Question

A 5 KVA transformer has 35 watt core loss and 40 watt cu loss at full load. It operates at rated KVA and
0.8 p.f lag for 6 hours. Half rated KVA and 0.5 p.f. lag for 12hrs and no load for 6 hrs.
Find all day efficiency.

Answer 1

. A single phase transformer is rated at 100 KVA. At full load its copper loss is 1200 Watts and its iron loss is 960 Watts. Calculate.

Answer 2

Answer:

All day efficiency = 98.7%.

Explanation:

kWh output in 24hr = $=P=6.25*6+5*12+6*0=97.5kWh$

Iron loss in 24hr = $P_i=\frac{35}{1000}*24 = 0.84kWh$

Copper loss in 24hr = $P_c=1^2*\frac{40}{1000} *6+(0.5)^2*\frac{40}{1000} *12+(0)*\frac{40}{1000}*6$

$=0.24+0.12+0=0.36kWh$

All day efficiency = $\eta_{eff} = \frac{P}{P_i+P_c}$

$\eta_{eff}=\frac{97.5}{97.5+0.84+0.36}=0.987$

$\eta_{eff}=98.7$%

All day efficiency = 98.7%.

#SPJ2