Page 593 - Basic Electrical Engineering

Example 6.18 Calculate the all-day efficiency of a 25 kVA distribution

transformer whose loading pattern is as follows:

15 kW at 0.8 power factor for 6 hours
12 kW at 0.7 power factor for 6 hours

10 kW at 0.9 power factor for 8 hours

Negligible load for 4 hours.

The core loss is 500 W and full-load copper loss is 800 W.

Solution:

Output of the transformer in 24 hours is calculated as

15 kW ×6 + 12kW ×6 + 10kW ×8 +0×4
= 242 kWh = output energy

We have to calculate the core loss and copper loss for 24 hours at different
loading conditions as Core loss remains constant at all loads. Therefore,

Core loss for 24 hours = 0.5 × 24 = 12 kWh
= Energy lost in the core

Copper loss varies as the square of the load. The loads on transformer have to
be calculated in terms of kVA.