Set 38 — Substation Earthing and Insulation
Twenty worked problems on the three questions a substation designer must answer together. How shall the neutral be earthed — because that single choice fixes the earth-fault current, the healthy-phase voltage rise and the relaying philosophy at once. How shall the buried grid be sized — because the fault current has to leave through the soil, and the soil is not an equipotential. And how shall the insulation be graded against the arrester — because the earth-fault factor set by the first decision propagates all the way to the transformer's BIL.
The arithmetic is unglamorous: a few reactance sums, one Laurent–Niemann expression, the IEEE 80 step-and-touch formulae, and a protective margin. What makes the set worth doing is that the three groups are not independent. Problem 20 closes the loop deliberately, and Problems 11 to 13 show a grid design failing and being rescued in three different ways at three different prices.
The neutral impedance appears tripled, and only in the zero-sequence path. \(I_f = 3E_a/(Z_1+Z_2+Z_0+3Z_n)\), from Set 23. The physical neutral carries \(3I_0\) while the per-phase zero-sequence network carries \(I_0\), so one ohm at the star point is three ohms in the network. It is the only element that changes earth-fault current without touching anything else.
An isolated neutral draws only charging current, and pays for it in volts. With phase \(a\) earthed the two healthy phases sit at full line voltage, and the current returning through the fault is \(I_C = 3\omega C V_{ph}\) — the same factor of three, and for the same reason: the three phase-to-earth capacitances are in parallel as seen from the neutral.
The Peterson coil cancels that current. Setting \(I_L = V_{ph}/\omega L\) equal to \(I_C\) gives \(L = 1/(3\omega^2C)\). Because \(C\) depends on how much line is connected, a fixed coil is detuned the moment a feeder is switched, which is why real coils have a plunger.
Grid resistance falls only as \(1/\sqrt{A}\). \(R_g = \frac{\rho}{4}\sqrt{\pi/A} + \rho/L\) (Laurent–Niemann), or Sverak's refinement including the burial depth \(h\). Land buys very little; adding conductor inside a fixed area buys almost nothing, because the second term is already small.
Safety is a difference, never an absolute. \(\mathrm{GPR}=I_GR_g\) may be kilovolts and the yard still be safe. What must be bounded is the step voltage between the feet and the touch voltage between hand and feet, and those are set by the shape of the surface potential, not by its height.
Dalziel fixes the tolerable body current as \(k/\sqrt{t_s}\). With \(1000\;\Omega\) of body and each foot a plate of \(3C_s\rho_s\) — in parallel for touch, in series for step — the limits are \(E_{\text{touch}}=(1000+1.5C_s\rho_s)k/\sqrt{t_s}\) and \(E_{\text{step}}=(1000+6C_s\rho_s)k/\sqrt{t_s}\), with \(k=0.116\) for 50 kg and \(0.157\) for 70 kg.
Crushed rock works through \(C_s\). The derating factor \(C_s = 1 - 0.09(1-\rho/\rho_s)/(2h_s+0.09)\) admits that a thin layer does not hide the native soil from the foot. It is the cheapest safety measure in a substation and it multiplies the tolerable touch voltage by roughly three.
Touch governs, always. The attained values are \(E_m = \rho K_mK_iI_G/L_M\) and \(E_s = \rho K_sK_iI_G/L_S\); the design passes when \(E_m \le E_{\text{touch}}\) and \(E_s \le E_{\text{step}}\). Because the feet are in parallel for touch and in series for step, the step check is loose by a factor of four or more and is almost never the binding one.
The arrester is rated from the temporary overvoltage, not from the surge. \(U_r \ge k_{\text{eff}}U_m/\sqrt3\), with \(k_{\text{eff}}\approx1.4\) when the system is effectively earthed (\(X_0/X_1\le3\), \(R_0/X_1\le1\)) and about \(1.73\) when it is not. The residual voltage scales with \(U_r\), so the earthing decision reappears as a protective level.
The margin is computed at the transformer, not at the arrester. \(V_p = U_{res} + L\,\mathrm{d}i/\mathrm{d}t + (2S/v)\,\mathrm{d}v/\mathrm{d}t\) with \(v\approx300\;\mathrm{m}/\mu\mathrm{s}\) and about \(1\;\mu\mathrm{H}\) per metre of lead; then \(\mathrm{PM} = (\mathrm{BIL}-V_p)/V_p \ge 20\%\).
Security is bought in breakers. \(n\) for a single bus or a ring, \(n+1\) for the sectionalised, transfer and double-bus schemes, \(1.5n\) for breaker-and-a-half, \(2n\) for double-bus-double-breaker. The knee of the curve is at \(1.5n\), which is why every large EHV station in the world looks the same.
At a \(33\;\mathrm{kV}\), 50 Hz bus the sequence reactances looking back from the fault point are \(X_1 = X_2 = j7.5\;\Omega\) and \(X_0 = j5.0\;\Omega\). For a solid single line-to-ground fault at that bus, find
- the fault current with the transformer neutral solidly earthed;
- the fault current with a \(12\;\Omega\) neutral resistor;
- the neutral resistance that limits the fault current to \(400\;\mathrm{A}\), and the resistor's rating for a 10 s duty;
- the highest healthy-phase voltage in each of cases (i) and (iii).
The one relation that does all the work. From Set 23, a single line-to-ground fault gives
with \(E_a = 33\,000/\sqrt3 = 19\,053\;\mathrm{V}\) and therefore \(3E_a = 57\,158\;\mathrm{V}\). Every part of this problem is that one expression with a different \(Z_n\).
(i) Solid earthing, \(Z_n = 0\). The three reactances add directly:
Compare the three-phase value, \(E_a/X_1 = 19\,053/7.5 = 2540\;\mathrm{A}\). The earth fault is the larger, by 12.5%, because \(X_0 < X_1\) — the usual situation at the terminals of a solidly earthed star-delta transformer, and the reason the breaker duty of Set 25 must be checked against the earth fault too.
(ii) A \(12\;\Omega\) resistor. It enters as \(3R_n = 36\;\Omega\), in quadrature with the reactances:
48.6% of the solid-earthed value. Note that the reduction is far less than \(36/20\) would suggest, because resistance and reactance add in quadrature; this is why the first few ohms of a neutral resistor are so much less effective than the beginner expects.
(iii) Sizing for \(400\;\mathrm{A}\). Invert the same expression:
The reactances have almost dropped out: at \(400\;\mathrm{A}\) the resistor dominates so completely that \(3R_n \approx 3E_a/I_f\) to within 1%. That approximation is safe whenever the limited current is below about a tenth of the solid-earthed value.
The resistor's duty. It carries the full fault current for as long as the fault lasts:
Seven and a half megawatts is not a continuous rating anybody would buy; it is a stack of stainless-steel grids sized thermally for ten seconds and cooled by its own mass. The 10 s figure is chosen to sit above the backup clearing time of Set 37, not above the primary one.
(iv) What the limiting costs in volts. Solving the sequence network for the healthy phases with \(Z_0 = j5\) and then with \(Z_0 = 3(47.16)+j5 = 141.5+j5\):
The second figure exceeds \(\sqrt3 = 1.732\), which surprises people who expect \(\sqrt3\) to be an upper bound. It is an upper bound only when \(Z_0\) is purely reactive; a large resistive \(Z_0\) rotates the neutral displacement out of line with the phase voltages and can push the healthy phase to about \(1.9\,V_{ph}\). The insulation must be rated for it.
A \(33\;\mathrm{kV}\), 50 Hz overhead network of total route length \(80\;\mathrm{km}\) has a capacitance to earth of \(0.005\;\mu\mathrm{F}\) per kilometre per phase, and its neutral is left isolated. A solid earth fault occurs on phase \(a\). Find the current in the fault, the voltage on the healthy phases, and state the two reasons this arrangement is no longer built.
The capacitance seen from the neutral. Each phase has its own capacitance to earth, and with no galvanic return the only path is through those capacitances:
What happens to the healthy phases. Phase \(a\) is pulled to earth potential, so the neutral point is displaced by \(V_{ph}\) and each healthy phase now stands at the full line voltage above earth:
Phase \(a\)'s own capacitance is short-circuited and carries nothing; the other two are driven by \(33\;\mathrm{kV}\) instead of \(19\;\mathrm{kV}\).
Adding the two charging currents. They are equal in magnitude and \(60^\circ\) apart, so their resultant is \(2\cos30^\circ\) times either one:
Seven amperes on a network whose load current is hundreds. The system can, and historically did, keep running with the fault on it — that is the whole attraction of the isolated neutral.
The first reason it was abandoned: the sustained voltage. The healthy phases sit at \(33\;\mathrm{kV}\) to earth for as long as the fault is left on, which may be hours. Every bushing, cable termination and insulator in the network must therefore be insulated for the line voltage to earth rather than the phase voltage — a \(\sqrt3\) penalty on the whole system's insulation to save a few relays.
The second and worse reason: the arcing ground. Seven amperes is far too little to hold an arc, so the arc extinguishes at a current zero. But it extinguishes with the line capacitance charged, and that trapped charge is still there when the voltage on the faulted phase swings back:
The escalation is stepwise and it does not stop at \(\sqrt3\): transient overvoltages of five to six times normal are measured, and they puncture insulation at a point in the network far from the original fault. An intermittent earth fault in an isolated system therefore destroys healthy plant, which is exactly the outcome earthing exists to prevent.
