The secondary of a 500 kVA, 4400/500 V, 50 Hz, single-phase transformer has 500 turns. Determine (i) emf per turn, (ii) primary turns, (iii) secondary full-load current, (iv) maximum flux, (v) gross cross-sectional area of the core for flux density of 1.2 tesla and iron factor is 0.92, and (vi) if

Question

The secondary of a 500 kVA, 4400/500 V, 50 Hz, single-phase transformer has 500 turns. Determine (i) emf per turn, (ii) primary turns, (iii) secondary full-load current, (iv) maximum flux, (v) gross cross-sectional area of the core for flux density of 1.2 tesla and iron factor is 0.92, and (vi) if the core is of square cross-section finds the width of the limb.

Accepted Answer

Here, Rating = 500 kVA; [latex]E_{1} = V_{1}[/latex] = 4400 V; [latex]E_{2} = V_{2}[/latex] = 500 V

f = 50 Hz; [latex]N_{2} = 500 ; B_{m} = 1.2 T ; k_{i} = 0.92[/latex]

(i) emf per turn [latex] = \frac{E_{2}}{N_{2}} = \frac{500}{500} = 1.0 V/turn[/latex]

(ii) emf per turn [latex] = \frac{E_{2}}{N_{2}} = \frac{E_{1}}{N_{1}} = \frac{4400}{N_{1}} = 1.0[/latex]

∴ Primary turns, [latex] N_{1} = \frac{4400}{1.0} = 4400[/latex]

(iii) Secondary full-load current, [latex]I_{2} = \frac{ ext{KVA} imes 1000}{V_{2}} = \frac{500 imes 1000}{500} = 1000 A[/latex]

(iv) Maximum flux, [latex] \phi_{m} = \frac{E_{2}}{4.44 imes N_{2} imes f} = \frac{500}{4.44 imes 500 imes 50} = 4.5 mWb[/latex]

(v) Iron area of the core, [latex] A_{i} = \frac{\phi_{m}}{B_{m}} = \frac{4.5 imes 10^{-3}}{1.2} = 37.54 imes 10^{-4} m^{2} = 37.54 cm^{2}[/latex]

Gross area of the core, [latex] A_{g} = \frac{A_{i}}{k_{i}} = \frac{37.54}{0.92} = 4.8 cm^{2}[/latex]

(vi) Width of squared limb [latex] \sqrt{A_{g}} = \sqrt{40.8} = 6.39 cm[/latex]

Question 10.6: The secondary of a 500 kVA, 4400/500 V, 50 Hz, single-phase ...

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