The following results were obtained on a 50 kVA transformer. Opencircuit test – primary voltage, 3300 V; secondary voltage, 400 V; primary power, 430 W. Short-circuit test – primary voltage, 124 V; primary current, 15.3 A; primary power, 525 W; secondary current, full-load value. Calculate: (a) the

Question

The following results were obtained on a 50 kVA transformer. Opencircuit test – primary voltage, 3300 V; secondary voltage, 400 V; primary power, 430 W. Short-circuit test – primary voltage, 124 V; primary current, 15.3 A; primary power, 525 W; secondary current, full-load value. Calculate:

(a) the efficiencies at full load and at half load for 0.7 power factor;

(b) the voltage regulations for power factor 0.7, (i) lagging, (ii) leading;

(c) the secondary terminal voltages corresponding to (i) and (ii).

Accepted Answer

(a) Core loss[latex]=430 \mathrm{~W}[/latex]

[latex]I^{2} R[/latex] loss on full load [latex]=525 \mathrm{~W}[/latex]

[latex] herefore \quad \quad \quad[/latex] Total loss on full load [latex]=955 \mathrm{~W}=0.955 \mathrm{~kW}[/latex]

and [latex]\quad \quad [/latex] Efficiency on full load [latex]=\frac{50 imes 0.7}{(50 imes 0.7)+0.955}[/latex]

[latex]\begin{aligned}=\left(1-\frac{0.955}{35.95} ight) &=\mathbf{0.973} \mathbf{per unit } \&=\mathbf{97.3} \mathbf{ per cent }\end{aligned}[/latex]

[latex]I^{2} R[/latex] loss on half load [latex]=525 imes(0.5)^{2}=131 \mathrm{~W}[/latex]

[latex] herefore \quad[/latex] Total loss on half load [latex]=430+131=561 \mathrm{~W}=0.561 \mathrm{~kW}[/latex]

and Efficiency on half load [latex]=\frac{25 imes 0.7}{(25 imes 0.7)+0.561}[/latex]

[latex]\begin{aligned}=\left(1-\frac{0.561}{18.06} ight) &=\mathbf{0 . 9 6 9} \mathbf { per unit } \&=\mathbf{9 6 . 9} \mathbf{per cent }\end{aligned}[/latex]

(b) [latex]\quad \quad \quad\quad \cos \phi_{\mathrm{e}}=\frac{525}{124 imes 15.3}=0.2765[/latex]

[latex] herefore \quad \quad \quad \quad \phi_{\mathrm{e}}=73^{\circ} 57^{\prime}[/latex]

For [latex]\cos \phi_{2}=0.7[/latex]

[latex]\phi_{2}=45^{\circ} 34^{\prime}[/latex]

From expression [32.22],

[latex] ext { Per-unit voltage regulation} = \frac{V_{sc} \cos \left(\phi_{e}-\phi_{2} ight)}{V_1}[/latex] [32.22]

for power factor [latex]0.7[/latex] lagging,

[latex]\begin{aligned} ext { Voltage regulation } &=\frac{124 \cos \left(73^\circ 57^{\prime}-45^{\circ} 34^{\prime} ight)}{3300} \&=\mathbf{0 . 0 3 3} \mathbf{per unit }=\mathbf{3 .3} \mathbf{ per cent }\end{aligned}[/latex]

For power factor [latex]0.7[/latex] leading,

[latex]\begin{aligned} ext { Voltage regulation } &=\frac{124 \cos \left(73^{\circ} 57^{\prime}+45^{\circ} 34^{\prime} ight)}{3300} \&=-\mathbf{0.0185} \mathbf{per unit } \&=-\mathbf{1.85} \mathbf{per cent }\end{aligned}[/latex]

(c) Secondary voltage on open circuit [latex]=400 \mathrm{~V}[/latex]. Therefore secondary voltage on full load, p.f. [latex]0.7[/latex] lagging, is

[latex]400(1 - 0.033)=\mathbf{3 8 7} \mathbf{V}[/latex]

and secondary voltage on full load, p.f. [latex]0.7[/latex] leading, is

[latex]400(1+0.0185)=\mathbf{407} \mathbf{~V}[/latex]

Question 32.8: The following results were obtained on a 50 kVA transformer....

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