--- title: "Reactor schemes in stations and substations" subject: "Power Plant Equipment" chapter: 4 key: "semester_7/power_plant_equipment:reactor-schemes" source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/reactor-schemes site: BCT Engineering --- # Reactor schemes in stations and substations Power Plant Equipment · Chapter 4: Substation Equipments - Asked in: 2070 Chaitra, 8 marks - Video: [5. Substation Components - Reactors](https://www.youtube.com/watch?v=AbNUhO0eV84) — CMTEQ - 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-schemes - Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-4 ## Question Explain different reactor schemes used in generating station and substations. ## Short answer Reactor schemes include generator reactors in series with each generator, feeder reactors in series with outgoing feeders, and bus-bar reactors that divide the bus into sections. Bus-bar schemes are ring and tie-bar; the tie-bar puts two reactors in the fault path, allowing smaller reactors. ## Answer **Current-limiting reactors** are placed at chosen points of a station so that the short-circuit current is limited and a fault in one part does not collapse the voltage of the whole station. The main schemes are below. #### 1. Generator reactor scheme Reactor in series with each generator, between the generator and the bus. ```text G1 G2 G3 [X] [X] [X] | | | =+===+===+===+===+= bus | | feeders ``` - Protects the generator against faults beyond it. - Disadvantage: full load current always flows, so constant voltage drop and loss; a feeder fault still lowers the whole bus voltage. Now seldom used, as modern machines have high reactance. #### 2. Feeder reactor scheme Reactor in series with each outgoing feeder. ```text G1 G2 | | =+===+===+===+= bus [X] [X] | | feeder feeder ``` - A feeder fault is limited by its own reactor; the bus voltage stays nearly normal, so other feeders are not disturbed. - Disadvantage: no protection against bus faults; constant drop and loss in each feeder. #### 3. Bus-bar reactor schemes The bus is divided into sections, each with a generator; sections are joined through reactors. In normal operation little current passes between sections, so losses are small. **(a) Ring system:** each section connected to the next through a reactor, forming a ring. ```text G1 G2 G3 =S1===[X]===S2===[X]===S3= | | +----------[X]-----------+ ``` A fault on one section is fed by the other generators only through reactors. Each section is fed from both sides, so it is more reliable than a straight in-line arrangement. **(b) Tie-bar system:** each section is connected through its own reactor to a common tie-bar. ```text =S1= =S2= =S3= sections [X] [X] [X] | | | =+=======+=======+= tie-bar ``` Fault current from any other section passes through **two** reactors in series, so each reactor can be about half the value needed in a ring system. New sections are added easily. It needs an additional bus-bar. #### Comparison | Scheme | Normal loss/drop | Main merit | Main demerit | |---|---|---|---| | Generator | High | Protects generator | Constant loss | | Feeder | High | Bus voltage held on feeder fault | No bus-fault protection | | Ring bus | Low | Fault confined to section | Unbalanced loads cause section voltage difference | | Tie-bar | Low | Two reactors in fault path; easy extension | Extra tie-bar cost | ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What do current-limiting reactors achieve in a station? A. Limit short-circuit current so a fault in one part does not collapse voltage of the whole station B. Raise generator output voltage C. Increase the fault MVA at the bus D. Remove the need for circuit breakers Answer: A. They add reactance at chosen points to limit fault current and localise the effect of a fault. (Key point: Purpose of current-limiting reactors) 2. Why is the ring bus-bar system more reliable than a straight in-line arrangement? A. Each section is fed from both sides B. It uses no reactors C. It has an extra tie-bar D. Each section has two generators Answer: A. In a ring every section has two neighbouring supply paths, so loss of one link still leaves the section fed. (Key point: Ring system reliability) 3. Why can each reactor in a tie-bar system be about half the value needed in a ring system? A. Fault current from another section passes through two reactors in series B. The tie-bar has zero reactance C. Only one generator feeds the fault D. Reactors in the tie-bar carry no current Answer: A. Two reactors are in series in the fault path, so each needs only about half the reactance for the same limiting. (Key point: Tie-bar reactor value) 4. Which is a disadvantage of the generator reactor scheme? A. Full load current always flows, giving constant drop and loss, and a feeder fault still lowers bus voltage B. It cannot be fitted to old machines C. It needs a tie-bar D. It increases fault current from the generator Answer: A. The reactor is in series with the generator all the time, and does not stop a feeder fault from pulling down the whole bus. 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