Question

A transformer is connected to a 2200 V, 40 Hz supply. the core loss is 800 Watts out of 600 Watts are due to hysteresis and the remaining eddy current losses. Determine the core loss if the supply voltage and frequency are 3300 V and 60 Hz respectively:
1. 450 Watts
2. 900 Watts
3. 1350 Watts
4. None of these

Answer 1

Correct Answer - Option 3 : 1350 Watts

Concept:

Separation of Iron loss in Transformer:

In transformer iron loss (Pi) has two components namely Hysteresis loss (Ph) and Eddy current loss (Pe).

Pi = Ph + Pe

\({P_i} = {k_h}fB_m^n + {k_e}{f^2}B_m^2\)  …. (1)

Where,

Kh and ke are loss coefficient constant and their value depends on the type of material

Bm is the maximum value of flux density

f is supply frequency

\({P_h} \propto f\)

\({P_h} = af\)

\({P_e} \propto {f^2}\)

\({P_e} = b{f^2}\)

Where a and b are constants.

\({P_i} = af + b{f^2}\)

\(\frac{{{P_i}}}{f} = a + bf\)

Note: For the separation of these two losses, the no-load test is performed on the transformer.

Calculation:

Given,

V1 = 2200 V

f1 = 40 Hz

V2 = 3300 V

f2 = 60 Hz

\(\frac{{{V_1}}}{{{f_1}}} = \frac{{{V_2}}}{{{f_2}}} = \frac{2200}{40}=\frac{3300}{60}=55\)

Hence, (V / f) ratio is constant,

At frequency 40 Hz

P_i = 800 W = P_h + P_e

P_h = 600 W = af

a = 600/40 = 15

Now,

P_e = 800 - 600 = 200 W = bf²

∴ b = \(\frac{200}{f^2}=\frac{200}{1600}=0.125\)

At, supply 3300 V and 60 Hz,

P_i = af + bf² = (15 × 60) + (0.125 × 60²) = 900 + 450

∴ P_i = 1350 W