A 4-pole, 50 Hz, 22 kV, 500 MV A synchronous generator having a synchronous reactance of 1.57 pu is feeding into a power system, which can be represented by a 22 kV infinite bus in series with a reactance of 0.4 Ω . The generator excitation is continually adjusted (by means of an automatic

Question

A 4-pole, 50 Hz, 22 kV, 500 MV A synchronous generator having a synchronous reactance of 1.57 pu is feeding into a power system, which can be represented by a 22 kV infinite bus in series with a reactance of [latex] 0.4 \Omega[/latex]. The generator excitation is continually adjusted (by means of an automatic voltage regulator) so as to maintain a terminal voltage of 22 kV independent of the load on the generator.

(a) Draw the phasor diagram, when the generator is feeding 250 MVA into the power system. Calculate the generator current, its power factor and real power fed by it. What is the excitation emf of the generator?

(b) Repeat part (a) when the generator load is 500 MVA.

Accepted Answer

The circuit diagram of the system is drawn in Fig. 8.114(a)

[latex] I_{a}( ext { rated })=\frac{500}{\sqrt{3} imes 22}=13.12 kA[/latex]

[latex] V_{t}( ext { rated })=22 / \sqrt{3}=12.7 kV[/latex]

[latex] Z_{b}=12.7 / 13.12=0.968 \Omega[/latex]

[latex] (M V A)_{B}=500,(M W)_{B}=500[/latex]

[latex] X_{s g}=1.57 pu[/latex] (given)

[latex] X_{b}=0.4 / 0.968=0.413 pu[/latex]

(a) Load = 250 MVA or 0.5 pu

[latex] V_{t}=1 pu , \quad I_{a}=0.5 pu[/latex]

[latex] V_{b}=1 pu[/latex]

[latex] X_{b} I_{a}=0.413 imes 0.5=0.207 pu[/latex]

The phasor diagram is drawn in Fig. 8.114(b)

[latex] \sin \phi=\frac{(0.207 / 2)}{1} ext { as } \bar{I}_{a} ext { is at } 90^{\circ} ext { to } \bar{I}_{a} X_{b}[/latex]

Or [latex] \phi=5.9^{\circ}[/latex]

[latex] p f=\cos \phi=0.995[/latex] lagging

[latex] P_{e}=0.5 imes 0.995=0.4975 pu \Rightarrow 124.4 MW[/latex]

[latex] \bar{E}_{g}=1 \angle 0^{\circ}+j 1.57 imes 0.5 \angle-5.9^{\circ}=1.333 \angle 35.9^{\circ}[/latex]

Or [latex] E_{g}=1.333 imes 22=29.32 kV[/latex] (line)

(b) MVA load = 1 pu, [latex] I_{a}=1 pu[/latex]

[latex] X_{b} I_{a}=0.413 imes 1=0.413 pu[/latex]

[latex] \sin \phi=\frac{(0.143 / 2)}{1} \Rightarrow \phi=11.9^{\circ}[/latex]

[latex] \cos \phi=0.98[/latex] lagging

[latex] P_{e}=1 imes 1 imes 0.98=0.98 pu[/latex] or 490 MW

[latex] \bar{E}_{g}=1 \angle 0^{\circ}+j 1.57 imes 1 \angle-11.9^{\circ}=1.324+j 1.536[/latex]

Or [latex] E_{g}=2.028 pu[/latex] or 44.62 kV (line)

Question 8.42: A 4-pole, 50 Hz, 22 kV, 500 MV A synchronous generator havin...

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