Question 6 of 7: Line-to-line fault at the mid-point of a line — fault current and phase voltages

Question 6: Line-to-line fault at the mid-point of a line — fault current and phase voltages (25 points)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. The positive-sequence network of Figure (4), stated to be identical to the negative-sequence network, on a 100 MVA base: a 1.00 pu source at the reference node $N^+$, a left-hand path of $j0.20+j0.15$ pu reaching bus W, a right-hand path of $j0.05+j0.15$ pu reaching bus V, and a $j0.20$ pu line joining W to V with a line-to-line fault at its mid-point.

Find. (a) the current flowing into the fault; (b) all three phase voltages at bus W and all three at bus V while the fault is on.

Approach. Reduce the positive-sequence network to its Thevenin reactance at F, connect the positive and negative networks in opposition (the line-to-line connection), take the sequence currents, then recover the phase currents and the phase voltages at each bus from the transfer reactances.

1. Read the network as a circuit. Each vertical reactance and the horizontal reactance beneath it are in series on the way from the reference to the bus, so $$X_{NW}=0.20+0.15=0.35\ \text{pu},\qquad X_{NV}=0.05+0.15=0.20\ \text{pu}$$ The fault splits the $j0.20$ pu line into two halves of $j0.10$ pu, so from F the reference is reachable by $0.35+0.10=0.45$ through W and by $0.20+0.10=0.30$ through V.
2. Reduce to the Thevenin reactance at F. The two paths are simply in parallel, and no delta–star transform is needed because there is no third branch bridging W and V: $$X_1=\frac{(0.45)(0.30)}{0.45+0.30}=\boxed{X_1=X_2=0.1800\ \text{pu}}$$ The transfer reactances that will be needed for part (b) come out of the same current division. Injecting unit current at F, the fraction $0.30/0.75=0.400$ returns through W and $0.45/0.75=0.600$ through V, so $$Z_{WF}=0.400\times0.35=j0.1400,\qquad Z_{VF}=0.600\times0.20=j0.1200\ \text{pu}$$ Both must reproduce $X_1$ when the remaining half-line is added, and they do: $0.1400+0.400(0.10)=0.1800$ and $0.1200+0.600(0.10)=0.1800$.
3. Connect the sequence networks for a line-to-line fault. Take the fault to be between phases b and c, with phase a healthy. The boundary conditions are $I_a=0$, $I_b=-I_c$ and $V_b=V_c$, which transform into $\hat I_{a0}=0$, $\hat I_{a2}=-\hat I_{a1}$ and $\hat V_{a1}=\hat V_{a2}$ — that is, the positive and negative networks joined in opposition at F with the zero-sequence network left entirely out of the problem. This is why the paper supplies no zero-sequence diagram: a phase-to-phase fault has no earth path, so no zero-sequence current can exist. Hence $$\begin{aligned} \hat I_{a1}&=\frac{1.0\angle0^\circ}{j(X_1+X_2)}=\frac{1.0}{j(2\times0.1800)} =-j2.77778\ \text{pu}\\ \hat I_{a2}&=+j2.77778\ \text{pu},\qquad \hat I_{a0}=0 \end{aligned}$$
4. Recover the phase currents. Applying the symmetrical-component synthesis with $a=1\angle120^\circ$, $$\hat I_a=\hat I_{a1}+\hat I_{a2}=0$$ $$\hat I_b=a^{2}\hat I_{a1}+a\hat I_{a2}=(a^{2}-a)\hat I_{a1}=-j\sqrt3\,\hat I_{a1}$$ so the answer to part (a) is $$\boxed{|\hat I_b|=|\hat I_c|=\sqrt3\,|\hat I_{a1}|=\sqrt3\,(2.77778)=4.81125\ \text{pu}}$$ with $\hat I_b=4.81125\angle180^\circ$ and $\hat I_c=4.81125\angle0^\circ$ — equal and opposite, as the boundary condition $I_b=-I_c$ demands, and both real because the network is purely reactive and the driving voltage is real. Phase a carries nothing. On the stated 100 MVA base this is 4.81125 times the base current at the fault location. The free check is the ratio to the three-phase current at the same point, $|I_{3\phi}|=1/X_1=5.5556$ pu, giving 0.86603 — exactly $\sqrt3/2$, which must hold whenever $X_1=X_2$.
5. Compute the sequence voltages at each bus. The positive-sequence network still contains its source, so its bus voltages fall from 1.0 by the transfer drop; the negative-sequence network has no source, so its bus voltages are purely the transfer drop: $$\hat V_{k1}=1.0-Z_{kF}\hat I_{a1},\qquad \hat V_{k2}=-Z_{kF}\hat I_{a2}=+Z_{kF}\hat I_{a1}$$ At bus W, with $Z_{WF}=j0.1400$, $$\hat V_{W1}=1.0-(j0.1400)(-j2.77778)=0.61111,\qquad \hat V_{W2}=0.38889$$ and at bus V, with $Z_{VF}=j0.1200$, $$\hat V_{V1}=0.66667,\qquad \hat V_{V2}=0.33333$$ Both sets are purely real because every impedance is a pure reactance and $\hat I_{a1}$ is purely imaginary.
6. Synthesise the phase voltages at bus W. With $\hat V_{a0}=0$, $$\hat V_a=\hat V_1+\hat V_2,\qquad \hat V_b=a^{2}\hat V_1+a\hat V_2,\qquad \hat V_c=a\hat V_1+a^{2}\hat V_2$$ At bus W, $$\hat V_a=0.61111+0.38889=\boxed{1.0000\angle0^\circ\ \text{pu}}$$ $$\hat V_b=0.53576\angle-158.948^\circ,\qquad \hat V_c=0.53576\angle158.948^\circ\ \text{pu}$$ The healthy phase holding exactly 1.0 pu is not a coincidence: since $\hat I_{a2}=-\hat I_{a1}$, the two transfer drops cancel in $\hat V_a=\hat V_1+\hat V_2$ at every bus. It is the single most useful check on this class of problem.
7. Synthesise the phase voltages at bus V, and interpret. Repeating with $\hat V_{V1}=0.66667$ and $\hat V_{V2}=0.33333$, $$\hat V_a=\boxed{1.0000\angle0^\circ},\qquad \hat V_b=0.57735\angle-150.0^\circ,\qquad \hat V_c=0.57735\angle--150.0^\circ\ \text{pu}$$ Bus V holds its faulted phases higher (0.57735 pu) than bus W does (0.53576 pu) because it is electrically stiffer — it sits only 0.20 pu from the reference against W's 0.35 pu. At the fault point itself the two sequence voltages are each 0.5 pu, so $\hat V_b=\hat V_c=-0.5$ pu there, equal as the boundary condition $V_b=V_c$ requires, while phase a remains at 1.0 pu. The faulted phase voltages at the two buses therefore bracket that value, which is the qualitative shape of every line-to-line fault.