The machine of Example 7.11 is connected to an infinite bus through a link with reactance of 0.2 p.u. The excitation voltage is 1.3 p.u. and the infinite-bus voltage is maintained at 1 p.u. For a power angle of 25 deg, compute the active and reactive power supplied to the bus.

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Accepted Answer

We have

[latex]\begin{array}{l}{{X_{d}=0.95+0.2=1.15\,\mathrm{p.u.}}}\ {{X_{q}=0.7+0.2=0.9\mathrm{p.u.}}}\end{array} \ E_{f}=1.3\qquad\delta=25\;\mathrm{deg}\qquad V_{t}=1[/latex]

As a result, we use Eq. (7.69) to obtain

[latex]P=\frac{V E_{f}}{X_{d}}\sin\delta+\frac{V^{2}}{2}\left(\frac{1}{X_{q}}-\frac{1}{X_{d}} ight)\sin2\delta[/latex] (7.69)

[latex]P={\frac{E_{f}V}{X_{d}}}\sin\delta+{\frac{V^{2}}{2}}\left({\frac{1}{X_{q}}}-{\frac{1}{X_{d}}} ight)\sin2\delta \ ={\frac{1.3(1)}{1.15}}\sin25\,\mathrm{deg}+{\frac{1}{2}}\left({\frac{1}{0.9}}-{\frac{1}{1.15}} ight)\sin50\,\mathrm{deg} \ =0.57\,{\mathrm{p.u}}.[/latex]

Equation (7.70) yields

[latex]Q=\frac{V E_{f}}{X_{d}}\cos\delta-V^{2}\biggl(\frac{\cos^{2}\!\delta}{X_{d}}+\frac{\sin^{2}\!\delta}{X_{q}}\biggr)[/latex] (7.70)

[latex]Q=\frac{E_{f}V}{X_{d}}\cos\delta-V^{2}\biggl(\frac{\cos^{2}\delta}{X_{d}}+\frac{\sin^{2}\delta}{X_{q}}\biggr) \ ={\frac{1.3(1)}{1.15}}\cos25\operatorname{deg}-\left({\frac{\cos^{2}25}{1.15}}+{\frac{\sin^{2}25}{0.9}} ight) \ =0.11[/latex]

which are the required answers.

Question 7.12: The machine of Example 7.11 is connected to an infinite bus ......

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