Power System Fault Calculation

A bolted three-phase fault is balanced, so only the positive-sequence reactance diagram is needed. Base: 100 MVA, 11 kV on the generator side and 132 kV on the line side. Pre-fault voltage is taken as 1.0 pu and pre-fault load current is neglected (usual assumption for fault level calculation). As described, both transformers feed a common 132 kV bus, and two parallel 200 km lines run from that bus to F.

Per-unit reactances

$X_{new} = X_{old}\dfrac{S_{new}}{S_{old}}$ (voltage bases match the ratings):

Line: $X_L = 0.2 \times 200 = 40\ \Omega$ per line.

$$\begin{aligned} Z_{base} &= \frac{132^2}{100} = 174.24\ \Omega \\ X_L &= \frac{40}{174.24} = 0.2296\ \text{pu} \end{aligned}$$

Reactance diagram

 E=1 --[G1 j0.15]--[T1 j0.10]--+
                               |   +--[L1]--+
                            132 kV |        |-- F
                               |   +--[L2]--+
 E=1 --[G2 j0.20]--[T2 j0.16]--+

Thevenin reactance at F

$$\begin{aligned} X_{G1T1} &= 0.15 + 0.10 = 0.25 \\ X_{G2T2} &= 0.20 + 0.16 = 0.36 \\ X_{gen} &= \frac{0.25 \times 0.36}{0.25 + 0.36} = 0.1475 \\ X_{lines} &= \frac{0.2296}{2} = 0.1148 \\ X_{th} &= 0.1475 + 0.1148 = 0.2623\ \text{pu} \end{aligned}$$

Fault current and fault MVA

$$\begin{aligned} I_f &= \frac{1}{X_{th}} = \frac{1}{0.2623} = 3.8121\ \text{pu} \\ I_{base} &= \frac{100 \times 10^6}{\sqrt{3} \times 132 \times 10^3} = 437.39\ \text{A} \\ I_f &= 3.8121 \times 437.39 = 1667.3\ \text{A} \\ \text{Fault MVA} &= \frac{S_{base}}{X_{th}} = \frac{100}{0.2623} = 381.21\ \text{MVA} \end{aligned}$$

Answer: total fault current $\approx 1.667$ kA (at 132 kV), fault level $\approx 381.2$ MVA.

Base: 100 MVA, 11 kV (generator side), 132 kV (line side), $f = 50$ Hz. Pre-fault voltage is taken as 1.0 pu and pre-fault load current is neglected (usual assumption for fault level calculation).

Per-unit reactances

• G1, G2: $X'' = 0.20$ pu each (already on 100 MVA, 11 kV).
• T1, T2: $X = 0.05$ pu each.
• So from each generator EMF to its 132 kV bus: $0.20 + 0.05 = 0.25$ pu.

Lines ($X = 2\pi fL$):

$$\begin{aligned} Z_{base} &= \frac{132^2}{100} = 174.24\ \Omega \\ X_{L1} &= 2\pi(50)(1.2 \times 10^{-3} \times 100) = 37.699\ \Omega = \frac{37.699}{174.24} = 0.2164\ \text{pu} \\ X_{L2} = X_{L3} &= 2\pi(50)(1.0 \times 10^{-3} \times 50) = 15.708\ \Omega = 0.0902\ \text{pu} \end{aligned}$$

Reactance diagram

  E=1                         E=1
   |                           |
 j0.25 (G1+T1)               j0.25 (G2+T2)
   |                           |
 bus3 ------- L1 j0.216 ------- bus4
    \                         /
  L3 j0.090            L2 j0.090
       \                   /
            bus 5 (F)

Thevenin reactance at bus 5

Both generator EMFs are equal (1 pu), and the two halves of the network are identical (0.25 pu to each source, $X_{L2} = X_{L3}$). So buses 3 and 4 are always at the same potential and line L1 carries no current; it can be removed. The two paths (0.25 + 0.0902) are then in parallel:

$$\begin{aligned} X_{th} &= \frac{0.25 + 0.0902}{2} = 0.1701\ \text{pu} \end{aligned}$$

(The same value is obtained as $Z_{55}$ of the bus impedance matrix.)

Fault current and fault MVA

$$\begin{aligned} I_f &= \frac{1}{0.1701} = 5.8797\ \text{pu} \\ I_{base} &= \frac{100 \times 10^6}{\sqrt{3}(132 \times 10^3)} = 437.39\ \text{A} \\ I_f &= 5.8797 \times 437.39 = 2571.7\ \text{A} \\ \text{Fault MVA} &= \frac{100}{0.1701} = 587.97\ \text{MVA} \end{aligned}$$

Answer: fault MVA $\approx 588.0$ MVA and fault current $\approx 2.572$ kA at 132 kV.

The short-circuit MVA that the feeder circuit breaker must interrupt is the fault level at bus 3, since the fault is just beyond the breaker and the feeder impedance up to F is negligible. Base: 100 MVA, 11 kV / 132 kV. Pre-fault $V_3 = 1$ pu and pre-fault current is zero, so all source EMFs are 1 pu.