A \(33\;\mathrm{kV}\), 50 Hz overhead distribution network consists of \(120\;\mathrm{km}\) of line whose capacitance to earth is \(0.0048\;\mu\mathrm{F}\) per kilometre per phase. Find the inductance of an arc-suppression coil giving complete compensation, the current it carries, its volt-ampere rating, and the residual current left if \(30\;\mathrm{km}\) of feeder is switched out with the coil left at its original setting.
The capacitance and the current to be cancelled.
Ten amperes into the fault — small, but quite enough to hold an arc across a punctured pin insulator and to keep it burning until somebody notices.
The tuning condition. Once the neutral has been displaced, the whole phase voltage stands across the coil, so it draws \(I_L = V_{ph}/\omega L\). Setting \(I_L = I_C\):
Note that \(V_{ph}\) cancels: the tuning condition is a property of the network's capacitance and the frequency alone, not of its voltage.
The coil current, checked both ways.
Equal to \(I_C\), as the condition demands, and in exact antiphase with it at the fault: the coil current lags the neutral voltage by \(90^\circ\), the capacitive current leads it by \(90^\circ\).
The rating. The coil sits at phase voltage carrying \(I_L\):
Short-time rated, because current flows only while an earth fault is present. Coils are specified for a stated duty — commonly 2 h or continuous on networks where faults are left on deliberately.
Switching out \(30\;\mathrm{km}\). The capacitance falls to \(90 \times 0.0048 = 0.432\;\mu\mathrm{F}\) and the capacitive current falls with it, but the coil is unchanged:
The residual is 33% of the remaining charging current, and it is now inductive rather than capacitive — the coil has become the dominant source. The setting that would restore the tune is \(L' = 1/(3\omega^2 \times 0.432\times10^{-6}) = 7.82\;\mathrm{H}\).
The network of Problem 3 is operated with anything between \(90\;\mathrm{km}\) and \(150\;\mathrm{km}\) of feeder in service. Define the degree of detuning, find the inductance range the coil must cover, and evaluate the worst residual current if a fixed coil tuned for \(120\;\mathrm{km}\) is used instead. Over what length range would a fixed coil hold the residual within 5% of the charging current?
Define the detuning. The residual current at the fault is the difference of two antiphase currents, so it is natural to normalise it on the capacitive one:
Positive \(\nu\) means undercompensation (the residual is capacitive, the coil too small); negative means overcompensation. Utilities normally run slightly overcompensated, because an undercompensated network can pass through exact series resonance as feeders are switched, and a resonance at the neutral is what the coil exists to avoid producing unintentionally.
The three operating points. With \(0.0048\;\mu\mathrm{F/km}\) and \(V_{ph} = 19\,053\;\mathrm{V}\):
The coil must therefore be adjustable from \(4.69\;\mathrm{H}\) to \(7.82\;\mathrm{H}\), a ratio of \(150/90 = 1.67:1\) — exactly the ratio of the line lengths, since \(L \propto 1/C \propto 1/\ell\).
The fixed coil at the two extremes. It carries \(I_L = 10.34\;\mathrm{A}\) whatever the network, because \(V_{ph}\) and \(\omega L\) are both fixed:
The residual is \(2.59\;\mathrm{A}\) at both extremes — necessarily, since it is \(3\omega V_{ph}|\Delta C|\) and \(\Delta C\) is \(0.144\;\mu\mathrm{F}\) either way. But the fraction differs, and the worse figure is the undercapacitance case at \(90\;\mathrm{km}\), because the residual is measured against a smaller \(I_C\).
The 5% band. Since \(I_C \propto \ell\) and \(I_L\) is fixed at the \(120\;\mathrm{km}\) value, the condition \(|I_C - I_L| \le 0.05\,I_C\) becomes \(0.95 \le \ell/120 \le 1.05\) in the sense that
A window of twelve kilometres out of a sixty-kilometre operating range. One feeder switched at either end of the day takes the network out of it.
Is a \(2.6\;\mathrm{A}\) residual actually dangerous? It is not lethal, and it will not damage the line. What it does is defeat the coil's purpose: an arc drawing two and a half amperes across a wet insulator restrikes reliably, so the fault does not self-extinguish and the network is left running with a permanent earth fault that the coil was supposed to have cleared. The healthy phases stay at \(\sqrt3 V_{ph}\) the whole time. Practice keeps the residual below about \(10\%\) of \(I_C\), and modern controllers hold it near \(2\%\).
A \(60\;\mathrm{MVA}\), \(11\;\mathrm{kV}\) generator has \(X_1 = j0.20\), \(X_2 = j0.18\) and \(X_0 = j0.06\) per unit on its own base. Find the neutral reactance, in ohms and in henries, that makes the single line-to-ground fault current at its terminals equal to the three-phase fault current, and state both currents in amperes. What would the earth-fault current be without it?
The three-phase fault sets the target. In per unit, with \(E = 1.0\),
Set the earth-fault current equal to it and solve for the total sequence impedance the fault must see:
Convert to ohms on the machine base.
A tenth of an ohm — a single-turn-per-layer air-cored reactor a metre across, not a large item. The per-unit method of Set 4 is doing the work here: the answer is set by the machine's own reactances, and the ohmic value is only the last line.
The currents in amperes.
Without the reactor, the small \(X_0\) lets the earth fault run away:
36% above the three-phase value. That is the whole reason for the reactor: a generator's circuit breaker, its bus ducting and its stator bracing are all specified against the three-phase fault, and there is no sense in letting a different fault type exceed the figure everything was designed to.
Why reactance and not resistance here. A resistor sized for the same current would dissipate \(I^2R\) at fifteen kiloamperes — hundreds of megawatts, absurd. Reactance limits with negligible loss. The price is that reactance earthing gives no damping, so \(X_n\) must be kept small enough that \(X_0/X_1\) stays within the effective-earthing bound; here the effective \(X_0\) becomes \(0.06+3(0.0533) = 0.22\), so \(X_0/X_1 = 1.1\) and the machine remains comfortably effectively earthed.
A \(132\;\mathrm{kV}\) bus has \(Z_1 = Z_2 = j0.15\) and \(Z_0 = 0.10 + j0.35\) per unit. Test it against the effective-earthing criterion, compute the earth-fault factor it actually produces, and repeat both with a neutral reactor of \(X_n = 0.20\) pu inserted. State the consequence for the arrester.
The criterion. A system is effectively earthed when
Both satisfied, so the bus qualifies and the standard earth-fault factor of \(1.4\) may be used for arrester selection.
What the earth-fault factor actually is. Solve the connected sequence networks for the healthy phases. With \(I_1=I_2=I_0=E/(Z_1+Z_2+Z_0)\) and \(V_{a1}=E-I_1Z_1\), \(V_{a2}=-I_2Z_2\), \(V_{a0}=-I_0Z_0\):
Comfortably inside \(1.4\). Notice that the two healthy phases are not equal — the resistive part of \(Z_0\) rotates the neutral displacement, so the fault is not symmetric about the faulted phase and one healthy phase always rises more than the other. It is the larger that matters.
Insert \(X_n = 0.20\) pu. It appears tripled, so \(Z_0 \to 0.10 + j(0.35+0.60) = 0.10+j0.95\):
The criterion fails and the computed factor confirms it. The reactor that was inserted to limit fault current has cost \(0.18\) pu of healthy-phase voltage.
Where the criterion stops being satisfied. The boundary is \(X_0 = 3X_1 = 0.45\), so
One-sixth of the reactor actually proposed. Anything larger, and the system must be treated as non-effectively earthed for insulation purposes even though the fault current is still substantial.
How conservative is the criterion? Evaluating the factor for a purely reactive \(Z_0\) gives the closed form \(\text{EFF} = \sqrt3\sqrt{k^2+k+1}/(k+2)\) with \(k = X_0/X_1\):
So with no resistance at all, \(k=3\) gives only \(1.25\), and \(1.4\) is not reached until \(k = 5.5\). The \(k\le3\) rule is deliberately conservative because it must also hold when \(R_0/X_1\) is at its own limit of 1, where \(k = 3\) gives \(1.35\) and the \(1.4\) boundary arrives at \(k = 4.3\).
The arrester consequence. With \(U_m = 145\;\mathrm{kV}\), \(U_r \ge k_{\text{eff}}U_m/\sqrt3 = k_{\text{eff}}(83.7)\):
The standard \(120\;\mathrm{kV}\) arrester of the \(132\;\mathrm{kV}\) class covers the first and not the second; the reactor forces the next rating up, and with it a residual voltage about 5% higher on every arrester in the station.
A substation earthing grid is a square of side \(80\;\mathrm{m}\) buried at \(h = 0.5\;\mathrm{m}\) in soil of resistivity \(\rho = 150\;\Omega\mathrm{\cdot m}\). It is meshed on an \(8\;\mathrm{m}\) square pitch — eleven conductors each way — so the buried length is \(L = 2(11)(80) = 1760\;\mathrm{m}\). Compute the resistance to remote earth by the Laurent–Niemann expression and by Sverak's refinement, and identify which term dominates.
