All-day efficiency is the ratio of energy (kWH) delivered in a 24 hour period divided by the energy (kWH) input in the same length of time.
Transformers used on residence-lighting circuits (and distribution circuits generally) are either idle or only lightly loaded during much of 24 hour period. However, they must at all times be connected to the line and ready to serve, so that the core losses are being supplied continually. It is therefore very important that such transformers be designed for minimum core loss. The copper losses of the time, distribution transformers are designed for relatively large full load copper loss and have their maximum power efficiencies at light loads. This design results in improved all day efficiency for these transformers. Power transformers, on the other hand are loaded more or less continuously and are designed for full load copper losses equal to about twice the no-load losses.
To calculate all day efficiency, it is necessary to know how the load on the transformer varies from hour to hour. The quotient obtained by dividing the energy output plus energy losses over a 24-hour period yields the efficiency expressed as a decimal fraction.
The use of a load factor facilitates practical calculations.
Example 48. A 15 kVA, 2000/200 V transformer has an iron loss of 250 W and full-load copper loss 350 W. During the day it is loaded as follows :
No. of hours |
Load |
Power factor |
9 |
¼ load |
0.6 |
7 |
Full load |
0.8 |
6 |
¾ load |
1.0 |
2 |
No-load |
- |
Calculate the all-day efficiency.
Solution. Rating of transformer = 15 kVA
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