Question 6 of 7: System Grounding of an Industrial Distribution Bus

Question 6: System Grounding of an Industrial Distribution Bus (20 marks)

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 systems are grounded. System grounding means deliberately connecting some point of the system — almost always a transformer or generator neutral — to earth, either solidly or through an impedance. An ungrounded system is not free of earth connection; it is capacitively coupled to earth through the distributed capacitance of every cable and winding, and that coupling is what makes it dangerous rather than safe.

The first reason is control of overvoltage. On an ungrounded system a single phase-to-earth fault raises the two healthy phases to full line-to-line voltage against earth, a $\sqrt{3}$ increase sustained for as long as the fault persists, which accelerates insulation ageing and frequently causes a second fault elsewhere. Worse, an intermittent arcing ground fault can resonate with the system charging capacitance and produce restrike transients of five to six times normal crest voltage. Grounding the neutral clamps the healthy-phase rise to about $1.25$–$1.4$ per unit and eliminates the arcing-ground mechanism entirely.

The second is reliable and selective protection. A grounded system delivers a definite, calculable ground-fault current, which residual and zero-sequence relays can measure and use to identify and isolate the faulted feeder. On an ungrounded system the only fault current is the system's own charging current — often a few amperes, below the pickup of any overcurrent device — so the fault is not cleared at all, merely alarmed, and must be hunted down by switching while the plant runs with degraded insulation.

The third is personnel and equipment safety. A defined ground return limits the touch and step potentials that appear on enclosures and around the grounding system, which is what CSA C22.1 (the Canadian Electrical Code) and IEEE Std 80 are written around. And the fourth is limitation of damage, which is where the choice of grounding impedance enters. Solid grounding maximises ground-fault current and hence relay sensitivity, but a high-current arcing fault at 4160 V burns iron and can destroy a machine or a switchgear cubicle. Resistance grounding, the near-universal choice in Canadian industrial plants at medium voltage, deliberately limits ground-fault current to a few hundred amperes — enough for selective relaying, far too little to cause burning damage. Parts (d) and (e) below quantify exactly that trade.

Given. All sequence reactances on a 5000 kVA base at 4160 V, read from Figure (4).

Find. (b) the SLG current with the utility alone; (c) the SLG current with the generator added; (d) the resistor in per unit; (e) the double line-to-ground current with the resistor in service.

