Question 6 of 7: System Grounding and Single Line-to-Ground Fault Currents

Problem 6: System Grounding and Single Line-to-Ground Fault Currents (20 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.

Find. Why systems are grounded at all; then the phase-A line-to-ground fault current on the plant bus for three successive configurations — utility alone, utility plus an ungrounded generator, and utility plus generator with a resistor in the utility transformer neutral — together with that resistor expressed in per unit.

Part (a) — why systems are grounded. Connecting the system neutral to earth, directly or through an impedance, serves four distinct purposes and no single one of them is the whole answer. The first is control of overvoltage. An ungrounded system has no reference to earth except its own distributed capacitance, so a single line-to-ground fault does not draw useful current but does displace the neutral by a full phase voltage, raising the two healthy phases to line-to-line voltage; worse, an intermittent (arcing) ground fault can resonate with that capacitance and produce transient overvoltages of five to six times normal, puncturing insulation anywhere on the system. Grounding pins the neutral and caps the healthy-phase rise at about 1.25 times normal for an effectively grounded system.

The second purpose is reliable fault detection and clearing. Ground faults are by far the commonest fault type, and unless the system is grounded there is no return path to give the overcurrent or ground-fault relays anything to measure. A grounded system turns an insulation failure into a current large enough to be selectively detected and cleared before it escalates into a phase-to-phase or three-phase fault. The third is personnel and equipment safety: bonding and grounding hold exposed metalwork near earth potential, limiting touch and step voltages, which is the basis of the equipment-grounding and bonding requirements of CSA C22.1, the Canadian Electrical Code, Part I (Section 10), and of CSA C22.3 No. 1 for outdoor distribution. The fourth is a defined path for lightning and switching surges, letting surge arresters discharge to a solid reference and allowing arresters of lower rating — and therefore lower protective level — to be used.

The choice of grounding method then trades these against each other, and part (f) of this question is a direct illustration. Solid grounding gives the best overvoltage control and the simplest relaying but the largest ground-fault current, with the attendant arc-flash energy and equipment damage. Low-resistance grounding, as introduced here, limits the current to a few hundred amperes — enough for selective relaying, small enough to avoid burning laminations and to cut incident energy sharply. High-resistance grounding limits it to a few amperes so the plant can keep running on the first fault, which is why it is common in continuous-process industry; and reactance or resonant (Petersen coil) grounding occupies the ground between.

Approach. For a single line-to-ground fault the three sequence networks are connected in series, so $I_{a1}=I_{a2}=I_{a0}=E/(Z_1+Z_2+Z_0)$ and $I_f=I_a=3I_{a1}$. The work in each configuration is therefore only to build the three sequence impedances correctly — and the decisive reading is which sources are admitted to the zero-sequence network.

1. Establish the per-unit bases before anything else. On a 5000 kVA, 4160 V three-phase base, $$\begin{aligned}Z_b&=\frac{V_b^{2}}{S_b}=\frac{(4160)^{2}}{5\times10^{6}}=3.4611\ \Omega\\I_b&=\frac{S_b}{\sqrt{3}\,V_b}=\frac{5\times10^{6}}{\sqrt{3}(4160)}=693.93\ \text{A}\end{aligned}$$ Every per-unit current below is converted to amperes with this base, and the resistor of part (e) is converted with $Z_b$.
2. Part (b) — build the three sequence impedances for the utility alone. Only one source is present, so each network contains one branch: $$X_1=X_2=0.06\ \text{pu},\qquad X_0=0.04\ \text{pu}$$ The utility transformer is drawn delta–wye with the wye solidly earthed on the plant side, which is exactly what gives the zero-sequence network a path to the reference at all.
3. Series-connect the networks and take the fault current. For an SLG fault on phase A with a bolted connection to earth, $$I_{a1}=\frac{E}{j\left(X_1+X_2+X_0\right)}=\frac{1.0}{j(0.06+0.06+0.04)}=\frac{1.0}{j0.16}=6.25\ \text{pu}$$ $$I_f=3I_{a1}=18.75\ \text{pu}\;\Longrightarrow\;\boxed{\ I_f=18.75\ \text{pu}=13\,011\ \text{A}\ }$$ Note the signature: $X_0=0.04$ is smaller than $X_1=0.06$, so the single-line-to-ground fault is more severe than a three-phase fault at the same bus, which would draw only $1/0.06=16.67$ pu. That inequality is the standard free check on a sequence network.
4. Part (d) — add the generator, and read its grounding off the figure. The generator is stated to be ungrounded, so although Figure (4-b) prints $X_0=0.1$ pu for it, that value is a decoy: with no neutral connection there is no path for zero-sequence current into the machine and it does not appear in the zero-sequence network at all. In the positive and negative networks it is fully in parallel with the utility: $$\begin{aligned}X_1=X_2&=\frac{(0.06)(0.15)}{0.06+0.15}=\frac{0.009}{0.21}=0.042857\ \text{pu}\\X_0&=0.04\ \text{pu (utility only)}\end{aligned}$$
5. Recompute the fault current with the generator in service. $$I_{a1}=\frac{1.0}{j\left(0.042857+0.042857+0.04\right)}=\frac{1.0}{j0.125714}=7.9545\ \text{pu}$$ $$I_f=3(7.9545)=23.864\ \text{pu}\;\Longrightarrow\;\boxed{\ I_f=23.86\ \text{pu}=16\,560\ \text{A}\ }$$ Adding a local generator has raised the ground-fault duty by 27 per cent even though it contributes nothing at all to the zero-sequence network — a point worth making explicitly, because plant switchgear is frequently found to be under-rated after on-site generation is added.
6. Part (e) — convert the neutral resistor to per unit. Using the base impedance already established, $$R_{n,\text{pu}}=\frac{R_n}{Z_b}=\frac{6}{3.4611}\;\Longrightarrow\;\boxed{\ R_n=1.7335\ \text{pu}\ }$$ It is worth pausing on the magnitude: the resistor is more than forty times the entire zero-sequence reactance of the system, so it will dominate everything that follows.
7. Part (f) — place the resistor in the zero-sequence network as $3R_n$. A neutral impedance carries $I_{a0}+I_{b0}+I_{c0}=3I_{a0}$ while the single-phase sequence network carries only $I_{a0}$, so to develop the same voltage drop it must be represented by three times its ohmic value: $$Z_0=3R_n+jX_{0,\text{utility}}=3(1.73354)+j0.04=5.2006+j0.04\ \text{pu}$$ The positive- and negative-sequence networks are unchanged at $j0.042857$ pu each, since the neutral resistor is invisible to balanced current.
8. Take the resistance-grounded fault current. Summing the three sequence impedances, $$Z_1+Z_2+Z_0=5.2006+j\left(0.042857+0.042857+0.04\right)=5.2006+j0.12571=5.2021\angle 1.385^{\circ}\ \text{pu}$$ $$I_f=\frac{3(1.0)}{5.2021}=0.5767\ \text{pu}\;\Longrightarrow\;\boxed{\ I_f=0.577\ \text{pu}=400.2\ \text{A}\ }$$ The resistor has cut the ground-fault current from 16.6 kA to 400 A, a reduction of more than forty to one, and the fault is now almost purely resistive — the arc energy falls with the square of the current, so incident energy drops by a factor of about 1700 while the 400 A remains ample for selective ground relaying. This is the textbook case for low-resistance grounding of an industrial distribution bus.