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Fault current: two 50 MVA generators, feeder reactor

Short answer

On 50 MVA, 11 kV base with negligible feeder reactance, parallel reactance is 0.01875 pu. Fault current = 1/0.01875 = 53.33 pu = 139.96 kA (G1: 52.49 kA, G2: 87.48 kA), fault power 2666.7 MVA. To reduce by 40% (to 1600 MVA), add series reactor of 0.0303 Ω (0.0125 pu).

Two generators each of capacities 50 MVA, 11 kV having their sub transient reactance of 5% and 3% respectively operating in parallel and supplying to a common load through a feeder. If a 3 phase to ground fault occurs on the outgoing feeder, calculate the fault current supplied by each generator and fault power on the outgoing feeder. Also calculate the value of reactor to be connected in series with the outgoing feeder to reduce the fault level by 40 %.

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

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

Fault current and fault power

Xeq=0.05×0.030.05+0.03=0.01875 puIf=10.01875=53.33 pu=53.33×2624.3=139,964 A≈139.96 kAFault MVA=500.01875=2666.7 MVA\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:

IG1=IfXG2XG1+XG2=139.96×0.030.08=52.49 kAIG2=IfXG1XG1+XG2=139.96×0.050.08=87.48 kA\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 =0.6×2666.7=1600= 0.6\times2666.7 = 1600 MVA.

Xeq,new=501600=0.03125 puXadded=0.03125−0.01875=0.0125 puXadded(Ω)=0.0125×2.42=0.03025 Ω\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: If≈139.96I_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).

✓ Answer checked by Claude Opus 5.5 (high effort)

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