A 13.8-kV, 50-MVA, 0.9-power-factor-lagging, 60-Hz, four-pole Y-connected synchronous generator has a direct-axis reactance of 2.5 Ω, a quadrature-axis reactance of 1.8 Ω, and an armature resistance of 0.2 Ω. Friction, windage, and stray losses may be assumed negligible. The generator’s open

Question

A 13.8-kV, 50-MVA, 0.9-power-factor-lagging, 60-Hz, four-pole Y-connected synchronous generator has a direct-axis reactance of 2.5 Ω, a quadrature-axis reactance of 1.8 Ω, and an armature resistance of 0.2 Ω. Friction, windage, and stray losses may be assumed negligible. The generator’s open-circuit characteristic is given by Figure P5-1.

Figure P4-1

(a) How much field current is required to make the terminal voltage [latex]V_T[/latex] (or line voltage [latex]V_T[/latex] ) equal to 13.8 kV when the generator is running at no load?
(b) What is the internal generated voltage of this machine when it is operating at rated conditions? How does this value of [latex]\mathbf{E}_A
[/latex] compare to that of Problem 4-2b?
(c) What fraction of this generator’s full-load power is due to the reluctance torque of the rotor?

Accepted Answer

(a) If the no-load terminal voltage is 13.8 kV, the required field current can be read directly from the open-circuit characteristic. It is 3.50 A.

(b) This generator is Y-connected, so [latex]I_L=I_A[/latex] . At rated conditions, the line and phase current in this generator is

[latex]I_A=I_L=\frac{P}{\sqrt{3} V_L}=\frac{50 \mathrm{\ MVA}}{\sqrt{3}(13800 \mathrm{~V})}=2092 \mathrm{~A} ext { at an angle of }-25.8^{\circ}[/latex]

The phase voltage of this machine is [latex]V_\phi=V_T / \sqrt{3}=7967 \mathrm{~V}[/latex] . The internal generated voltage of the machine is

[latex]\begin{aligned}& \mathbf{E}_A^{\prime \prime}=\mathbf{V}_\phi+R_A \mathbf{I}_A+j X_q \mathbf{I}_A \& \mathbf{E}_A^{\prime \prime}=7967 \angle 0^{\circ}+(0.20 \ \Omega)\left(2092 \angle-25.8^{\circ} \mathrm{A} ight)+j(1.8 \ \Omega)\left(2092 \angle-25.8^{\circ} \mathrm{A} ight) \& \mathbf{E}_A^{\prime \prime}=10485 \angle 17.8^{\circ} \mathrm{V}\end{aligned}[/latex]

Therefore, the torque angle [latex]\delta[/latex] is 17.8°. The direct-axis current is

[latex]\begin{aligned}& \mathbf{I}_d=I_A \sin ( heta+\delta) \angle \delta-90^{\circ} \& \mathbf{I}_d=(2092 \mathrm{~A}) \sin \left(43.6^{\circ} ight) \angle-72.2^{\circ} \& \mathbf{I}_d=1443 \angle-72.2^{\circ} \mathrm{A}\end{aligned}[/latex]

The quadrature-axis current is

[latex]\begin{aligned}& \mathbf{I}_q=I_A \cos ( heta+\delta) \angle \delta \& \mathbf{I}_q=(2092 \mathrm{~A}) \cos \left(43.6^{\circ} ight) \angle 17.8^{\circ} \& \mathbf{I}_q=1515 \angle 17.8^{\circ} \mathrm{A}\end{aligned}[/latex]

Therefore, the internal generated voltage of the machine is

[latex]\begin{aligned}& \mathbf{E}_A=\mathbf{V}_\phi+R_A \mathbf{I}_A+j X_d \mathbf{I}_d+j X_q \mathbf{I}_q \& \mathbf{E}_A=7967 \angle 0^{\circ}+(0.20 \ \Omega)\left(2092 \angle-25.8^{\circ} ight)+j(2.5 \ \Omega)\left(1443 \angle-72.2^{\circ} ight)+j(1.8 \ \Omega)\left(1515 \angle 17.8^{\circ} ight) \& \mathbf{E}_A=11496 \angle 17.8^{\circ} \mathrm{\ V}\end{aligned}[/latex]

[latex]\left|\mathbf{E}_A ight|[/latex] is approximately the same magnitude here as in Problem 4-2b, but the angle δ is quite different.

(c) The power supplied by this machine is given by the equation

[latex]\begin{aligned}& P=\frac{3 V_\phi E_A}{X_d} \sin \delta+\frac{3 V_\phi^2}{2}\left\lgroup\frac{X_d-X_q}{X_d X_q} ight group \sin 2 \delta \& P=\frac{3(7967)(11496)}{2.5} \sin 17.8^{\circ}+\frac{3(7967)^2}{2}\left\lgroup\frac{2.5-1.8}{(2.5)(1.8)} ight group \sin 35.6^{\circ}\end{aligned}[/latex]

P = 33.6 MW + 8.6 MW = 42.2 MW

The cylindrial rotor term is 33.6 MW, and the reluctance term is 8.6 MW, so the reluctance torque accounts for about 20% of the power in this generator.

Question C.1: A 13.8-kV, 50-MVA, 0.9-power-factor-lagging, 60-Hz, four-pol...

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