In some generators, a very slight current results even when the terminals of the machine are actually short circuited; that is, due to residual magnetism in the pole pieces, the lower portion of the curve often terminates, not exactly at zero, but at a point some distance along the current line.
The working portion of the curve is from A to C, at which time the machine is supplying a fairly constant voltage. From C to E shows a critical condition of affairs, while the straight portion D O represents the unstable part of the curve caused by the field current being below its proper value.
The position of the point C determines the maximum power the machine is capable of developing, being in this case (47.5 × 25) ÷ 746 = 1.59 horse power.
Ques. How may the commercial efficiency of a generator be determined?
Ans. To obtain the commercial efficiency, the input and output must be found for different loads.
The input may be found by running the generator as a motor at its rated speed, loading it by means of a Prony brake. The generator must be stripped of all belting or other mechanical connections, supplied with its normal voltage and full load current, and the pressure of the Prony brake upon its armature shaft or pulley adjusted until the rated speed of the armature is obtained. The data thus obtained is substituted in the formula.
| 2π L W R | ||
| input in brake horse power = | —————— | (1) |
| 33,000 |
in which
L = length of Prony brake lever;
W = pounds pull at end of lever;
R = revolutions per minute.
The output or electrical horse power for the same load is easily calculated from the formula
| amperes × volts | ||
| output in electrical horse power = | ———————— | (2) |
| 746 |