---
title: "Mid-line fault cleared at 60°, E = 1.04"
subject: "Power System Analysis II"
chapter: 6
key: "semester_5/power_system_analysis_ii:mid-line-fault-50mva-cleared-60"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/mid-line-fault-50mva-cleared-60
site: BCT Engineering
---

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

Power System Analysis II · Chapter 6: Power System Stability

- Asked in: 2071 Chaitra, 12 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/mid-line-fault-50mva-cleared-60
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-6

## Question

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°]

## Short answer

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

## Answer

**Data:** $E = 1.04$, $V = 1$, $X_s = 0.2$, each line $0.4$ pu, $P_m = 1.2$ pu. The fault is at the midpoint of one line and is cleared by opening that line.

#### Power-angle curves

| Condition | Transfer $X$ (pu) | $P_{max} = EV/X$ (pu) |
|---|---|---|
| Pre-fault | $0.2 + 0.4/2 = 0.4$ | 2.600 |
| During fault | 1.0 (star-delta) | 1.040 |
| Post-fault | $0.2 + 0.4 = 0.6$ | 1.733 |

Transfer reactance during the fault. Star at node A: $E$–A $0.2$, A–$V$ $0.4$ (healthy line), A–ground $0.2$ (half of the faulted line):

$$
X_{II} = \frac{0.2(0.4) + 0.4(0.2) + 0.2(0.2)}{0.2} = 1.0\ \text{pu}
$$

Angles:

$$
\begin{aligned}
\delta_0 &= \sin^{-1}\frac{1.2}{2.6} = 27.49^\circ = 0.4797\ \text{rad} \\
\delta_{max} &= 180^\circ - \sin^{-1}\frac{1.2}{1.733} = 136.19^\circ = 2.3769\ \text{rad}
\end{aligned}
$$

#### (a) Fault cleared at $\delta_c = 60^\circ$ (1.0472 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 $A_{2,max} > A_1$, **the system is stable**. Equating the areas gives a maximum rotor swing of about $94.3^\circ$.

#### (b) Critical clearing angle

$$
\begin{aligned}
\cos\delta_{cr} &= \frac{P_m(\delta_{max}-\delta_0) + P_{max3}\cos\delta_{max} - P_{max2}\cos\delta_0}{P_{max3}-P_{max2}} \\
&= \frac{1.2(2.3769-0.4797) + 1.7333(-0.7215) - 1.04(0.8871)}{1.7333-1.04} \\
&= \frac{2.2766 - 1.2506 - 0.9226}{0.6933} = 0.1489 \\
\delta_{cr} &= 81.44^\circ
\end{aligned}
$$

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

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which Pmax values apply pre-fault, during fault and post-fault?
   A. 2.600, 0 and 1.733 pu
   B. 1.733, 1.040 and 2.600 pu
   C. 1.040, 2.600 and 1.733 pu
   D. 2.600, 1.040 and 1.733 pu
   Answer: D. X = 0.4, 1.0 and 0.6 pu give EV/X = 2.6, 1.04 and 1.733 for E = 1.04. (Key point: Power-angle table)

2. Why is the fault transfer reactance not zero-power here?
   A. The fault is mid-line, so some power still flows through the healthy line
   B. A mid-line fault never reduces power
   C. The faulted line is open immediately
   D. The fault is at the infinite bus
   Answer: A. Unlike a bus fault, a mid-line fault leaves a path through the healthy line; star-delta gives X = 1.0 pu. (Key point: Mid-line fault reactance)

3. What is the stability check at clearing angle 60 degrees?
   A. Compare Pmax1 with Pmax3
   B. Compare A1 = 0.278 with A2,max = 0.522; A2,max larger means stable
   C. A1 must equal zero
   D. Compare delta_0 with 60 degrees
   Answer: B. A2,max > A1 so the rotor is decelerated before delta_max = 136.19 degrees. (Key point: Stability at 60 degrees)

4. Which formula gives the critical clearing angle?
   A. cos dcr = Pm/Pmax3
   B. cos dcr = (Pmax3 - Pmax2)/Pm
   C. cos dcr = [Pm(dmax - d0) + Pmax3 cos dmax - Pmax2 cos d0]/(Pmax3 - Pmax2)
   D. cos dcr = [Pm(dmax - d0)]/Pmax1
   Answer: C. General equal-area result with non-zero fault curve; gives 81.44 degrees. (Key point: Critical angle formula)

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