Laurent and Niemann. Their approximation adds a hemispherical-plate term, representing the area of soil the grid engages, to a conductor-length term:
Sverak's refinement replaces the plate term with one that recognises the grid is buried at a depth \(h\) rather than lying on the surface:
The two agree to 0.4%, which is the usual outcome for a shallow grid of ordinary proportions. Sverak's expression differs materially only when the grid is deep — the bracketed factor \((1 + 1/(1+h\sqrt{20/A}))\) runs from 2 at \(h = 0\) down towards 1 for a very deep grid, so burying deeper genuinely helps, but at \(0.5\;\mathrm{m}\) the factor is 1.97 and almost nothing has been gained.
Which term dominates. Split the Laurent–Niemann result:
Nine-tenths of the resistance is set by how much ground the grid covers. Doubling the copper inside the same fence — from \(1760\) to \(3520\;\mathrm{m}\) — would remove only \(0.043\;\Omega\), a 4.7% improvement for twice the material.
And the resistivity multiplies everything. Both terms are proportional to \(\rho\), so a soil survey that reports \(150\;\Omega\mathrm{\cdot m}\) in the monsoon and \(400\;\Omega\mathrm{\cdot m}\) in May is reporting a grid resistance that varies from \(0.92\) to \(2.4\;\Omega\) over the year. Design is done at the dry-season value.
The grid of Problem 7 carries a grid current of \(I_G = 8\;\mathrm{kA}\). Find the ground potential rise. Then determine how much land would be needed to halve the grid resistance, first with the conductor length held at \(1760\;\mathrm{m}\) and then with the \(8\;\mathrm{m}\) mesh spacing held constant so that the conductor length grows with the area. Comment.
The rise itself. Using Sverak's value,
The entire grid — and every fence, structure, enclosure and cable screen bonded to it — sits at seven and a quarter kilovolts above remote earth for the duration of the fault. Nothing about that figure is by itself unsafe, because a person standing on the grid is lifted with it.
Halving \(R_g\) with the copper fixed. The target is \(0.456\;\Omega\), and the conductor term already accounts for \(0.085\;\Omega\) of it:
Five times the area — 3.21 hectares instead of 0.64 — to halve one number.
Halving it with the mesh spacing kept. Now \(L = 0.275\,A\) grows with the area, and the conductor term helps a little:
Still 3.7 times the land, and now with \(6480\;\mathrm{m}\) of buried copper instead of \(1760\;\mathrm{m}\). The extra conductor has bought a reduction from 5.0 to 3.7 area-multiples — real, but not a change of kind.
Why the arithmetic comes out this way. The dominant term is proportional to \(A^{-1/2}\), so
The factor of four is the floor; the conductor term's presence pushes the honest answer above it in the first case, and additional copper pulls it back towards four in the second.
What this means for design. The GPR is essentially not adjustable once the site boundary is fixed. It follows that the safety of a substation yard can never be established by driving \(R_g\) down; it must be established by controlling the surface potential gradient, which is the subject of Problems 9 to 12. The only genuinely effective lever on the GPR is \(I_G\) itself — that is, the earthing method of Problems 1 to 6.
For the station of Problems 7 and 8 the fault clearing time is \(t_s = 0.4\;\mathrm{s}\) and the native soil is \(150\;\Omega\mathrm{\cdot m}\). Find the tolerable step and touch voltages for a \(50\;\mathrm{kg}\) person (a) on bare soil and (b) under a \(0.15\;\mathrm{m}\) layer of crushed rock of \(\rho_s = 3000\;\Omega\mathrm{\cdot m}\). Repeat (b) for a \(70\;\mathrm{kg}\) person and say which figure a designer should use.
The body current first. Dalziel's fibrillation threshold for a shock lasting \(t_s\) seconds is \(I_B = k/\sqrt{t_s}\):
183 mA for four-tenths of a second. The \(\sqrt{t_s}\) is the whole of the physiology: fibrillation depends on the energy the heart absorbs, so a shorter shock is tolerable at a higher current.
The circuit the current flows through. Body \(1000\;\Omega\); each foot a plate of \(3C_s\rho_s\). For touch the two feet are in parallel (\(1.5C_s\rho_s\)); for step they are in series (\(6C_s\rho_s\)):
(a) Bare soil. With no surface layer \(\rho_s = \rho\) and \(C_s = 1\):
(b) With the rock layer. The derating factor accounts for the layer being thin enough that the foot still sees the soil beneath:
The \(70\;\mathrm{kg}\) figures use the same resistances with the larger \(k\):
35% higher throughout, since \(0.157/0.116 = 1.353\).
Which to use. The \(50\;\mathrm{kg}\) criterion, without exception, unless access to the yard is provably restricted to a workforce whose lightest member exceeds \(70\;\mathrm{kg}\) — a condition no Indian utility can demonstrate. Designing to the \(70\;\mathrm{kg}\) limit would let a grid pass at \(1120\;\mathrm{V}\) of touch voltage that a \(50\;\mathrm{kg}\) person would not survive.
What the rock bought.
Fifteen centimetres of stone, at a few hundred rupees a square metre, has multiplied the touch limit by nearly four. Nothing else in earthing design offers that return, and the step limit improves twice as fast because the feet are in series and the \(6C_s\rho_s\) term dominates.
For the soil and rock of Problem 9 (\(\rho = 150\), \(\rho_s = 3000\;\Omega\mathrm{\cdot m}\), \(t_s = 0.4\;\mathrm{s}\), \(50\;\mathrm{kg}\)), tabulate the derating factor and the tolerable touch voltage for rock thicknesses of \(0.05\), \(0.10\), \(0.15\), \(0.20\) and \(0.30\;\mathrm{m}\). What thickness should be specified, and why is more not better?
The factor as a function of thickness alone. With \(\rho/\rho_s = 0.05\) fixed,
A hyperbola in \(h_s\), rising towards 1 but never reaching it. The \(0.09\;\mathrm{m}\) is the foot-plate radius scale that Dalziel's two-layer model contributes; below that thickness the layer is essentially transparent.
The table.
Read the last column. Going from \(0.05\) to \(0.10\;\mathrm{m}\) buys 20%; doubling again from \(0.15\) to \(0.30\;\mathrm{m}\) buys 9.5% for twice the stone. The asymptote is \(C_s = 1\), giving \(E_{\text{touch}} = (1000+4500)(0.1834) = 1009\;\mathrm{V}\), so even an infinitely thick layer is only 22% better than \(0.15\;\mathrm{m}\).
Why more is not better, for three reasons that have nothing to do with the formula:
The third is decisive over a station's life. A layer whose \(\rho_s\) has fallen from \(3000\) to \(1000\;\Omega\mathrm{\cdot m}\) through contamination gives \(C_s = 0.802\) and \(E_{\text{touch}} = 404\;\mathrm{V}\) — less than half the design value, and a grid that passed on paper now fails. Periodic replacement of the top layer is a maintenance item, not an optional one.
Specify \(0.10\) to \(0.15\;\mathrm{m}\), which is exactly the range IEEE 80 quotes, and specify the wet resistivity of the stone rather than its dry value, since rain is when people work outdoors and when \(\rho_s\) is lowest.
The grid of Problems 7 to 10 — \(80\;\mathrm{m}\) square, \(8\;\mathrm{m}\) mesh pitch, \(L_M = L_S = 1760\;\mathrm{m}\), \(\rho = 150\;\Omega\mathrm{\cdot m}\), \(I_G = 8\;\mathrm{kA}\) — has geometric factors \(K_m = 0.811\), \(K_s = 0.395\) and irregularity factor \(K_i = 2.272\). Compute the attained mesh and step voltages and decide whether the design passes.
The two attained voltages. IEEE 80 gives them as the product of the soil resistivity, the geometry and the current, divided by the effective buried length:
\(E_m\) is the worst touch voltage in the yard, found at the centre of a corner mesh where the surface potential dips furthest below the grid. \(E_s\) is the largest step voltage, just outside the perimeter where the profile is steepest.
The mesh voltage.
The step voltage.
The verdict, against Problem 9's limits.
The design fails, and it fails only in touch. The mesh voltage is 52% above the tolerable value; the step voltage has a margin of 350%.
Why touch always governs. Two independent factors conspire:
Together a factor of about seven separates the two checks. In thirty years of practice the step check has essentially never been the binding one, which is why a designer who is short of time computes \(E_m\) first and \(E_s\) only to write it down.
Note what the GPR did not tell us. Problem 8's ground potential rise of \(7.30\;\mathrm{kV}\) is nine times the tolerable touch voltage, and by itself proves nothing at all — a station with a far higher GPR and a finer mesh would pass. The GPR fixes the transferred-potential problem and the insulation of any circuit leaving the yard; it does not fix the safety of the yard.
Rescue the failing grid of Problem 11 by refining the mesh. For a square grid of side \(\ell\) with \(n\) parallel conductors each way at pitch \(D\), buried at depth \(h\) with conductor diameter \(d\), IEEE 80 gives
with \(K_{ii}=1\) and \(K_h=\sqrt{1+h}\) for \(h\) in metres. Take \(d = 10\;\mathrm{mm}\). Evaluate \(D = 5\;\mathrm{m}\) and \(D = 4\;\mathrm{m}\), and say what else could be done instead.
Set up the two candidates. With \(\ell = 80\;\mathrm{m}\), \(n = \ell/D + 1\) and \(L_c = 2n\ell\):
Evaluate \(K_m\) at \(D = 5\;\mathrm{m}\). With \(h = 0.5\), \(d = 0.01\), \(K_h = \sqrt{1.5} = 1.2247\):
And the mesh voltage it gives:
Still fails — by 3.9%. Nearly a kilometre of extra copper has taken the excess from 52% to 4%, but a fail is a fail, and a margin of \(-4\%\) is exactly the region in which a designer must not be tempted.