Per-unit reactances (100 MVA base)

• G1, G2: 0.10 pu each; T1, T2: 0.05 pu each, so each source branch is $0.10 + 0.05 = 0.15$ pu.
• Lines 1-2, 1-3, 2-3: 0.10 pu each.

  E=1                 E=1
   |                   |
 j0.15               j0.15
   |                   |
 bus1 ---- j0.1 ---- bus2
     \               /
    j0.1          j0.1
        \         /
          bus 3 --CB-- F

Thevenin reactance at bus 3

The network is symmetrical about bus 3: buses 1 and 2 are always at the same potential, so line 1-2 carries no current. Each path is $0.15 + 0.10 = 0.25$ pu, and the two are in parallel:

$$X_{th} = \frac{0.25 \times 0.25}{0.25 + 0.25} = 0.125\ \text{pu}$$

Short-circuit MVA of the breaker

$$\begin{aligned} I_f &= \frac{V_3}{X_{th}} = \frac{1}{0.125} = 8\ \text{pu} \\ \text{SC MVA} &= \frac{S_{base}}{X_{th}} = \frac{100}{0.125} = 800\ \text{MVA} \\ I_f &= 8 \times \frac{100 \times 10^6}{\sqrt{3} \times 132 \times 10^3} = 8 \times 437.39 = 3499.1\ \text{A} \end{aligned}$$

Answer: the circuit breaker should be rated for a short-circuit capacity of 800 MVA, with a fault current of about 3.50 kA at 132 kV. (For the 11/110 kV variant the per-unit values and the 800 MVA result are the same; only the current in amperes changes, to 4.20 kA at 110 kV.)

Use the internal (sub-transient) EMFs of both machines found from the pre-fault load. Base: 100 MVA, 11 kV.

Per-unit values

• Generator: $X''_g = 0.25$ pu.
• Motor: $X''_m = 0.2 \times \dfrac{100}{50} = 0.4$ pu.
• Line: $X_L = 0.05$ pu.
• Motor terminal voltage: $V_m = \dfrac{10.95}{11} = 0.9955\angle 0^\circ$ pu (reference).
• Motor input: $P = 40/100 = 0.4$ pu at 0.8 pf leading.

 Eg --j0.25--(Vt)--j0.05--(Vm)--j0.4-- Em
             |
             F (fault at generator terminals)

Pre-fault current

$$\begin{aligned} |I| &= \frac{P}{V_m\cos\phi} = \frac{0.4}{0.9955 \times 0.8} = 0.5023\ \text{pu} \\ I &= 0.5023\angle 36.87^\circ = 0.4018 + j0.3014\ \text{pu}\ \text{(leading)} \end{aligned}$$

Internal EMFs

$$\begin{aligned} V_t &= V_m + jX_LI = 0.9955 + j0.05(0.4018 + j0.3014) = 0.9804 + j0.0201\ \text{pu} \\ E''_g &= V_t + jX''_gI = 0.9804 + j0.0201 + j0.25(0.4018 + j0.3014) = 0.9050 + j0.1205 = 0.9130\angle 7.59^\circ\ \text{pu} \\ E''_m &= V_m - jX''_mI = 0.9955 - j0.4(0.4018 + j0.3014) = 1.1160 - j0.1607 = 1.1275\angle -8.20^\circ\ \text{pu} \end{aligned}$$

Currents during the fault

With the generator terminals shorted, each machine feeds the fault through its own reactance:

$$\begin{aligned} I''_g &= \frac{E''_g}{jX''_g} = \frac{0.9050 + j0.1205}{j0.25} = 0.4822 - j3.6202 = 3.6521\angle -82.41^\circ\ \text{pu} \\ I''_m &= \frac{E''_m}{j(X''_m + X_L)} = \frac{1.1160 - j0.1607}{j0.45} = -0.3572 - j2.4800 = 2.5056\angle -98.20^\circ\ \text{pu} \\ I_f &= I''_g + I''_m = 0.1250 - j6.1002 = 6.1015\angle -88.83^\circ\ \text{pu} \end{aligned}$$

Base current at 11 kV: $I_{base} = \dfrac{100 \times 10^6}{\sqrt{3} \times 11 \times 10^3} = 5248.6$ A.

Check: $I_f = V_t\left(\frac{1}{j0.25} + \frac{1}{j0.45}\right) = 0.1250 - j6.1002$ pu, the same.

Answer: generator current $\approx 19.17$ kA, motor current $\approx 13.15$ kA, total fault current $\approx 32.02$ kA.

Base: 30 MVA, 13.2 kV (the machine ratings). The motor terminal voltage is the reference. Internal EMFs are found from the pre-fault load, then each machine's contribution to the fault at the motor terminals is found separately.

Per-unit data

• $X''_g = X''_m = 0.20$ pu, line $X_L = 0.08$ pu.
• $V_f = \dfrac{12.8}{13.2} = 0.9697\angle 0^\circ$ pu.
• Motor load: $P = 20/30 = 0.6667$ pu at 0.8 pf leading.

