Skip to main content

Fault level: adding a third generator

Short answer

Fault level is 400 MVA, fault current ≈ 20.99 kA. After adding G3, equivalent reactance drops to 0.04444 pu; to restore 0.05 pu total, add a feeder reactor of 0.00556 pu (0.0336 Ω).

Following figure shows two generators operating parallel. i) If a 3-phase to ground fault occurs on the outgoing feeder, calculate the fault current and fault level on the feeder. ii) If a 3rd generator (G3: 10MVA, 11kV, X3=0.2 pu) is added at point 'C', calculate the value of reactor to be added on the outgoing feeder so that the fault level remains same as before. [Figure: G1 20 MVA, 11 kV, X1 = 0.1 pu (path-1) and G2 20 MVA, 11 kV, X2 = 0.1 pu (path-2) connected to a common busbar at points A and B; busbar extends to point C; outgoing feeder from the busbar through a C.B., fault point 'F' on the outgoing feeder]

Fault MVA =MVAb/Xeq=\text{MVA}_b/X_{eq}. The feeder has no reactance at first (none given).

Base: 20 MVA, 11 kV; Zbase=112/20=6.05 ΩZ_{base} = 11^2/20 = 6.05\ \Omega; Ibase=20×1063×11×103=1049.7 AI_{base} = \dfrac{20\times10^6}{\sqrt3\times11\times10^3} = 1049.7\ \text{A}.

 G1 20MVA   G2 20MVA   G3 10MVA (added)
  0.1pu      0.1pu      0.2pu
   |A         |B         |C
 ==+==========+==========+=== 11 kV
         |
        C.B.--[X]-- F

(i) Fault current and fault level (two generators)

Xeq=0.12=0.05 puFault MVA=200.05=400 MVAIf=10.05×1049.7=20,995 A≈20.99 kA\begin{aligned} X_{eq} &= \frac{0.1}{2} = 0.05\ \text{pu} \\ \text{Fault MVA} &= \frac{20}{0.05} = 400\ \text{MVA} \\ I_f &= \frac{1}{0.05}\times1049.7 = 20{,}995\ \text{A} \approx 20.99\ \text{kA} \end{aligned}

(ii) Reactor after adding G3

G3 on 20 MVA base: X3=0.2×2010=0.4 puX_3 = 0.2\times\dfrac{20}{10} = 0.4\ \text{pu}.

1Xeq′=10.1+10.1+10.4=22.5Xeq′=0.04444 pu  ⇒  Fault MVA=450 MVA (too high)\begin{aligned} \frac{1}{X_{eq}'} &= \frac{1}{0.1}+\frac{1}{0.1}+\frac{1}{0.4} = 22.5 \\ X_{eq}' &= 0.04444\ \text{pu} \;\Rightarrow\; \text{Fault MVA} = 450\ \text{MVA (too high)} \end{aligned}

To keep 400 MVA the total must stay at 0.05 pu:

Xreactor=0.05−0.04444=0.005556 puXreactor(Ω)=0.005556×6.05=0.0336 Ω\begin{aligned} X_{reactor} &= 0.05 - 0.04444 = 0.005556\ \text{pu} \\ X_{reactor}(\Omega) &= 0.005556\times6.05 = 0.0336\ \Omega \end{aligned}

Answer: (i) Fault level = 400 MVA, If≈I_f \approx 20.99 kA. (ii) Feeder reactor ≈ 0.00556 pu = 0.0336 Ω keeps the fault level at 400 MVA after G3 is added.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Power Plant Equipment chapter 4 questions →