Substation Equipments

The fire fighting system of a power station detects a fire early, raises an alarm and puts it out quickly, so that costly equipment (transformers, generators, cable galleries, oil stores) is saved and people are safe.

Main fire risks

• Transformer and switchgear insulating oil
• Generator windings (insulation, hydrogen in large units)
• Cable trenches and galleries
• Fuel oil and coal stores, battery rooms

Components of the system

 +--------+   +-------------+   +-----------------+
 | Fire   |-->| Fire alarm  |-->| Alarm (bells,   |
 |detector|   | control pnl |   | lamps) + trip   |
 +--------+   +------+------+   +-----------------+
                     | opens deluge valve
 +-------+  +------+ v  +-------+   +-----------+
 | Water |->|Pumps |--->|Deluge |-->| Spray     |
 | tank  |  |(el+DG)|    | valve |   | nozzles   |
 +-------+  +------+    +-------+   | (on xfmr) |
                                     +-----------+

1. Detection: smoke, heat (rate-of-rise) and flame detectors; for transformers, quartz bulb or linear heat detectors placed around the tank.
2. Alarm panel: receives detector signals, sounds alarm, shows the zone, trips the equipment and starts the extinguishing system.
3. Extinguishing systems:
   • Hydrant system: ring main of pipes with hydrant points and hoses, kept pressurised by jockey pump; main electric pump and diesel standby pump.
   • Water spray (deluge) / mulsifire system: for transformers; high-velocity water spray forms an emulsion on the oil surface, cools it and cuts oxygen.
   • CO$_2$ flooding system: for generators, control rooms and cable vaults; CO$_2$ cylinders discharge automatically and dilute oxygen. Used where water would damage equipment.
   • Foam system: for oil tanks and fuel stores.
   • Portable extinguishers: CO$_2$, dry chemical powder and foam types near equipment.
4. Passive measures: oil soak pits filled with gravel under transformers, fire walls between transformers, fire-resistant cable coating and fire barriers.

Requirements

• Automatic operation with manual back-up.
• Reliable, separate water supply and standby diesel pump.
• Regular testing of detectors, pumps and cylinders.

SCADA (Supervisory Control And Data Acquisition) is a computer-based system that collects real-time data from power stations and substations, shows it to operators at a control centre, and lets them control equipment remotely.

Architecture

      +-------------------------------+
      | Master station (control ctr)  |
      | SCADA/EMS servers, HMI, data  |
      | historian, alarm printer      |
      +---------------+---------------+
                      | communication
       (fibre, PLCC, microwave, radio, GSM)
         +------------+------------+
         |            |            |
     +---+---+    +---+---+    +---+---+
     | RTU 1 |    | RTU 2 |    | RTU 3 |
     +---+---+    +---+---+    +---+---+
         |            |            |
   CT, PT, relays, CB aux contacts, OLTC,
   transducers in plant / substations

Components

1. Field devices: CTs, PTs, transducers, breaker status contacts, tap-position indicators, intelligent electronic devices (IEDs, numerical relays).
2. Remote Terminal Unit (RTU): installed at each substation/plant; acquires analog values (V, I, MW, MVAr, f) and digital status (CB open/closed, alarms), converts them to digital data and executes control commands from the master.
3. Communication system: carries data between RTUs and master using protocols such as IEC 60870-5-101/104, DNP3, Modbus, IEC 61850. Media: optical fibre (OPGW), PLCC, microwave, radio.
4. Master station: servers and software that poll RTUs, process data, raise alarms, store history and run applications.
5. HMI: operator consoles with single-line diagrams, trends and alarm lists.

Implementation steps

1. Survey the plants/substations and list the points (analog, digital, control) to be monitored.
2. Install transducers/IEDs and RTUs in each station and wire them to CTs, PTs, CBs and isolators.
3. Build the communication network (e.g. OPGW fibre with PLCC as back-up) and choose the protocol.
4. Set up the master station with redundant servers, database of all points and HMI displays.
5. Test point-to-point, then commission and train operators.

Functions in a power system

• Data acquisition: voltage, current, power, frequency, energy.
• Supervisory control: open/close CBs, change transformer taps, start/stop units.
• Alarm and event recording with time stamps (sequence of events).
• Trend and historical data for analysis.
• EMS applications: automatic generation control (load–frequency), economic dispatch, state estimation, load shedding.
• Distribution automation: fault location, isolation and restoration.

Advantages

• Fast, accurate information and quick fault response.
• Fewer staff at unmanned substations.
• Better reliability, economy and security of supply.

In Nepal, the NEA Load Dispatch Centre at Siuchatar uses SCADA to monitor and control the integrated grid.

A reactor is a coil of large inductive reactance and very small resistance. Its main function in a power network is to limit the short-circuit (fault) current to a value that circuit breakers and equipment can safely handle.

• It raises the reactance between the source and the fault, so $I_f = V/X$ falls and the fault MVA falls.
• It localises the fault so that the voltage on healthy parts of the system does not collapse.
• Shunt reactors also absorb reactive power on long lightly loaded lines to control over-voltage (Ferranti effect).

Work in per unit on a common base. Fault level $= \dfrac{\text{Base MVA}}{X_{pu}}$ and a reactor of $X$ ohms has $X_{pu} = X\cdot\dfrac{\text{MVA}_b}{\text{kV}_b^2}$.

