A 20kw,200V shunt generator is operated at rated load. if the driving engine is developing 30bhp (break horse power), determine the iron and friction losses.PLease use all the given data above..thanksAnswer is IFL=1039W.Thanks a lot.
The brake horsepower is the power delivered to the load by the engine under specified operating conditions. The engine delivers 30 X 745.7 = 22371 watts to the generator. A machine described as a 20 kW, 200 V shunt generator would be assumed to be a DC generator with a nominal rating of 20 kW at 200 V dc. It would be capable or delivering 20 kW, but the actual delivered power is not stated. You could assume that the generator is operating at full load. If that is the case, the total losses, copper, iron and friction would then be 22371 - 20000 = 2371 watts. If you have been given that the iron and friction losses are 1039 watts, the copper losses must be 2371 - 1039 = 1331 watts. There is no information given in the problem, such as armature and field resistance, that could be used to calculate the copper losses. If you had that information, you could calculate that the copper losses are 1331 watts and from that calculate the iron and friction losses. Are you sure that you are not supposed use machine specifications provided in another problem?
This is the answer I gave 3 other askers of the same question. brake horse power of an engine is the actual amount of horse power an engine can put out. Using 745.5 watts as a horse power, the engine will supply the equivalent of 30 times 745.7 watts at 30 brake horse power. This is 22,371watts. The generator is supplying 20 Kw or 20,000 watts. The difference in the engine output and the generator output is 22,371 minus 20,000 or 2.371 watts. I don't see how the answer is 1039 watts, if my calculations are correct. Perhaps someone else will have the answer you seek. I may be missing something in your question
