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Mid-line fault cleared at 60°, E = 1.04

Short answer

When cleared at 60°, the system is stable. The critical clearing angle is δcr = 81.44°.

For the system shown in figure, the numerical values for different quantities are: E = 1.04 pu, V = 1 pu, Xs = 0.2 pu, and reactances of each line is 0.4 pu. The generator is delivering a power of 1.2 pu to the infinite bus. If a 3-phase short circuit fault occurs at the mid point of one the transmission lines, perform the following: a) If the fault is cleared (switching out of the faulted line) when power angle is 60°, check whether the system is stable or not; b) Determine the critical clearing angle. [Figure: generator 50 MVA, E∠δ, Xs = 0.2; two parallel lines Line-1 (X = 0.4) and Line-2 (X = 0.4) with breakers at both ends; infinite bus V∠0°]

Data: E=1.04E = 1.04, V=1V = 1, Xs=0.2X_s = 0.2, each line 0.40.4 pu, Pm=1.2P_m = 1.2 pu. The fault is at the midpoint of one line and is cleared by opening that line.

Power-angle curves

ConditionTransfer XX (pu)Pmax=EV/XP_{max} = EV/X (pu)
Pre-fault0.2+0.4/2=0.40.2 + 0.4/2 = 0.42.600
During fault1.0 (star-delta)1.040
Post-fault0.2+0.4=0.60.2 + 0.4 = 0.61.733

Transfer reactance during the fault. Star at node A: EE–A 0.20.2, A–VV 0.40.4 (healthy line), A–ground 0.20.2 (half of the faulted line):

XII=0.2(0.4)+0.4(0.2)+0.2(0.2)0.2=1.0 puX_{II} = \frac{0.2(0.4) + 0.4(0.2) + 0.2(0.2)}{0.2} = 1.0\ \text{pu}

Angles:

δ0=sin⁡−11.22.6=27.49∘=0.4797 radδmax=180∘−sin⁡−11.21.733=136.19∘=2.3769 rad\begin{aligned} \delta_0 &= \sin^{-1}\frac{1.2}{2.6} = 27.49^\circ = 0.4797\ \text{rad} \\ \delta_{max} &= 180^\circ - \sin^{-1}\frac{1.2}{1.733} = 136.19^\circ = 2.3769\ \text{rad} \end{aligned}

(a) Fault cleared at δc=60∘\delta_c = 60^\circ (1.0472 rad)

A1=Pm(δc−δ0)−Pmax2(cos⁡δ0−cos⁡δc)=1.2(1.0472−0.4797)−1.04(0.8871−0.5)=0.2784 pu⋅radA2,max=Pmax3(cos⁡δc−cos⁡δmax)−Pm(δmax−δc)=1.7333(0.5+0.7215)−1.2(2.3769−1.0472)=0.5218 pu⋅rad\begin{aligned} A_1 &= P_m(\delta_c-\delta_0) - P_{max2}(\cos\delta_0 - \cos\delta_c) \\ &= 1.2(1.0472-0.4797) - 1.04(0.8871-0.5) = 0.2784\ \text{pu·rad} \\ A_{2,max} &= P_{max3}(\cos\delta_c - \cos\delta_{max}) - P_m(\delta_{max}-\delta_c) \\ &= 1.7333(0.5+0.7215) - 1.2(2.3769-1.0472) = 0.5218\ \text{pu·rad} \end{aligned}

Since A2,max>A1A_{2,max} > A_1, the system is stable. Equating the areas gives a maximum rotor swing of about 94.3∘94.3^\circ.

(b) Critical clearing angle

cos⁡δcr=Pm(δmax−δ0)+Pmax3cos⁡δmax−Pmax2cos⁡δ0Pmax3−Pmax2=1.2(2.3769−0.4797)+1.7333(−0.7215)−1.04(0.8871)1.7333−1.04=2.2766−1.2506−0.92260.6933=0.1489δcr=81.44∘\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{1.2(2.3769-0.4797) + 1.7333(-0.7215) - 1.04(0.8871)}{1.7333-1.04} \\ &= \frac{2.2766 - 1.2506 - 0.9226}{0.6933} = 0.1489 \\ \delta_{cr} &= 81.44^\circ \end{aligned}

Answer: (a) stable when cleared at 60° (maximum swing ≈ 94.3°); (b) critical clearing angle ≈ 81.4°.

✓ Answer checked by Claude Sonnet 5.5

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