 Eg --j0.2--+--j0.08--+--j0.2-- Em
          (gen)       (Vf) motor terminals
                       |
                       F

Pre-fault current

$$\begin{aligned} |I_L| &= \frac{0.6667}{0.9697 \times 0.8} = 0.8594\ \text{pu} \\ I_L &= 0.8594\angle 36.87^\circ = 0.6875 + j0.5156\ \text{pu} \end{aligned}$$

Internal voltages

$$\begin{aligned} E''_g &= V_f + j(X_L + X''_g)I_L = 0.9697 + j0.28(0.6875 + j0.5156) = 0.8253 + j0.1925\ \text{pu} \\ E''_m &= V_f - jX''_mI_L = 0.9697 - j0.2(0.6875 + j0.5156) = 1.0728 - j0.1375\ \text{pu} \end{aligned}$$

Sub-transient currents with the fault at the motor terminals

$$\begin{aligned} I''_g &= \frac{E''_g}{j(X''_g + X_L)} = \frac{0.8253 + j0.1925}{j0.28} = 0.6875 - j2.9476\ \text{pu} \\ I''_m &= \frac{E''_m}{jX''_m} = \frac{1.0728 - j0.1375}{j0.2} = -0.6875 - j5.3641\ \text{pu} \\ I''_f &= I''_g + I''_m = -j8.3117\ \text{pu} \end{aligned}$$

Base current: $I_{base} = \dfrac{30 \times 10^6}{\sqrt{3} \times 13.2 \times 10^3} = 1312.2$ A.

Check (Thevenin): $I''_f = V_f\left(\frac{1}{j0.28} + \frac{1}{j0.2}\right) = -j8.3117$ pu, the same.

Answer: $I''_g \approx 3972$ A, $I''_m \approx 7096$ A, $I''_f \approx 10906$ A.

Base: 500 MVA, 20 kV. The terminal voltage is $V_t = 1\angle 0^\circ$ pu.

Pre-fault load current

The load is purely resistive, so the current is in phase with $V_t$:

$$I_L = \frac{P}{V_t} = \frac{400/500}{1} = 0.8\angle 0^\circ\ \text{pu}$$

Internal (sub-transient) voltage

$$\begin{aligned} E'' &= V_t + jX''_dI_L = 1 + j0.2(0.8) \\ &= 1 + j0.16 = 1.0127\angle 9.09^\circ\ \text{pu} \end{aligned}$$

Fault current

When all three load phases are shorted, the terminals are at zero voltage and the generator sees only $X''_d$:

$$\begin{aligned} I'' &= \frac{E''}{jX''_d} = \frac{1 + j0.16}{j0.2} = 0.8000 - j5.0000\ \text{pu} \\ |I''| &= \sqrt{0.8^2 + 5^2} = 5.0636\ \text{pu} \end{aligned}$$

In amperes: $I_{base} = \dfrac{500 \times 10^6}{\sqrt{3} \times 20 \times 10^3} = 14433.8$ A, so $|I''| = 5.0636 \times 14433.8 \approx 73.09$ kA.

Answer: initial symmetrical rms current $= 5.0636$ pu (about 73.1 kA).

Base: 30 MVA (transformer rating), 11 kV on the generator side and 33 kV on the line side (ratio 1:3). Pre-fault voltage is taken as 1.0 pu and pre-fault load current is neglected (usual assumption for fault level calculation).

Per-unit impedances

$Z_{base}$ at 33 kV $= 33^2/30 = 36.3\ \Omega$.

 E=1--jX_G1--+
             +--j0.05--(3+j5 ohm)--F
 E=1--jX_G2--+

Total impedance up to F

$$\begin{aligned} X_G &= \frac{0.225 \times 0.300}{0.225 + 0.300} = 0.1286\ \text{pu} \\ Z_{th} &= jX_G + j0.05 + Z_L = j0.1286 + j0.05 + (0.0826 + j0.1377) \\ &= 0.0826 + j0.3163 = 0.3269\angle 75.36^\circ\ \text{pu} \end{aligned}$$

Fault current and fault level

$$\begin{aligned} I_f &= \frac{1}{|Z_{th}|} = \frac{1}{0.3269} = 3.0588\ \text{pu} \\ I_{base} &= \frac{30 \times 10^6}{\sqrt{3} \times 33 \times 10^3} = 524.86\ \text{A} \\ I_f &= 3.0588 \times 524.86 = 1605.4\ \text{A} \\ \text{Fault level} &= \frac{S_{base}}{|Z_{th}|} = \frac{30}{0.3269} = 91.76\ \text{MVA} \end{aligned}$$

Answer: fault current $\approx 1605$ A at 33 kV, fault level $\approx 91.8$ 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: $0.1 + 0.12 = 0.22$ pu.
• Bus 3 has no source, so lines 1-3 and 3-2 are simply in series: $0.06 + 0.10 = 0.16$ pu.
• This path is in parallel with line 1-2 (0.08 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

$$\begin{aligned} I_f &= \frac{1}{0.2733} = 3.6585\ \text{pu} \\ \text{SC MVA} &= \frac{S_{base}}{X_{th}} = \frac{100}{0.2733} = 365.85\ \text{MVA} \end{aligned}$$

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

Base: 30 MVA, with base voltages 6.6 kV, 66 kV and 3.81 kV in the three zones. Loads are neglected and the pre-fault voltage is 1.0 pu (usual for fault level), so all generator EMFs are 1 pu. The fault is on the 3.81 kV bus of load B.

