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Mid-line fault, cleared at 60°

Short answer

For a sustained fault, Pe_max during fault (1.2 pu) is below Pm (1.5 pu), so stable operation is impossible. Clearing at 60° yields a stable case with maximum swing about 105.9°; the critical clearing angle is 68.5°.

A balanced 3-phase fault occurs at the middle point of line 2 when power transfer is 1.5 pu in the system shown in the figure below. Given: E = 1.2, V = 1, X'd = 0.2, X1 = X2 = 0.4 pu. i) Determine whether the system is stable for a sustained fault. ii) The fault is cleared at δ = 60°. Is the system stable? If so find the maximum rotor swing iii) Find the critical clearing angle. [Figure: generator E∠δ behind X'd feeding two parallel lines X1 and X2 (circuit breakers at both ends) to the infinite bus V∠0; fault F at the middle of line 2]

Data: E=1.2E = 1.2, V=1V = 1, Xd′=0.2X_d' = 0.2, X1=X2=0.4X_1 = X_2 = 0.4 pu, Pm=1.5P_m = 1.5 pu.

Power-angle curves

Pre-fault:

XI=0.2+0.2=0.4 pu  ⇒  Pmax1=1.20.4=3.0 pu,δ0=sin⁡−11.53.0=30∘X_I = 0.2 + 0.2 = 0.4\ \text{pu} \;\Rightarrow\; P_{max1} = \frac{1.2}{0.4} = 3.0\ \text{pu}, \qquad \delta_0 = \sin^{-1}\frac{1.5}{3.0} = 30^\circ

During the fault (midpoint of line 2): the network is a star with EE to node A: 0.20.2; A to VV: 0.40.4 (line 1); A to ground: 0.20.2 (half of line 2). The other half of line 2, from the fault to VV, only shunts the infinite bus. Star-delta conversion gives the transfer reactance:

XII=0.2×0.4+0.4×0.2+0.2×0.20.2=0.200.2=1.0 puPmax2=1.21.0=1.2 pu\begin{aligned} X_{II} &= \frac{0.2\times0.4 + 0.4\times0.2 + 0.2\times0.2}{0.2} = \frac{0.20}{0.2} = 1.0\ \text{pu} \\ P_{max2} &= \frac{1.2}{1.0} = 1.2\ \text{pu} \end{aligned}

Post-fault (line 2 open):

XIII=0.2+0.4=0.6 pu  ⇒  Pmax3=2.0 pu,δmax=180∘−sin⁡−10.75=131.41∘X_{III} = 0.2 + 0.4 = 0.6\ \text{pu} \;\Rightarrow\; P_{max3} = 2.0\ \text{pu}, \qquad \delta_{max} = 180^\circ - \sin^{-1}0.75 = 131.41^\circ

(i) Sustained fault

During the fault, Pmax2=1.2<Pm=1.5P_{max2} = 1.2 < P_m = 1.5. The electrical output can never equal the input, so the rotor keeps accelerating and no decelerating area exists. The system is unstable for a sustained fault.

(ii) Fault cleared at δc=60∘\delta_c = 60^\circ

Accelerating area:

A1=Pm(δc−δ0)−Pmax2(cos⁡δ0−cos⁡δc)=1.5(1.0472−0.5236)−1.2(0.8660−0.5)=0.3462 pu⋅rad\begin{aligned} A_1 &= P_m(\delta_c-\delta_0) - P_{max2}(\cos\delta_0 - \cos\delta_c) \\ &= 1.5(1.0472-0.5236) - 1.2(0.8660-0.5) = 0.3462\ \text{pu·rad} \end{aligned}

Decelerating area available:

A2,max=Pmax3(cos⁡δc−cos⁡δmax)−Pm(δmax−δc)=2.0(0.5+0.6614)−1.5(2.2935−1.0472)=0.4534 pu⋅rad\begin{aligned} A_{2,max} &= P_{max3}(\cos\delta_c - \cos\delta_{max}) - P_m(\delta_{max}-\delta_c) \\ &= 2.0(0.5 + 0.6614) - 1.5(2.2935 - 1.0472) = 0.4534\ \text{pu·rad} \end{aligned}

Since A2,max>A1A_{2,max} > A_1, the system is stable. For the maximum swing δ1\delta_1, set 2.0(cos⁡60∘−cos⁡δ1)−1.5(δ1−1.0472)=0.34622.0(\cos60^\circ - \cos\delta_1) - 1.5(\delta_1 - 1.0472) = 0.3462. Solving numerically gives δ1=105.89∘\delta_1 = 105.89^\circ.

(iii) Critical clearing angle

cos⁡δcr=Pm(δmax−δ0)+Pmax3cos⁡δmax−Pmax2cos⁡δ0Pmax3−Pmax2=1.5(2.2935−0.5236)+2.0(−0.6614)−1.2(0.8660)2.0−1.2=0.3660δcr=68.53∘\begin{aligned} \cos\delta_{cr} &= \frac{P_m(\delta_{max}-\delta_0) + P_{max3}\cos\delta_{max} - P_{max2}\cos\delta_0}{P_{max3}-P_{max2}} \\ &= \frac{1.5(2.2935-0.5236) + 2.0(-0.6614) - 1.2(0.8660)}{2.0-1.2} \\ &= 0.3660 \\ \delta_{cr} &= 68.53^\circ \end{aligned}

Answer: (i) unstable for a sustained fault; (ii) stable, maximum swing ≈ 105.9°; (iii) δcr≈68.5∘\delta_{cr} \approx 68.5^\circ.

✓ Answer checked by Claude Sonnet 5.5

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