Base: 40 MVA, 11 kV.

$$Z_{base} = \frac{11^2}{40} = 3.025\ \Omega$$

 G1 40MVA,5%   G2 40MVA,4%
     |             |
 ====+=====+=======+==== 11 kV bus
           |
          X=2 ohm feeder
           |
           F  (3-ph fault)

Fault level

$$\begin{aligned} X_{G1} &= 0.05\ \text{pu},\quad X_{G2} = 0.04\ \text{pu} \\ X_{G1}\parallel X_{G2} &= \frac{0.05\times0.04}{0.05+0.04} = 0.02222\ \text{pu} \\ X_{feeder} &= \frac{2}{3.025} = 0.66116\ \text{pu} \\ X_{eq} &= 0.02222 + 0.66116 = 0.68338\ \text{pu} \\ \text{Fault MVA} &= \frac{40}{0.68338} = 58.53\ \text{MVA} \end{aligned}$$

Fault current: $I_{base} = \dfrac{40\times10^6}{\sqrt3\times 11\times10^3} = 2099.5\ \text{A}$, so $I_f = 2099.5/0.68338 = 3072\ \text{A}$.

Reactor to reduce fault level by 40 %

New fault level $= 0.6\times58.53 = 35.12$ MVA.

$$\begin{aligned} X_{eq,new} &= \frac{40}{35.12} = \frac{0.68338}{0.6} = 1.13897\ \text{pu} \\ X_{added} &= 1.13897 - 0.68338 = 0.45559\ \text{pu} \\ X_{added}(\Omega) &= 0.45559\times3.025 = 1.378\ \Omega \end{aligned}$$

Answer: Fault level = 58.53 MVA (about 3.07 kA). A reactor of about 1.378 Ω (0.4556 pu on 40 MVA) in series with the 2 Ω feeder reduces it by 40 % to 35.12 MVA.

A power transformer transfers power between two voltage levels by mutual induction. Besides the core and windings it has many accessories for cooling, protection and voltage control.

     HV bushing   LV bushing   conservator
         |  |        |  |     +---------+--breather
         |  |        |  |     |   oil   |  (silica gel)
  vent  _|__|________|__|_____+----+----+
  ===> |      oil-filled tank      | Buchholz relay
       |  +--------------------+   |
       |  | core + HV/LV wndgs |   |=== radiators
       |  +--------------------+   |
       |   tap changer   OTI WTI   |
       +---------------------------+
          drain valve     earthing

Parts and functions

Power Line Carrier Communication (PLCC) uses the high-voltage transmission line itself as the medium to send speech, data, telemetry and protection signals between two substations, by superimposing a high-frequency carrier (about 30–500 kHz) on the 50 Hz power.

 Substation A                           Substation B
  bus                                         bus
   |                                           |
  [WT]=========== HV line ================[WT]
   |                                           |
  [CC]                                        [CC]
   |                                           |
  [LMU]                                      [LMU]
   |                                           |
 PLC terminal                           PLC terminal
 (tx/rx: speech, data, teleprotection)

Components and working

1. PLC terminal (transmitter/receiver): modulates the speech/data/protection signal on to an HF carrier at the sending end and demodulates it at the receiving end.
2. Coupling capacitor (CC): high-voltage capacitor that passes the HF carrier to the line but blocks the 50 Hz power frequency.
3. Line matching unit (LMU): matches the impedance of the coaxial cable to the line and, with the CC, forms a high-pass filter; it also provides surge protection and earthing switch.
4. Wave trap (line trap): a parallel LC circuit in series with the line, tuned to the carrier frequency. It offers high impedance to the carrier, so the signal does not leak into the substation bus, but negligible impedance to 50 Hz power.

The carrier travels along the line conductors (phase-to-earth or phase-to-phase coupling) to the other substation, where it is picked up through its own CC and LMU.

Uses

• Telephone (speech) between substations and load dispatch centre.
• Teleprotection: carrier-aided distance and inter-tripping.
• SCADA data and telemetering.

Merits

• No separate communication line needed; the line is strong and reliable.
• Economical over long distances.

The feeder reactance is not given, so it is taken as negligible: the fault is close to the bus. Fault MVA $= \text{MVA}_b/X_{pu}$.

Base: 50 MVA, 11 kV; $Z_{base} = 11^2/50 = 2.42\ \Omega$; $I_{base} = \dfrac{50\times10^6}{\sqrt3\times11\times10^3} = 2624.3\ \text{A}$.

 G1 50MVA,5%   G2 50MVA,3%
     |             |
 ====+=====+=======+==== 11 kV bus
           |  outgoing feeder
           F

Fault current and fault power

$$\begin{aligned} X_{eq} &= \frac{0.05\times0.03}{0.05+0.03} = 0.01875\ \text{pu} \\ I_f &= \frac{1}{0.01875} = 53.33\ \text{pu} = 53.33\times2624.3 = 139{,}964\ \text{A} \approx 139.96\ \text{kA} \\ \text{Fault MVA} &= \frac{50}{0.01875} = 2666.7\ \text{MVA} \end{aligned}$$

Current from each generator

Current divides inversely as the reactances:

$$\begin{aligned} I_{G1} &= I_f\frac{X_{G2}}{X_{G1}+X_{G2}} = 139.96\times\frac{0.03}{0.08} = 52.49\ \text{kA} \\ I_{G2} &= I_f\frac{X_{G1}}{X_{G1}+X_{G2}} = 139.96\times\frac{0.05}{0.08} = 87.48\ \text{kA} \end{aligned}$$

Reactor for 40 % reduction

New fault MVA $= 0.6\times2666.7 = 1600$ MVA.

$$\begin{aligned} X_{eq,new} &= \frac{50}{1600} = 0.03125\ \text{pu} \\ X_{added} &= 0.03125 - 0.01875 = 0.0125\ \text{pu} \\ X_{added}(\Omega) &= 0.0125\times2.42 = 0.03025\ \Omega \end{aligned}$$

Answer: $I_f \approx 139.96$ kA (G1: 52.49 kA, G2: 87.48 kA), fault power = 2666.7 MVA; series reactor = 0.0125 pu ≈ 0.0303 Ω (fault level becomes 1600 MVA).

i) Conservator tank

A small cylindrical tank mounted above the main tank and connected to it by a pipe (with the Buchholz relay in between).

