A single-phase, 500 V, 50 Hz alternator produces a short-circuit current of 170 A and an open circuit emf of 425 V when a field current of 15A passes through its field winding. If its armature has an effective resistance of 0.2 ohm, determine its full-load regulation at unity pf and at 0.8 pf

Question

A single-phase, 500 V, 50 Hz alternator produces a short-circuit current of 170 A and an open circuit emf of 425 V when a field current of 15A passes through its field winding. If its armature has an effective resistance of 0.2 ohm, determine its full-load regulation at unity pf and at 0.8 pf lagging.

Accepted Answer

Here, [latex]Rated power = 50 kVA = 50 \times 10^{3} VA[/latex]

Terminal voltage, [latex]V = 500 V[/latex]; Armature resistance, [latex]R = 0.2 ohm[/latex]

Short circuit current, [latex]I_{sc} = 170 A[/latex]; Open circuit emf, [latex]E = 425 V[/latex]

Synchronous impedance, [latex]Z_{s} = \frac{E}{I_{SC}} = \frac{425}{170} = 2.5 ohm[/latex]

Synchronous reactance, [latex]X_{s} = \sqrt{\left(Z_{s}\right)^{2} - \left(R\right)^{2} } = \sqrt{\left(2.5\right)^{2} - \left(0.2\right)^{2} } = 2.492 ohm[/latex]

Full load current, [latex]I = \frac{50 \times 10^{3}}{500} = 100 A[/latex]

When [latex] p.f. \cos \phi = 1; \sin \phi = 0[/latex]

[latex] E_{0} = \sqrt{\left(V \cos \phi + IR\right)^{2} + \left(IX_{S}\right)^{2} } \[0.5cm] \quad \; = \sqrt{\left(500 \times 1 + 100 \times 0.2\right)^{2} + \left(100 \times 2.492\right)^{2}} \[0.5cm] \quad \; = 576.63 V[/latex]

[latex]\% Reg. = \frac{E_{0} - V}{V} \times 100 = \frac{576.63 - 500}{500} \times 100 \[0.5cm] \quad \quad \quad = \mathbf{15.326 \%} [/latex]

When [latex]p.f. \cos \phi = 0.8 lagging; \sin \phi = \sin \cos^{–1} 0.8 = 0.6 [/latex]

[latex] E_{0} = \sqrt{\left(V \cos \phi + IR\right)^{2} + \left(V \sin \phi + IX_{S}\right)^{2} } \[0.5cm] \quad \; = \sqrt{\left(500 \times 0.8 + 100 \times 0.2\right)^{2} + \left(500 \times 0.6 + 100 \times 2.492\right)^{2}} \[0.5cm] \quad \; = \sqrt{\left(420\right)^{2} + \left(549.2\right)^{2}} = 671.62 V[/latex]

[latex]\% Reg. = \frac{E_{0} - V}{V} \times 100 = \frac{671.62 - 500}{500} \times 100 \[0.5cm] \quad \quad \quad = \mathbf{34.35 \%} [/latex]

Question 6.18: A single-phase, 500 V, 50 Hz alternator produces a short-cir...

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