Base impedances

$$\begin{aligned} Z_{b,6.6} &= \frac{6.6^2}{30} = 1.452\ \Omega \\ Z_{b,66} &= \frac{66^2}{30} = 145.2\ \Omega \\ Z_{b,3.81} &= \frac{3.81^2}{30} = 0.48387\ \Omega \end{aligned}$$

Per-unit reactances

 G1--+
     +--Tr1--line--Tr2--+-- 3.81 kV bus (F, load B)
 G2--+                  |
                        G3

Thevenin reactance at load B

$$\begin{aligned} X_{G1\|G2} &= \frac{0.4511 \times 0.9022}{0.4511 + 0.9022} = 0.3007 \\ X_{left} &= 0.3007 + 0.1000 + 0.1198 + 0.1000 = 0.6206 \\ X_{th} &= \frac{0.6206 \times 0.3001}{0.6206 + 0.3001} = 0.2023\ \text{pu} \end{aligned}$$

Short-circuit MVA and current

$$\begin{aligned} \text{SC MVA} &= \frac{30}{0.2023} = 148.32\ \text{MVA} \\ I_{base} &= \frac{30 \times 10^6}{\sqrt{3} \times 3.81 \times 10^3} = 4546.1\ \text{A} \\ I_f &= \frac{1}{0.2023} \times 4546.1 = 4.9439 \times 4546.1 = 22475\ \text{A} \end{aligned}$$

Answer: short-circuit MVA $\approx 148.3$ MVA, short-circuit current $\approx 22.48$ kA at 3.81 kV.

Base: 100 MVA, 11 kV (generators) and 220 kV (lines). Pre-fault voltage is taken as 1.0 pu and pre-fault load current is neglected (usual assumption for fault level calculation).

Per-unit reactances (100 MVA)

 G1-j0.15-T1-j0.10-+          +-j42 ohm-+
                   +-220 kV --+         +-- F
 G2-jX_G2-T2-jX_T2-+          +-j42 ohm-+

Thevenin reactance at F

$$\begin{aligned} X_a &= 0.15 + 0.10 = 0.25, \quad X_b = 0.1667 + 0.1067 = 0.2733 \\ X_{gen} &= \frac{0.25 \times 0.2733}{0.25 + 0.2733} = 0.1306 \\ X_{lines} &= 0.0868/2 = 0.0434 \\ X_{th} &= 0.1306 + 0.0434 = 0.1740\ \text{pu} \end{aligned}$$

Total fault current and fault level

$$\begin{aligned} I_f &= \frac{1}{0.1740} = 5.7484\ \text{pu} \\ I_{base,220} &= \frac{100 \times 10^6}{\sqrt{3} \times 220 \times 10^3} = 262.43\ \text{A} \\ I_f &= 5.7484 \times 262.43 = 1508.6\ \text{A} \\ \text{Fault level} &= \frac{100}{0.1740} = 574.84\ \text{MVA} \end{aligned}$$

Current supplied by each generator

The fault current divides between the two generator branches in inverse ratio of their reactances:

$$\begin{aligned} I_{G1} &= I_f\frac{X_b}{X_a + X_b} = 5.7484 \times \frac{0.2733}{0.5233} = 3.0023\ \text{pu} \\ I_{G2} &= I_f\frac{X_a}{X_a + X_b} = 5.7484 \times \frac{0.25}{0.5233} = 2.7460\ \text{pu} \end{aligned}$$

On the 11 kV side, $I_{base} = \dfrac{100 \times 10^6}{\sqrt{3} \times 11 \times 10^3} = 5248.6$ A:

Answer: total fault current $\approx 1509$ A at 220 kV, fault level $\approx 574.8$ MVA; G1 supplies about 15.76 kA and G2 about 14.41 kA at 11 kV.

Base: 30 MVA, 11 kV / 132 kV. Pre-fault voltage is taken as 1.0 pu and pre-fault load current is neglected (usual assumption for fault level calculation). The load on the HV bus is neglected.

Per-unit reactances (30 MVA base)

 G1--j0.75--T1 j0.075--+
                       +-- HV bus (F)
 G2--j0.50--T2 j0.05--[X]--+

(i) Fault at the HV bus

$$\begin{aligned} X_1 &= 0.75 + 0.075 = 0.825, \quad X_2 = 0.50 + 0.05 = 0.55 \\ X_{th} &= \frac{0.825 \times 0.55}{0.825 + 0.55} = 0.33\ \text{pu} \\ \text{SC MVA} &= \frac{30}{0.33} = 90.91\ \text{MVA} \\ I_{base} &= \frac{30 \times 10^6}{\sqrt{3} \times 132 \times 10^3} = 131.22\ \text{A} \\ I_f &= \frac{1}{0.33} \times 131.22 = 3.0303 \times 131.22 = 397.6\ \text{A} \end{aligned}$$

