An elementary book on electricity and magnetism and their applications . tual voltage curve at no load. 150 kv-a. and 60 cycles. 249. Efficiency of alternators. The efficiency of any machinemeans the ratio of the power delivered by the machine to thopower received by it. Efficiency = <^ In the case of a large alternator it is generally impracticableto measure directly the input and so we determine the outputand the losses and calculate the efficiency thus: Efficiency = ^PSL^ Output + total losses The results of tests on alternators of varying sizes from 100kv-a. to 10,000 kv-a. sh
An elementary book on electricity and magnetism and their applications . tual voltage curve at no load. 150 kv-a. and 60 cycles. 249. Efficiency of alternators. The efficiency of any machinemeans the ratio of the power delivered by the machine to thopower received by it. Efficiency = <^ In the case of a large alternator it is generally impracticableto measure directly the input and so we determine the outputand the losses and calculate the efficiency thus: Efficiency = ^PSL^ Output + total losses The results of tests on alternators of varying sizes from 100kv-a. to 10,000 kv-a. show that the efficiencies run from about91 per cent to 97 per cent. It is, of course, possible to getvery high efficiency by increasing the first cost of the experience has shown that sometimes it does not pay to build machines for the highest possible efficiency. Hencewe must distinguish between efficiency and economy. 366 ELECTRICITY AND MAGNETISM If we know the efficiency of an alternator, we can computethe size of the engine required to drive it. Engirt. Fig. 249. — Size of engine required to drive three-phase alternator. For example, what horse power would be required at the pulley infigure 249 when the alternator is impressing 2300 volts on the externalcircuit and delivering a current of 25 amperes, it being assumed thatthe power factor of the load is 80% and the efficiency of the generator is 92%? Solution. — Let x = horse power supplied at pulley,746 x = input in watts, output in watts, X746 x = X2300 X25 , x = = Problems 1. What is the kilovolt-ampere output of a three-phase alternatorwhen it is delivering 54 amperes in each of the three-phase wires andimpressing on the external circuit a voltage of 2200 ? 2. Find the kilowatt capacity of the machine in problem 1 whendelivering a load at 80% power factor. 3. If it requires 715 horse power to drive a three-phase generatordelivering 600 kv-a.
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