--- title: "RRRV from short circuit test (132 kV)" subject: "Switchgear and Protection" chapter: 6 key: "semester_6/switchgear_and_protection:rrrv-132kv-test" source: https://www.bctengineer.com/q/semester_6/switchgear_and_protection/rrrv-132kv-test site: BCT Engineering --- # RRRV from short circuit test (132 kV) Switchgear and Protection · Chapter 6: Circuit Breakers - Asked in: 2072 Asoj, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/switchgear_and_protection/rrrv-132kv-test - Whole chapter: https://www.bctengineer.com/md/semester_6/switchgear_and_protection/chapter-6 ## Question In a short circuit test on a 132 kV three phase system, the breaker gave the following results: pf of the fault is 0.4, recovery voltage is 0.9 of full line value and the natural frequency is 16 kHz. Determine the rate of rise of restriking voltage in following cases: i) Assume that symmetrical fault is grounded. ii) Assume that symmetrical fault is not grounded. ## Short answer With sinφ=0.9165 and V_m=107.78 kV, the average RRRV is 5.69 kV/μs for a grounded fault and 8.53 kV/μs for an ungrounded fault because the first-pole-to-clear factor is 1.5. ## Answer The **rate of rise of restriking voltage (RRRV)** is the slope of the transient voltage across the CB contacts after current zero; it decides whether the arc re-ignites. #### Method - Peak phase voltage: *V_m* = √2 × *V_L*/√3. - At current zero the instantaneous system voltage is *V_m* sin φ (fault current lags by φ, cos φ = fault p.f.). - Active recovery voltage: *V_r* = *k* × sin φ × *V_m*, where *k* is the recovery voltage factor (0.9 of full line value). - For an **ungrounded** symmetrical fault the first pole to clear sees 1.5 times this (first-pole-to-clear factor 1.5). - Restriking voltage peak = 2 *V_r*, reached at *t* = 1/(2*f_n*). - RRRV (average up to the peak, the usual textbook value) = 2*V_r* / *t* = 4 *f_n* *V_r*. The maximum instantaneous rate is *ω_n V_r*. #### Data *V_L* = 132 kV, cos φ = 0.4, *k* = 0.9, *f_n* = 16 kHz. ```text sin φ = √(1 − 0.4²) = 0.9165 V_m = √2 × 132/√3 = 107.78 kV t = 1/(2 × 16000) = 31.25 μs ``` #### i) Fault grounded ```text V_r = 0.9 × 0.9165 × 107.78 = 88.90 kV Peak = 2 × 88.90 = 177.80 kV RRRV = 177.80 kV / 31.25 μs = 5.69 kV/μs (max instantaneous = ω_n V_r = 2π×16000×88.90 kV = 8.94 kV/μs) ``` #### ii) Fault not grounded (first-pole factor 1.5) ```text V_r = 1.5 × 88.90 = 133.35 kV Peak = 2 × 133.35 = 266.71 kV RRRV = 266.71 kV / 31.25 μs = 8.53 kV/μs (max instantaneous = 13.41 kV/μs) ``` **Answer: RRRV ≈ 5.69 kV/μs with a grounded fault and ≈ 8.53 kV/μs with an ungrounded fault (average up to the first peak).** An ungrounded fault therefore puts 50% more stress on the breaker. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the active recovery voltage V_r obtained? A. k x sin(phi) x V_m, with V_m the peak phase voltage B. k x cos(phi) x V_m C. V_L / sqrt(3) only D. 2 x V_m x cos(phi) Answer: A. At current zero the instantaneous voltage is V_m sin(phi), multiplied by the recovery voltage factor k = 0.9. (Key point: Active recovery voltage formula) 2. For the grounded fault, the RRRV (average up to first peak) is about A. 8.53 kV/us B. 11.4 kV/us C. 2.85 kV/us D. 5.69 kV/us Answer: D. V_r = 88.9 kV, peak 177.8 kV in 31.25 us, giving 5.69 kV/us. (Key point: Grounded fault result) 3. Why is the RRRV higher for an ungrounded fault? A. The natural frequency is 1.5 times higher B. The first pole to clear sees 1.5 times the recovery voltage C. The power factor changes D. The breaker has no capacitance Answer: B. First-pole-to-clear factor of 1.5 multiplies V_r, giving 133.35 kV and 8.53 kV/us. (Key point: Ungrounded fault factor) 4. The average RRRV up to the peak is A. 2 V_r x f_n B. V_r / f_n C. 4 f_n V_r (peak 2 V_r reached at 1/(2 f_n)) D. pi f_n V_r Answer: C. 2 V_r / (1/(2 f_n)) = 4 f_n V_r. (Key point: RRRV expression) ## Close variants set in other exams - [RRRV from short circuit test (220 kV)](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Arrrv-220kv-test) — Switchgear and Protection, chapter 6, asked 2× ## Related in Switchgear and Protection - [Recovery, restriking voltage, RRRV and derivation](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Arestriking-voltage-derivation) — Switchgear and Protection, chapter 6 - [Specifying breaker on 132/66 kV transformer](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Abreaker-x-specification) — Switchgear and Protection, chapter 6 - [Resistance switching numerical (132 kV, 350 Ω)](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Aresistance-switching-132kv) — Switchgear and Protection, chapter 6, asked 2× ## Related in other subjects - [Air clearances for 132 kV double circuit line](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Aair-clearance-132kv-double-circuit) — Transmission and Distribution System Design, chapter 4 - [Current chopping overvoltage in 132 kV breaker](https://www.bctengineer.com/question.md?key=semester_8%2Fhigh_voltage_engineering%3Achopping-overvoltage-10a) — High Voltage Engineering, chapter 3 - [Fault at receiving end of 132 kV lines](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Afault-132kv-receiving-end-2071) — Power System Analysis II, chapter 3 - [Nominal-T 100 km line, 50 MVA at 132 kV](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Anominal-t-100km-50mva) — Power System Analysis I, chapter 6