Question

A 300 KVA transformer has 95% efficiency at full load 0.8 pf lagging
and 96% efficiency at half load, unity pf.w hat is the maximum efficiency (in%) at unity pf load?​

Answer 1

Answer:

Wc = 8.51, Wi = 4.12

Explanation:

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Answer 2

Concept

A transformer is a passive part that transfers electrical energy between circuits, whether they be single circuits or numerous circuits.

Given

A 300 KVA transformer has 95% efficiency at full load 0.8 pf lagging and 96% efficiency at half load, unity pf is given

Find

We have to find the maximum efficiency (in%) at unity pf load

Solution

The steps are as follow:

• Let,

η = Efficiency

pf = power factor

• For 1st condition for full load

η = (kVA x pf)/(kVA x pf x Wcu x Wi)

95 % = (300 x 0.8)/(300 x 0.8 x Wcu x Wi)

Wcu + Wi = 12.63 ---------- (1)

• For 2nd condition

96 % = (300 x 0.5)/(300 x 0.5 x Wcu x Wi)

0.25 Wcu + 0.96 Wi = 6.25 ---------------(2)

• By solving the equation (1) and equation (2) we will get

Wcu = 8.51

Wi = 4.118

Hence the maximum efficiency (in%) at unity pf load will be 8.51

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