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L-L fault through Zf with two motors

Short answer

On 25 MVA base, positive/negative Thevenin impedance at the motor bus is Z1 = Z2 = j0.1597 pu. With Zf = j0.1 pu, Ia1 = -j2.384 pu, and L-L fault current is √3 × 2.384 = 4.13 pu = 5418 A.

For the power system network shown below, if Line to Line fault occurs at point 'F' then find the fault current, take fault impedance Zf = 0.1 p.u. For Generator: X1 = X2 = 20%, X0 = 5%; For Both Motor: X1 = X2 = 25%, X0 = 10%; For Both Transformer: X1 = X2 = X0 = 10%. The neutral reactance of generator G1 and motor M2 are in p.u. [Figure: G1 25 MVA, 11 kV, Y grounded through Xn = 0.1 - T1 30 MVA, 10.8/121 kV, Δ/Y grounded - line X1 = X2 = 100 Ω, X0 = 250 Ω - T2 30 MVA, 121/10.8 kV, Y grounded/Δ - motor bus with M1 (15 MVA, 10 kV, Y ungrounded) and M2 (7.5 MVA, 10 kV, Y grounded through Xn = 0.036); fault F at the motor bus]

Assumptions: base 25 MVA, 11 kV in the generator circuit; pre-fault voltage at F is 1 pu with no load current; Zf=j0.1Z_f = j0.1 pu. An L-L fault involves only positive and negative sequence networks, so the neutral reactances and zero-sequence data are not needed.

Base voltages

Vb,line=11×12110.8=123.24 kV,Vb,motor=123.24×10.8121=11 kVV_{b,line} = 11 \times \frac{121}{10.8} = 123.24\ \text{kV}, \quad V_{b,motor} = 123.24 \times \frac{10.8}{121} = 11\ \text{kV}

Per-unit reactances (25 MVA)

ElementCalculationX1=X2X_1 = X_2 (pu)
G1given0.20
T1, T20.1×2530×(10.811)20.1 \times \frac{25}{30} \times (\frac{10.8}{11})^20.0803 each
Line100/123.24225100 / \frac{123.24^2}{25}0.1646
M10.25×2515×(1011)20.25 \times \frac{25}{15} \times (\frac{10}{11})^20.3444
M20.25×257.5×(1011)20.25 \times \frac{25}{7.5} \times (\frac{10}{11})^20.6887

Positive (and negative) sequence network at F

 Eg-j0.2-T1 j0.0803-line j0.1646-T2 j0.0803-(F)
                                             |
                               +-------------+
                               |             |
                            j0.3444       j0.6887
                              Em1           Em2
Xgen side=0.2+0.0803+0.1646+0.0803=0.5253Xmotors=0.3444∥0.6887=0.2296Z1=Z2=j(0.5253∥0.2296)=j0.1597 pu\begin{aligned} X_{gen\ side} &= 0.2 + 0.0803 + 0.1646 + 0.0803 = 0.5253 \\ X_{motors} &= 0.3444 \| 0.6887 = 0.2296 \\ Z_1 = Z_2 &= j(0.5253 \| 0.2296) = j0.1597\ \text{pu} \end{aligned}

L-L fault through ZfZ_f

Ia1=−Ia2=VfZ1+Z2+Zf=1j(0.1597+0.1597+0.1)=1j0.4195=−j2.384 pu∣If∣=∣Ib∣=∣Ic∣=3×2.384=4.129 pu\begin{aligned} I_{a1} = -I_{a2} &= \frac{V_f}{Z_1 + Z_2 + Z_f} = \frac{1}{j(0.1597 + 0.1597 + 0.1)} = \frac{1}{j0.4195} = -j2.384\ \text{pu} \\ |I_f| = |I_b| = |I_c| &= \sqrt3 \times 2.384 = 4.129\ \text{pu} \end{aligned}

Base current at the motor bus (11 kV):

Ibase=25×1063×11×103=1312.2 AI_{base} = \frac{25\times10^6}{\sqrt3 \times 11\times10^3} = 1312.2\ \text{A} If=4.129×1312.2=5418 AI_f = 4.129 \times 1312.2 = 5418\ \text{A}

Answer: L-L fault current at F ≈ 4.13 pu ≈ 5.42 kA.

✓ Answer checked by Claude Sonnet 5.5

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