A 40KVA, single phase transformer has core loss of 450 Watts and full load copper loss 850 Watts. If the power factor of the load is 0.8. Calculate i) Full load efficiency ii) Maximum efficiency at UPF iii) Load for maximum efficiency. * 8 points i) 96.09% ii) 97.26% i) 97.09% ii) 98.26% i) 98.09% ii) 99.26%?

Question

Answer

Solution:

Given data:

KVA rating (S) = 40KVA

Core loss (Wc) = 450W

Full load copper loss (Wcu) = 850W

Power factor (pf) = 0.8

i) Full load efficiency:

The efficiency of the transformer is given by:

Efficiency (η) = Output power / Input power

Output power (Pout) = S x pf

Input power (Pin) = Pout + Wc + Wcu

Substituting the given values, we get:

Efficiency (η) = (40 x 0.8) / (40 x 0.8 + 450 + 850)

η = 0.9609 or 96.09%

ii) Maximum efficiency at UPF:

At unity power factor (UPF), the efficiency of the transformer is maximum.

Efficiency (η) = Output power / Input power

Output power (Pout) = S x 1

Input power (Pin) = Pout + Wc + Wcu

Substituting the given values, we get:

Efficiency (η) = (40 x 1) / (40 x 1 + 450 + 850)

η = 0.9726 or 97.26%

iii) Load for maximum efficiency:

The load for maximum efficiency can be found using the following formula:

Load for maximum efficiency = √(Wcu / (pf x S))

Substituting the given values, we get:

Load for maximum efficiency = √(850 / (0.8 x 40))

Load for maximum efficiency = 18.75KVA

Therefore, the load for maximum efficiency is 18.75KVA.

Hence, the correct answer is:

i) Full load efficiency: 96.09%

ii) Maximum efficiency at UPF: 97.26%

iii) Load for maximum efficiency: 18.75KVA.

Answer

A

Answer

Answer

Efficiency at any load, η=(x•S•pf)/(x•S•pf+x^2•Wc+Wi), x=1 For full load.
Just substitute the values here, you'll get η at full load =96% (B)..

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