Test: Electrical Machines- 2 - Electrical Engineering (EE) MCQ

Assuming the 110-V side 1 and the 220-V side to be side 2, N₁/N₂ = 0.5

Total leakage reactance referred to side 1

E₁ = 4.44 B_mNAf
⇒ 400 = 4.44 × Bm × 50 × 10⁻⁴ × 50 × 350
⇒ B_m = 1.03T

No load current I₀ = 1.3 A
Iron loss component I_W = 0.5 A

Magnetizing component of no load current

EMT equation of Transformer is given by 
Where f is the flux density, given as ϕ = B × A
Now, the given radius is reduced by half, i.e.  r → r/2,
So, the area will be reduced by 1/4 i.e. to maintain no load current constant (in turn emf (EE)) to be constant, number of turns will be 4 times

Since maximum efficiency occurs at 80% of full load at 0.8 P.F
Output at η_max = (100 × 0.8) × 0.8 = 64 kW

∴ Total loss = 65.3 – 64 = 1.3 kW
This loss is divided equally between cu and iron
∴ Cu loss at 80% of full load = 1.3/2 = 0.65 kW
Cu loss at full load = 0.65/82 = 1 kW

∴ % Regulation = [1 × 0.8 + 5 × 0.6] = 3.8%

Salving equs (1) and (2) we get
P_i = 11.846
P_c = 19.734
at 3/4 th full load.

Eddy current loss = 150 W
No load loss W₀ = 300 W

Now W₀ = W_h + W_e
300 = W_h + 150
W_h = 150 W
W_h ∝ f

W_h ∝ f²

∴ Total no load loss = W_e2 + W_h2 = 600 + 300 = 900 W

P.F. = Cas^-1­ 0.8 = 36.86°
Full load voltage
V₂ = E₂ – I₂ ∠-ϕ (R₂ + jX₂) = 440 – 10∠-36.86 (0.12 + j2.42)
= 424.93 ∠-2.51° V

Given that no load losses components are equally divided
W_h = W_e = 10W
Initially test is conducted on LV side
Now V/f ratio is 100/50 = 2
In HV side, applied voltage is 160 V, this voltage on LV side is equal to 80 V.
Now V/f ratio is constant, W^h ∝ f, W_e ∝ f²

In SC test
I(HV side) = 5 A
I(LV side) = 10 A
As the SC tests were conducted at rated current on both sides, the copper losses are same.

At 25 Hz, x_e = 0.2615Ω, r_e = 0.1067Ω
⇒ z_e = 0.2822Ω