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Fault at receiving end of 132 kV lines

Short answer

Assuming 200 km lines (40 Ω each), on100 MVA base X_th =0.255 pu. Fault level =392 MVA, fault current =1.71 kA at132 kV. At11 kV, G1 supplies12.55 kA, G2 supplies8.02 kA.

Figure below shows power system fed by two generators. The rating and reactances of the equipments are shown. A 3-phase balanced short circuit fault occurs at the receiving end bus of 132 kV. Find (a) Fault level and fault current at receiving end (b) Fault current supplied by generators G1 and G2. Note that reactance of machines are based on their own rating. [Figure: G1 100 MVA, 11 kV, 15% - T1 100 MVA, 11/132 kV, X = 8%; G2 50 MVA, 11 kV, 10% - T2 50 MVA, 11/132 kV, X = 8%; both transformers feed the 132 kV bus; Line-1 and Line-2, each X = 0.2 Ω/km/phase, in parallel to the receiving end bus F (line length not given)]

Fault level at a bus is MVAbase/Xth\text{MVA}_{base}/X_{th} (pu), where XthX_{th} is the total reactance from the source EMFs (1 pu) to the fault.

Assumption: the line length is not given in the question, so a length of 200 km is assumed. Each line then has X=0.2×200=40 ΩX = 0.2 \times 200 = 40\ \Omega. G2 is 50 MVA, 11 kV, 10%; T2 is 50 MVA, 11/132 kV, 8%.

Step 1: Per unit values on 100 MVA base

Base: 100 MVA, 11 kV (generators), 132 kV (lines). Zb,132=1322/100=174.24 ΩZ_{b,132} = 132^2/100 = 174.24\ \Omega.

ElementCalculationX (pu)
G10.15 (own base = 100 MVA)0.15
T10.080.08
G20.10×100/500.10 \times 100/500.20
T20.08×100/500.08 \times 100/500.16
Each line40/174.2440/174.240.2296

Step 2: Reactance diagram and reduction

 G1 0.15  T1 0.08
 (~)--------||---+               Line-1 0.2296
                 |-- 132 kV bus ====+=========+--- F
 (~)--------||---+               Line-2 0.2296
 G2 0.20  T2 0.16
X1=0.15+0.08=0.23,X2=0.20+0.16=0.36Xgen=0.23×0.360.23+0.36=0.1403Xline=0.22962=0.1148Xth=0.1403+0.1148=0.2551 pu\begin{aligned} X_1 &= 0.15 + 0.08 = 0.23, \quad X_2 = 0.20 + 0.16 = 0.36 \\ X_{gen} &= \frac{0.23 \times 0.36}{0.23 + 0.36} = 0.1403 \\ X_{line} &= \frac{0.2296}{2} = 0.1148 \\ X_{th} &= 0.1403 + 0.1148 = 0.2551\ \text{pu} \end{aligned}

(a) Fault level and fault current at the receiving end

If=10.2551=3.920 puFault level=1000.2551=392.0 MVAIb,132=100×1063×132×103=437.4 AIf=3.920×437.4=1714 A=1.714 kA\begin{aligned} I_f &= \frac{1}{0.2551} = 3.920\ \text{pu} \\ \text{Fault level} &= \frac{100}{0.2551} = 392.0\ \text{MVA} \\ I_{b,132} &= \frac{100\times10^6}{\sqrt3 \times 132\times10^3} = 437.4\ \text{A} \\ I_f &= 3.920 \times 437.4 = 1714\ \text{A} = 1.714\ \text{kA} \end{aligned}

(b) Currents supplied by G1 and G2

The fault current divides between the two generator branches inversely as their reactances:

IG1=If X2X1+X2=3.920×0.360.59=2.392 puIG2=If X1X1+X2=3.920×0.230.59=1.528 pu\begin{aligned} I_{G1} &= I_f\,\frac{X_2}{X_1+X_2} = 3.920 \times \frac{0.36}{0.59} = 2.392\ \text{pu} \\ I_{G2} &= I_f\,\frac{X_1}{X_1+X_2} = 3.920 \times \frac{0.23}{0.59} = 1.528\ \text{pu} \end{aligned}

Base current on the 11 kV side: Ib,11=100×106/(3×11×103)=5248.6I_{b,11} = 100\times10^6/(\sqrt3\times 11\times10^3) = 5248.6 A.

IG1=2.392×5248.6=12.55 kA,IG2=1.528×5248.6=8.02 kAI_{G1} = 2.392 \times 5248.6 = 12.55\ \text{kA}, \qquad I_{G2} = 1.528 \times 5248.6 = 8.02\ \text{kA}

Answer (200 km lines): fault level ≈ 392 MVA, fault current ≈ 1.71 kA at 132 kV; G1 supplies ≈ 12.55 kA and G2 ≈ 8.02 kA at 11 kV. For any other line length ll km, only Xline=0.2l/(2×174.24)X_{line} = 0.2l/(2\times174.24) pu changes in Step 2.

✓ Answer checked by Claude Sonnet 5.5

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