Assume that, at the sending-end of one of the transmission lines in the system shown in Fig. 8.7, a three-phase fault occurs. Develop and solve the swing equation of the system, if the fault reactance is 0.07 pu. The inertia constant of the generator is H = 0.5 s and the frequency f = 60 Hz.

Question

Assume that, at the sending-end of one of the transmission lines in the system shown in Fig. 8.7, a three-phase fault occurs. Develop and solve the swing equation of the system, if the fault reactance is 0.07 pu. The inertia constant of the generator is H = 0.5 s and the frequency f = 60 Hz. Initially the generator delivers a 0.8 pu power with a transient EMF of 1.22 pu.

Accepted Answer

The power-angle characteristic prior to the fault is

[latex]P_e=\frac{1.22\cdot 1}{0.6} \sin \delta =2.03\sin \delta, [/latex]

where [latex]E^{\prime}_d=1.22 and X_{tot}=0.6 [/latex] (see Fig. 8.7(b)).

Performing Y→ Δ transformation in the given circuit we obtain

[latex]Y_{12} = \frac{Y_1Y_2}{Y_1+Y_2+Y_3}=\frac{(1/0.3)(1/0.3)}{1/0.3+1/0.3+1/0.07} =0.528. [/latex]

The electrical power output of the sending-end at the fault is

[latex]P_e=1.22·1·0.528 \sin \delta =0.644 \sin \delta . [/latex]

Therefore, with equation 8.15 the swing equation is

[latex]\frac{2H}{\omega_r}\frac{d^2\delta }{dt^2}=P_{an^\bullet } [/latex] (8.15)

[latex]\frac{d^2\delta }{dt^2}=\frac{\omega_r}{2H}(P_m-P_e) , [/latex]

or

[latex] \frac{d^2\delta }{dt^2} =\frac{377}{10}(0.8−0.644 \sin \delta ) = 37.7(0.8−0.644 \sin \delta ). [/latex]

The solution of this nonlinear equation is obtained by the MATCAD program, which for the initial conditions of [latex]\delta_0 = \sin^{− 1}(0.8/2.03) = 23.2° (0.405 rad) and \omega(0) = \omega_0 − \omega_r =0[/latex] is shown in Fig. 8.8. As can be seen, the power angle increases indefinitely and the system is unstable.

Assume now that the fault is cleared in 0.2 s by opening the faulted line.
The equivalent circuit is now shown in Fig. 8.7(c), in which the total reactance is

[latex]X_{12} = 0.3 + 0.6 = 0.9[/latex] pu,

and the transient power will be

[latex]P_e=\frac{1.22}{0.9} \sin \delta = 1.36 \sin \delta . [/latex]

Thus, the swing equation is

[latex]\frac{d^2\delta }{dt^2}=37.7(0.8−1.36 \sin \delta ). [/latex]

The initial values of d and v are calculated with the previous solution for time 0.2 s, which gives [latex]\delta_{0.2} = 0.788 rad and \omega_{0.2} = 3.56[/latex], and the time solution is now shown in Fig. 8.9. (Note that for the new solution [latex]\delta_{0.2} \equiv \delta_0 and \omega_{0.2} \equiv \omega_0.[/latex]) As can be seen the time change of the power angle is oscillatory and the first peak of about [latex]\delta_{\max} =1.26 [/latex] rad is reached at t = 0.238 s after which [latex]\delta[/latex] is decreased until it reaches a minimum value of about [latex]\delta_{\min} = 0.085 rad at t = 0.756 [/latex] s and the oscillations of the rotor angle continue until they decay due to the damping effect. For the system under study and for the given fault the synchronism is not lost and the machine is stable.

Question 8.3: Assume that, at the sending-end of one of the transmission l...

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