Question 6 of 7: Sequence networks and a double line-to-ground fault

Question 6: Sequence networks and a double line-to-ground fault

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 machine, transformer and line data listed in the question, with a 25 MVA, 11 kV base chosen on the generator side. Reading the connections off Figure (3-a): transformer T₂ is delta on the generator side and grounded-wye on the line side; transformer T₁ is grounded-wye on the line side and delta on the motor side; the generator and motor 2 are wye-connected and grounded through 2.5 Ω reactors; motor 1 is wye-connected and ungrounded. The fault point f is on the line side of T₁; point g is on its delta (motor) side.

Find. Confirmation of every per-unit reactance printed on the sequence networks of Figures (3-b) and (3-c), then the phase currents flowing at point g when a double line-to-ground fault occurs at point f.

Approach. Convert every impedance to the common base, checking each against the printed figure. Then reduce the positive-, negative- and zero-sequence networks to their Thevenin values at f, connect them in the double line-to-ground configuration, and finally divide the sequence currents between the generator branch and the motor branch — remembering that the delta winding of T₁ blocks zero-sequence current from ever reaching g.

Part (a) — verification of the sequence networks

1. Establish the three voltage bases. The generator-side base is 11 kV, so the line-side base follows the transformer ratio, $$\begin{aligned} V_{b,\text{line}} &= 11\left(\frac{121}{10.8}\right)=123.24\ \text{kV} \\ V_{b,\text{motor}} &= 123.24\left(\frac{10.8}{121}\right)=11.0\ \text{kV} \end{aligned}$$ and the impedance bases are $Z_b = V_b^{2}/S_b$, that is 4.840 Ω on the machine sides and 607.53 Ω on the line side.
2. Convert the rotating machines. The generator is already on base, so $X_g'' = 0.20$ pu. The motors are converted by $X_{\text{new}} = X_{\text{old}}\left(S_{\text{new}}/S_{\text{old}}\right)\left(V_{\text{old}}/V_{\text{new}}\right)^{2}$: $$\begin{aligned} X_{m1} &= 0.25\left(\frac{25}{15}\right)\left(\frac{10}{11}\right)^{2}=0.3444\ \text{pu} \\ X_{m2} &= 0.25\left(\frac{25}{7.5}\right)\left(\frac{10}{11}\right)^{2}=0.6887\ \text{pu} \end{aligned}$$ matching the j0.345 and j0.69 printed on Figure (3-b).
3. Convert the transformers and the line. Both transformers give $$X_T=0.10\left(\frac{25}{30}\right)\left(\frac{10.8}{11}\right)^{2}=0.08033\ \text{pu}$$ against the j0.0805 printed, and the line gives $X_{\text{line}} = 100/607.53 = 0.16460$ pu against j0.164 printed. Both agree to the third decimal, the small residual being the exam's own rounding.
4. Convert the zero-sequence quantities. The line contributes $X_{0,\text{line}} = 300/607.53 = 0.4938$ pu (printed j0.494). The motors' zero-sequence reactances, referred from their own ratings, are $0.06(25/15)(10/11)^{2} = 0.0826$ pu (printed j0.082) and $0.06(25/7.5)(10/11)^{2} = 0.1653$ pu (printed j0.164). Each neutral reactor enters the zero-sequence network as three times its ohmic value, $$3X_n = \frac{3(2.5)}{4.840}=1.550\ \text{pu}$$ which is the j1.548 printed against the generator and motor 2. Every value on Figures (3-b) and (3-c) is therefore confirmed.
5. Confirm the zero-sequence topology, which is where the marks really are. A delta winding provides a circulating path for zero-sequence current but no path out of it, so it appears as an open circuit on the line side of the transformer arm. T₂ carries its delta on the generator side, so the generator branch $j(1.548+0.06)$ terminates at node d and never connects to the line; T₁ carries its delta on the motor side, so the motor branches terminate at g and are likewise isolated. The only paths from the reference bus to the line are the two grounded-wye transformer neutrals. The tabulated generator and motor zero-sequence reactances are therefore decoys for this fault — they are drawn on Figure (3-c) but carry no current.

