On a 25 MVA base, how is the fault level at the bus obtained from the equivalent reactance?

Fault MVA = base MVA / X_eq (pu)

Fault MVA = base MVA x X_eq (pu)

Fault MVA = sqrt(3) x 11 x X_eq

With the reactor X between B and C, which description of the network seen by the fault is correct?

G3 and G4 feed directly (0.08 pu) and G1, G2 feed through (0.08 + X)

All four generators are in series with X

Only G1 and G2 feed the fault, G3 and G4 are isolated

G1 and G2 feed directly and G3, G4 feed through X

2073 Shrawan · 5 marks

Fault level: four 25 MVA generators, bus reactor

Short answer

On 25 MVA base, four 0.16 pu generators in parallel give equivalent reactance 0.04 pu, so fault level is 625 MVA. To reduce to 375 MVA, a bus-bar reactor between B and C must be 0.32 pu, about 1.549 Ω.

The figure below shows four identical generators, each rated 11 kV, 25 MVA and each having sub-transient reactance of 16% on its own rating. Find 3-ϕ fault level at one of the outgoing feeder. Also calculate the value of reactance to be connected in the bus bar between "B" and "C" so that fault level reduces by 40%. [Figure: four generators connected to a busbar at points A, B, C, D; outgoing feeders leave the busbar, with a 3-ϕ fault on the feeder between C and D]

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.

✓ Answer checked by Claude Opus 5.5 (high effort)

Fault current: two 50 MVA generators, feeder reactor · Power Plant Equipment, chapter 4 · asked 3×
Fault level: two 40 MVA generators, 2 Ω feeder · Power Plant Equipment, chapter 4 · asked 5×
Fault level: three generators, feeder reactor · Power Plant Equipment, chapter 4
Bus-bar reactor scheme with four generators · Power Plant Equipment, chapter 4 · asked 2×
Fault level: two 25 MVA generators, 50% reduction · Power Plant Equipment, chapter 4
Fault level: two 50 MVA generators, 50% reduction · Power Plant Equipment, chapter 4
Reactance diagram: 20 MVA generator, 15 kV/25 MVA base · Power System Analysis I, chapter 3
Reactance diagram: 50 MVA, 13.8 kV base, three generators · Power System Analysis I, chapter 3
HV bus fault and limiting reactor · Power System Analysis II, chapter 3
Fault MVA at bus 5 with inductive lines · Power System Analysis II, chapter 3 · asked 2×

All Power Plant Equipment chapter 4 questions →

Fault level: four 25 MVA generators, bus reactor

Fault level without reactor

Reactor between B and C for 40 % reduction

Related questions