---
title: "Critical clearing time, H = 8 MJ/MVA"
subject: "Power System Analysis II"
chapter: 6
key: "semester_5/power_system_analysis_ii:critical-clearing-time-h8"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/critical-clearing-time-h8
site: BCT Engineering
---

# Critical clearing time, H = 8 MJ/MVA

Power System Analysis II · Chapter 6: Power System Stability

- Asked in: 2078 Bhadra, 12 marks
- Video: [Numerical on Transient stability](https://www.youtube.com/watch?v=yJT_z61r9aA) — Crash Course in Electrical Engineering
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/critical-clearing-time-h8
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-6

## Question

In the power system network shown below, if the generator delivers 1 pu real power at infinite bus 2. A 3-phase bolted fault occurs at the end of the 1st transmission line. The fault is isolated by simultaneous opening of circuit breakers on the both end of the line. Find the critical fault clearing time so that transient stability of the machine is maintained. Inertia constant of the generator is 8 MJ/MVA. [Figure: generator E∠δ with X'd = 0.3 - transformer j0.2 - bus 1; two parallel lines L1 (j0.3) and L2 (j0.3) with breakers at both ends between bus 1 and bus 2; infinite bus 1∠0° at bus 2]

## Short answer

Pre-fault Pmax=1.5385 pu, delta0=40.54°. Post-fault Pmax=1.25 pu, delta_max=126.87°. Equal area gives delta_cr=52.74°. Then t_cr = 0.147 s (assuming E=1 pu).

## Answer

**Assumption:** the internal emf magnitude is not given in the figure, so $|E| = 1.0$ pu and $f = 50$ Hz are assumed. The method is the same for any other $E$.

**Data:** $X_d' = 0.3$, $X_T = 0.2$, $L_1 = L_2 = 0.3$ pu, $P_m = 1$ pu, $H = 8$ MJ/MVA, $V = 1\angle0^\circ$.

#### Step 1: Pre-fault

$$
\begin{aligned}
X_I &= 0.3 + 0.2 + \frac{0.3}{2} = 0.65\ \text{pu} \\
P_{max1} &= \frac{1\times1}{0.65} = 1.5385\ \text{pu} \\
\delta_0 &= \sin^{-1}\frac{1}{1.5385} = 40.54^\circ = 0.7076\ \text{rad}
\end{aligned}
$$

#### Step 2: During the fault

The fault is at the end of line 1, either at bus 1 or at the infinite bus 2. In both cases one of the buses on the power path is at zero voltage, so $P_{max2} = 0$ and the power transferred is zero.

#### Step 3: Post-fault (line 1 removed)

$$
\begin{aligned}
X_{III} &= 0.3 + 0.2 + 0.3 = 0.8\ \text{pu} \\
P_{max3} &= \frac{1}{0.8} = 1.25\ \text{pu} \\
\delta_{max} &= 180^\circ - \sin^{-1}\frac{1}{1.25} = 126.87^\circ = 2.2143\ \text{rad}
\end{aligned}
$$

Since $P_{max3} > P_m$, the system can be stable if the fault is cleared in time.

#### Step 4: Critical clearing angle

$$
\begin{aligned}
\cos\delta_{cr} &= \frac{P_m(\delta_{max}-\delta_0) + P_{max3}\cos\delta_{max}}{P_{max3}} \\
&= \frac{1(2.2143-0.7076) + 1.25\cos 126.87^\circ}{1.25} \\
&= \frac{1.5067 - 0.75}{1.25} = 0.6054 \\
\delta_{cr} &= 52.74^\circ = 0.9206\ \text{rad}
\end{aligned}
$$

#### Step 5: Critical clearing time

With $P_e = 0$ during the fault, $\delta = \delta_0 + \dfrac{\pi f P_m}{2H}t^2$. So

$$
\begin{aligned}
t_{cr} &= \sqrt{\frac{2H(\delta_{cr}-\delta_0)}{\pi f P_m}} \\
&= \sqrt{\frac{2\times 8\times(0.9206-0.7076)}{\pi\times 50\times 1}} \\
&= 0.147\ \text{s}
\end{aligned}
$$

**Answer (with $E = 1.0$ pu): $\delta_{cr} = 52.74^\circ$, critical clearing time $t_{cr} \approx 0.147$ s (about 7.4 cycles).**

## Check yourself

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

1. After line 1 is removed, the total reactance and Pmax3 are (E = V = 1):
   A. 0.65 pu and 1.538 pu
   B. 0.8 pu and 1.25 pu
   C. 0.5 pu and 2.0 pu
   D. 0.3 pu and 3.33 pu
   Answer: B. X = 0.3 + 0.2 + 0.3 = 0.8 pu, so Pmax3 = 1/0.8 = 1.25 pu. (Key point: Post-fault network)

2. The initial angle delta0 with Pmax1 = 1.5385 pu and Pm = 1 pu is about:
   A. 30 degrees
   B. 65 degrees
   C. 52.7 degrees
   D. 40.54 degrees
   Answer: D. delta0 = asin(1/1.5385) = 40.54 degrees. (Key point: Pre-fault angle)

3. Which expression gives cos(delta_cr) when Pmax2 = 0?
   A. [Pm(dmax - d0) + Pmax3 cos(dmax)] / Pmax3
   B. [Pm(dmax + d0) - Pmax3 cos(dmax)] / Pmax3
   C. Pm/Pmax3
   D. [Pmax3 cos(d0)] / Pm
   Answer: A. Equating A1 = Pm(dcr - d0) and A2 gives this result; here cos(dcr) = 0.6054 and dcr = 52.74 degrees. (Key point: Critical clearing angle)

4. The critical clearing time is about:
   A. 0.047 s
   B. 0.147 s
   C. 0.5 s
   D. 1.47 s
   Answer: B. t = sqrt(2 x 8 x 0.2130 / (pi x 50)) = 0.147 s. (Key point: Clearing time)

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