Repeat at \(D = 4\;\mathrm{m}\).
A margin of 17%. The step voltage at this spacing is \(E_s = 628\;\mathrm{V}\) against \(2761\;\mathrm{V}\), still irrelevant.
Why the improvement is so much slower than the copper. Track the three factors from \(D=8\) to \(D=4\):
Doubling the copper halved \(K_m\)'s contribution and doubled \(L_c\), but \(K_i\) — which penalises a grid for having many parallel conductors and therefore an uneven current distribution among them — rose by two-thirds and gave back most of the gain. The net is \(0.650\times1.651/1.909 = 0.562\). Halving the mesh pitch does not halve the mesh voltage.
The alternatives, in the order IEEE 80 recommends trying them:
Remedy 1 alone rescues the \(D=5\;\mathrm{m}\) design: \(889 \gt 860\), a margin of 3.4%. Combining remedy 1 with \(D=5\;\mathrm{m}\) costs \(960\;\mathrm{m}\) of copper plus \(0.10\;\mathrm{m}\) more stone over \(6400\;\mathrm{m^2}\); the \(D=4\;\mathrm{m}\) design costs \(1600\;\mathrm{m}\) of copper and no stone. At current prices the stone is far cheaper — but it is also the remedy that degrades with time, as Problem 10 showed.
A colleague proposes to rescue the original \(8\;\mathrm{m}\) grid of Problem 11 without touching it, by shortening the fault clearing time instead. Find the clearing time that would make the design pass, express it in cycles at 50 Hz, and say whether the proposal is sound.
Where \(t_s\) enters. Only through Dalziel's threshold — the resistances of body, feet and rock are unaffected:
Set it equal to the attained mesh voltage of \(1256\;\mathrm{V}\) and invert:
Check the scaling. The limit varies as \(t_s^{-1/2}\), so going from \(0.4\;\mathrm{s}\) to \(0.174\;\mathrm{s}\) multiplies it by \(\sqrt{0.4/0.174} = 1.52\) — precisely the factor by which Problem 11's design was failing. Consistent.
Is 8.7 cycles achievable? As a primary clearing time, easily: a distance relay from Set 37 operating in one cycle plus a \(2\)-cycle breaker clears in three, and a modern \(220\;\mathrm{kV}\) feeder clears in about \(60\;\mathrm{ms}\).
But that is the wrong time to use, and this is the point of the problem. IEEE 80 requires the clearing time in the shock-duration formula to be the time a person may actually be exposed, which is the backup clearing time — because if the primary protection fails, the person is standing there for the whole of the backup zone's delay:
The \(0.4\;\mathrm{s}\) of Problem 9 is already a backup figure. To justify \(0.174\;\mathrm{s}\) one would have to show that every credible backup path clears within nine cycles, which requires breaker-failure protection with a very short timer and a demonstration that the timer is never the limiting element. It is possible on a modern station and it is not free.
A second objection. A shorter clearing time on the same network usually goes with a larger grid current, because the fastest clearing schemes are used on the strongest parts of the system, and \(E_m \propto I_G\) directly while the limit improves only as \(t_s^{-1/2}\). If the tighter protection settings came with a system reinforcement raising \(I_G\) from \(8\) to \(10\;\mathrm{kA}\), the gain would be entirely eaten.
A \(400\;\mathrm{kV}\) effectively earthed system has a highest equipment voltage \(U_m = 420\;\mathrm{kV}\) and an earth-fault factor of \(1.4\). Gapless metal-oxide arresters are offered in \(6\;\mathrm{kV}\) steps, with a residual voltage at \(10\;\mathrm{kA}\) of \(2.3\,U_r\) and \(U_c = 0.8\,U_r\). Select the rated voltage, state the continuous and residual voltages, and verify that the arrester will not conduct on the healthy system.
Selection starts from the temporary overvoltage, not from the surge. The arrester must survive the power-frequency voltage that appears on a healthy phase during an earth fault elsewhere, for the ten seconds the fault might last:
Round up to a catalogue rating. In \(6\;\mathrm{kV}\) steps the first available value is \(342\;\mathrm{kV}\), but utilities customarily specify \(360\;\mathrm{kV}\) at this class:
The extra headroom pays for an earth-fault factor that turns out slightly worse than assumed, for a load-rejection overvoltage on a long line (Set 14), and for the fact that a metal-oxide arrester which conducts on a temporary overvoltage does not trip — it heats, and if it heats past its thermal stability point it fails short-circuit.
The continuous operating voltage.
19% above the highest continuous phase voltage the arrester will ever see, so it draws only its leakage current — well under a milliampere, and mostly capacitive — for the whole of its thirty-year life.
The residual voltage, which is the number the insulation actually cares about:
Note the structure: \(U_{res}\) is fixed as a multiple of \(U_r\), and \(U_r\) was fixed by the earthing. Choosing the safer \(360\;\mathrm{kV}\) rating over \(342\;\mathrm{kV}\) has cost \(41\;\mathrm{kV}\) of protective level. Nothing in arrester selection is free in both directions.
The check that it does not conduct. The crest of the normal phase voltage is
and far below the knee of the \(V\)–\(I\) curve. Because the zinc-oxide exponent is \(\alpha = 25\) to \(50\), the resistance between \(U_c\) and the discharge region changes by six orders of magnitude, which is why no series gap is needed and why there is no follow current to interrupt.
A \(33\;\mathrm{kV}\) network (\(U_m = 36\;\mathrm{kV}\)) is resonant-earthed by the Peterson coil of Problem 3. Select the arrester rating, compare it with the rating the same network would need if it were solidly earthed, and evaluate the protective margin in each case against the standard BIL of \(170\;\mathrm{kV}\). Take \(U_{res} = 2.4\,U_r\) and neglect lead and separation effects.
The phase voltage the rating is built on.
Resonant earthing gives the worst possible earth-fault factor. Problem 2 showed the healthy phases sit at the full line voltage, and the Peterson coil does nothing to change that — it removes the current, not the voltage:
And the fault is deliberately not cleared on a resonant-earthed network, so the arrester holds this voltage not for ten seconds but for hours. The rating must be checked against the arrester's continuous, not its 10 s, capability.
Solidly earthed, the same network needs less.
A 20% lower rating, exactly the \(1.73/1.4\) ratio rounded to the catalogue.
What that does to the protective level.
Both pass the \(20\%\) criterion with enormous room, because at \(33\;\mathrm{kV}\) the standard BIL is set by air clearance and creepage, not by the arrester. The resonant system has lost 39 percentage points of margin and still has five times what it needs.
So why does the earthing choice matter at all here? Not for the BIL, but for two other things:
The second is the real constraint. A \(36\;\mathrm{kV}\) arrester has \(U_c = 0.8(36) = 28.8\;\mathrm{kV}\), which covers the normal phase voltage of \(20.78\;\mathrm{kV}\) but not the \(\sqrt3(20.78) = 36.0\;\mathrm{kV}\) that a persistent earth fault leaves on the healthy phases. On a network where faults are deliberately left on, the arrester must satisfy \(U_c \ge 36\;\mathrm{kV}\).
The corrected selection, therefore, for a genuinely resonant-earthed network on which faults persist:
Still acceptable, and now honest. This is the standard trap in arrester selection on non-effectively-earthed systems: the \(U_r\) formula assumes the temporary overvoltage is temporary.
The \(360\;\mathrm{kV}\) arrester of Problem 14 protects a \(400\;\mathrm{kV}\) transformer of \(\mathrm{BIL} = 1425\;\mathrm{kV}\). It is connected by \(4\;\mathrm{m}\) of lead in total and stands \(15\;\mathrm{m}\) from the transformer bushing. The incoming surge has a steepness of \(800\;\mathrm{kV}/\mu\mathrm{s}\) and the discharge current rises at \(10\;\mathrm{kA}/\mu\mathrm{s}\). Find the voltage at the transformer and the protective margin, and say how the margin would change if the lower standard BIL of \(1300\;\mathrm{kV}\) were bought instead.
The arrester's own contribution.
The lead inductance. Connecting leads carry the discharge current at high \(\mathrm{d}i/\mathrm{d}t\), and stray inductance is about \(1\;\mu\mathrm{H}\) per metre:
"Total lead" means both the line-side lead and the earth-side lead down to the grid, because the current flows through both and the transformer is outside the loop.
The separation effect. The wave arriving at the transformer keeps rising for the round-trip travel time before the arrester's clamping reaches it:
with \(v \approx 300\;\mathrm{m}/\mu\mathrm{s}\), the propagation velocity on an overhead conductor. Fifteen metres of busbar is a hundred nanoseconds during which the transformer has no protection at all.
The voltage the transformer sees.
The margin.
Comfortably above the \(20\%\) criterion.
Where the margin went. Against the bare residual voltage the margin would have been \((1425-828)/828 = 72.1\%\):
Eight points to four metres of lead and fourteen more to fifteen metres of spacing. Quoting the protective margin from the catalogue residual voltage overstates it by 22 percentage points here, and by far more at lower voltages.
With the lower BIL. Standard practice offers two levels at each \(U_m\):
Still well above \(20\%\), so buying the cheaper transformer is legitimate here — provided the layout stays as assumed. It is exactly this kind of economy that later fails when somebody relocates the arrester during a bay extension.
