A 22 kV, 3-phase, star-connected turbo-alternator with a synchronous impedance of $\text { j } 1.2 \Omega / \text { phase }$ is delivering 230 MW at UPF to 22 kV grid. With the turbine power remaining constant, the alternator excitation is increased by 30%. Determine machine current and power factor based upon linearity assumption.

At the new excitation, the turbine power is now increased till the machine delivers 275 MW. Calculate the new current and power factor

Question

A 22 kV, 3-phase, star-connected turbo-alternator with a synchronous impedance of $\text { j } 1.2 \Omega / \text { phase }$ is delivering 230 MW at UPF to 22 kV grid. With the turbine power remaining constant, the alternator excitation is increased by 30%. Determine machine current and power factor based upon linearity assumption.

At the new excitation, the turbine power is now increased till the machine delivers 275 MW. Calculate the new current and power factor

Accepted Answer

230 MW at UPF

[latex] I_{a}=\frac{230}{\sqrt{3} imes 22 imes 1}=6.04 kA , \bar{I}_{a}=6.04 \angle 0^{\circ} kA[/latex]

[latex] \bar{V}_{t}=(22 / \sqrt{3}) \angle 0^{\circ}=12.7 \angle 0^{\circ} kV[/latex]

With reference to Fig. 8.112(a)

[latex] \bar{E}_{f}=12.7 \angle 0^{\circ}+j 1.2 imes 6.04 \angle 0^{\circ}=12.7+j 7.25[/latex]

Or [latex] E_{f}=14.62 kV [/latex]

Excitation increased by 30%; turbine power input constant at 230 MW, refer Fig. 8.112(b)

[latex] E_{f}^{\prime}=14.62 imes 1.3=19 kV[/latex]

[latex] \frac{230}{3}=\frac{19 imes 12.7}{1.2} \sin \delta^{\prime}[/latex]

Or [latex] \delta^{\prime}=22.4^{\circ}[/latex]

[latex] \bar{I}_{a}^{\prime}=\frac{19 \angle 22.4^{\circ}-12.7 \angle 0^{\circ}}{j 1.2}=7.275 \angle-33.9^{\circ} kA[/latex]

[latex] I_{a}=7.275 kA[/latex]

[latex] p_{f}=\cos 33.9^{\circ}=0.83 [/latex] lagging

Excitation constant at new value; turbine power increased to 275 MW

[latex] \frac{275}{3}=\frac{19 imes 12.7}{1.2} \sin \delta^{\prime \prime} [/latex]

Or [latex] \delta^{\prime \prime}=27.1^{\circ} [/latex]

[latex] \bar{I}_{a}^{\prime \prime}=\frac{19 \angle 27.1-12.7 \angle 0^{\circ}}{j 1.2}=8.03 \angle-25.9^{\circ} kA [/latex]

[latex] I_{a}=8.03 kA , pf =\cos 25.9^{\circ}=0.9 ext { lagging } [/latex]

Question 8.40: A 22 kV, 3-phase, star-connected turbo-alternator with a syn...

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