Part (b) — the double line-to-ground fault at f

1. Reduce the positive-sequence network to f. The generator reaches f through its own reactance, T₂ and the line; the two motors are in parallel and reach f through T₁: $$X_{\text{gen path}}=0.20+0.0805+0.164=0.4445\ \text{pu}$$ $$X_{\text{motor path}}=\frac{(0.345)(0.69)}{0.345+0.69}+0.0805=0.230+0.0805=0.3105\ \text{pu}$$ $$X_1=\frac{(0.4445)(0.3105)}{0.4445+0.3105}=0.18280\ \text{pu}$$ The negative-sequence network is identical in topology and value, so $X_2 = 0.18280$ pu.
2. Reduce the zero-sequence network to f. From f there are exactly two paths to the reference bus: straight through the T₁ arm, or along the line and through the T₂ arm. $$X_0=\frac{(0.0805)(0.494+0.0805)}{0.0805+0.494+0.0805}=\frac{(0.0805)(0.5745)}{0.655}$$ $$\boxed{X_0 = 0.07061\ \text{pu}}$$ Because $X_0$ is well below $X_1$, this fault location will produce ground-fault currents that exceed the three-phase value — the usual signature of a solidly earthed transformer close to the fault.
3. Connect the networks for a double line-to-ground fault. For a fault on phases b and c to ground, the positive network drives the parallel combination of the negative and zero networks: $$I_{a1}=\frac{E}{j\left[X_1+\dfrac{X_2X_0}{X_2+X_0}\right]} =\frac{1.0}{j\left[0.18280+0.05093\right]}=\frac{1.0}{j0.23374}$$ $$I_{a1}=4.2783\angle-90^\circ\ \text{pu}$$
4. Find the sequence voltage and the other two sequence currents. All three sequence voltages are equal at the fault point: $$V_{a1}=V_{a2}=V_{a0}=E-jX_1I_{a1}=1.0-(j0.18280)(4.2783\angle-90^\circ)=0.21791\ \text{pu}$$ $$\begin{aligned} I_{a2} &= -\frac{V_{a2}}{jX_2}=1.1920\angle 90^\circ\ \text{pu} \\ I_{a0} &= -\frac{V_{a0}}{jX_0}=3.0863\angle 90^\circ\ \text{pu} \end{aligned}$$ and $I_{a1}+I_{a2}+I_{a0}=0$, confirming that the healthy phase a carries no current, which is the defining condition of this fault.
5. Divide the positive- and negative-sequence currents at the fault. Each network splits its current between the generator path and the motor path in inverse proportion to their reactances. The motor share is $$k_m=\frac{0.4445}{0.4445+0.3105}=0.58874$$ so the currents arriving at f from the motor side are $$\begin{aligned} I_{a1}^{(g)} &= 0.58874(4.2783\angle-90^\circ)=2.5188\angle-90^\circ\ \text{pu} \\ I_{a2}^{(g)} &= 0.58874(1.1920\angle 90^\circ)=0.7018\angle 90^\circ\ \text{pu} \end{aligned}$$
6. Apply the delta blocking to the zero sequence. Point g lies on the delta winding of T₁. Zero-sequence current circulates inside that delta but cannot leave it, so $$I_{a0}^{(g)}=0$$ This single fact is what makes the currents at g quite different in character from those at f, and it is the reason the whole of the generator and motor zero-sequence data plays no part in the answer.
7. Transform to phase currents at g. With $a = 1\angle120^\circ$ and only two sequence components present, $$I_a=I_{a1}^{(g)}+I_{a2}^{(g)}=1.8170\angle-90^\circ\ \text{pu}$$ $$I_b=a^{2}I_{a1}^{(g)}+aI_{a2}^{(g)}=2.9334\angle 161.96^\circ\ \text{pu}$$ $$I_c=aI_{a1}^{(g)}+a^{2}I_{a2}^{(g)}=2.9334\angle 18.04^\circ\ \text{pu}$$ The three sum to zero, as they must in a delta-side winding, and phase a is no longer current-free: on the delta side the healthy phase carries 1.817 pu.
8. Convert to amperes. The base current at g, on the 11 kV motor-side base, is $$I_b^{\text{base}}=\frac{25\times10^{6}}{\sqrt3\,(11\,000)}=1312.2\ \text{A}$$ so $$\boxed{\begin{aligned} |I_b| = |I_c| &= 2.9334\ \text{pu} = 3849\ \text{A} \\ |I_a| &= 1.8170\ \text{pu} = 2384\ \text{A} \end{aligned}}$$ For comparison, at the fault point itself the base current is 117.1 A and the faulted phases carry $6.6238$ pu $= 776$ A each, with $3I_{a0} = 9.2588$ pu $= 1084$ A returning through ground.