Question

A 230V, 2.5 kVA single phase transformer has an
iron loss of 100 watt at 40 Hz and 70 Watt at 30
Hz. Find the hysteresis and eddy current losses at
50 Hz.​

Answer 1

Explanation:

How do you find the conjugate of a cubed root?

Since we have a cube root in the numerator, we need to multiply by the cube root of an expression that will give us a perfect cube under the cube root in the numerator. Step 3: Simplify the fraction if needed. Step 1: Find the conjugate of the denominator. In general the conjugate of a + b is a - b and vice versa

Answer 2

The hysteresis losses is 90 Watt

The eddy current losses is 42.5 Watt

Explanation:

Given:

A 230 V, 2.5 KVA single phase transformer has an iron loss of 100 watt at 40 Hz and 70 watt at 30 Hz

To find out:

The hysteresis and eddy current losses at 50 Hz

Solution:

Hysteresis loss is proportional to frequency $f$ and eddy loss is proportional to square of the frequency

Thus,

Hysteresis loss

$W_h=K_hf$

And

Eddy loss

$W_e=K_ef^2$

Total iron loss

$W=W_h+W_e$

$\implies W=K_hf+K_ef^2$

When $W=100$ Watt,

$f=40$ Hz

$100=K_h\times 40+K_e\times 40^2$

or, K_h + 40K_e = 2.5                    (1)

When $W=70$ Watt,

$f=30$ Hz

$70=K_h\times 30+K_e\times 30^2$

or, 3K_h + 90 K_e = 7                     (2)

Multiplying equation (1) by 3 and subtracting eq (2) from eq (1)

$30K_e=7.5-7$

$\imples K_e=0.5/30=0.017$

Putting the value of $K_e$ in eq (1)

$K_h+40\times 0.017=2.5$

$\implies K_h=1.8$

Thus,

At 50 Hz

$W_h=1.8\times 50=90$ Watt

$W_e=0.017\times 50^2$

$\implies W_e=42.5$ Watt

Hope this answer is helpful.