--- title: "RRRV from short circuit test (220 kV)" subject: "Switchgear and Protection" chapter: 6 key: "semester_6/switchgear_and_protection:rrrv-220kv-test" source: https://www.bctengineer.com/q/semester_6/switchgear_and_protection/rrrv-220kv-test site: BCT Engineering --- # RRRV from short circuit test (220 kV) Switchgear and Protection · Chapter 6: Circuit Breakers - Asked in: 2078 Chaitra, 6 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-220kv-test - Whole chapter: https://www.bctengineer.com/md/semester_6/switchgear_and_protection/chapter-6 ## Question In a short circuit test on a 220 kV 3-phase system, the breaker gave the following results: p.f. of the fault 0.45, recovery voltage 0.9 of full line value; the breaking current is symmetrical and the restriking transient had a natural frequency of 18 kHz. Determine the rate of rise of restriking voltage. Assume that the fault is grounded. ## Short answer For the grounded fault, V_r=144.37 kV and peak restriking voltage is 288.75 kV reached at 27.78 μs, so average RRRV is 10.39 kV/μs; maximum instantaneous RRRV is 16.33 kV/μs. ## Answer The **RRRV** is the rate at which the voltage across the CB contacts rises after the arc is extinguished at current zero. #### 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 would see 1.5 times this (first-pole-to-clear factor 1.5); here the fault is grounded, so the factor is 1. - 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* = 220 kV, cos φ = 0.45, *k* = 0.9, *f_n* = 18 kHz, fault grounded (factor 1). ```text sin φ = √(1 − 0.45²) = 0.8930 V_m = √2 × 220/√3 = 179.63 kV V_r = 0.9 × 0.8930 × 179.63 = 144.37 kV Peak restriking = 2 V_r = 288.75 kV t = 1/(2 f_n) = 1/(2×18000) = 27.78 μs RRRV = 288.75 kV / 27.78 μs = 10.39 kV/μs ``` Maximum instantaneous rate (at *t* = 1/(4*f_n*) = 13.9 μs): ```text ω_n V_r = 2π × 18000 × 144.37 kV = 16.33 kV/μs ``` **Answer: RRRV ≈ 10.39 kV/μs (average up to the first peak of 288.75 kV); maximum instantaneous RRRV ≈ 16.33 kV/μs.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For a 220 kV system the peak phase voltage V_m is A. 220 kV B. 311 kV C. 179.6 kV D. 127 kV Answer: C. V_m = sqrt(2) x 220/sqrt(3) = 179.63 kV. (Key point: Peak phase voltage) 2. With pf 0.45 and k = 0.9, V_r is A. 144.4 kV B. 72.7 kV C. 161.7 kV D. 289 kV Answer: A. sin(phi) = 0.893, V_r = 0.9 x 0.893 x 179.63 = 144.37 kV. (Key point: Active recovery voltage) 3. With f_n = 18 kHz, the restriking voltage peak of 2 V_r is reached at A. 55.6 us B. 13.9 us C. 27.8 us D. 6.9 us Answer: C. t = 1/(2 f_n) = 27.78 us. (Key point: Time to peak) 4. The average RRRV for the grounded fault is A. 10.39 kV/us B. 16.33 kV/us C. 5.2 kV/us D. 15.6 kV/us Answer: A. 288.75 kV / 27.78 us = 10.39 kV/us; 16.33 kV/us is the maximum instantaneous rate w_n V_r. (Key point: RRRV result) ## Close variants set in other exams - [RRRV from short circuit test (132 kV)](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Arrrv-132kv-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 - [Restriking voltage vs recovery voltage](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Arestriking-vs-recovery-voltage) — Switchgear and Protection, chapter 6 - [Restriking voltage numerical (6.6 kV, 8 mH)](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Arestriking-6-6kv-8mh) — Switchgear and Protection, chapter 6, asked 2× - [Restriking voltage numerical (11 kV, 10 mH)](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Arestriking-11kv-10mh) — Switchgear and Protection, chapter 6, asked 2× - [Breaking MVA and making kA (800 A, 33 kV)](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Acb-breaking-making-800a) — Switchgear and Protection, chapter 6 - [Resistance switching numerical (220 kV, 600 Ω)](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Aresistance-switching-220kv) — Switchgear and Protection, chapter 6, asked 2× ## Related in other subjects - [Tower dimensions for 220 kV single circuit line](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Atower-dimensions-220kv) — Transmission and Distribution System Design, chapter 4, asked 2× - [Tower dimensions for 220 kV double circuit line](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Atower-dimensions-220kv-double-circuit) — Transmission and Distribution System Design, chapter 4, asked 2× - [220 kV line, 200 MW: sending end from ABCD](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Aabcd-220kv-200mw) — Power System Analysis I, chapter 6 - [Air clearances for 220 kV single circuit line](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Aair-clearances-220kv-single-circuit) — Transmission and Distribution System Design, chapter 4