• Oil expands and contracts with load and temperature; the conservator takes up this change, so the main tank always stays completely full of oil.
• Only the small oil surface in the conservator touches air, which reduces oxidation and moisture absorption.
• Its oil level indicator shows the oil level.

ii) Explosion vent

A bent pipe on top of the main tank, closed by a thin diaphragm (bakelite/glass), or a modern spring-loaded pressure relief valve.

• During a severe internal fault, arcing decomposes oil into gas and pressure rises suddenly.
• The diaphragm bursts and releases the pressure, preventing the tank from bursting or exploding.

iii) Breather

A small container filled with silica gel, with an oil seal cup, fitted on the conservator air pipe.

• When the oil contracts, air is drawn into the conservator through the breather.
• Silica gel absorbs moisture (blue colour turns pink when saturated); the oil cup traps dust.
• This keeps the oil dry and maintains its dielectric strength.

Convert all reactances to one base, combine the three parallel generators, and use Fault MVA $=\text{MVA}_b/X_{eq}$. The feeder reactance is not given, so it is taken as zero (fault just beyond the C.B.).

Base: 100 MVA, 11 kV. $X_{new} = X_{old}\dfrac{\text{MVA}_{new}}{\text{MVA}_{old}}$; $Z_{base} = 11^2/100 = 1.21\ \Omega$.

 G1 20MVA   G2 30MVA   G3 40MVA
  0.1pu      0.2pu      0.3pu
   |A         |B         |C
 ==+==========+==========+=== 11 kV
              |
             C.B.
              |---- F

Reactances on 100 MVA base

$$\begin{aligned} \frac{1}{X_{eq}} &= \frac{1}{0.5}+\frac{1}{0.6667}+\frac{1}{0.75} = 2+1.5+1.3333 = 4.8333 \\ X_{eq} &= 0.20690\ \text{pu} \end{aligned}$$

Fault level and fault current

$$\begin{aligned} \text{Fault MVA} &= \frac{100}{0.2069} = 483.33\ \text{MVA} \\ I_f &= \frac{483.33\times10^6}{\sqrt3\times11\times10^3} = 25{,}368\ \text{A} \approx 25.37\ \text{kA} \end{aligned}$$

Feeder reactor for 40 % reduction

New fault MVA $= 0.6\times483.33 = 290$ MVA.

$$\begin{aligned} X_{eq,new} &= \frac{100}{290} = 0.34483\ \text{pu} \\ X_{reactor} &= 0.34483 - 0.20690 = 0.13793\ \text{pu} \\ X_{reactor}(\Omega) &= 0.13793\times1.21 = 0.1669\ \Omega \end{aligned}$$

Answer: Fault level = 483.33 MVA, fault current ≈ 25.37 kA; feeder reactor ≈ 0.167 Ω (0.1379 pu on 100 MVA) brings the level down to 290 MVA.

Generators in parallel feed the fault through the bus. Fault MVA $=\text{MVA}_b/X_{eq}$. Feeder reactance (not given) is taken as zero before adding the reactor.

Base: 50 MVA, 11 kV; $Z_{base} = 11^2/50 = 2.42\ \Omega$. Both generators are already on 50 MVA.

 G1 50MVA,0.15   G2 50MVA,0.3
      |               |
 =====+=======+=======+==== 11 kV
              |  feeder (+X)
              F

Fault level

$$\begin{aligned} X_{eq} &= \frac{0.15\times0.3}{0.15+0.3} = 0.1\ \text{pu} \\ \text{Fault MVA} &= \frac{50}{0.1} = 500\ \text{MVA} \\ I_f &= \frac{500\times10^6}{\sqrt3\times11\times10^3} = 26.24\ \text{kA} \end{aligned}$$

Reactance for 50 % reduction

New fault MVA = 250 MVA, so

$$\begin{aligned} X_{eq,new} &= \frac{50}{250} = 0.2\ \text{pu} \\ X &= 0.2 - 0.1 = 0.1\ \text{pu} \\ X(\Omega) &= 0.1\times2.42 = 0.242\ \Omega \end{aligned}$$

(Halving the fault level needs the total reactance to double, so the added reactance equals the existing $X_{eq}$.)

Answer: Fault level = 500 MVA; feeder reactance = 0.1 pu = 0.242 Ω (fault level becomes 250 MVA).

Fault MVA $=\text{MVA}_b/X_{eq}$. The feeder has no reactance at first (none given).

Base: 20 MVA, 11 kV; $Z_{base} = 11^2/20 = 6.05\ \Omega$; $I_{base} = \dfrac{20\times10^6}{\sqrt3\times11\times10^3} = 1049.7\ \text{A}$.

