Protective Relaying

The pilot loop. The current goes out and comes back, so the conductor length is twice the route length:

A resistance of \(6.88\) mΩ per metre of cable is worth memorising for \(2.5\) mm² copper; it turns a pilot run into ohms in one multiplication.

The relay burden in ohms. A volt-ampere rating is quoted at rated secondary current, so

The same \(2.5\) VA relay on a \(1\) A CT would be \(2.5\) Ω — twenty-five times as much. The current rating of the secondary matters more than the VA number.

(i) The total loop.

Forty-one volt-amperes, of which the relay is \(2.5\) and the pilot cable is \(31.0\). On a \(5\) A secondary the cable is the burden and the relay is a rounding error.

(ii) The knee-point voltage. The CT must hold its accuracy to twenty times rating, so the core must be able to generate the emf that drives \(20\times5 = 100\) A through \(1.638\) Ω:

Specify the next standard core above this — \(V_k = 180\) V or \(200\) V — never the value itself, because \(R_{ct}\) rises with temperature and the pilot run is rarely the length the drawing claims.

(iii) The volt-amperes at the limit.

A device described on its nameplate as a \(41\) VA CT has to behave as a \(16\) kVA transformer for a few cycles. That is the whole reason a protection core is physically large, and the whole reason it saturates when the estimate of the loop resistance is optimistic.

The effective accuracy limit factor is the knee-point formula solved for ALF instead of for \(V_k\):

The nameplate says \(5\mathrm{P}20\), and in this installation the CT is a \(5\mathrm{P}17\). The nameplate figure is quoted at the CT's rated burden; it is a property of the CT and the burden together, never of the CT alone.

The saturation current in primary terms:

A \(20.5\) kA symmetrical fault. On a \(132\) kV bus with a \(31.5\) kA switchgear rating that is not comfortable, and Problem 3 shows that the DC offset destroys the margin entirely.

Route one — shorten the pilot run. For \(\mathrm{ALF} = 20\) the whole loop must not exceed

The panel would have to move \(17\) m closer to the switchyard, which in a substation already built is not a route at all.

Route two — upsize the cable, which is the one that gets used:

Going from \(2.5\) mm² to \(4\) mm² buys back \(8\) kA of saturation-free fault current for the price of a drum of cable. CT circuits are routinely wired in \(4\) or \(6\) mm² for no reason other than this.

The symmetrical requirement. The secondary current at the fault is

By this measure the \(140\) V core passes: \(140 > 122.9\), consistent with Problem 2's saturation current of \(20.5\) kA being above \(18\) kA. It is also the wrong measure.

The DC offset. A fault initiated at voltage zero carries a unidirectional component that drives the flux one way for as long as it lasts, so the flux demand is not the peak of a sine but the integral of a decaying exponential added to it. The transient dimensioning factor accounts for the worst case:

One thousand eight hundred volts against one hundred and forty. The core actually installed is a factor of \(13.2\) too small to hold accuracy through a fully offset fault.

The time constant tells you how long the problem lasts:

Rather more than two cycles, so the offset is still at \(37\%\) of its initial value when a zone 1 distance element is trying to make its decision, and at \(11\%\) when a graded overcurrent relay finally operates.

What the installed core will do. It will saturate, and it will saturate early — within the first half cycle for a fully offset fault. The consequences depend entirely on which relay is behind it:

The third is the dangerous one, and it is why busbar and generator differential schemes are the applications that get fully transient-rated TPX, TPY or TPZ cores while ordinary feeder overcurrent does not.

What is done instead. A \(1843\) V core for a feeder CT is not economic — it is a physically different device, several times the volume. Two things are done. The burden is reduced, which scales \(V_k\) directly: replacing the pilot cable as in Problem 2 and moving to a numerical relay of \(0.1\) Ω takes the loop to \(1.174\) Ω and the requirement to \(15\times75\times1.174 = 1321\) V. And the relay algorithm is made saturation-tolerant, using the undistorted portion of each cycle rather than the whole of it — something a numerical relay can do and an induction disc cannot.

What the residual lead carries. The three secondaries share a return, so the current in that return is the phasor sum:

This is Chapter 22's zero-sequence definition executed in copper, with no relay and no arithmetic. Set 21's transformation is being computed by three wires meeting at a point.

(i) Balanced load, 340 A. A balanced set sums to zero by construction:

Each phase relay carries \(340/400 = 0.85\) A; the residual relay carries nothing. In practice a few tens of milliamperes of CT-error residual appear, which is why the earth-fault setting is \(10\)–\(20\%\) rather than \(1\%\) on a star group.

(ii) Phase-to-phase fault, 2400 A between B and C. The current in B is the negative of that in C, and A carries nothing:

A \(2400\) A fault, and the earth-fault relay does not stir. That is the point of the residual connection: it is a filter that is blind to everything except an earth path.

(iii) Single-line-to-earth fault, 1800 A in phase A.

A multiple of thirty on a relay whose pickup is \(60\) A — one sixth of the load the feeder is carrying at the time.

Why a phase relay cannot be set there. A phase relay carries load current, so its pickup must sit above the maximum load the feeder will ever draw, including cold-load pickup:

The phase relay's pickup is seven times the earth relay's, because the phase relay measures a quantity that load produces and the earth relay measures one that load cannot. A \(150\) A high-resistance earth fault is invisible to the first and gives the second a multiple of \(2.5\).

