Question 6 of 7: System Grounding and Ground Faults on a Plant Bus

Problem 6: System Grounding and Ground Faults on a Plant Bus (5 + 4 + 8 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.

Part (a) — Why power systems are grounded

System grounding means deliberately connecting a point of the system — almost always the neutral of a wye winding — to earth, either solidly or through an impedance. It is done for four distinct reasons, and confusing it with equipment grounding (bonding the non-current-carrying metal) is the commonest conceptual error.

First, overvoltage control. On an ungrounded system a single phase-to-ground fault does not draw enough current to operate protection, so the system keeps running with the faulted phase at earth potential and the two healthy phases at full line-to-line voltage to ground, a permanent 73 per cent overvoltage on their insulation. Worse, an intermittent or arcing ground can resonate with the system capacitance and produce restriking transients of five to six times normal crest, which puncture insulation elsewhere in the plant. Grounding the neutral clamps the healthy-phase voltage rise and eliminates the arcing-ground mechanism.

Second, reliable relay operation. Ground faults are the most common fault type by a wide margin. A grounded neutral provides a defined low-impedance return path, so a ground fault produces a current large enough and predictable enough for overcurrent and ground-fault relays to detect selectively and clear quickly. This is what turns an undetectable insulation failure into a cleared, located fault.

Third, personnel safety and equipment protection. A bounded, fast-cleared fault current limits the duration of touch and step potentials around the affected equipment, limits the arc-flash incident energy to which a worker is exposed (energy scales with clearing time), and limits the thermal and magnetic damage to the faulted machine or cable. Fourth, insulation co-ordination and economy: a system whose neutral is effectively grounded (the usual criteria being $X_0/X_1 \le 3$ and $R_0/X_1 \le 1$) permits surge arresters and apparatus rated at about 80 per cent of line-to-line voltage instead of 100 per cent, a real saving at transmission voltages.

The choice of method follows from these aims. Solid grounding gives the largest fault current and the best relaying, and is standard on utility transmission and on 600-V and lower plant systems. Low-resistance grounding, which limits the ground-fault current to a few hundred amperes, is common on medium-voltage industrial systems such as the 4160-V bus in this problem, because it cuts the damage at the fault while still allowing selective relaying. High-resistance grounding limits the current to a few amperes so that a first ground fault can be alarmed rather than tripped — valuable in a continuous process — but it requires a strict find-and-fix discipline. In Canada these arrangements are governed by CSA C22.1, the Canadian Electrical Code, Section 10 (grounding and bonding), and by IEEE Std 142 for industrial practice. Note that the generator added in part (c) of this problem is explicitly ungrounded, which is a common industrial choice made precisely to keep the machine's winding out of the ground-fault path.

Parts (b) and (c) — Double line-to-ground fault calculations

Given.

Find. The magnitude of the fault current in phase B, first with the utility alone and then with the ungrounded generator added in parallel.

Approach. For a B-C-to-ground fault the three sequence networks are connected in parallel — all three fault points tied together and all three neutral points tied together. Solve for the sequence currents, transform back to phase quantities, and convert to amperes at the end.

1. Part (b) — Establish the base current. Everything is on a 5000-kVA, 4160-V base, so $$I_{base}=\frac{5000\times 10^{3}}{\sqrt{3}\,(4160)}=693.93\ \text{A}$$ This is the only conversion needed at the end; the whole calculation is done in per unit.
2. Reduce the parallel network and find the positive-sequence current. With the negative and zero networks in parallel across the positive network, $$\begin{aligned}X_2\parallel X_0&=\frac{(0.05)(0.03)}{0.05+0.03}=0.018750 \\ X_{th}&=X_1+X_2\parallel X_0=0.05+0.018750=0.068750\end{aligned}$$ $$I_{a1}=\frac{1.0}{j(0.068750)}=14.5455\angle -90.00^{\circ}\ \text{pu}$$
3. Recover the negative- and zero-sequence currents. The voltage that appears at the common fault point drives the two parallel branches: $$V_{a1}=1.0-I_{a1}(jX_1)=1.0-(14.5455)(0.05)=0.272727\ \text{pu}$$ $$\begin{aligned}I_{a2}&=\frac{-V_{a1}}{jX_2}=5.4545\angle 90.00^{\circ} \\ I_{a0}&=\frac{-V_{a1}}{jX_0}=9.0909\angle 90.00^{\circ}\ \text{pu}\end{aligned}$$ Their sum with $I_{a1}$ is zero, which is the defining condition of this fault type — phase A is healthy and carries no current — and it is the free check on the three sequence values.
4. Transform to the phase-B current. With $a=1\angle 120^{\circ}$, $$I_B=I_{a0}+a^{2}I_{a1}+aI_{a2}$$ $$\begin{aligned} a^{2}I_{a1}&=-12.5967+j7.2727\\ aI_{a2}&=-4.7238-j2.7273\\ I_{a0}&=-0.0000+j9.0909 \end{aligned}$$ $$I_B=-17.3205+j13.6364=22.0443\angle 141.79^{\circ}\ \text{pu}$$ $$\boxed{|I_B|=(22.0443)(693.93)=15{,}297\ \text{A}}$$ For reference the total current returning through the ground is $3I_{a0}=27.2727$ pu, or 18,925 A, and by symmetry $|I_C| = |I_B|$.
5. Part (c) — Recognise which reactances change. The added generator is in parallel with the utility for the positive and negative sequences: $$X_1'=X_2'=\frac{(0.05)(0.14)}{0.05+0.14}=0.036842\ \text{pu}$$ Its zero-sequence reactance of 0.08 pu, however, is a decoy: the machine is stated to be ungrounded, so its zero-sequence network has no connection to the reference bus and no zero-sequence current can flow in it. The zero-sequence network is therefore unchanged at $X_0=0.03$, supplied only by the utility's grounded-wye transformer neutral.
6. Repeat the reduction with the new reactances. $$\begin{aligned}X_2'\parallel X_0&=\frac{(0.036842)(0.03)}{0.036842+0.03}=0.016535 \\ X_{th}'&=0.036842+0.016535=0.053378\end{aligned}$$ $$\begin{aligned}I_{a1}&=\frac{1.0}{j(0.053378)}=18.7345\angle -90.00^{\circ} \\ V_{a1}&=0.309783\end{aligned}$$ $$\begin{aligned}I_{a2}&=8.4084\angle 90.00^{\circ} \\ I_{a0}&=10.3261\angle 90.00^{\circ}\ \text{pu}\end{aligned}$$
7. Transform and interpret. Applying the same phase-B transformation, $$I_B=I_{a0}+a^{2}I_{a1}+aI_{a2}=-23.5064+j15.4891=28.1507\angle 146.62^{\circ}\ \text{pu}$$ $$\boxed{|I_B|=(28.1507)(693.93)=19{,}535\ \text{A}}$$ Adding the generator raises the phase-B fault duty by 27.7 per cent, from 15,297 to 19,535 A, entirely through the positive- and negative-sequence paths. This is exactly the calculation that governs the interrupting rating of the plant switchgear, and it is why adding on-site generation to an existing industrial bus so often forces a switchgear replacement even when the generator is small.

Final results.