 G1 20MVA   G2 20MVA   G3 10MVA (added)
  0.1pu      0.1pu      0.2pu
   |A         |B         |C
 ==+==========+==========+=== 11 kV
         |
        C.B.--[X]-- F

(i) Fault current and fault level (two generators)

$$\begin{aligned} X_{eq} &= \frac{0.1}{2} = 0.05\ \text{pu} \\ \text{Fault MVA} &= \frac{20}{0.05} = 400\ \text{MVA} \\ I_f &= \frac{1}{0.05}\times1049.7 = 20{,}995\ \text{A} \approx 20.99\ \text{kA} \end{aligned}$$

(ii) Reactor after adding G3

G3 on 20 MVA base: $X_3 = 0.2\times\dfrac{20}{10} = 0.4\ \text{pu}$.

$$\begin{aligned} \frac{1}{X_{eq}'} &= \frac{1}{0.1}+\frac{1}{0.1}+\frac{1}{0.4} = 22.5 \\ X_{eq}' &= 0.04444\ \text{pu} \;\Rightarrow\; \text{Fault MVA} = 450\ \text{MVA (too high)} \end{aligned}$$

To keep 400 MVA the total must stay at 0.05 pu:

$$\begin{aligned} X_{reactor} &= 0.05 - 0.04444 = 0.005556\ \text{pu} \\ X_{reactor}(\Omega) &= 0.005556\times6.05 = 0.0336\ \Omega \end{aligned}$$

Answer: (i) Fault level = 400 MVA, $I_f \approx$ 20.99 kA. (ii) Feeder reactor ≈ 0.00556 pu = 0.0336 Ω keeps the fault level at 400 MVA after G3 is added.

Function of reactor

A current-limiting reactor is a coil with high inductive reactance and negligible resistance. It limits the short-circuit current to a value that circuit breakers can interrupt, localises the fault so the rest of the bus keeps its voltage, and reduces mechanical and thermal stress on equipment.

Ring type bus-bar reactor scheme

The bus-bar is split into sections, one generator per section, and adjacent sections are joined through reactors so that the sections form a closed ring. Feeders are taken from each section.

      G1            G2            G3
      |             |             |
  ==[S1]===[X]===[S2]===[X]===[S3]==
    |  |          |  |          |  |
    |  feeders    feeders     feeders
    |                              |
    +-------------[X]--------------+
         (reactor closing the ring)

Working:

• In normal operation each generator mainly feeds the load on its own section. Only small balancing currents flow through the reactors, so their voltage drop and power loss are small.
• When a fault occurs on a feeder of, say, section S2, generator G2 feeds it directly, but G1 and G3 feed it only through the reactors. The fault current from the other sections is therefore limited.
• Since each section is fed from two sides, the current through each reactor is lower than in a simple tie-bar scheme with the same reactor value.

Advantages:

• Good fault-current limitation with low losses in normal running.
• A fault is confined to one section; the voltage on other sections stays fairly high.
• Each section is supplied from both directions, so supply is more reliable than in a plain series (in-line) scheme.

Disadvantages:

• More reactors and more complex layout than the simple series bus-bar scheme.
• Load sharing between sections depends on the loads being balanced.

Fault MVA $=\text{MVA}_b/X_{eq}$, with $E = 1$ pu. Feeder reactance (not given) is taken as zero before adding the reactor.

Base: 25 MVA, 11 kV; $Z_{base} = 11^2/25 = 4.84\ \Omega$.

 G1 25MVA,0.15   G2 25MVA,0.3
      |               |
 =====+=======+=======+==== 11 kV
              |
             C.B.--[X]-- F

Fault level

$$\begin{aligned} X_{eq} &= \frac{0.15\times0.3}{0.15+0.3} = 0.1\ \text{pu} \\ \text{Fault MVA} &= \frac{25}{0.1} = 250\ \text{MVA} \\ I_f &= \frac{250\times10^6}{\sqrt3\times11\times10^3} = 13.12\ \text{kA} \end{aligned}$$

Reactance for 50 % reduction

New fault MVA = 125 MVA.

$$\begin{aligned} X_{eq,new} &= \frac{25}{125} = 0.2\ \text{pu} \\ X &= 0.2 - 0.1 = 0.1\ \text{pu} \\ X(\Omega) &= 0.1\times4.84 = 0.484\ \Omega \end{aligned}$$

Answer: Fault level = 250 MVA; feeder reactance = 0.1 pu = 0.484 Ω (fault level becomes 125 MVA).

The bus-bar is split into two sections joined by a reactor $X$: G1 on bus 1 and G2 on bus 2. For a fault on bus 1, G1 feeds directly and G2 feeds through $X$. Fault MVA $=\text{MVA}_b/X_{eq}$.

Base: 100 MVA.

$$X_{G1} = 0.10\ \text{pu},\qquad X_{G2} = 0.08\times\frac{100}{200} = 0.04\ \text{pu}$$

  G1 100MVA,10%        G2 200MVA,8%
       |                     |
  ===[Bus 1]=====[ X ]=====[Bus 2]===
       |
       F (fault on bus 1)

Without reactor: $X_{eq} = 0.1\parallel0.04 = 0.02857$ pu, fault MVA $= 100/0.02857 = 3500$ MVA (too high).

With reactor: required

$$X_{eq} = \frac{100}{1500} = 0.06667\ \text{pu}$$

The two paths are in parallel:

$$\begin{aligned} \frac{1}{X_{eq}} &= \frac{1}{X_{G1}} + \frac{1}{X_{G2}+X} \\ 15 &= 10 + \frac{1}{0.04+X} \\ 0.04 + X &= \frac{1}{5} = 0.2 \\ X &= 0.16\ \text{pu} \end{aligned}$$

Answer: Bus-bar reactor $X$ = 0.16 pu (16 %) on 100 MVA base. (If the bus voltage $V$ kV is known, $X_\Omega = 0.16\,V^2/100$; e.g. at 11 kV, 0.194 Ω.)

