---
title: "Fault cleared at 30°, Pe = 2.1 sin δ"
subject: "Power System Analysis II"
chapter: 6
key: "semester_5/power_system_analysis_ii:fault-cleared-thirty-degrees"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/fault-cleared-thirty-degrees
site: BCT Engineering
---

# Fault cleared at 30°, Pe = 2.1 sin δ

Power System Analysis II · Chapter 6: Power System Stability

- Asked in: 2072 Kartik, 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/fault-cleared-thirty-degrees
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-6

## Question

A synchronous machine is connected to an infinite bus and the following condition exists during normal operation and the following relationship is valid during that period: Pe = 2.1 sinδ, Pm = 1.0 pu. Following a three phase symmetrical fault at the bus of synchronous generator, and fault is cleared at δ = 30°. Check the stability. Calculate the maximum fault clearing angle so as to maintain synchronism.

## Short answer

The fault at the generator bus reduces electrical power to zero. Clearing at 30° leaves a very small accelerating area, so the system is stable. The maximum (critical) clearing angle to maintain synchronism is about 81.7°.

## Answer

**Data:** $P_e = 2.1\sin\delta$ (normal and, assuming a temporary fault, also post-fault), $P_m = 1.0$ pu. The fault is at the generator bus, so $P_e = 0$ during the fault.

#### Step 1: Initial and maximum angles

$$
\begin{aligned}
\delta_0 &= \sin^{-1}\frac{1.0}{2.1} = 28.44^\circ = 0.4963\ \text{rad} \\
\delta_{max} &= 180^\circ - 28.44^\circ = 151.56^\circ
\end{aligned}
$$

#### Step 2: Stability check for clearing at $30^\circ$

$$
\begin{aligned}
A_1 &= P_m(\delta_c - \delta_0) = 1.0(0.5236 - 0.4963) = 0.0273\ \text{pu·rad} \\
A_{2,max} &= 2.1(\cos30^\circ - \cos151.56^\circ) - 1.0(2.6453 - 0.5236) = 1.5436\ \text{pu·rad}
\end{aligned}
$$

Since $A_{2,max} \gg A_1$, **the system is stable**. Clearing occurs only $1.56^\circ$ after the fault.

#### Step 3: Maximum (critical) clearing angle

With $P_{max2} = 0$ and $P_{max3} = P_{max1}$:

$$
\begin{aligned}
\cos\delta_{cr} &= (\pi - 2\delta_0)\sin\delta_0 - \cos\delta_0 \\
&= (3.1416 - 0.9926)(0.4762) - 0.8793 \\
&= 0.1440 \\
\delta_{cr} &= 81.72^\circ
\end{aligned}
$$

**Answer: Stable when cleared at 30°; maximum (critical) clearing angle ≈ 81.7°.**

## Check yourself

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

1. How is delta_0 obtained from Pe = 2.1 sin(delta) and Pm = 1.0?
   A. sin^-1(2.1) 
   B. cos^-1(1/2.1)
   C. 30 degrees
   D. sin^-1(1/2.1) = 28.44 degrees
   Answer: D. In normal operation Pe = Pm, so sin(delta_0) = 1.0/2.1. (Key point: Initial angle)

2. What happens to Pe during a 3-phase fault at the generator bus?
   A. Pe = 0, so Pmax2 = 0
   B. Pe stays 2.1 sin(delta)
   C. Pe doubles
   D. Pe becomes 1.0 pu
   Answer: A. Bus voltage is zero, so no power is delivered and the accelerating area is Pm(dc - d0). (Key point: Fault at generator bus)

3. How is stability confirmed when cleared at 30 degrees?
   A. A1 is bigger than A2,max
   B. A1 = 0.027 is much less than A2,max = 1.544
   C. 30 degrees is below 28.44 degrees
   D. Because Pm is less than 2.1
   Answer: B. Only a tiny acceleration occurs, far less than the decelerating area available up to 151.56 degrees. (Key point: Stability check)

4. Which formula gives the critical clearing angle here?
   A. cos dcr = sin d0 + cos d0
   B. cos dcr = (pi - d0) cos d0
   C. cos dcr = (pi - 2 d0) sin d0 - cos d0
   D. dcr = 2 d0
   Answer: C. With Pmax2 = 0 and Pmax3 = Pmax1, delta_max = pi - d0 and the equal area condition reduces to this, giving 81.7 degrees. (Key point: Critical clearing angle formula)

## Related in Power System Analysis II

- [Critical clearing angle/time, fault at infinite bus](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Acritical-clearing-infinite-bus-h5) — Power System Analysis II, chapter 6
- [L-G fault and 3-phase fault, 30 MVA generator](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Alg-fault-30mva-voltages-3phase) — Power System Analysis II, chapter 5
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- [Critical clearing, fault at infinite bus, H = 6](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Acritical-clearing-fault-infinite-bus-h6) — Power System Analysis II, chapter 6
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- [Mid-line fault, cleared at 60°](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Amid-line-fault-cleared-60) — Power System Analysis II, chapter 6

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