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Bus-bar reactor scheme: 30 MVA and 50 MVA

Short answer

On 50 MVA base, G1 reactance is 0.16667 pu, G2 is 0.1 pu, bus reactor 0.08264 pu. Equivalent reactance is 0.087145 pu, so fault current is 30.11 kA and fault MVA is 573.76 MVA; G1 supplies 15.75 kA and G2 supplies 14.37 kA.

Figure below shows a bus-bar reactor scheme. If a 3 phase to ground fault occurs on the outgoing feeder, calculate the fault current (in kA) supplied by each generator and fault MVA required for the circuit breaker (CB) on the feeder. Rating of generators are given as follow: G1: 30MVA, 11kV, Xg1" = 0.1 pu; G2: 50MVA, 11kV, Xg2" = 0.1 pu. [Figure: G1 (0.1 pu) and G2 (0.1 pu) on two bus-bar sections joined by a bus-bar reactor X = 0.2 ohm; the feeder leaves the G1 section through the CB]

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 =MVAb/Xeq=\text{MVA}_b/X_{eq}.

Base: 50 MVA, 11 kV; Zbase=112/50=2.42 ΩZ_{base} = 11^2/50 = 2.42\ \Omega; Ibase=50×1063×11×103=2624.3 AI_{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

XG1=0.1×5030=0.16667 puXG2=0.1 puXR=0.22.42=0.08264 puXG2+XR=0.18264 pu\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

Xeq=0.16667×0.182640.16667+0.18264=0.087145 puIf=2624.30.087145=30,114 A=30.11 kAFault MVA=500.087145=573.76 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

IG1=30.114×0.182640.34931=15.75 kAIG2=30.114×0.166670.34931=14.37 kA\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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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