(ii) Reactor to reduce the fault level by 25%

"Limit the fault level by 25%" is read as reducing it by 25%, i.e. to 75% of the value above. (Reducing it to 25% is impossible here, because the G1 branch alone gives $30/0.825 = 36.4$ MVA, which is 40% of the present level.)

$$\begin{aligned} \text{New fault level} &= 0.75 \times 90.91 = 68.18\ \text{MVA} \\ X_{th,new} &= \frac{30}{68.18} = 0.44\ \text{pu} \end{aligned}$$

With reactor $X$ (pu) in series with T2 on its 132 kV (secondary) side:

$$\begin{aligned} \frac{1}{0.44} &= \frac{1}{0.825} + \frac{1}{0.55 + X} \\ \frac{1}{0.55 + X} &= 2.2727 - 1.2121 = 1.0606 \\ 0.55 + X &= 0.9429 \\ X &= 0.3929\ \text{pu} \end{aligned}$$

In ohms on the 132 kV side, $Z_{base} = 132^2/30 = 580.8\ \Omega$:

$$X = 0.3929 \times 580.8 = 228.2\ \Omega$$

Answer: (i) SC MVA $= 90.91$ MVA, fault current $\approx 397.6$ A at 132 kV. (ii) A reactor of about 228 Ω per phase (0.393 pu on 30 MVA) on the 132 kV side of T2 reduces the fault level to 68.2 MVA.

Base: 15 MVA (transformer rating), 11 kV on the generator side and 33 kV on the line side. Pre-fault voltage is taken as 1.0 pu and pre-fault load current is neglected (usual assumption for fault level calculation).

Per-unit reactances (15 MVA)

 G1 j0.150--+
               +--T1 j0.06--line j0.275--F (load end)
 G2 j0.225--+

Thevenin reactance

$$\begin{aligned} X_G &= \frac{0.150 \times 0.225}{0.150 + 0.225} = 0.090 \\ X_{th} &= 0.090 + 0.06 + 0.2755 = 0.4255\ \text{pu} \end{aligned}$$

Short-circuit MVA and fault current

$$\begin{aligned} \text{SC MVA} &= \frac{15}{0.4255} = 35.25\ \text{MVA} \\ I_f &= \frac{1}{0.4255} = 2.3503\ \text{pu} \\ I_{base,33} &= \frac{15 \times 10^6}{\sqrt{3} \times 33 \times 10^3} = 262.43\ \text{A} \Rightarrow I_f = 616.8\ \text{A at 33 kV} \end{aligned}$$

Current supplied by each generator

On the 11 kV side, $I_{base,11} = \dfrac{15 \times 10^6}{\sqrt{3} \times 11 \times 10^3} = 787.30$ A, so the fault current there is $2.3503 \times 787.30 = 1850.4$ A. It divides inversely as the generator reactances:

$$\begin{aligned} I_{G1} &= 1850.4 \times \frac{0.225}{0.150 + 0.225} = 1110.2\ \text{A} \\ I_{G2} &= 1850.4 \times \frac{0.150}{0.150 + 0.225} = 740.1\ \text{A} \end{aligned}$$

Answer: SC MVA $\approx 35.25$ MVA; fault current $\approx 617$ A at 33 kV; G1 supplies $\approx 1110$ A and G2 $\approx 740$ A (at 11 kV).

A three-phase fault is balanced, so only the positive-sequence network is used; the transformer connections (Y/Δ) and grounding do not affect it. Pre-fault voltage is 1.0 pu and load current is neglected.

Positive-sequence network

• Source 1 to bus 1: $0.1 + 0.1 = 0.2$ pu.
• Source 2 to bus 2: $0.2 + 0.2 = 0.4$ pu.
• Lines: $X_{12} = 0.8$, $X_{13} = 0.4$, $X_{23} = 0.4$ pu (a delta between buses 1, 2, 3).

Delta-star conversion of the line delta

$$\begin{aligned} X_1 &= \frac{X_{12}X_{13}}{X_{12} + X_{13} + X_{23}} = \frac{0.8 \times 0.4}{1.6} = 0.20 \\ X_2 &= \frac{X_{12}X_{23}}{1.6} = \frac{0.8 \times 0.4}{1.6} = 0.20 \\ X_3 &= \frac{X_{13}X_{23}}{1.6} = \frac{0.4 \times 0.4}{1.6} = 0.10 \end{aligned}$$

 E=1-j0.2-bus1-j0.2-+
                    N--j0.1--bus3 (F)
 E=1-j0.4-bus2-j0.2-+

Thevenin reactance at bus 3

$$\begin{aligned} X_{path1} &= 0.2 + 0.20 = 0.40, \quad X_{path2} = 0.4 + 0.20 = 0.60 \\ X_{th} &= \frac{0.40 \times 0.60}{0.40 + 0.60} + 0.10 = 0.24 + 0.10 = 0.34\ \text{pu} \end{aligned}$$

(The same value is $Z_{33}$ of the bus impedance matrix.)

