Answer :

Solution :

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

(P_e). P_i = P_h + P_e

$P_i=k_{h}f{B}_m^n+k_e{f}^2{B}_m^2$

Where,

K_h and k_e are loss coefficient constant and their value depends on the type of material

B_m is the maximum value of flux density

f is supply frequency

P _h ∝ f

P _h = a f

P _e ∝ f ²

P _e = b f ²

Where a and b are constants.

P _i = a f + 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 = 440 V

f₁ = 50 Hz

V₂ = 220 V

f₂ = 25 Hz

$\frac{V_1}{f_1}=\frac{V_2}{f_2}=8.8$

Hence, (V / f) ratio is constant,

P₁ = 2500 W

P₂ = 850 W

From the above concept,

P_i = af + bf²

2500 = 50a + 2500b .... (1)

850 = 25a + 625b .... (2)

From equation (1) and (2),

a = 18 and b = 0.64 Iron loss at 50 Hz,

Hysteresis loss (P_h) = af = 18 × 50 = 900 W

Eddy current loss (P_e) = bf² = 0.64 × 2500 = 1600 W