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Fault current: 50 MW and 40 MW generators

Short answer

Fault current ≈2903 A, fault power ≈55.32 MVA; G1 supplies ≈1405 A, G2 ≈1499 A. For equal currents, add a reactor of 0.01 pu (0.0242 Ω, ≈0.077 mH) in series with G2.

Below figure show 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 = 40 MW, 11kV, X2 = 0.12 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 outgoing feeder leaves the busbar through XL] a) If a three-phase to ground fault occurs at outgoing feeder, calculate the fault current and fault power at the outgoing feeder and fault current supplied by each generator. b) Calculate the value of inductor (in mH) of the reactor to be connected in series with X2 so that both generators delivers equal amp of fault current during 3 phase to ground fault on the feeder.

Generator ratings are given in MW; they are taken as MVA (unity power factor assumed). Fault MVA =MVAb/Xeq=\text{MVA}_b/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 40MW,0.12
      |               |
     X1              X2 (+ Xr in part b)
      |               |
  ====+=======+=======+==== 11 kV bus
              |
             XL = 2 ohm
              |
              F (3-ph to ground)

a) Fault current, fault power and generator currents

X1=0.16 pu,X2=0.12×5040=0.15 puX1∥X2=0.16×0.150.31=0.077419 puXL=22.42=0.82645 puXeq=0.077419+0.82645=0.90387 puIf=2624.30.90387=2903.4 AFault power=500.90387=55.32 MVA\begin{aligned} X_1 &= 0.16\ \text{pu},\quad X_2 = 0.12\times\frac{50}{40} = 0.15\ \text{pu} \\ X_1\parallel X_2 &= \frac{0.16\times0.15}{0.31} = 0.077419\ \text{pu} \\ X_L &= \frac{2}{2.42} = 0.82645\ \text{pu} \\ X_{eq} &= 0.077419 + 0.82645 = 0.90387\ \text{pu} \\ I_f &= \frac{2624.3}{0.90387} = 2903.4\ \text{A} \\ \text{Fault power} &= \frac{50}{0.90387} = 55.32\ \text{MVA} \end{aligned}

Current divides inversely as the branch reactances:

IG1=2903.4×0.150.31=1404.9 AIG2=2903.4×0.160.31=1498.5 A\begin{aligned} I_{G1} &= 2903.4\times\frac{0.15}{0.31} = 1404.9\ \text{A} \\ I_{G2} &= 2903.4\times\frac{0.16}{0.31} = 1498.5\ \text{A} \end{aligned}

b) Reactor in series with X2X_2 for equal currents

Both generators feed the same bus, so they deliver equal current only when their branch reactances are equal:

X2+Xr=X1  ⇒  Xr=0.16−0.15=0.01 puXr(Ω)=0.01×2.42=0.0242 ΩL=Xr2πf=0.02422π×50=7.70×10−5 H=0.0770 mH\begin{aligned} X_2 + X_r &= X_1 \;\Rightarrow\; X_r = 0.16 - 0.15 = 0.01\ \text{pu} \\ X_r(\Omega) &= 0.01\times2.42 = 0.0242\ \Omega \\ L &= \frac{X_r}{2\pi f} = \frac{0.0242}{2\pi\times50} = 7.70\times10^{-5}\ \text{H} = 0.0770\ \text{mH} \end{aligned}

Check: new Xeq=0.08+0.82645=0.90645X_{eq} = 0.08 + 0.82645 = 0.90645 pu, If=2895.2I_f = 2895.2 A, each generator supplies 1447.6 A.

Answer: (a) IfI_f ≈ 2903 A, fault power ≈ 55.32 MVA, IG1I_{G1} ≈ 1405 A, IG2I_{G2} ≈ 1499 A. (b) LL ≈ 0.077 mH (0.0242 Ω).

✓ Answer checked by Claude Opus 5.5 (high effort)

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