Fault current

$$\begin{aligned} I_f &= \frac{V_f}{jX_{th}} = \frac{1}{j0.34} = -j2.9412\ \text{pu} \\ \text{Fault MVA} &= \frac{100}{0.34} = 294.1\ \text{MVA} \end{aligned}$$

Answer: $I_f = 2.941$ pu (fault level 294.1 MVA on the 100 MVA base). The current in amperes is $I_f \times I_{base}$ once the bus voltage base is known.

The fault MVA is found by converting every reactance to one common MVA base, reducing the network to a single Thevenin reactance seen from F, and using $\text{Fault MVA} = \text{Base MVA}/X_{th}$ with prefault voltage 1 pu.

Step 1: Choose base and convert reactances

Base: $S_b = 20$ MVA, $V_b = 11$ kV. Use $X_{new} = X_{old}\,\dfrac{S_b}{S_{rated}}$ (same voltage, so no voltage correction).

Step 2: Reactance diagram

All generator EMFs are equal (1 pu) and their neutral is the reference, so for a 3-phase fault every source branch ends on the reference bus.

   F (G1 bus)        R1=0.1       B (G2 bus)  R2=0.1
 ---o----------------/\/\/\---------o-------/\/\/\---+
    |                               |                |
  G1 0.2                         G2 0.15          G3 0.15
    |                               |                |
 ===+===========  reference  =======+================+===

Step 3: Reduce the network as seen from F

$$\begin{aligned} X_{R2+G3} &= 0.10 + 0.15 = 0.25 \\ X_B &= 0.15 \parallel 0.25 = \frac{0.15 \times 0.25}{0.40} = 0.09375 \\ X_{R1+B} &= 0.10 + 0.09375 = 0.19375 \\ X_{th} &= 0.20 \parallel 0.19375 = \frac{0.20 \times 0.19375}{0.39375} = 0.09841\ \text{pu} \end{aligned}$$

Step 4: Fault current and fault MVA

$$\begin{aligned} I_f &= \frac{1}{X_{th}} = \frac{1}{0.09841} = 10.16\ \text{pu} \\ I_b &= \frac{S_b}{\sqrt{3}\,V_b} = \frac{20\times 10^6}{\sqrt{3}\times 11\times 10^3} = 1049.7\ \text{A} \\ I_f &= 10.16 \times 1049.7 = 10\,667\ \text{A} \approx 10.67\ \text{kA} \\ \text{Fault MVA} &= \frac{S_b}{X_{th}} = \frac{20}{0.09841} = 203.2\ \text{MVA} \end{aligned}$$

Check: $\sqrt{3} \times 11 \times 10.667 = 203.2$ MVA.

The result does not depend on the base chosen; with a 10 MVA base all pu values double and $10/0.1968$ gives the same 203.2 MVA.

Answer: Fault current ≈ 10.67 kA (10.16 pu on 20 MVA base); fault MVA ≈ 203.2 MVA.

The fault current is found from the Thevenin (driving point) impedance at bus 2, $I_f = V_f/(Z_{22}+Z_f)$, with prefault voltage 1 pu. Bus voltages during the fault then give the generator currents.

Assumptions: all values are reactances (lines 1-2 = j0.75 and 3-2 = j1.0 pu), $Z_f = j0.088$ pu, no prefault load, prefault voltage 1 pu everywhere. Base: 20 MVA, 11 kV (generator side), so 110 kV on the line side.

Step 1: Reactance network

Generator + transformer branches: bus 1 to reference $= 0.15 + 0.10 = 0.25$; bus 3 to reference $= 0.15 + 0.225 = 0.375$.

      (ref)               (ref)
        |                   |
      j0.25               j0.375
        |                   |
  bus 1 o------j0.5---------o bus 3
         \                 /
        j0.75           j1.0
           \             /
            o---- bus 2 (fault, Zf=j0.088)

Step 2: Bus admittance matrix and Z-bus (values of $-j$ / $j$ dropped, all reactive)

$$Y_{bus} = -j\begin{bmatrix} 4+2+1.333 & -1.333 & -2 \\ -1.333 & 1.333+1 & -1 \\ -2 & -1 & 2.667+2+1 \end{bmatrix} = -j\begin{bmatrix} 7.333 & -1.333 & -2 \\ -1.333 & 2.333 & -1 \\ -2 & -1 & 5.667 \end{bmatrix}$$

Inverting:

$$Z_{bus} = j\begin{bmatrix} 0.1884 & 0.1473 & 0.0925 \\ 0.1473 & 0.5788 & 0.1541 \\ 0.0925 & 0.1541 & 0.2363 \end{bmatrix}$$

So the Thevenin impedance at bus 2 is $Z_{22} = j0.5788$ pu.

Step 3: Fault current

$$\begin{aligned} I_f &= \frac{1}{Z_{22}+Z_f} = \frac{1}{j(0.5788+0.088)} = -j1.4998\ \text{pu} \\ I_{b,110} &= \frac{20\times10^6}{\sqrt3 \times 110\times10^3} = 104.97\ \text{A} \\ |I_f| &= 1.4998 \times 104.97 = 157.4\ \text{A} \end{aligned}$$

Step 4: Bus voltages during fault ($V_i = 1 - Z_{i2} I_f$)

$$\begin{aligned} V_1 &= 1 - 0.1473 \times 1.4998 = 0.7791\ \text{pu} \\ V_2 &= 1 - 0.5788 \times 1.4998 = 0.1320\ \text{pu} \\ V_3 &= 1 - 0.1541 \times 1.4998 = 0.7689\ \text{pu} \end{aligned}$$

Check: $V_2 = Z_f I_f = 0.088 \times 1.4998 = 0.132$ pu.

