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

# Mid-line fault, cleared at 60°

Power System Analysis II · Chapter 6: Power System Stability

- Asked in: 2073 Chaitra, 8 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-cleared-60
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-6

## Question

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]

## 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°.

## Answer

**Data:** $E = 1.2$, $V = 1$, $X_d' = 0.2$, $X_1 = X_2 = 0.4$ pu, $P_m = 1.5$ pu.

#### Power-angle curves

**Pre-fault:**

$$
X_I = 0.2 + 0.2 = 0.4\ \text{pu} \;\Rightarrow\; P_{max1} = \frac{1.2}{0.4} = 3.0\ \text{pu}, \qquad \delta_0 = \sin^{-1}\frac{1.5}{3.0} = 30^\circ
$$

**During the fault (midpoint of line 2):** the network is a star with $E$ to node A: $0.2$; A to $V$: $0.4$ (line 1); A to ground: $0.2$ (half of line 2). The other half of line 2, from the fault to $V$, only shunts the infinite bus. Star-delta conversion gives the transfer reactance:

$$
\begin{aligned}
X_{II} &= \frac{0.2\times0.4 + 0.4\times0.2 + 0.2\times0.2}{0.2} = \frac{0.20}{0.2} = 1.0\ \text{pu} \\
P_{max2} &= \frac{1.2}{1.0} = 1.2\ \text{pu}
\end{aligned}
$$

**Post-fault (line 2 open):**

$$
X_{III} = 0.2 + 0.4 = 0.6\ \text{pu} \;\Rightarrow\; P_{max3} = 2.0\ \text{pu}, \qquad \delta_{max} = 180^\circ - \sin^{-1}0.75 = 131.41^\circ
$$

#### (i) Sustained fault

During the fault, $P_{max2} = 1.2 < P_m = 1.5$. 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 $\delta_c = 60^\circ$

Accelerating area:

$$
\begin{aligned}
A_1 &= P_m(\delta_c-\delta_0) - P_{max2}(\cos\delta_0 - \cos\delta_c) \\
&= 1.5(1.0472-0.5236) - 1.2(0.8660-0.5) = 0.3462\ \text{pu·rad}
\end{aligned}
$$

Decelerating area available:

$$
\begin{aligned}
A_{2,max} &= P_{max3}(\cos\delta_c - \cos\delta_{max}) - P_m(\delta_{max}-\delta_c) \\
&= 2.0(0.5 + 0.6614) - 1.5(2.2935 - 1.0472) = 0.4534\ \text{pu·rad}
\end{aligned}
$$

Since $A_{2,max} > A_1$, **the system is stable**. For the maximum swing $\delta_1$, set $2.0(\cos60^\circ - \cos\delta_1) - 1.5(\delta_1 - 1.0472) = 0.3462$. Solving numerically gives $\delta_1 = 105.89^\circ$.

#### (iii) Critical clearing angle

$$
\begin{aligned}
\cos\delta_{cr} &= \frac{P_m(\delta_{max}-\delta_0) + P_{max3}\cos\delta_{max} - P_{max2}\cos\delta_0}{P_{max3}-P_{max2}} \\
&= \frac{1.5(2.2935-0.5236) + 2.0(-0.6614) - 1.2(0.8660)}{2.0-1.2} \\
&= 0.3660 \\
\delta_{cr} &= 68.53^\circ
\end{aligned}
$$

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

## Check yourself

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

1. What are Pmax1 and delta_0 before the fault?
   A. Pmax1 = 2.0 pu, delta_0 = 48.6 degrees
   B. Pmax1 = 6.0 pu, delta_0 = 14.5 degrees
   C. Pmax1 = 1.2 pu, delta_0 = 90 degrees
   D. Pmax1 = 3.0 pu (X = 0.4), delta_0 = 30 degrees
   Answer: D. X = 0.2 + 0.4/2 = 0.4, so Pmax1 = 1.2/0.4 = 3.0 and sin d0 = 1.5/3.0. (Key point: Pre-fault angle)

2. Why is the system unstable for a sustained fault?
   A. Pmax2 = 1.2 pu is less than Pm = 1.5 pu, so Pe can never balance Pm
   B. Pmax2 is greater than Pm
   C. Because the fault is at the midpoint
   D. Because delta_0 is above 90 degrees
   Answer: A. If the fault curve peak is below Pm the rotor accelerates continuously with no decelerating area. (Key point: Sustained fault)

3. Why is the fault-period transfer reactance 1.0 pu?
   A. Series sum 0.2 + 0.4 + 0.4
   B. Star-delta conversion of the network with the faulted node shunted to ground (0.2 pu)
   C. Parallel of the two lines
   D. Because half of line 2 is out of service
   Answer: B. The fault at mid-line gives a star with arms 0.2, 0.4 and 0.2 to ground, so X = (sum of products)/0.2 = 1.0 pu. (Key point: Fault transfer reactance)

4. Why is the system stable when cleared at 60 degrees?
   A. A1 is greater than A2,max
   B. Because 60 degrees is greater than delta_0
   C. A2,max (0.453) exceeds A1 (0.346)
   D. Because Pmax3 is greater than Pmax1
   Answer: C. The available decelerating area is larger than the accelerating area, so the rotor swings back (max about 105.9 degrees). (Key point: Stability when cleared at 60 degrees)

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