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Reactor schemes in stations 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.

Explain different reactor schemes used in generating station and substations.

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.

  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.

  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.

  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.

 =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

SchemeNormal loss/dropMain meritMain demerit
GeneratorHighProtects generatorConstant loss
FeederHighBus voltage held on feeder faultNo bus-fault protection
Ring busLowFault confined to sectionUnbalanced loads cause section voltage difference
Tie-barLowTwo reactors in fault path; easy extensionExtra tie-bar cost

✓ Answer checked by Claude Opus 5.5 (high effort)

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