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Bus-bar reactor scheme with four generators

Short answer

Using a 50 MVA, 11 kV base and assuming G1/G2=30 MVA, G3/G4=50 MVA, the total fault current is 35.53 kA and CB fault MVA is 677 MVA. Each G1/G2 contributes 7.87 kA; each G3/G4 contributes 9.89 kA.

In figure below a bus-bar reactor scheme with four generators. 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/G2: [?]MVA, 11kV, Xg1" = 0.2 pu; G3/G4: [?]MVA, 11kV, Xg2" = 0.1 pu. [Figure: G1 (0.2 pu) and G2 (0.2 pu) on one bus-bar section, G3 (0.1 pu) and G4 (0.1 pu) on the other, the two sections joined by a bus-bar reactor X = 0.2 ohms; the feeders leave the G1/G2 section through the CB] (Generator MVA ratings are unreadable in the scan.)

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; Zbase=2.42 ΩZ_{base} = 2.42\ \Omega; Ibase=2624.3 AI_{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

XG1=XG2=0.2×5030=0.33333 pu  ⇒  X12=0.16667 puXG3=XG4=0.1 pu  ⇒  X34=0.05 puXR=0.22.42=0.08264 puX34+XR=0.13264 pu\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

Xeq=0.16667×0.132640.16667+0.13264=0.073861 puFault MVA=500.073861=676.95 MVAIf=2624.30.073861=35,531 A=35.53 kA\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

IG1+G2=35.53×0.132640.29931=15.75 kA⇒IG1=IG2=7.87 kAIG3+G4=35.53×0.166670.29931=19.78 kA⇒IG3=IG4=9.89 kA\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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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