For the arrester and transformer of Problem 16, find the greatest separation distance at which the protective margin still reaches \(20\%\), for each of the two standard BIL values. Then find the separation limit if the surge steepness were the \(1200\;\mathrm{kV}/\mu\mathrm{s}\) characteristic of a nearby back-flashover rather than \(800\).
Invert the coordination criterion. \(\mathrm{PM} \ge 20\%\) is \(\mathrm{BIL} \ge 1.2V_p\), so
With \(\mathrm{BIL} = 1425\;\mathrm{kV}\). The fixed contributions are \(828\) of residual and \(40\) of lead:
Sixty metres — larger than any \(400\;\mathrm{kV}\) bay is long, so the coordination is not in practice separation-limited at this BIL.
With \(\mathrm{BIL} = 1300\;\mathrm{kV}\).
A third less, and now within the range a real layout could reach if the arrester were placed at the line entrance rather than at the transformer. This is the concrete cost of the cheaper transformer: it does not reduce the margin much at the design layout, but it removes two-thirds of the layout freedom.
With a steeper front. A back-flashover on the first tower launches a much steeper wave than a distant stroke, because it has not been attenuated by corona along the span (Set 15):
The limit is inversely proportional to the steepness, so the assumed waveshape is as much a design input as the BIL. A study that assumes the incoming surge has already travelled several spans, when the station is at the end of a line whose first tower has a high footing resistance, will overstate \(S_{\max}\) by a third.
And the reason low-voltage stations look different. The available voltage \(V_p^{\max}-U_{res}-L\,\mathrm{d}i/\mathrm{d}t\) scales roughly with \(U_m\), while the separation term does not scale at all — it is metres times steepness. That is why \(400\;\mathrm{kV}\) arresters sit on their own structures a bay away and \(33\;\mathrm{kV}\) arresters are bolted to the transformer tank. Problem 18 works the extreme case.
A \(66\;\mathrm{kV}\) effectively earthed substation (\(U_m = 72.5\;\mathrm{kV}\)) has transformers of \(\mathrm{BIL} = 325\;\mathrm{kV}\). The arrester has \(U_{res} = 2.4\,U_r\). It is proposed to mount the arresters on a separate structure \(20\;\mathrm{m}\) from the transformer with \(3\;\mathrm{m}\) of lead, exactly as at the \(400\;\mathrm{kV}\) station. Take \(\mathrm{d}v/\mathrm{d}t = 1000\;\mathrm{kV}/\mu\mathrm{s}\) and \(\mathrm{d}i/\mathrm{d}t = 10\;\mathrm{kA}/\mu\mathrm{s}\). Evaluate the proposal, find the greatest permissible separation, and state what must be done instead.
The arrester rating.
Which against the \(325\;\mathrm{kV}\) BIL looks magnificent: a bare margin of \((325-144)/144 = 126\%\).
Now add the layout. The lead and separation terms do not know what voltage class they are in:
The margin.
The proposal fails. A margin of 126% has collapsed to 5.8% — below a quarter of the requirement — purely because the same steel arrangement was copied from a station of six times the voltage.
The greatest permissible separation.
So \(20\;\mathrm{m}\) is not merely marginal, it is 38% beyond the limit.
What is done instead: tank mounting. With the arrester bolted to the transformer, \(1\;\mathrm{m}\) of lead and \(2\;\mathrm{m}\) of separation:
Ninety-four per cent, from the identical arrester and the identical transformer. The whole of the difference is eighteen metres of busbar and two metres of lead.
Why the effect scales this way. Compare with Problem 16 term by term:
The layout terms are absolute voltages set by metres and microseconds; the arrester's residual voltage scales with the system. Below about \(132\;\mathrm{kV}\) the layout dominates, and above about \(220\;\mathrm{kV}\) it is a correction. The design rule "mount the arrester as close as the layout allows" is not advice at \(66\;\mathrm{kV}\) — it is the design.
A \(220\;\mathrm{kV}\) substation has six circuits: two incoming lines, two outgoing lines and two transformers. For each of the seven standard busbar arrangements, count the breakers, state what a bus fault costs and whether a breaker can be maintained with its circuit in service, and compare the capital cost at \(₹2.5\) crore per \(220\;\mathrm{kV}\) breaker bay. If a bus fault occurs once in twenty years and costs a four-hour outage of \(2000\;\mathrm{MW}\) valued at \(₹5\) per kWh, is the breaker-and-a-half scheme worth its extra breakers over a ring?
The counts, for \(n = 6\).
Read the ladder. The first extra breaker — sectionalising — is the best value in the whole table: \(₹2.5\) crore converts a whole-station bus fault into a half-station one. The next four schemes all cost \(n+1\) and differ only in which weakness they remove, which is why the choice among them is made on operating philosophy rather than on cost.
The breaker-and-a-half arrangement, for six circuits, is three diameters of three breakers each strung between two buses, with a circuit tapped either side of each shared centre breaker:
Each circuit is fed from both buses. A bus fault trips only the breakers on that bus, and every circuit continues through its centre breaker. Any breaker can be withdrawn without disconnecting anything.
The pairing rule inside a diameter. Losing a whole diameter — a rare but real event, on a busbar-protection maloperation — takes out both its circuits together. So the two must be chosen so that losing them together is survivable:
Here the three diameters would be (line 1, transformer 1), (line 2, transformer 2), (line 3, line 4) — never both transformers on one diameter.
Now the economics. The ring and the breaker-and-a-half schemes both survive a bus fault, so the ring's weakness is different: while the ring is open — after any fault, or during any breaker maintenance — a second event splits the station. Value the bus-fault event first:
Against the extra capital of \(₹7.5\) crore (9 breakers against 6):
On the bus-fault event alone the ring wins — which is exactly why ring buses exist. The breaker-and-a-half scheme is justified by what the count above cannot show: the ring is open for every breaker maintenance (several weeks a year across six breakers), it cannot be extended beyond about six circuits without becoming a chain of single points, and each of its circuits is fed through two breakers in series so that any breaker failure loses two circuits. Add those exposures and the payback falls under fifteen years.
And the last step buys nothing. Going from \(9\) to \(12\) breakers costs another \(₹7.5\) crore and removes no outage that the breaker-and-a-half scheme has not already removed — only the very rare double contingency of a diameter loss. That is why \(1.5n\) is the knee of the curve and why every large EHV station in the world looks the same.
A \(132\;\mathrm{kV}\) substation (\(U_m = 145\;\mathrm{kV}\), BIL \(650\;\mathrm{kV}\)) is solidly earthed and passes a grid current of \(12\;\mathrm{kA}\) into a grid of \(R_g = 0.5\;\Omega\). Its mesh voltage is \(1.05\) times the tolerable touch voltage — a marginal fail. An engineer proposes to fix the earthing grid by resistance-earthing the transformer neutral to limit the earth fault to \(1000\;\mathrm{A}\). Trace the consequences through the ground potential rise, the mesh voltage, the arrester rating, the residual voltage and the protective margin, and give a verdict. Take \(3\;\mathrm{m}\) of arrester lead, \(10\;\mathrm{m}\) of separation, \(\mathrm{d}v/\mathrm{d}t = 1000\;\mathrm{kV}/\mu\mathrm{s}\), \(\mathrm{d}i/\mathrm{d}t = 10\;\mathrm{kA}/\mu\mathrm{s}\) and \(U_{res} = 2.4U_r\).
Link 1 — the ground potential rise falls by the current ratio.
Link 2 — the mesh voltage falls in the same proportion, since \(E_m = \rho K_mK_iI_G/L_M\) is linear in \(I_G\) and nothing else has changed:
A margin of over eleven hundred per cent. The grid problem is not merely solved, it is obliterated — and the proposal was made for exactly that reason.
Link 3 — the earth-fault factor. With \(3R_n \approx 3E_a/I_f = 3(76\,210)/1000 = 229\;\Omega\) against sequence reactances of tens of ohms, the system is emphatically no longer effectively earthed:
Link 4 — the arrester rating rises. With \(U_m/\sqrt3 = 83.7\;\mathrm{kV}\):
Link 5 — the protective margin. The layout contributions are unchanged at \(30\;\mathrm{kV}\) of lead and \((2\times10/300)(1000) = 66.7\;\mathrm{kV}\) of separation:
Both still comfortably above \(20\%\). On the insulation coordination alone, the proposal survives.
Now the objections the chain does not show.
The first is decisive. A \(76\;\mathrm{MW}\) neutral resistor at \(132\;\mathrm{kV}\) is a yard of steel grids the size of the transformer it protects, and it is the reason resistance earthing is a distribution-voltage technique. Add that the transformer must be re-specified without graded insulation — a real cost on a \(100\;\mathrm{MVA}\) unit — and the proposal is an expensive way to avoid buying copper.
The verdict, and the correct remedy. Reject the earthing change. A \(5\%\) touch-voltage overshoot is at the very bottom of IEEE 80's remedy ladder: \(0.05\;\mathrm{m}\) more crushed rock, or one extra pass of the mesh in the corners, would clear it for a few lakhs. Reducing \(I_G\) is the last remedy on that ladder precisely because it reaches back into decisions that six other calculations depend on.
The general shape of the chain, which is what Chapter 37 is about:
One decision, two chains, pulling in opposite directions: less earth-fault current makes the yard safer and the insulation more expensive. Every real substation design is that trade iterated until it closes.