Step 5: Generator currents

$$\begin{aligned} I_{G1} &= \frac{1 - V_1}{0.25} = \frac{0.2209}{0.25} = 0.8834\ \text{pu} \\ I_{G2} &= \frac{1 - V_3}{0.375} = \frac{0.2311}{0.375} = 0.6163\ \text{pu} \end{aligned}$$

Check: $0.8834 + 0.6163 = 1.4997 \approx I_f$.

Generator currents flow on the 11 kV side, base current $I_{b,11} = 20\times10^6/(\sqrt3 \times 11\times10^3) = 1049.7$ A:

$$I_{G1} = 0.8834 \times 1049.7 = 927.4\ \text{A}, \qquad I_{G2} = 0.6163 \times 1049.7 = 647.0\ \text{A}$$

Answer: Fault current at bus 2 = 1.50 pu = 157.4 A (110 kV side); G1 supplies 0.883 pu = 927 A and G2 supplies 0.616 pu = 647 A (11 kV side), all lagging the prefault voltage by 90°.

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

Assumption: the line length is not given on the paper; the same network in other IOE papers uses 200 km, so each line has $X = 0.2 \times 200 = 40\ \Omega$. 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). $Z_{b,132} = 132^2/100 = 174.24\ \Omega$.

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

$$\begin{aligned} X_1 &= 0.15 + 0.08 = 0.23, \quad X_2 = 0.20 + 0.16 = 0.36 \\ X_{gen} &= \frac{0.23 \times 0.36}{0.23 + 0.36} = 0.1403 \\ X_{line} &= \frac{0.2296}{2} = 0.1148 \\ X_{th} &= 0.1403 + 0.1148 = 0.2551\ \text{pu} \end{aligned}$$

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

$$\begin{aligned} I_f &= \frac{1}{0.2551} = 3.920\ \text{pu} \\ \text{Fault level} &= \frac{100}{0.2551} = 392.0\ \text{MVA} \\ I_{b,132} &= \frac{100\times10^6}{\sqrt3 \times 132\times10^3} = 437.4\ \text{A} \\ I_f &= 3.920 \times 437.4 = 1714\ \text{A} = 1.714\ \text{kA} \end{aligned}$$

(b) Currents supplied by G1 and G2

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

$$\begin{aligned} I_{G1} &= I_f\,\frac{X_2}{X_1+X_2} = 3.920 \times \frac{0.36}{0.59} = 2.392\ \text{pu} \\ I_{G2} &= I_f\,\frac{X_1}{X_1+X_2} = 3.920 \times \frac{0.23}{0.59} = 1.528\ \text{pu} \end{aligned}$$

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

$$I_{G1} = 2.392 \times 5248.6 = 12.55\ \text{kA}, \qquad I_{G2} = 1.528 \times 5248.6 = 8.02\ \text{kA}$$

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 $l$ km, only $X_{line} = 0.2l/(2\times174.24)$ pu changes in Step 2.

For a bolted 3-phase fault, $I_f = 1/X_{th}$ pu, where $X_{th}$ is the parallel combination of the reactances from the fault point back to each generator (prefault voltage 1 pu, load current neglected).

Data and per unit values

Base: 50 MVA; 132 kV on the line, hence 6.6 kV on the generator buses (from the 6.6/132 kV transformers). $Z_{b,line} = 132^2/50 = 348.48\ \Omega$.

Base current at 132 kV: $I_b = \dfrac{50\times10^6}{\sqrt3 \times 132\times10^3} = 218.7$ A.

 G1      T1      P    Line     Q     T2     G2
(~)-0.1033-0.10--o--0.1435--o--0.10-0.1033-(~)
 |                                          |
 +============= reference (EMF 1 pu) =======+

i. Fault at point P

$$\begin{aligned} X_{left} &= X_{G1} + X_{T1} = 0.1033 + 0.10 = 0.2033 \\ X_{right} &= X_{line} + X_{T2} + X_{G2} = 0.1435 + 0.10 + 0.1033 = 0.3468 \\ X_{th} &= \frac{0.2033 \times 0.3468}{0.2033 + 0.3468} = 0.1282\ \text{pu} \\ I_f &= \frac{1}{0.1282} = 7.80\ \text{pu} \\ I_f &= 7.80 \times 218.7 = 1706\ \text{A} \end{aligned}$$

Contribution from G1 side $= 7.80 \times 0.3468/0.5501 = 4.92$ pu, from G2 side $= 2.88$ pu. Fault MVA $= 50/0.1282 = 390$ MVA.

ii. Fault at point Q

$$\begin{aligned} X_{left} &= X_{G1} + X_{T1} + X_{line} = 0.3468 \\ X_{right} &= X_{T2} + X_{G2} = 0.2033 \\ X_{th} &= \frac{0.3468 \times 0.2033}{0.5501} = 0.1282\ \text{pu} \\ I_f &= 7.80\ \text{pu} = 1706\ \text{A} \end{aligned}$$

Now G2 side supplies 4.92 pu and G1 side 2.88 pu.

