Question

a 500 kva transformer has full load efficiency of 95% at upf. it gives the same efficiency at 60% of full load and at upf . the iron loss of the transformer is

Answer 1

ANSWER

At maximum efficiency copper losses equal to iron losses by using efficiency formula u can find out iron losses and substitute that value ,to find efficiency at half load.

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Answer 2

26.31 kW is the total loss in the iron transformer.

Given:

Efficiency = 95% for full load and a 500 kva transformer.

To Find:

The iron loss of the transformer

Solution:

Let X be the load of the transformer.

Efficiency η = $\frac{x * S *cos\alpha }{({x * S *cos\alpha)}+ p_{i} +x^{2} p_{cu} } * 100 %$

For full load x= 1

$0.95 =\frac{1 * 500 *1 }{({1 * 500 *1)}+ p_{i} +1^{2} p_{cu} } * 100 %$

⇒$p_{i} + p_{cu} = 26.315$ --------------(1)

At 60% efficiency , x= 1.6

$0.95 =\frac{1 * 300 *1 }{({1 * 300 *1)}+ p_{i} +0.36 p_{cu} } * 100 %$

⇒$p_{i} + 0.36p_{cu} = 15.789$ --------------(2)

From Equations (1) and (2)

On solving

0.64 $p_{cu}$ = 10.526

$p_{cu}$ = 16.44 KW

$p_{i}$ = 26.315 - 16.44 = 9.86 KW

The input to the transformer must be (1/0.95) X 500 kW= 526.315 kW

The difference between input and output power

=  526.31 - 500 kW

= 26.31 kW

26.31 kW is the total loss in the iron transformer.

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