If it be desired to separate the load losses from the true I2R loss, the resistances can be measured, and the I2R loss calculated and subtracted from the wattmeter reading. The losses being known, the efficiency at any load is readily found by taking the core loss as constant and the copper loss as varying proportionally to the square of the load. Thus,
| output | |||
| efficiency | = | × 100 | |
| output + losses |
All Day Efficiency of Transformers.—This denotes the ratio of the total watt hour output of a transformer to the total watt hour input taken over a working day. To compute this efficiency it is necessary to know the load curve of the transformer over a day. Suppose that this is equivalent to 5 hours at full load, and 19 hours at no load. Then, if W1 be the core loss in watts, W2 the copper loss at rated load, and W the rated output,
- output = 5 × W,
- losses = 5 (W1 + W2) + 19 W1,
- input = 5 (W + W1 + W2) + 19 W1,
and the all day efficiency is equal to
| 5W × 100 | |
| per cent. | |
| 5(W + W1 + W2) + 19W1 |
Commercial or all day efficiency is a most important point in a good transformer. The principal factor in securing a high all day efficiency is to keep the core loss as low as possible. The core loss is constant—it continues while current is supplied to the primary, while copper loss takes place only when the secondary is delivering energy.
In general, if a transformer is to be operated at light loads the greater part of the day, it is much more economical to use one designed for a small iron loss than for a small full load copper loss.