Work in per unit on a common base. Fault MVA $=\text{MVA}_b/X_{eq}$; ohms to pu: $X_{pu} = X_\Omega\,\text{MVA}_b/\text{kV}^2$.

Base: 40 MVA, 11 kV; $Z_{base} = 11^2/40 = 3.025\ \Omega$.

 G1 40MVA,5%   G2 30MVA,4%
     |             |
 ====+=====+=======+==== 11 kV
           |
          2 ohm feeder + X
           |
           F

Present fault level

$$\begin{aligned} X_{G1} &= 0.05\ \text{pu},\quad X_{G2} = 0.04\times\frac{40}{30} = 0.05333\ \text{pu} \\ X_{G1}\parallel X_{G2} &= \frac{0.05\times0.05333}{0.10333} = 0.025806\ \text{pu} \\ X_{feeder} &= \frac{2}{3.025} = 0.66116\ \text{pu} \\ X_{eq} &= 0.025806 + 0.66116 = 0.68696\ \text{pu} \\ \text{Fault MVA} &= \frac{40}{0.68696} = 58.23\ \text{MVA} \end{aligned}$$

Reactor for 50 % reduction

New fault MVA $= 29.11$ MVA, so the total reactance must double:

$$\begin{aligned} X_{eq,new} &= 2\times0.68696 = 1.37393\ \text{pu} \\ X &= 1.37393 - 0.68696 = 0.68696\ \text{pu} \\ X(\Omega) &= 0.68696\times3.025 = 2.078\ \Omega \end{aligned}$$

Answer: Present fault level = 58.23 MVA; series reactor = 0.687 pu ≈ 2.078 Ω (fault level becomes 29.11 MVA).

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: $\%X_{new} = \%X\times\dfrac{\text{kVA}_b}{\text{kVA}}$.

   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

$A\parallel B$: $\dfrac{32\times120}{152} = 25.263\ \%$, 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.

Generator ratings are given in MW; they are taken as MVA (unity power factor assumed). Fault MVA $=\text{MVA}_b/X_{eq}$; ohms to pu: $X_{pu} = X_\Omega\,\text{MVA}_b/\text{kV}^2$.

Base: 50 MVA, 11 kV. $Z_{base} = 11^2/50 = 2.42\ \Omega$; $I_{base} = \dfrac{50\times10^6}{\sqrt3\times11\times10^3} = 2624.3\ \text{A}$.

  G1 50MW,0.16    G2 40MW,0.12
      |               |
     X1              X2 (+ Xr in part b)
      |               |
  ====+=======+=======+==== 11 kV bus
              |
             XL = 2 ohm
              |
              F (3-ph to ground)

a) Fault current, fault power and generator currents

$$\begin{aligned} X_1 &= 0.16\ \text{pu},\quad X_2 = 0.12\times\frac{50}{40} = 0.15\ \text{pu} \\ X_1\parallel X_2 &= \frac{0.16\times0.15}{0.31} = 0.077419\ \text{pu} \\ X_L &= \frac{2}{2.42} = 0.82645\ \text{pu} \\ X_{eq} &= 0.077419 + 0.82645 = 0.90387\ \text{pu} \\ I_f &= \frac{2624.3}{0.90387} = 2903.4\ \text{A} \\ \text{Fault power} &= \frac{50}{0.90387} = 55.32\ \text{MVA} \end{aligned}$$

Current divides inversely as the branch reactances:

$$\begin{aligned} I_{G1} &= 2903.4\times\frac{0.15}{0.31} = 1404.9\ \text{A} \\ I_{G2} &= 2903.4\times\frac{0.16}{0.31} = 1498.5\ \text{A} \end{aligned}$$

b) Reactor in series with $X_2$ for equal currents

Both generators feed the same bus, so they deliver equal current only when their branch reactances are equal:

$$\begin{aligned} X_2 + X_r &= X_1 \;\Rightarrow\; X_r = 0.16 - 0.15 = 0.01\ \text{pu} \\ X_r(\Omega) &= 0.01\times2.42 = 0.0242\ \Omega \\ L &= \frac{X_r}{2\pi f} = \frac{0.0242}{2\pi\times50} = 7.70\times10^{-5}\ \text{H} = 0.0770\ \text{mH} \end{aligned}$$

Check: new $X_{eq} = 0.08 + 0.82645 = 0.90645$ pu, $I_f = 2895.2$ A, each generator supplies 1447.6 A.

Answer: (a) $I_f$ ≈ 2903 A, fault power ≈ 55.32 MVA, $I_{G1}$ ≈ 1405 A, $I_{G2}$ ≈ 1499 A. (b) $L$ ≈ 0.077 mH (0.0242 Ω).

Work in per unit on a common base. Fault MVA $=\text{MVA}_b/X_{eq}$; ohms to pu: $X_{pu} = X_\Omega\,\text{MVA}_b/\text{kV}^2$.

Base: 40 MVA, 11 kV; $Z_{base} = 11^2/40 = 3.025\ \Omega$.

