Ans. Slightly less than one.
Ques. Why is it desirable to keep the power factor near unity?
Ans. Because with a low power factor, while the alternator may be carrying its full load and operating at a moderate temperature, the consumer is paying only for the actual watts which are sent over the line to him.
Fig. 1,366.—Diagram illustrating power factor test, when on non-inductive and inductive circuits. The instruments are connected as shown and by means of the double throw switch can be put on either the non-inductive or inductive circuit. First turn switch to left so that current passes through the lamps; for illustration, the following readings are assumed: ammeter 10, voltmeter 110, and wattmeter 1,100. The power factor then is wattmeter reading ÷ volts × amperes = 1,100 actual watts ÷ 1,100 apparent watts = 1, that is, on non-inductive circuit the power factor is unity. Now throwing the switch to the right connecting instruments with the inductive circuits, then for illustration the following readings may be assumed: ammeter 8, voltmeter 110, and wattmeter 684. Now, as before, power factor = wattmeter reading ÷ volts × amperes = 684 ÷ (8 × 110) = 684 ÷ 880 = .78.
For instance, if a large alternator supplying 1,000 kilowatts at 6,600 volts in a town where a number of induction motors are used on the line be operating with a power factor of say .625 during a great portion of the time, the switchboard instruments connected to the alternator will give the following readings:
Voltmeter 6,600 volts; ammeter 242.4 amperes; power factor meter .625.
The apparent watts would equal 1,600,000 watts or 1,600 kilowatts, which, if multiplied by the power factor .625 would give 100,000,000 watts or 1,000 kilowatts which is the actual watts supplied. The alternator and line must carry 242.4 amperes instead of 151 amperes and the difference 242.4 - 151 = 91.4 amperes represents a wattless current flowing in the circuit which causes useless heating of the alternator.
The mechanical power which is required to drive the alternator is equivalent to the actual watts produced, since that portion of the current which lags, is out of phase with the pressure and therefore requires no energy.