Question 3.27: Calculate the all-day efficiency of a 25 kVA distribution tr...
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.
Output of the transformer in 24 hours is calculated as
15 kW × 6 + 12 kW × 6 + 10 kW × 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.
15 kW at 0.8 p.f. =\frac{15 kW}{0.8} = 18.75 kVA
12 kW at 0.7 p.f. =\frac{12 kW}{0.7} = 17.14 kVA
10 kW at 0.9 p.f. =\frac{10 kW}{0.9} = 11.11 kVA
at 25 kVA load copper loss = 0.8 kW.
at 18.75 kVA load copper loss = 0.8 × \left( \frac{18.75}{25}\right)^{2} = 450 W =0.45 kW
at 17.14 kVA load copper loss = 0.8 × \left( \frac{17.14}{25}\right)^{2} = 376 W =0.376 kW
at 11.11 kVA load copper loss = 0.8 × \left( \frac{11.11}{25}\right)^{2} = 157 W
=0.157 kW
All day efficiency
Thus \eta _{all-day} = 0.929 = 92.9 per cent