 G1 40MVA,5%   G2 30MVA,4%
     |             |
 ====+=====+=======+==== 11 kV
           |
          2 ohm feeder + X
           |
           F

Present fault level

$$\begin{aligned} X_{G1} &= 0.05\ \text{pu},\quad X_{G2} = 0.04\times\frac{40}{30} = 0.05333\ \text{pu} \\ X_{G1}\parallel X_{G2} &= \frac{0.05\times0.05333}{0.10333} = 0.025806\ \text{pu} \\ X_{feeder} &= \frac{2}{3.025} = 0.66116\ \text{pu} \\ X_{eq} &= 0.68696\ \text{pu} \\ \text{Fault MVA} &= \frac{40}{0.68696} = 58.23\ \text{MVA} \end{aligned}$$

Reactor for 40 % reduction

New fault MVA $= 0.6\times58.23 = 34.94$ MVA.

$$\begin{aligned} X_{eq,new} &= \frac{0.68696}{0.6} = 1.14494\ \text{pu} \\ X &= 1.14494 - 0.68696 = 0.45798\ \text{pu} \\ X(\Omega) &= 0.45798\times3.025 = 1.385\ \Omega \end{aligned}$$

Answer: Present fault level = 58.23 MVA; series reactor ≈ 0.458 pu = 1.385 Ω (fault level becomes 34.94 MVA).

Work in ohms at 11 kV. Phase voltage $V_{ph} = 11000/\sqrt3 = 6350.85$ V. Fault current $I_f = V_{ph}/X_{total}$.

$$\begin{aligned} Z_{base} &= \frac{11^2}{1.6} = 75.625\ \Omega \\ X_g &= 0.2\times75.625 = 15.125\ \Omega \end{aligned}$$

  G1      G2       G3      G4
  |Xg     |Xg      |Xg     |Xg
  A--[5]--B----+---C--[5]--D   11 kV bus
               |       ([5] only in part ii)
              CB
               |
            X=10 ohm
               |
               F

(i) Without bus reactors

All four generators are directly in parallel:

$$\begin{aligned} X_{par} &= \frac{15.125}{4} = 3.781\ \Omega \\ X_{total} &= 3.781 + 10 = 13.781\ \Omega \\ I_f &= \frac{6350.85}{13.781} = 460.83\ \text{A} \end{aligned}$$

Each generator supplies $460.83/4 = 115.21$ A. (Fault MVA $=\sqrt3\times11\times0.46083 = 8.78$ MVA.)

(ii) With 5 Ω reactors between A–B and C–D

G1 and G4 now reach the feeder through 5 Ω; G2 and G3 connect directly.

$$\begin{aligned} X_{G1} = X_{G4} &= 15.125 + 5 = 20.125\ \Omega \\ \frac{1}{X_{par}} &= \frac{2}{20.125} + \frac{2}{15.125} = 0.23161 \\ X_{par} &= 4.3176\ \Omega \\ X_{total} &= 4.3176 + 10 = 14.3176\ \Omega \\ I_f &= \frac{6350.85}{14.3176} = 443.57\ \text{A} \end{aligned}$$

The bus voltage behind the reactors is $I_f X_{par} = 443.57\times4.3176 = 1915.2$ V, so

$$\begin{aligned} I_{G1} = I_{G4} &= \frac{1915.2}{20.125} = 95.16\ \text{A} \\ I_{G2} = I_{G3} &= \frac{1915.2}{15.125} = 126.62\ \text{A} \end{aligned}$$

Check: $2(95.16) + 2(126.62) = 443.57$ A.

Answer: (i) $I_f$ = 460.83 A, each generator 115.21 A. (ii) $I_f$ = 443.57 A; G1 and G4 supply 95.16 A each, G2 and G3 supply 126.62 A each.

All four generators are identical, so work on their own base. Fault MVA $=\text{MVA}_b/X_{eq}$.

Base: 25 MVA, 11 kV; $X_g = 0.16$ pu; $Z_{base} = 11^2/25 = 4.84\ \Omega$.

  G1      G2         G3      G4
  |       |          |       |
  A-------B---[X]----C-------D   bus
                         |
                         F (feeder between C and D)

Fault level without reactor

$$\begin{aligned} X_{eq} &= \frac{0.16}{4} = 0.04\ \text{pu} \\ \text{Fault MVA} &= \frac{25}{0.04} = 625\ \text{MVA} \end{aligned}$$

($I_f = 625/(\sqrt3\times11) = 32.80$ kA.)

Reactor between B and C for 40 % reduction

New fault MVA $= 0.6\times625 = 375$ MVA, so $X_{eq,new} = 25/375 = 0.06667$ pu.

With the reactor, G3 and G4 feed the fault directly ($0.16\parallel0.16 = 0.08$ pu) and G1, G2 feed it through $X$ ($0.08 + X$):

$$\begin{aligned} \frac{1}{0.06667} &= \frac{1}{0.08} + \frac{1}{0.08+X} \\ 15 &= 12.5 + \frac{1}{0.08+X} \\ 0.08 + X &= 0.4 \\ X &= 0.32\ \text{pu} \end{aligned}$$ $$X(\Omega) = 0.32\times4.84 = 1.549\ \Omega$$

Answer: Fault level = 625 MVA; bus-bar reactor between B and C = 0.32 pu (32 %) on 25 MVA ≈ 1.549 Ω, giving 375 MVA.

Reactors are coils with large inductive reactance and very small resistance. They are used in power systems mainly to limit short-circuit current so that circuit breakers and equipment are not over-stressed, and to localise faults.

Why reactors are used

• Fault current $I_f = V/X$; adding reactance cuts $I_f$ and the fault MVA, so cheaper breakers can be used.
• Keep the voltage on healthy sections from collapsing during a fault.
• Reduce mechanical forces and heating in windings during short circuits.
• Shunt reactors absorb surplus VAr on long, lightly loaded lines.