1. Establish the bases first. Everything downstream needs these two numbers: $$Z_{base} = \frac{V_{base}^2}{S_{base}} = \frac{(4160)^2}{5\times10^6} = 3.46112\ \Omega, \qquad I_{base} = \frac{S_{base}}{\sqrt{3}\,V_{base}} = \frac{5\times10^6}{\sqrt{3}(4160)} = 693.93\ \text{A}$$
2. Part (b) — single line-to-ground fault, utility alone. For an SLG fault the three sequence networks connect in series, so $I_{a1} = I_{a2} = I_{a0}$ and $$I_f = 3I_{a1} = \frac{3E}{X_1 + X_2 + X_0} = \frac{3(1.0)}{0.15 + 0.15 + 0.10} = \frac{3}{0.40}$$ $$\boxed{I_f = 7.500\ \text{pu} = 7.500 \times 693.93 = 5204\ \text{A}}$$
Figure 6.1 — single line-to-ground fault: the three sequence networks in series, so I_a1 = I_a2 = I_a0 and the phase-A fault current is 3 I_a1.
3. Check the answer against the three-phase duty. At the same bus the three-phase current is $1/X_1 = 6.667$ pu, or 4626 A. The SLG current exceeds it, which is the correct signature when $X_0 < X_1$: the ground-fault loop here is stiffer than the phase loop because the transformer neutral is solidly earthed close to the bus. Whenever the inequality comes out the other way with $X_0 < X_1$, a sequence network has been built wrongly.
4. Part (c) — add the ungrounded generator. The generator is drawn as an ungrounded wye, so it offers no path for zero-sequence current: its printed $X_0 = 0.04$ pu is a decoy and never enters any answer. It does parallel the utility in the positive- and negative-sequence networks: $$X_1 = X_2 = \frac{(0.15)(0.06)}{0.15 + 0.06} = 0.042857\ \text{pu}, \qquad X_0 = 0.10\ \text{pu (unchanged)}$$ $$I_f = \frac{3}{2(0.042857) + 0.10} = \frac{3}{0.185714}$$ $$\boxed{I_f = 16.154\ \text{pu} = 11\,210\ \text{A}}$$ Adding local generation has more than doubled the ground-fault duty, from 5204 A to 11 210 A, while contributing nothing whatever to the zero-sequence network — and the switchgear must be rated for the larger figure. Note also that the inequality has now reversed: the three-phase current is $1/0.042857 = 23.333$ pu (16 192 A) and exceeds the SLG current, exactly as it must once $X_0 > X_1$.
5. Part (d) — the grounding resistor in per unit. A resistance is converted on the same base as any other impedance: $$R_{pu} = \frac{R_\Omega}{Z_{base}} = \frac{8}{3.46112}$$ $$\boxed{R = 2.3114\ \text{pu}}$$ It enters the zero-sequence network as three times its ohmic value, because the physical neutral carries $3I_{a0}$ while the single-phase sequence network carries only $I_{a0}$: $$3R = 6.9342\ \text{pu}$$ That figure dwarfs every reactance in the network, which is the whole design intent.
6. Part (e) — double line-to-ground fault on B and C, resistor in service. For a DLG fault the positive-sequence network is in series with the parallel combination of the negative- and zero-sequence networks. With the generator retained (Figure 4-c draws it) the sequence impedances are $$Z_1 = Z_2 = j0.042857\ \text{pu}, \qquad Z_0 = 3R + jX_0 = 6.93417 + j0.10\ \text{pu}$$ $$I_{a1} = \frac{E}{Z_1 + \dfrac{Z_2Z_0}{Z_2 + Z_0}} = 11.667\angle -89.82^\circ\ \text{pu}$$ $$I_{a2} = -I_{a1}\frac{Z_0}{Z_2+Z_0} = 11.666\angle 89.82^\circ, \qquad I_{a0} = -I_{a1}\frac{Z_2}{Z_2+Z_0} = 0.0721\angle 179.00^\circ$$
Figure 6.2 — double line-to-ground fault with the 8 Ω neutral resistor in place. The resistor enters the zero-sequence network as 3R = 6.934 pu, which swamps every reactance in it.
7. Synthesise the phase currents. With $a = 1\angle 120^\circ$, $$I_B = I_{a0} + a^2I_{a1} + aI_{a2}, \qquad I_C = I_{a0} + aI_{a1} + a^2I_{a2}$$ $$\boxed{I_B = 20.315\angle 180.0^\circ\ \text{pu} = 14\,097\ \text{A}}$$ $$\boxed{I_C = 20.099\angle 0.0^\circ\ \text{pu} = 13\,947\ \text{A}}$$ and the current returning through the resistor and earth is $$\boxed{3I_{a0} = 0.2163\ \text{pu} = 150.1\ \text{A}}$$ The healthy phase carries $I_A = I_{a0} + I_{a1} + I_{a2} = 0$ exactly, which is the standard closure check on any DLG solution.
8. Read what the resistor actually bought. Repeating part (e) with the neutral solidly grounded gives $I_B = 21.130$ pu and a ground current of $3I_{a0} = 12.353$ pu, or 8572 A. The resistor therefore reduces the phase currents by only 4 % — they are set almost entirely by the positive- and negative-sequence networks, which the resistor does not touch — but it cuts the earth current by a factor of 57, from 8572 A to 150 A. That is precisely the design objective: the switchgear interrupting duty is unchanged, so nothing has to be re-rated, while arcing damage at the fault and touch potentials around the grounding system collapse. The same resistor reduces a plain single line-to-ground fault from 16.15 pu to $|3/(Z_1+Z_2+Z_0)| = 0.4325$ pu, i.e. 300 A — still ten times the pickup of an ordinary ground relay, and far below the level that burns iron.