Ques. How are alternators rated by manufacturers in order to avoid disputes?
Ans. They usually rate their alternators as producing so many kilovolt amperes instead of kilowatts.
Fig. 1,367.—Ayrton and Sumpner method of alternating current power measurement. Three voltmeters are required, and accordingly the method is sometimes called the three voltmeter method. It is a good method where the voltage can be regulated to suit the load. In the figure, let the non-inductive resistance R be placed in series with the load AB. Measure the following voltages: V across the terminals of R, V1 across the load AB, and V2 across both, that is from A to C. Then, true watts = (V22 - V12 - V2) ÷ 2R. The best conditions are when V = V1, and, if R = ½ ohm, then W = V22 - V12 - V2.
Ques. What is a kilovolt ampere (kva)?
Ans. A unit of apparent power in an alternating current circuit which is equal to one kilowatt when the power factor is equal to one.
The machine mentioned on page [1,120] would be designed to carry 151 amperes without overheating and also carry slight overloads for short periods. It would be rated as 6.6 kilo volts and 151 amperes which would equal approximately 1,000 kilowatts when the power factor is 1 or unity, and it should operate without undue heating. Now the lower the power factor becomes, the greater the heating trouble will be in trying to produce the 1,000 actual kilowatts.
Fig. 1,368—Curves illustrating power factor. In a circuit having no capacity or inductance, the power is given by the product of the respective readings of the voltmeter and ammeter, as in the case of a direct current. In the case of a circuit having capacity or inductance, this product is higher than the true value as found by a wattmeter, and is known as the apparent watts. The ratio true watts ÷ apparent watts is known as the power factor. The current flowing in an inductive circuit, such as the primary of a transformer, is really made up of two components, as already explained, one of which (the load or active component), is in phase with the pressure, while the other the magnetizing component, is at right angles to it, that is, it attains its crest value when the other is at zero, and vice versa. To illustrate, take a complete cycle divided into 360 degrees and lay out on it the current required to correspond to a given load on the secondary of a transformer, say a crest value of 100 amperes, and at right angles to this lay out the current required for exciting the magnetic circuit of the transformer, giving A, merely for purposes of illustration, a crest value of 25 amperes. Combining these curves, the dotted curve in the figure is obtained and which represents the resultant current that would be indicated by an ammeter placed in the primary circuit of the transformer. It will be noted that this current attains its maximum at a point 14° 2´ later than the load current, giving the angle of lag. Multiplying the apparent watts by the cosine of the angle of lag gives the true watts. Now assuming the diagram to show the full load condition of the transformer, the angle of lag being 14° 2´, the power factor at full load is .97 (.97 being the value of the natural cosine of 14° 2´ as obtained from table, such as on page 451). With no external load on the transformer, the load component of the current is that necessary to make up the core losses. For instance, at 5 amperes, while the magnetizing current remains as before at 25 amperes, the angle of lag becomes 78° 41´ and the power factor .196. It is thus seen that in transformers, induction motors, etc., the power factor is a function of the load.