Types by construction

Types by location (reactor schemes)

1. Generator reactor: in series with each generator. Protects the generator, but is in circuit all the time, so causes constant voltage drop and loss. Modern generators have enough internal reactance, so it is seldom used.
2. Feeder reactor: in series with each outgoing feeder. A fault on one feeder does not disturb the bus voltage much. But it gives no protection for bus faults, and causes drop and loss on each feeder.
3. Bus-bar reactor: connects sections of the bus.
   • Ring system: sections joined in a ring through reactors.
   • Tie-bar system: each section connects through a reactor to a common tie-bar; fault current from other sections passes through two reactors.

Shunt reactor

Connected between line and earth at the end of long EHV lines to absorb charging current and limit the rise of receiving-end voltage (Ferranti effect).

Work in ohms. Phase voltage $V_{ph} = 11000/\sqrt3 = 6350.85$ V, and $I_f = V_{ph}/X_{total}$.

$$\begin{aligned} Z_{base} &= \frac{11^2}{0.5} = 242\ \Omega \\ X_{G1} = X_{G2} &= 0.15\times242 = 36.3\ \Omega \end{aligned}$$

   G1 500kVA        G2 500kVA
    15%              15%
     |                |
    CB               CB
     |                |
  ===A==[10]===B======C=== 11 kV
              |   ([10] only in part ii)
             CB
              |
            X=20 ohm
              |
              F

(i) Without reactor between A and B

$$\begin{aligned} X_{par} &= \frac{36.3}{2} = 18.15\ \Omega \\ X_{total} &= 18.15 + 20 = 38.15\ \Omega \\ I_f &= \frac{6350.85}{38.15} = 166.47\ \text{A} \end{aligned}$$

Each alternator supplies $166.47/2 = 83.24$ A.

(ii) With 10 Ω between A and B

$$\begin{aligned} X_{G1,path} &= 36.3 + 10 = 46.3\ \Omega \\ X_{par} &= \frac{46.3\times36.3}{46.3+36.3} = 20.347\ \Omega \\ X_{total} &= 20.347 + 20 = 40.347\ \Omega \\ I_f &= \frac{6350.85}{40.347} = 157.40\ \text{A} \end{aligned}$$

Sharing (inversely as path reactance):

$$\begin{aligned} I_{G1} &= 157.40\times\frac{36.3}{82.6} = 69.17\ \text{A} \\ I_{G2} &= 157.40\times\frac{46.3}{82.6} = 88.23\ \text{A} \end{aligned}$$

Answer: (i) $I_f$ = 166.47 A, each alternator 83.24 A. (ii) $I_f$ = 157.40 A; G1 supplies 69.17 A, G2 supplies 88.23 A.

A reactor is an inductive coil with high reactance and negligible resistance. In generating stations and substations it is used to limit short-circuit current, localise faults and keep bus voltage up on healthy sections. Air-core (dry, concrete-supported) reactors are common because their reactance does not fall by saturation at high fault current.

1. Generator reactors

  G1      G2      G3
  |       |       |
 [X]     [X]     [X]
  |       |       |
 =+=======+=======+===  bus
     |       |
  feeders

• Merits: protects each generator from heavy fault current; useful for old machines with low reactance.
• Demerits: always carries full load current, so constant voltage drop and $I^2R$ loss; a bus or feeder fault still affects all generators; gives no protection between feeders.
• Application: rarely used now, since modern generators have enough leakage reactance.

2. Feeder reactors

  G1      G2
  |       |
 =+===+===+===+===  bus
      |       |
     [X]     [X]
      |       |
   feeder1  feeder2

• Merits: a fault on one feeder hardly affects bus voltage, so other feeders keep supplying; breaker ratings of feeders reduced.
• Demerits: no protection for bus-bar faults; constant drop and loss on every feeder; poor regulation.
• Application: distribution substations with many radial feeders.

3. Bus-bar reactors

(a) Ring system

  G1          G2          G3
  |           |           |
 =S1===[X]===S2===[X]===S3=
  |                        |
  +----------[X]-----------+

(b) Tie-bar system

  G1      G2      G3
  |       |       |
 =S1=    =S2=    =S3=   sections
  |       |       |
 [X]     [X]     [X]
  |       |       |
 =+=======+=======+=  tie-bar

• Merits: in normal operation little current flows through bus reactors, so loss and drop are small; a fault is confined to one section; tie-bar system lets sections be added easily, and fault current from other sections passes through two reactors in series.
• Demerits: a fault on a section gets full fault current from its own generator; tie-bar needs an extra bus; large voltage difference between sections if loads are unbalanced.
• Application: large generating stations with several sections and many generators.

Summary

G1 and G2 are on two bus sections joined by a bus-bar reactor. The feeder is on the G1 section; its reactance is not given, so the fault is taken just beyond the CB. G1 feeds directly; G2 feeds through the bus reactor. Fault MVA $=\text{MVA}_b/X_{eq}$.

Base: 50 MVA, 11 kV; $Z_{base} = 11^2/50 = 2.42\ \Omega$; $I_{base} = \dfrac{50\times10^6}{\sqrt3\times11\times10^3} = 2624.3\ \text{A}$.

