A 100-MVA, 14.4-kV, 0.8-PF-lagging, 50-Hz, two-pole, Y-connected synchronous generator has a perunit synchronous reactance of 1.1 and a per-unit armature resistance of 0.011. (a) What are its synchronous reactance and armature resistance in ohms? (b) What is the magnitude of the internal generated

Question

A 100-MVA, 14.4-kV, 0.8-PF-lagging, 50-Hz, two-pole, Y-connected synchronous generator has a perunit synchronous reactance of 1.1 and a per-unit armature resistance of 0.011.

(a) What are its synchronous reactance and armature resistance in ohms?

(b) What is the magnitude of the internal generated voltage [latex] { E }_{ A } [/latex] at the rated conditions? What is its torque angle [latex] \delta [/latex] at these conditions?

(c) Ignoring losses in this generator, what torque must be applied to its shaft by the prime mover at full load?

Accepted Answer

The base phase voltage of this generator is [latex]V_{\phi, ext { base }}=14,400 / \sqrt{3}=8314 \mathrm{~V}[/latex] . Therefore, the base impedance of the generator is

[latex]Z_{ ext {base }}=\frac{3 V_{\phi, ext { base }}{ }^2}{S_{ ext {base }}}=\frac{3(8314 \mathrm{~V})^2}{100,000,000 \mathrm{\ VA}}=2.074 \ \Omega[/latex]

(a) The generator impedance in ohms are:

[latex]\begin{aligned}& R_A=(0.011)(2.074 \ \Omega)=0.0228 \ \Omega \& X_S=(1.1)(2.074 \ \Omega)=2.281 \ \Omega\end{aligned}[/latex]

(b) The rated armature current is

[latex]I_A=I_L=\frac{S}{\sqrt{3} V_T}=\frac{100 \mathrm{\ MVA}}{\sqrt{3}(14.4 \mathrm{\ kV})}=4009 \mathrm{~A}[/latex]

The power factor is 0.8 lagging, so [latex]\mathbf{I}_A=4009 \angle-36.87^{\circ} \mathrm{A}[/latex] . Therefore, the internal generated voltage is

[latex]\begin{aligned}& \mathbf{E}_A=\mathbf{V}_\phi+R_A \mathbf{I}_A+j X_S \mathbf{I}_A \& \mathbf{E}_A=8314 \angle 0^{\circ}+(0.0228 \ \Omega)\left(4009 \angle-36.87^{\circ} \mathrm{A} ight)+j(2.281 \ \Omega)\left(4009 \angle-36.87^{\circ} \mathrm{A} ight) \& \mathbf{E}_A=15,660 \angle 27.6^{\circ} \mathrm{V}\end{aligned}[/latex]

Therefore, the magnitude of the internal generated voltage [latex] { E }_{ A } [/latex] = 15,660 V, and the torque angle [latex] \delta [/latex] = 27.6°.

(c) Ignoring losses, the input power would equal the output power. Since

[latex]P_{ ext {OUT }}=(0.8)(100 \mathrm{\ MVA})=80 \mathrm{\ MW}[/latex]

and

[latex]n_{\mathrm{sync}}=\frac{120 f_{s e}}{P}=\frac{120(50 \mathrm{~Hz})}{2}=3000 \mathrm{\ r} / \mathrm{min}[/latex]

the applied torque would be

[latex] au_{ ext {app }}= au_{ ext {ind }}=\frac{80,000,000 \mathrm{~W}}{(3000 \mathrm{\ r} / \mathrm{min})(2 \ \pi \mathrm{\ rad} / \mathrm{r})(1 \mathrm{~min} / 60 \mathrm{~s})}=254,700 \mathrm{~N}.\mathrm{m}[/latex]

Question 4.7: A 100-MVA, 14.4-kV, 0.8-PF-lagging, 50-Hz, two-pole, Y-conne...

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