Testing of d.c.Machines:Main Disadvantages

Main Disadvantages

1. No account is taken of the change in iron losses from no-load to full-load. At full-load, due to armature reaction, flux is distorted which increases the iron losses in some cases by as much as 50%.

2. As the test is on no-load, it is impossible to know whether commutation would be satisfactory at full-load and whether the temperature rise would be within the specified limits.

Example 31.3. A 220 V, d.c. shunt motor at no load takes a current of 2.5 A. The resistances of the armature and shunt field are 0.8 W and 200 W respectively. Estimate the efficiency of the motor when the input current is 20 A. State precisely the assumptions made.

(Electrical Technology, Kerala Univ. 1986)

Solution. No-load input = 220 ´ 2.5 = 550 W

This input meets all kinds of no-load losses i.e. armature Cu loss and constant losses.

Ish = 220/200 = 1.1 A. No-load arm current, Ia0 = 2.5 – 1.1 = 1.4 A

Example 31.8. In a test on a d.c. shunt generator whose full-load output is 200 kW at 250 V, the following figures were obtained :

(a) When running light as a motor at full speed, the line current was 36 A, the field current 12 A, and the supply voltage 250.

(b) With the machine at rest, a p.d. of 6 V produced a current of 400 A through the armature circuit. Explain how these results may be utilised to obtain the efficiency of generator at full-load and half-load. Neglect brush voltage drop.