Practice Problems
Work each on paper before opening the answer. The answer is given so you can check yourself; the method is deliberately not. Take \(f = 50\;\mathrm{Hz}\) throughout.
P1. An \(11\;\mathrm{kV}\) bus has \(X_1 = X_2 = j0.9\;\Omega\) and \(X_0 = j0.6\;\Omega\), solidly earthed. Find the single line-to-ground and three-phase fault currents.
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\(I_{LG} = \mathbf{7939\;\mathrm{A}}\), \(I_{3\phi} = \mathbf{7057\;\mathrm{A}}\) — the earth fault is 12.5% larger. Problem 1.P2. A \(66\;\mathrm{kV}\) network of \(60\;\mathrm{km}\) has \(0.008\;\mu\mathrm{F/km}\) per phase to earth. Find the Peterson coil inductance, current and rating.
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\(C = 0.48\;\mu\mathrm{F}\), \(L = \mathbf{7.04\;\mathrm{H}}\), \(I_L = \mathbf{17.24\;\mathrm{A}}\), \(S = \mathbf{657\;\mathrm{kVA}}\). Problem 3.P3. Soil at \(200\;\Omega\mathrm{\cdot m}\) carries \(0.10\;\mathrm{m}\) of crushed rock at \(2500\;\Omega\mathrm{\cdot m}\). Find \(C_s\) and \(C_s\rho_s\).
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\(C_s = \mathbf{0.714}\), \(C_s\rho_s = \mathbf{1786\;\Omega\mathrm{\cdot m}}\). Problems 9 and 10.P4. A square grid of side \(60\;\mathrm{m}\) uses \(600\;\mathrm{m}\) of conductor at \(0.5\;\mathrm{m}\) depth in \(200\;\Omega\mathrm{\cdot m}\) soil. Find \(R_g\) both ways.
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\(\mathbf{1.810\;\Omega}\) (Laurent–Niemann), \(\mathbf{1.797\;\Omega}\) (Sverak). Problem 7.P5. That grid carries \(6\;\mathrm{kA}\). Find the ground potential rise, and say whether it alone condemns the design.
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\(\mathrm{GPR} = \mathbf{10.9\;\mathrm{kV}}\). It condemns nothing — only \(E_m\) and \(E_s\) do. Problems 8 and 11.P6. Select the arrester rated voltage for a \(132\;\mathrm{kV}\) effectively earthed system, \(U_m = 145\;\mathrm{kV}\).
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\(U_r \ge 1.4(83.7) = 117.2\;\mathrm{kV}\), so the standard \(\mathbf{120\;\mathrm{kV}}\) unit. Problem 14.P7. A \(220\;\mathrm{kV}\) transformer has \(\mathrm{BIL} = 950\;\mathrm{kV}\) and sees \(V_p = 620\;\mathrm{kV}\). Find the protective margin.
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\(\mathbf{53.2\%}\), comfortably above the \(20\%\) requirement. Problem 16.P8. How many breakers does a breaker-and-a-half scheme need for eight circuits, and how many diameters?
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\(1.5(8) = \mathbf{12}\) breakers in four diameters of three. Problem 19.P9. A generator has \(X_1 = X_2 = j0.25\) and \(X_0 = j0.10\) pu. Find \(X_n\) that equalises the earth and three-phase faults.
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Need \(\sum = 3/4 = 0.75\); present sum \(0.60\), so \(3X_n = 0.15\) and \(X_n = \mathbf{0.05}\) pu. Problem 5.P10. Bare soil of \(100\;\Omega\mathrm{\cdot m}\), clearing time \(0.5\;\mathrm{s}\). Find the tolerable step and touch voltages for a \(70\;\mathrm{kg}\) person.
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\(E_{\text{touch}} = \mathbf{255\;\mathrm{V}}\), \(E_{\text{step}} = \mathbf{355\;\mathrm{V}}\). Problem 9.P11. A quarter of a resonant-earthed network is switched out and the coil is not retuned. Express the residual current as a fraction of the remaining charging current.
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\(I_{\text{res}}/I_C' = 0.25/0.75 = \mathbf{33\%}\), inductive — for Problem 3's coil, \(2.59\;\mathrm{A}\). Problems 3 and 4.P12. A grid in \(150\;\Omega\mathrm{\cdot m}\) soil has \(K_m = 0.75\), \(K_i = 2.2\), \(L_M = 1500\;\mathrm{m}\) and \(I_G = 6\;\mathrm{kA}\). Find \(E_m\).
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\(E_m = \mathbf{990\;\mathrm{V}}\). Whether it passes depends entirely on the surface layer. Problem 11.
Challenge Problems
Three problems in which the formula is not the difficulty. Each needs a judgement about which quantity to change and what else moves when it does.
C1 — Costing the remedy ladder. The grid of Problem 11 attains \(E_m = 1256\;\mathrm{V}\) against a tolerable \(828\;\mathrm{V}\). Evaluate every remedy IEEE 80 offers, quantify what each achieves, and recommend one. Include perimeter ground rods, which Problem 12 did not.
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The requirement. \(E_m\) must fall by a factor of \(828/1256 = 0.659\), or \(E_{\text{touch}}\) must rise by \(1.52\), or the two must meet somewhere between.
Remedy A — thicker crushed rock. Raising \(h_s\) from \(0.15\) to \(0.25\;\mathrm{m}\) gives \(C_s = 0.855\) and \(E_{\text{touch}} = 889\;\mathrm{V}\) — a rise of only 7.4%. Even an infinitely thick layer reaches \(1009\;\mathrm{V}\), a rise of 22%. Insufficient alone. The alternative within this remedy is a better stone: granite at \(5000\;\Omega\mathrm{\cdot m}\) at \(0.15\;\mathrm{m}\) gives \(C_s = 0.776\), \(C_s\rho_s = 3881\) and \(E_{\text{touch}} = 1251\;\mathrm{V}\) — which alone clears the fail.
Remedy B — perimeter ground rods. Twenty \(3\;\mathrm{m}\) rods add \(L_R = 60\;\mathrm{m}\), and IEEE 80 weights rod length heavily in the effective length:
\[ L_M = L_c + \left[1.55+1.22\frac{L_r}{\sqrt{L_x^{2}+L_y^{2}}}\right]L_R = 1760 + 1.582(60) = 1855\;\mathrm{m} \]\[ E_m \to 1256 \times \frac{1760}{1855} = 1192\;\mathrm{V} \]A 5% improvement — insufficient. Rods earn their place when they reach a lower-resistivity stratum, changing the effective \(\rho\) rather than the length; in uniform soil they are almost decorative for touch voltage, though they do help the step voltage at the fence line, which was never the problem.
Remedy C — finer mesh. From Problem 12: \(D = 5\;\mathrm{m}\) gives \(860\;\mathrm{V}\) (still a fail), \(D = 4\;\mathrm{m}\) gives \(707\;\mathrm{V}\) (a 17% margin) for \(1600\;\mathrm{m}\) of extra copper.
Remedy D — enlarge the grid. Extending the fence changes \(R_g\) (Problem 8) and, at constant mesh pitch, adds conductor. But \(K_i = 0.644+0.148n\) grows with the conductor count and \(K_m\) barely moves at constant \(D\), so \(E_m\) improves roughly as \(K_i/L_c\) — about \(A^{-1/2}\). Halving \(E_m\) would need four times the land. Absurd for this purpose.
Remedy E — reduce \(I_G\). Directly proportional, and directly attached to the earthing method, the arrester rating and the transformer specification. Problem 20 shows what it costs.
The recommendation. Combine the two cheapest partial remedies rather than buying one complete one:
\[ \begin{array}{lccc} \text{Option} & E_m\ (\mathrm{V}) & E_{\text{touch}}\ (\mathrm{V}) & \text{Margin} \\ \hline D=4\;\mathrm{m}\ \text{alone} & 707 & 828 & +17\% \\ D=5\;\mathrm{m} + 0.25\;\mathrm{m}\ \text{rock} & 860 & 889 & +3.4\% \\ D=5\;\mathrm{m} + \text{granite at }5000 & 860 & 1251 & +45\% \\ D=4\;\mathrm{m} + \text{rods} & 687 & 828 & +21\% \end{array} \]Choose \(D = 4\;\mathrm{m}\) alone, and specify the granite surface layer as a second, independent margin rather than as part of the safety case. The reasoning is the one that runs through the whole of earthing design: the buried copper cannot be degraded by anybody, and a margin that depends on a maintainable item should never be the margin the design needs.
C2 — Changing an \(11\;\mathrm{kV}\) system from resistance to solid earthing. An industrial \(11\;\mathrm{kV}\) system with \(X_1 = X_2 = j0.9\;\Omega\), \(X_0 = j0.6\;\Omega\) is presently earthed through a resistor limiting the earth fault to \(400\;\mathrm{A}\). Solid earthing is proposed. List every consequence, quantified, and state which save money and which cost it.
Show answer
The present resistor. \(3E_a/400 = 47.63\;\Omega\), so \(3R_n = \sqrt{47.63^2-2.4^2} = 47.57\) and \(R_n = 15.86\;\Omega\), dissipating \((400)^2(15.86) = 2.54\;\mathrm{MW}\) while the fault lasts.
