Skip to main content

Critical clearing angle, Pmax 2, 0.5, 1.5

Short answer

Pre-fault Pmax=2 pu gives delta0=30°, post-fault Pmax=1.5 pu gives delta_max=138.19°. Using equal area, delta_cr = 70.3°.

A 50 Hz generator with inertia constant 5 MJ/MVA is delivering 1 pu power to an infinite bus. When a fault occurs, the maximum power transferable reduces to 0.5 pu. The maximum power transferable before the occurrence of fault was 2 pu. The maximum power after clearance of fault is 1.5 pu. Compute the critical fault clearing angle.

Data: Pm=1P_m = 1 pu, Pmax1=2P_{max1} = 2 pu (pre-fault), Pmax2=0.5P_{max2} = 0.5 pu (during fault), Pmax3=1.5P_{max3} = 1.5 pu (post-fault), H=5H = 5 MJ/MVA, f=50f = 50 Hz. HH is not needed for the angle.

Step 1: Initial and maximum angles

δ0=sin⁡−1PmPmax1=sin⁡−112=30∘=0.5236 radδmax=180∘−sin⁡−1PmPmax3=180∘−41.81∘=138.19∘=2.4119 rad\begin{aligned} \delta_0 &= \sin^{-1}\frac{P_m}{P_{max1}} = \sin^{-1}\frac{1}{2} = 30^\circ = 0.5236\ \text{rad} \\ \delta_{max} &= 180^\circ - \sin^{-1}\frac{P_m}{P_{max3}} = 180^\circ - 41.81^\circ = 138.19^\circ = 2.4119\ \text{rad} \end{aligned}

Step 2: Equal area condition

Accelerating area (from δ0\delta_0 to δcr\delta_{cr} on curve 2) = decelerating area (from δcr\delta_{cr} to δmax\delta_{max} on curve 3):

∫δ0δcr(Pm−Pmax2sin⁡δ) dδ=∫δcrδmax(Pmax3sin⁡δ−Pm) dδ\int_{\delta_0}^{\delta_{cr}}(P_m - P_{max2}\sin\delta)\,d\delta = \int_{\delta_{cr}}^{\delta_{max}}(P_{max3}\sin\delta - P_m)\,d\delta

This gives

cos⁡δcr=Pm(δmax−δ0)+Pmax3cos⁡δmax−Pmax2cos⁡δ0Pmax3−Pmax2\cos\delta_{cr} = \frac{P_m(\delta_{max}-\delta_0) + P_{max3}\cos\delta_{max} - P_{max2}\cos\delta_0}{P_{max3}-P_{max2}}

Step 3: Substitute

cos⁡δcr=1(2.4119−0.5236)+1.5cos⁡138.19∘−0.5cos⁡30∘1.5−0.5=1.8883−1.1180−0.43301.0=0.3372δcr=70.29∘\begin{aligned} \cos\delta_{cr} &= \frac{1(2.4119-0.5236) + 1.5\cos138.19^\circ - 0.5\cos30^\circ}{1.5-0.5} \\ &= \frac{1.8883 - 1.1180 - 0.4330}{1.0} \\ &= 0.3372 \\ \delta_{cr} &= 70.29^\circ \end{aligned}

Answer: Critical clearing angle δcr≈70.3∘\delta_{cr} \approx 70.3^\circ.

✓ Answer checked by Claude Sonnet 5.5

Related questions

All Power System Analysis II chapter 6 questions →