Skip to main content

Short circuit MVA at bus 2

Short answer

Using Thevenin reactance at bus 2 on 100 MVA base, the short-circuit MVA is 365.9 MVA.

In the network given below, compute the short circuit MVA for a fault at bus 2. [Figure: generator j0.1 - transformer j0.12 - bus 1; line 1-2 j0.08; line 1-3 j0.06; line 2-3 j0.1; Sbase = 100 MVA]

Base: 100 MVA. Pre-fault voltage is taken as 1.0 pu and pre-fault load current is neglected (usual assumption for fault level calculation). Only one source (the generator) feeds the network.

 E=1--j0.1--j0.12--bus1
                  /    \
             j0.08      j0.06
                /          \
            bus2 --j0.1-- bus3
             |
             F

Thevenin reactance at bus 2

  • Source to bus 1: 0.1+0.12=0.220.1 + 0.12 = 0.22 pu.
  • Bus 3 has no source, so lines 1-3 and 3-2 are simply in series: 0.06+0.10=0.160.06 + 0.10 = 0.16 pu.
  • This path is in parallel with line 1-2 (0.08 pu):
X1−2=0.08×0.160.08+0.16=0.0533 puXth=0.22+0.0533=0.2733 pu\begin{aligned} X_{1-2} &= \frac{0.08 \times 0.16}{0.08 + 0.16} = 0.0533\ \text{pu} \\ X_{th} &= 0.22 + 0.0533 = 0.2733\ \text{pu} \end{aligned}

Short-circuit MVA

If=10.2733=3.6585 puSC MVA=SbaseXth=1000.2733=365.85 MVA\begin{aligned} I_f &= \frac{1}{0.2733} = 3.6585\ \text{pu} \\ \text{SC MVA} &= \frac{S_{base}}{X_{th}} = \frac{100}{0.2733} = 365.85\ \text{MVA} \end{aligned}

Answer: short-circuit MVA at bus 2 ≈365.9\approx 365.9 MVA (fault current 3.659 pu).

✓ Answer checked by Claude Sonnet 5.5

Related questions

All Power System Analysis II chapter 3 questions →