Consequence 1 — fault current, ×19.8.
\[ I_{LG}: 400\;\mathrm{A} \longrightarrow \frac{3(6351)}{2.4} = 7939\;\mathrm{A}, \qquad I_{3\phi} = 7057\;\mathrm{A} \]The earth fault now exceeds the three-phase fault by 12.5%, so the switchgear rating of Set 25 must be checked against \(7.94\;\mathrm{kA}\) rather than \(7.06\;\mathrm{kA}\). Costs money if the existing boards are rated at the three-phase value.
Consequence 2 — damage at the fault point, ×394. Arc energy goes as \(I^2t\):
\[ \left(\frac{7939}{400}\right)^{2} = 394 \]This is the reason resistance earthing exists. A \(400\;\mathrm{A}\) stator earth fault burns a hole; a \(7.9\;\mathrm{kA}\) one welds the laminations, and the repair changes from a rewind to a re-core. Costs money, and this consequence alone normally settles the question for a system with rotating machines on it.
Consequence 3 — the earthing grid, ×19.8 on \(E_m\). A grid designed for \(400\;\mathrm{A}\) of grid current has a mesh voltage twenty times too small and would need complete replacement — the largest single item in the whole change. Costs money, heavily.
Consequence 4 — earth-fault relaying. At \(400\;\mathrm{A}\) the system needs sensitive core-balance CTs and settings well below full-load current; at \(7.9\;\mathrm{kA}\) ordinary residual connection of the phase CTs suffices, and grading against downstream devices becomes easy. Saves money, and improves selectivity.
Consequence 5 — arrester rating, \(-17\%\). With \(U_m = 12\;\mathrm{kV}\) and \(U_m/\sqrt3 = 6.93\;\mathrm{kV}\):
\[ \begin{array}{lccc} & k_{\text{eff}} & U_r\ (\mathrm{kV}) & U_{res} = 2.4U_r \\ \hline \text{Resistance} & 1.73 & 12 & 28.8 \\ \text{Solid} & 1.4 & 10 & 24.0 \end{array} \]With tank-mounted arresters (\(1\;\mathrm{m}\) lead, \(10\;\mathrm{kV}\)) the margin against a \(75\;\mathrm{kV}\) BIL improves from \((75-38.8)/38.8 = 93\%\) to \((75-34)/34 = 121\%\). Saves money, though only a little at this voltage.
Consequence 6 — transformer insulation. Solid earthing permits graded insulation, the winding near the neutral being wound with less insulation than the line end. Saves money on new plant; saves nothing on plant already installed.
Consequence 7 — the resistor itself. A \(2.54\;\mathrm{MW}\), \(15.9\;\Omega\) grid bank is deleted. Saves money, both capital and the periodic testing it requires.
Consequence 8 — healthy-phase voltage, from about \(1.7V_{ph}\) down to below \(1.4V_{ph}\). Genuinely beneficial for cable and motor insulation life.
The verdict. On a distribution system of cables and transformers, solid earthing is the right answer and the savings are real. On a system with directly connected motors or generators, consequence 2 dominates everything else and the change should be refused: \(₹2.5\) crore of switchgear and grid work would be spent to make every future stator fault four hundred times more destructive.
C3 — The grid that passed and the yard that was not safe. A design report states \(E_m = 780\;\mathrm{V}\) against a tolerable touch voltage of \(1171\;\mathrm{V}\), a margin of 50%. Two years later a fitter is injured at a structure inside the yard during a line fault. The grid is intact and matches the drawing. Identify the five errors the report is most likely to contain, quantify each, and state the true margin.
Show answer
Error 1 — the primary clearing time was used. The report took \(t_s = 0.2\;\mathrm{s}\), the distance relay's zone-1 time. IEEE 80 requires the time a person can actually be exposed, which is the backup: \(0.5\;\mathrm{s}\). Since the limit varies as \(t_s^{-1/2}\),
\[ E_{\text{touch}}: 1171 \longrightarrow 1171\sqrt{0.2/0.5} = 741\;\mathrm{V} \]The design already fails on this correction alone. Confirm by reading the breaker-failure timer setting.
Error 2 — soil resistivity measured in the monsoon. \(150\;\Omega\mathrm{\cdot m}\) in August against \(400\;\Omega\mathrm{\cdot m}\) in May. \(E_m\) is directly proportional to \(\rho\):
\[ E_m \times \frac{400}{150} = \times 2.67 \]The tolerable value improves slightly, because a higher native \(\rho\) raises \(C_s\): \(C_s = 0.800\) instead of \(0.781\), giving \(755\;\mathrm{V}\) at \(t_s = 0.5\;\mathrm{s}\). A 2% gain against a 167% loss. Confirm by the date on the Wenner survey.
Error 3 — the split factor was assumed and then invalidated. The report took \(S_f = 0.6\), assuming 40% of the fault current returns through the overhead earth wires and the cable sheaths rather than through the soil. If the incoming circuits were later converted to cable with single-point-bonded sheaths, or the earth wire bonds have corroded, \(S_f \to 1.0\):
\[ I_G = S_fI_f: \quad \times \frac{1.0}{0.6} = \times 1.67 \]Confirm by a staged earth-fault test with a current measurement in the earth wires — the only way to establish \(S_f\) honestly.
Error 4 — the decrement factor was omitted. The asymmetrical component of the fault current in the first cycles raises the rms equivalent by \(D_f = 1.05\) to \(1.2\) for short clearing times; take \(1.1\). \(\times 1.10\). Confirm by the \(X/R\) ratio at the bus.
The true margin. Compounding the three multipliers on \(E_m\):
\[ E_m = 780 \times 2.67 \times 1.67 \times 1.10 = 3813\;\mathrm{V} \quad\text{against}\quad E_{\text{touch}} = 755\;\mathrm{V} \]\[ \frac{E_m}{E_{\text{touch}}} = 5.05 \]Not a marginal fail but a factor of five. And note that each individual error looked defensible in isolation: a 10% factor here, a seasonal measurement there. They multiply.
Error 5 — transferred potential, which is not in the formula at all. If the fitter was in contact with anything referenced to remote earth — a water pipe, a telephone pair, a low-voltage supply neutral brought in from outside, or a fence that continues past the grid — the exposure is not the mesh voltage but the entire ground potential rise, here \(I_GR_g\) of several kilovolts. No amount of grid refinement addresses this; it is dealt with by isolating transformers, non-conducting fence sections and insulated pipe joints at the boundary. Confirm by establishing what the injured person was touching.
The general lesson. Every one of the five errors makes the report look better, and none of them is a mistake in the formulae. Earthing calculations fail through their inputs — clearing time, season, split factor, decrement, and the things outside the fence — which is why IEEE 80 spends more pages on choosing \(I_G\) and \(t_s\) than on computing \(E_m\).
Multiple-Choice Questions
MCQ 1. A neutral impedance \(Z_n\) appears in the sequence networks as:
(a) \(Z_n\) in all three (b) \(3Z_n\) in the zero-sequence network only (c) \(Z_n/3\) in the zero-sequence network (d) \(3Z_n\) in all threeShow answer
(b). The physical neutral carries \(3I_0\) while the per-phase zero-sequence network carries \(I_0\), so the same voltage drop requires three times the impedance. Positive and negative sequence currents sum to zero in the neutral, so it is invisible to them. Problem 1.MCQ 2. The Peterson coil tuning condition is:
(a) \(L = 1/(\omega^2C)\) (b) \(L = 3/(\omega^2C)\) (c) \(L = 1/(3\omega^2C)\) (d) \(L = 1/(\sqrt3\,\omega^2C)\)Show answer
(c). The three phase-to-earth capacitances appear in parallel from the neutral, so \(I_C = 3\omega CV_{ph}\); equating to \(V_{ph}/\omega L\) gives the factor of three in the denominator. Option (a) is the answer for a single phase and is the commonest wrong choice. Problem 3.MCQ 3. On an isolated-neutral system the healthy phases during an earth fault reach:
(a) \(V_{ph}\) (b) \(1.4V_{ph}\) (c) \(\sqrt3V_{ph}\) (d) \(2V_{ph}\)Show answer
(c), and sustained, which is why the whole network must be insulated for line voltage to earth. Option (b) is the effectively-earthed value. Problem 2.MCQ 4. A system is effectively earthed when:
(a) \(Z_n = 0\) (b) \(X_0/X_1 \le 3\) and \(R_0/X_1 \le 1\) (c) every neutral is earthed (d) \(X_0 = X_1\)Show answer
(b). Note that (a) is neither necessary nor sufficient: a solidly earthed transformer at the end of a long line can still fail the ratio test, and a small neutral reactor can pass it. Problem 6.MCQ 5. Doubling the area of an earthing grid reduces its resistance by about:
(a) 50% (b) 30% (c) 10% (d) 75%Show answer
(b). The dominant term goes as \(A^{-1/2}\), so doubling the area multiplies it by \(1/\sqrt2 = 0.707\). Answer (a) would require the resistance to go as \(1/A\). Problems 7 and 8.MCQ 6. The tolerable touch voltage uses \(1.5C_s\rho_s\) and the step voltage \(6C_s\rho_s\) because:
(a) the step exposure lasts longer (b) the feet are in parallel for touch and in series for step (c) the current path is longer for step (d) of the \(1\;\mathrm{m}\) spacingShow answer
(b). Each foot is a plate of \(3C_s\rho_s\); hand-to-both-feet puts them in parallel (\(1.5\)), foot-to-foot puts them in series (\(6\)). This factor of four is most of the reason the step check never governs. Problems 9 and 11.MCQ 7. The derating factor \(C_s\) is less than 1 because:
(a) the rock is wet (b) the layer is thin enough that the foot still sees the soil beneath (c) of contact resistance (d) the rock resistivity is uncertainShow answer
(b). \(C_s \to 1\) as \(h_s \to \infty\) and \(C_s \to \rho/\rho_s\)-limited behaviour as \(h_s \to 0\). It is a two-layer field correction, not a safety factor. Problem 10.MCQ 8. An earthing grid design normally fails on:
(a) grid resistance (b) GPR (c) mesh (touch) voltage (d) step voltageShow answer
(c). The step check has both a looser limit (factor 3.3) and a smaller attained value (\(K_s \lt K_m\)), giving about a factor of seven of extra room. The GPR is not a pass/fail criterion at all. Problem 11.MCQ 9. Halving the mesh spacing of a square grid reduces the mesh voltage by about:
(a) 75% (b) 50% (c) 44% (d) 10%Show answer
(c) — from \(1256\) to \(707\;\mathrm{V}\) in Problem 12, a reduction of 44%. Doubling the copper does not halve the mesh voltage because \(K_i = 0.644+0.148n\) rises with the conductor count and gives back most of the gain.MCQ 10. A surge arrester's rated voltage \(U_r\) is chosen from:
(a) the lightning surge magnitude (b) the temporary power-frequency overvoltage (c) the BIL (d) the discharge currentShow answer
(b). \(U_r \ge k_{\text{eff}}U_m/\sqrt3\) — the arrester must not conduct on the healthy-phase rise during an earth fault elsewhere. The BIL is an output of the selection, not an input. Problem 14.MCQ 11. The voltage at a transformer \(20\;\mathrm{m}\) from its arrester exceeds the residual voltage because of:
(a) lead inductance only (b) separation only (c) both, with separation usually larger (d) neither, if the arrester is correctly ratedShow answer
(c). In Problem 18 the lead added \(30\;\mathrm{kV}\) and the separation \(133\;\mathrm{kV}\). Together they turned a 126% margin into 5.8%. Option (d) is the error that produces most failed \(66\;\mathrm{kV}\) installations.MCQ 12. For six circuits, the breaker-and-a-half scheme needs:
(a) 6 breakers (b) 7 (c) 9 (d) 12Show answer
(c) — three diameters of three, i.e. \(1.5n\). Option (a) is the ring or single bus, (b) the sectionalised or double-bus schemes, (d) double-bus-double-breaker. Problem 19.
