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Fault current: two 50 MW generators, reactor inductance

Short answer

On a 50 MVA, 11 kV base, the feeder fault current is 2895 A. To reduce fault current by 40%, add a series reactor of 1.462 ohm, which is 4.65 mH at 50 Hz.

Figure below shows the single line diagram of parallel operated generators. Their ratings are: G1: capacity = 50 MW, 11kV, X1 = 0.16 pu based on its rating; G2: capacity = 50MW, 11kV, X2 = 0.16 pu based on its rating; Xl = 2 ohms. [Figure: G1 (through X1) and G2 (through X2), both star-grounded, connected to a common busbar; the feeder leaves the busbar through Xl] Calculate the fault current in the feeder during 3 phases to ground fault on the feeder. Calculate the value of inductance (in mH) of the reactor to be connected in series with Xl so that the fault current decreases by 40%.

Ratings in MW are taken as MVA (unity power factor). Fault current If=Ibase/XeqI_f = I_{base}/X_{eq}; ohms to pu: Xpu=XΩ MVAb/kV2X_{pu} = X_\Omega\,\text{MVA}_b/\text{kV}^2.

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

Fault current

X1∥X2=0.162=0.08 puXl=22.42=0.82645 puXeq=0.08+0.82645=0.90645 puIf=2624.30.90645=2895.2 A\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= 50/0.90645 = 55.16 MVA.)

Reactor for 40 % reduction in fault current

New If=0.6×2895.2=1737.1I_f = 0.6\times2895.2 = 1737.1 A, so the total reactance must become Xeq/0.6X_{eq}/0.6:

Xeq,new=0.906450.6=1.51074 puXadd=1.51074−0.90645=0.60430 puXadd(Ω)=0.60430×2.42=1.4624 ΩL=1.46242π×50=4.655×10−3 H\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: IfI_f ≈ 2895 A; series reactor ≈ 1.462 Ω, i.e. L ≈ 4.65 mH (fault current falls to about 1737 A).

✓ Answer checked by Claude Opus 5.5 (high effort)

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