--- title: "Reactor to protect circuit breaker of small station" subject: "Power Plant Equipment" chapter: 4 key: "semester_7/power_plant_equipment:reactor-protect-circuit-breaker" source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/reactor-protect-circuit-breaker site: BCT Engineering --- # Reactor to protect circuit breaker of small station Power Plant Equipment · Chapter 4: Substation Equipments - Asked in: 2076 Chaitra, 5 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/power_plant_equipment/reactor-protect-circuit-breaker - Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-4 ## Question A small generating station has two alternators of 2500KVA and 500KVA with percentage reactance of 8 & 6 percentage respectively. The circuit breakers are rated at 150,000kVA. Due to increase in system load it is intended to add a third generator of 10,000kVA rating and 7.5% reactance. If the system voltage is 3300volts, find the reactance X necessary to protect the C.B. [Figure: alternators A and B connected directly to the busbar, which feeds an outgoing circuit through the CB rated 150,000 kVA; the third generator C (10,000 kVA, 7.5%) is connected to the busbar through the reactor X] ## Short answer On 10,000 kVA base, A and B parallel give 25.26%, and C is 7.5%. To keep fault level ≤150,000 kVA, total reactance must be 6.667%, so add a reactor of 1.56% (≈0.017 Ω) in series with the new generator. ## Answer The C.B. must not see a fault level above its rating (150,000 kVA). G3 is connected to the bus through reactor $X$. For a fault on the outgoing circuit just beyond the C.B., all three machines feed the fault. **Base:** 10,000 kVA, 3.3 kV; $Z_{base} = \dfrac{3.3^2}{10} = 1.089\ \Omega$. Percentage reactance on new base: $\%X_{new} = \%X\times\dfrac{\text{kVA}_b}{\text{kVA}}$. ```text A 2500kVA B 500kVA C 10000kVA,7.5% 8% 6% | | | [X] ====+===========+===========+==== 3.3 kV | CB 150,000 kVA |--- F ``` #### Reactances on 10,000 kVA base | Machine | Calculation | %X | |---|---|---| | A | $8\times10000/2500$ | 32 % | | B | $6\times10000/500$ | 120 % | | C | given | 7.5 % | $A\parallel B$: $\dfrac{32\times120}{152} = 25.263\ \%$, giving fault kVA $= 10000\times100/25.263 = 39{,}583$ kVA (safe). If C were connected directly: $25.263\parallel7.5 = 5.783\ \%$, fault kVA $= 172{,}917$ kVA > 150,000 kVA. So a reactor is needed. #### Required reactor Allowed total reactance: $$ \%X_{eq} = \frac{10000\times100}{150000} = 6.667\ \% $$ $$ \begin{aligned} \frac{1}{6.667} &= \frac{1}{25.263} + \frac{1}{7.5+X} \\ 0.15 - 0.039583 &= 0.110417 = \frac{1}{7.5+X} \\ 7.5 + X &= 9.0566 \\ X &= 1.5566\ \% \end{aligned} $$ In ohms: $$ X_\Omega = \frac{1.5566}{100}\times1.089 = 0.01695\ \Omega $$ **Answer:** Reactor $X$ ≈ **1.557 % on 10,000 kVA base ≈ 0.017 Ω** in series with the new generator keeps the fault level at 150,000 kVA. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. On 10,000 kVA base, what are the reactances of A (2500 kVA, 8%) and B (500 kVA, 6%)? A. 32% and 120% B. 8% and 6% C. 2% and 0.3% D. 320% and 1200% Answer: A. %X x 10000/kVA gives 32% and 120%. (Key point: Common base) 2. Why is a reactor required? A. The fault level is already below the breaker rating B. With C directly connected, the fault level is about 172,917 kVA, above the 150,000 kVA breaker rating C. A and B alone exceed the rating D. The breaker has no rating Answer: B. A parallel B is 25.26%, parallel C gives 5.78%, so 172,917 kVA exceeds the breaker. (Key point: Why a reactor is needed) 3. What total % reactance equals 150,000 kVA on a 10,000 kVA base? A. 15% B. 0.667% C. 6.667% D. 66.7% Answer: C. %X = 10000 x 100/150000 = 6.667%. (Key point: Target reactance) 4. Where is the reactor and what is its value? A. In series with A: X about 5% B. In parallel with C: X about 1.557% C. In series with the CB: X about 7.5% D. In series with C: X about 1.557% (0.017 ohm) Answer: D. 7.5 + X = 9.057%, so X = 1.557%, times 1.089 ohm = 0.017 ohm. 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