Key Formulas
| Statement | Relation | Notes |
|---|---|---|
| Earth-fault current | \(I_f = 3E_a/(Z_1+Z_2+Z_0+3Z_n)\) | \(Z_n\) tripled, zero sequence only |
| Charging current at the fault | \(I_C = 3\omega CV_{ph}\) | Isolated or resonant neutral |
| Peterson coil | \(L = 1/(3\omega^2C)\), \(I_L = V_{ph}/\omega L\) | Rating \(S = V_{ph}I_L\), short time |
| Degree of detuning | \(\nu = (I_C-I_L)/I_C = 1-1/(3\omega^2LC)\) | Hold below 5–10% |
| Effective earthing | \(X_0/X_1 \le 3\), \(R_0/X_1 \le 1\) | Earth-fault factor \(\le 1.4\) |
| Earth-fault factor, \(R=0\) | \(\sqrt3\sqrt{k^2+k+1}/(k+2)\), \(k=X_0/X_1\) | \(1.00,\,1.25,\,1.73\) at \(k=1,\,3,\,\infty\) |
| Grid resistance (Laurent–Niemann) | \(R_g = \frac{\rho}{4}\sqrt{\pi/A} + \rho/L\) | Area term dominates, \(\propto A^{-1/2}\) |
| Grid resistance (Sverak) | \(R_g = \rho\left[\frac1L + \frac{1}{\sqrt{20A}}\left(1+\frac{1}{1+h\sqrt{20/A}}\right)\right]\) | Includes burial depth \(h\) |
| Ground potential rise | \(\mathrm{GPR} = I_GR_g\) | Not a pass/fail criterion |
| Grid current | \(I_G = S_fD_fI_f\) | Split and decrement factors |
| Tolerable voltages, 50 kg | \(E_{\text{touch}} = (1000+1.5C_s\rho_s)\,0.116/\sqrt{t_s}\) | \(6C_s\rho_s\) and \(0.157\) for step, 70 kg |
| Surface derating factor | \(C_s = 1 - 0.09(1-\rho/\rho_s)/(2h_s+0.09)\) | Saturates above \(h_s \approx 0.15\;\mathrm{m}\) |
| Attained mesh and step | \(E_m = \rho K_mK_iI_G/L_M\), \(E_s = \rho K_sK_iI_G/L_S\) | Safe when both are below tolerable |
| Irregularity factor | \(K_i = 0.644+0.148n\) | Penalises a finer mesh |
| Arrester rating | \(U_r \ge k_{\text{eff}}U_m/\sqrt3\), \(U_c \approx 0.8U_r\) | \(k_{\text{eff}} = 1.4\) or \(1.73\) |
| Voltage at the equipment | \(V_p = U_{res}+L\,\mathrm{d}i/\mathrm{d}t+(2S/v)\,\mathrm{d}v/\mathrm{d}t\) | \(1\;\mu\mathrm{H/m}\), \(v = 300\;\mathrm{m}/\mu\mathrm{s}\) |
| Protective margin | \(\mathrm{PM} = (\mathrm{BIL}-V_p)/V_p \ge 20\%\) | 15% against switching impulse |
| Maximum separation | \(S_{\max} = \frac{v}{2}\left(\mathrm{BIL}/1.2-U_{res}-L\,\mathrm{d}i/\mathrm{d}t\right)/(\mathrm{d}v/\mathrm{d}t)\) | Governs layout below \(132\;\mathrm{kV}\) |
| Busbar breaker counts | \(n\), \(n+1\), \(1.5n\), \(2n\) | Knee of the curve at \(1.5n\) |
Common Mistakes
Writing \(Z_n\) instead of \(3Z_n\), or putting it in all three networks. It is tripled and it belongs to the zero-sequence path alone. Every earth-fault current in Problems 1, 5 and 20 is wrong by a large factor if this slips.
Adding a neutral resistance arithmetically to the reactances. \(3R_n\) and \(X_0+X_1+X_2\) are in quadrature. In Problem 1 the difference between \(|36+j20| = 41.2\) and \(56\) is 36% of the answer.
Using \(L = 1/(\omega^2C)\) for the Peterson coil. The missing three makes the coil three times too large and leaves two-thirds of the charging current uncompensated — Problem 3.
Treating a Peterson coil as a fixed component. Its correct value is a function of the switching state; a quarter of the feeders out leaves a 33% residual — Problem 4.
Believing that a low ground potential rise proves the yard is safe, or that a high one condemns it. The GPR lifts everybody together; only the differences \(E_m\) and \(E_s\) matter. Problem 11's yard sits at \(7.3\;\mathrm{kV}\) and fails by \(430\;\mathrm{V}\).
Trying to fix a mesh-voltage failure by adding conductor for the resistance's sake. Resistance responds to area and safety responds to spacing. In Problem 7 the conductor term is 9% of \(R_g\); in Problem 12 the same copper moves \(E_m\) by 44%.
Assuming that halving the mesh pitch halves the mesh voltage. \(K_i\) grows with the conductor count and returns most of the gain; the true factor in Problem 12 is \(0.562\), not \(0.5\).
Using the primary clearing time in \(E_{\text{touch}}\). IEEE 80 wants the backup time, because that is how long a person can actually be in the circuit. Problem 13 and Challenge C3 both turn on it.
Designing to the 70 kg body criterion. It is 35% more permissive and can only be justified where access is provably restricted — Problem 9.
Quoting the protective margin as \(\mathrm{BIL}/U_{res}\). The margin is computed at the transformer with the lead and separation terms included. In Problem 18 the two figures are 126% and 5.8%.
Forgetting that the earthing method sets the arrester rating. Non-effective earthing raises \(k_{\text{eff}}\) from \(1.4\) to \(1.73\), and \(U_{res}\) follows \(U_r\) — Problems 6, 15 and 20.
Assuming a three-phase fault is always the worst. With \(X_0 \lt X_1\) at a solidly earthed transformer terminal, the earth fault exceeds it — by 12.5% in Problem 1 and 36% in Problem 5.
The substation is now specified end to end: a neutral earthing method that fixes the earth-fault current, a buried grid whose mesh spacing has been argued down to \(4\;\mathrm{m}\) to hold the touch voltage, an arrester whose rating followed from the earth-fault factor, and a transformer whose BIL exceeds the voltage at its bushings by half. Every one of those numbers moved when any other did, which is the point Chapter 37 exists to make.
Set 39 turns to the one substation this chapter has not described. A converter station terminates an HVDC link, and almost nothing in Part 8 transfers to it unchanged: there is no earth-fault factor because there is no power-frequency neutral to displace, the harmonics that a delta tertiary traps here are generated deliberately there, and the valve hall's insulation coordination is against a d.c. stress that redistributes itself through the insulation over minutes rather than microseconds. What does transfer is the method — one chain of decisions, iterated until it closes.