The network is symmetrical about the middle of the line, so the fault current magnitude is the same at P and Q; only the sharing between the generators changes.

Answer: At P and at Q, $I_f$ = 7.80 pu ≈ 1706 A (≈ 1.71 kA at 132 kV), fault level ≈ 390 MVA.

For a bolted fault at the generator terminal, both the generator and the synchronous motor (acting as a source) feed the fault. $I_f = V_f / X_{th}$ with $V_f = 1$ pu.

Per unit values

Base: 50 MVA, 11 kV at the generator, so 132 kV on the line and 6.6 kV at the motor (voltage bases follow the transformer ratios, so no voltage correction is needed). $Z_{b,line} = 132^2/50 = 348.48\ \Omega$.

        F
 G      o   T1     Line     T2     M
(~)-0.1-+-0.025--0.0717--0.04--0.06-(~)
 |                                   |
 +====== reference (EMF = 1 pu) =====+

Fault current

Reactance on the motor side seen from F:

$$X_m = 0.025 + 0.0717 + 0.04 + 0.06 = 0.1967\ \text{pu}$$ $$\begin{aligned} X_{th} &= \frac{0.10 \times 0.1967}{0.10 + 0.1967} = 0.0663\ \text{pu} \\ I_f &= \frac{1}{0.0663} = 15.08\ \text{pu} \\ I_b &= \frac{50\times10^6}{\sqrt3 \times 11\times10^3} = 2624.3\ \text{A} \\ I_f &= 15.08 \times 2624.3 = 39\,582\ \text{A} \approx 39.6\ \text{kA} \end{aligned}$$

Contributions: generator $1/0.1 = 10$ pu (26.24 kA); motor side $1/0.1967 = 5.08$ pu (13.34 kA).

Fault MVA at the generator terminal

$$\text{Fault MVA} = \frac{S_b}{X_{th}} = \frac{50}{0.0663} = 754.1\ \text{MVA}$$

Check: $\sqrt3 \times 11 \times 39.58 = 754.1$ MVA.

Answer: Fault current ≈ 15.08 pu ≈ 39.6 kA; fault MVA ≈ 754 MVA.

A short circuit fault is an abnormal low-impedance connection between phases or between phase and earth, which makes very large currents flow.

Effects of short circuit faults

• Heavy current: fault current may be 5 to 20 times full-load current; $I^2R$ heating can damage windings, cables and insulation.
• Mechanical forces: electromagnetic force is proportional to $I^2$, so bus bars, transformer and generator windings can be bent or broken.
• Voltage dip: voltage near the fault falls sharply, so motors may stall and sensitive loads trip.
• Loss of stability: electrical output of generators falls while mechanical input stays the same, so machines may lose synchronism.
• Fire and arcing: arcs at the fault can cause fire and explosion, and danger to people.
• Unbalance: unsymmetrical faults produce negative sequence currents that overheat rotors of machines.
• Interruption of supply to consumers when the faulty section is disconnected.

Importance of fault calculations

1. Circuit breaker selection: the breaking capacity (MVA, kA) and making capacity of breakers and fuses are chosen from the maximum fault level.
2. Relay setting and coordination: the minimum and maximum fault currents are needed to set protective relays so they operate selectively.
3. Equipment rating: bus bars, CTs, cables and switchgear must withstand the thermal and mechanical stress of the fault current.
4. Design of reactors and grounding: to limit fault level by current-limiting reactors or neutral grounding impedance.
5. Stability studies: fault conditions are used to check transient stability and critical clearing time.
6. Planning: to check the effect of new generators or interconnections on the fault level of existing substations.

Fault analysis is the calculation of the currents and voltages in a power network during abnormal (fault) conditions such as short circuits and open conductors.

Purpose of fault analysis

• To find the maximum fault current and fault MVA so circuit breakers and fuses of the right breaking capacity can be chosen.
• To find minimum and maximum fault currents for setting and coordinating protective relays.
• To check that bus bars, cables, CTs and other equipment can withstand the thermal and mechanical stresses.
• To size current-limiting reactors and neutral grounding impedances.
• To provide fault conditions for transient stability studies.

Types of faults

1. Symmetrical (balanced) fault

• Three-phase fault (L-L-L) or three-phase to ground fault (L-L-L-G). All phases remain balanced; about 2–5% of faults, but usually the most severe. Analysed with the per phase (positive sequence) network.

2. Unsymmetrical (shunt) faults — analysed with symmetrical components:

3. Series (open conductor) faults

• One conductor open or two conductors open (e.g. broken conductor, blown fuse in one phase, single-pole breaker operation). They cause unbalance but not heavy current.

Faults can also be classed as temporary (e.g. flashover due to lightning, cleared by reclosing) or permanent (e.g. broken conductor, insulator failure).