  G1 30MVA,0.1        G2 50MVA,0.1
      |                    |
 ===[Sec 1]====[X=0.2 ohm]====[Sec 2]===
      |
     CB
      |--- F

Reactances on 50 MVA base

$$\begin{aligned} X_{G1} &= 0.1\times\frac{50}{30} = 0.16667\ \text{pu} \\ X_{G2} &= 0.1\ \text{pu} \\ X_R &= \frac{0.2}{2.42} = 0.08264\ \text{pu} \\ X_{G2}+X_R &= 0.18264\ \text{pu} \end{aligned}$$

Fault current and fault MVA

$$\begin{aligned} X_{eq} &= \frac{0.16667\times0.18264}{0.16667+0.18264} = 0.087145\ \text{pu} \\ I_f &= \frac{2624.3}{0.087145} = 30{,}114\ \text{A} = 30.11\ \text{kA} \\ \text{Fault MVA} &= \frac{50}{0.087145} = 573.76\ \text{MVA} \end{aligned}$$

Current from each generator

$$\begin{aligned} I_{G1} &= 30.114\times\frac{0.18264}{0.34931} = 15.75\ \text{kA} \\ I_{G2} &= 30.114\times\frac{0.16667}{0.34931} = 14.37\ \text{kA} \end{aligned}$$

Answer: G1 supplies ≈ 15.75 kA, G2 ≈ 14.37 kA (total 30.11 kA); the feeder CB must be rated for about 574 MVA.

The generator MVA ratings are not readable, so assume G1 = G2 = 30 MVA and G3 = G4 = 50 MVA, all 11 kV (the method is the same for any rating). The feeder reactance is not given, so the fault is taken just beyond the CB. G1 and G2 feed directly; G3 and G4 feed through the bus-bar reactor.

Base: 50 MVA, 11 kV; $Z_{base} = 2.42\ \Omega$; $I_{base} = 2624.3\ \text{A}$.

  G1     G2            G3     G4
  0.2    0.2           0.1    0.1
  |      |             |      |
 ==[Section 1]==[X=0.2 ohm]==[Section 2]==
       |
      CB
       |--- F

Reactances on 50 MVA base

$$\begin{aligned} X_{G1} = X_{G2} &= 0.2\times\frac{50}{30} = 0.33333\ \text{pu} \;\Rightarrow\; X_{12} = 0.16667\ \text{pu} \\ X_{G3} = X_{G4} &= 0.1\ \text{pu} \;\Rightarrow\; X_{34} = 0.05\ \text{pu} \\ X_R &= \frac{0.2}{2.42} = 0.08264\ \text{pu} \\ X_{34} + X_R &= 0.13264\ \text{pu} \end{aligned}$$

Fault MVA and fault current

$$\begin{aligned} X_{eq} &= \frac{0.16667\times0.13264}{0.16667+0.13264} = 0.073861\ \text{pu} \\ \text{Fault MVA} &= \frac{50}{0.073861} = 676.95\ \text{MVA} \\ I_f &= \frac{2624.3}{0.073861} = 35{,}531\ \text{A} = 35.53\ \text{kA} \end{aligned}$$

Current from each generator

$$\begin{aligned} I_{G1+G2} &= 35.53\times\frac{0.13264}{0.29931} = 15.75\ \text{kA} \Rightarrow I_{G1} = I_{G2} = 7.87\ \text{kA} \\ I_{G3+G4} &= 35.53\times\frac{0.16667}{0.29931} = 19.78\ \text{kA} \Rightarrow I_{G3} = I_{G4} = 9.89\ \text{kA} \end{aligned}$$

Answer (with the assumed ratings): G1, G2 ≈ 7.87 kA each; G3, G4 ≈ 9.89 kA each; total 35.53 kA; CB fault rating ≈ 677 MVA.

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

Ratings in MW are taken as MVA (unity power factor). Fault current $I_f = I_{base}/X_{eq}$; ohms to pu: $X_{pu} = X_\Omega\,\text{MVA}_b/\text{kV}^2$.

Base: 50 MVA, 11 kV; $Z_{base} = 11^2/50 = 2.42\ \Omega$; $I_{base} = \dfrac{50\times10^6}{\sqrt3\times11\times10^3} = 2624.3\ \text{A}$.

  G1 50MW,0.16   G2 50MW,0.16
      |              |
  ====+======+=======+==== 11 kV
             |
           Xl=2 ohm + new L
             |
             F

Fault current

$$\begin{aligned} X_1\parallel X_2 &= \frac{0.16}{2} = 0.08\ \text{pu} \\ X_l &= \frac{2}{2.42} = 0.82645\ \text{pu} \\ X_{eq} &= 0.08 + 0.82645 = 0.90645\ \text{pu} \\ I_f &= \frac{2624.3}{0.90645} = 2895.2\ \text{A} \end{aligned}$$

(Fault MVA $= 50/0.90645 = 55.16$ MVA.)

Reactor for 40 % reduction in fault current

New $I_f = 0.6\times2895.2 = 1737.1$ A, so the total reactance must become $X_{eq}/0.6$:

$$\begin{aligned} X_{eq,new} &= \frac{0.90645}{0.6} = 1.51074\ \text{pu} \\ X_{add} &= 1.51074 - 0.90645 = 0.60430\ \text{pu} \\ X_{add}(\Omega) &= 0.60430\times2.42 = 1.4624\ \Omega \\ L &= \frac{1.4624}{2\pi\times50} = 4.655\times10^{-3}\ \text{H} \end{aligned}$$

Answer: $I_f$ ≈ 2895 A; series reactor ≈ 1.462 Ω, i.e. L ≈ 4.65 mH (fault current falls to about 1737 A).