NOTE.—Checking up of a recording wattmeter. This may conveniently be done by noting the deflections at short intervals on an ammeter connected in circuit, and also the readings on the dial of the recording wattmeter during this period. If this test be continued for an appreciable time, the product of the pressure in volts, the current in amperes, and the time in hours, should equal the number of watthours recorded on the counters of the dial.
NOTE.—Transformer testing. In the early days of transformer building, before the commercial wattmeter had been perfected, leakage or exciting current was the criterion of good design. After the introduction of the wattmeter, core loss became the all important factor, and for a long time the question of leakage current was lost sight of. With the introduction of silicon steel, leakage or exciting current again assumed prominence. Keeping in mind the fact that all characteristics of a transformer are of more or less importance, it is essential that the user of such apparatus have at hand the necessary facilities for making tests of all such variable quantities. The tests which all users of transformers should make, are given in this chapter.
Ques. Can direct current be measured by an alternating current voltmeter?
Ans. Yes.
Fig. 2,899.—Transformer copper loss by wattmeter measurement and impedance. At first glance, this method would seem better than the calculation of loss after measurement of the resistance. However, it should be noted that the wattmeter is, in itself, subject to considerable error under the low power factor that will exist in this test. The secondary of the transformer is short circuited, and a voltage applied to the primary which is just sufficient to cause full load primary current. If full current pass through the primary of the transformer with the secondary short circuited, the secondary will also carry full load current. With connections as shown, and with the full load current, the voltmeter indicates the impedance volts of the transformer. This divided by the rated voltage gives what is called the per cent. impedance of the transformer. In a commercial transformer of 5 kw., this should be approximately 3 per cent. The iron loss of the transformer under approximately 3 per cent. of the normal voltage will be negligible, and the losses measured will be the sum of the primary and secondary copper losses. As in the discussion of the core loss measurements, the wattmeter readings must be corrected for the loss in its pressure coil, the method of correction being the same as that discussed under the core loss measurement. If the impedance volts, as measured, be divided by the primary current, the impedance of the transformer is obtained. The reciprocal of this quantity is known by the term "admittance." When two or more transformers are connected in parallel they divide the load in proportion to their admittance. It is, therefore, important that the users of transformers know the impedance of the apparatus used, in order to determine whether two or more transformers will operate satisfactorily in parallel. For discussion of wattmeter error on low power factor, see note on page 2,075. For accurate measurement of impedance, the voltmeter should be connected directly across the terminals of the transformer rather than as shown in the diagram.
NOTE.—Transformer copper loss test. The usual and best method of obtaining copper losses is to separately measure the primary and secondary resistance and calculate from these the primary and secondary copper losses. For general diagram of connections and discussion of the drop method, [see fig. 2,875]. The current should be kept well within the load current of the transformer to avoid temperature rise during the test. In other words, the resistance of the coil is the voltage across its terminals divided by the current. The resistance of the primary coil can be measured similarly. The copper loss in watts in each coil will then be the product of the resistance and the square of the rated current for that coil. The total copper loss will be the sum.
Ques. What would be the effect of placing a direct voltmeter across an alternating current circuit, and why?
Ans. There would be no deflection of the pointer owing to the rapid reversals of the alternating current.