Examine the voltage stability of a non-static load supplied from a 275 kV infinite bus bar through a line of reactance 50 Ω per phase. The load consists of a constant active power of 200 MW and 200 MVAr rating reactive power, which is related to the voltage by the equation (in pu)

Question

Examine the voltage stability of a non-static load supplied from a 275 kV infinite bus bar through a line of reactance 50 Ω per phase. The load consists of a constant active power of 200 MW and 200 MVAr rating reactive power, which is related to the voltage by the equation (in pu) [latex]Q = 5(V−0.7)^2 + 0.8.[/latex]

Accepted Answer

With [latex]S_b = 200 MVA and V_b = 275 kV[/latex] the pu value of the line reactance is

[latex]X=\frac{50·200}{275^2 } =0.132 pu.[/latex]

The load voltage can then be found from the equation

[latex]E=V+\frac{QX}{V}. [/latex]

Thus,

[latex]E=V+\frac{0.132}{V} [5(V−0.7)^2+0.8]. [/latex]

Since E = 1 after simplification we have

[latex]V^2−1.159V + 0.258=0.[/latex]

Thus, the roots are

[latex] V_{1,2} = 0.858; 0.301.[/latex]

Taking the upper value (which is suitable to a physical reality) we obtain

[latex]Q=5(0.858−0.7)^2+0.8=0.925, [/latex]

and

[latex]\frac{dQ}{dV} =10(V−0.7)=1.58. [/latex]

Then,

[latex]\frac{dE}{dV}=1+\left(1.58-\frac{0.925}{0.858} \right)\frac{0.132}{0.858} =1.077. [/latex]

Since the result is positive, the system is stable. (Note that PX/V = 1·0.132/0.858 = 0.15, which is much less than [latex](V^2 + QX)/V= (0.8582 + 0.924·0.132)/0.858 \cong 1.)[/latex]
The reader can now convince himself that, if one of the two parallel lines is lost and the reactance changed to 0.264, dE/dV becomes negative and the system is unstable, i.e., the voltage collapses.

Question 8.5: Examine the voltage stability of a non-static load supplied ...

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