Question

A single phase transformer working at unity power factor has an efficiency of 90% at both half load and at full
load of 500 W. Determine the efficiency at 75% full load and the maximum efficiency

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

Explanation:

Answer 2

Given:

single-phase transformer working at unity power factor has an efficiency of 90%

load = 500 W

efficiency at 75% full load

We count on that there are best 2 critical additives of losses*)

which upload as much as approximately 11.111111 % below the cited conditions; examine the subsequent equations.

Vcu(100) + Vfe = 55.555 W

Vcu(50) + Vfe = 27.778 W

Vcu(100) = four x Vcu(50)

First of all,

we will remove Vfe with the aid of using subtracting the primary 2 equations and the result is

Vcu(100) - Vcu(50) = 27.778 W

Well,

now we update Vcu(100) with four x Vcu(50) and the result is:

three x Vcu(50) = 27.778 W and

Vcu(50) = (27.778 / 3) W = 9.259 W

Vfe = 2 x Vcu(50) = 2/three x 27.778 W = 18.519 W

For to reply to the unique query we want to understand losses at 75 % of 500 W

that's 150 % of 1/2 of load (= 250 W)

identical to 375 W.

Vcu(75) = (3/2)^2 x Vcu(50)

= 9/4 x 9.259 W = 20.833 W

Now, for general losses at 375 W load we must upload the middle loss:

Vcu(75) + Vfe = 20.833 W + 18.519 W = 39.352 W

Efficiency equals 1 minus those ensuing losses divided with the aid of using the sum of 375 W and those losses.

1 - ((39.352 W / (375 W + 39.352 W))

= 1 - (39.352 W / 414.352 W) = 90.5027 %

So,

the performance at 75 % of the complete load is ready at 90.50 %.

And those are the values for entering and output:

If you get 375 W out of that transformer enter is ready 414.352 W with losses of approximately 39.352 W.

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