200K/240 = 833.33A b) volatgeat no load = 125V at no load no current passes through armature So Volatge developed at no load = back emf =125V as speed is kep constant, E (backemf) is also constant loading of generator is usually siad in terms of armature current at rated load implies at rated current in armature output voltage = E-IR = 125-(0.02*833.33) = 108.33V C) As said while loading speed is kept constant E will remain constant hence output voltage is dependent on arature resistance and also armature current to maintain voltage at 240V 1. armature currant can be adjusted by adjusting torque 2.A variable restance is connected in series with armature and can be varied to get rq uotput voltage D) efficiency is 85% 0.85=(output)/inputinput power = 200kW output power = *0.85 = 170KW = *0.00134HP =