The lower bound on the pickup. A feeder relay is set at about \(1.3\) times full load, to clear the load, the tolerance and the cold-load pickup that follows a restoration:

Cold-load pickup is the reason for the margin, not relay tolerance. When a feeder is re-energised after an outage every thermostatic load in the area starts together, and the first few seconds can draw twice the normal maximum.

Convert the available plugs to primary amperes, which is the only form in which the constraint can be checked:

Adopt \(125\%\): the smallest plug that clears the load constraint, because every step upward lengthens every operating time on the curve (Problem 8).

The upper bound, which must also be checked and usually is not:

Comfortably above unity, so the relay will certainly operate on the smallest fault it must see. Since \(t\to\infty\) as \(M\to1\), a multiple below about \(1.5\) is useless however generous the TMS.

The PSM, both ways, as the arithmetic check every settings sheet should carry:

They agree, as they must — the CT ratio cancels. Disagreement means a factor of the ratio has been applied twice or not at all, which is the single commonest arithmetic error in the subject.

The operating time. On the IEC standard-inverse characteristic,

The quantity \(0.14/(M^{0.02}-1) = 4.980\) is the relay's time at \(\mathrm{TMS} = 1\), and it is worth carrying separately: every grading calculation in Problem 9 is a division by it.

Check the second multiple first.

A current ratio of \(15/4 = 3.75\) between the two faults. Everything that follows is about how much time each curve can make out of that ratio.

Evaluate all four at \(M = 4\):

At a modest multiple the four are within a few tens of per cent of each other — except long-time inverse, which is eight times slower and is a motor-thermal curve, not a feeder curve.

Evaluate all four at \(M = 15\):

Now the family has separated completely. Extremely inverse is seven times faster than standard inverse at the same TMS and the same pickup, on the same fault.

The time separation each curve extracts from the current ratio \(3.75\):

Standard inverse, with its exponent of \(0.02\), is very nearly flat: a fourfold current change buys a factor of two in time. Extremely inverse buys a factor of fifteen. When fault current varies strongly along a feeder, the steeper curve turns that variation into grading margin for free.

Which to choose against a fuse. An HRC fuse melts on \(I^2t\), so its time–current curve falls as \(1/I^{2}\). Only one IEC characteristic has that shape:

Choose extremely inverse. Matching the shape means the margin between relay and fuse stays roughly constant over the whole fault range instead of closing at one end; a standard-inverse relay graded against a fuse at \(M = 4\) will cross it somewhere above \(M = 15\) and lose selectivity exactly where the fault is largest.

Separate the curve from the multiplier. The characteristic factorises into a shape term that depends only on \(M\) and a scale term that is the TMS:

\(t_1(M)\) is the operating time at \(\mathrm{TMS}=1\). Every inverse problem in this sheet is one evaluation of \(t_1\) followed by one multiplication or one division.

Evaluate the shape term.

The \(0.02\) exponent makes the denominator a small difference of numbers close to one, so carry six figures through this step. Rounding \(6.5^{0.02}\) to \(1.038\) shifts the answer by \(0.4\%\); rounding it to \(1.04\) shifts it by \(4.6\%\).

Solve and round.

Nine milliseconds slower than required, and that is the correct direction.

Why the rounding is always up. The \(0.45\) s came from a downstream time plus a grading margin, and that margin was assembled from four physical contributions — breaker time, relay overshoot, CT and relay tolerance, safety allowance — each of which is already a minimum:

Rounding to the nearest step would have chosen \(0.125\) here anyway, which is exactly why the rule is stated as a rule: it must hold when the computed value is \(0.1249\) as well as when it is \(0.1226\). A grading margin lost to rounding is invisible until the day two breakers trip together.

The same relay at a smaller multiple. Fixing the TMS fixes the whole curve, so the time at any other fault follows:

Halving the multiple has not doubled the time — it has multiplied it by \(1.72\). That is the standard-inverse curve's gentleness, and it is why a chain of SI relays grades smoothly but accumulates time slowly rather than sharply.

Before: plug 100%.

After: plug 150%.

The relay has become \((0.646-0.540)/0.540 = 19.6\%\) slower for a fault that has not changed at all. Nobody touched the time multiplier.

The TMS that restores the original time.

Available steps do not permit an exact restoration, which is usual; \(0.565\) s is \(25\) ms slow rather than \(106\) ms slow.

What that costs. The TMS scales the whole curve, so restoring the time at one fault changes it at every other fault, and the relay below is graded at a different fault:

The relay has been made faster at low multiples — which is precisely where it is acting as backup for the relay downstream, and where a grading margin of \(0.3\) s was checked. A margin computed before this change is no longer valid.

The general statement. Plug setting and time multiplier are not independent knobs. Raising the plug moves the whole curve to the right in current, so at any fixed fault current the relay operates at a smaller multiple and therefore takes longer; lowering the TMS to compensate moves the whole curve down in time, which changes the operating time at every fault, not the one you were looking at. Any change to either setting invalidates the grading study and it must be redone from the load end.

Plug settings first, in primary amperes. Take \(100\%\) on each relay:

Each pickup is \(1.25\) times the load on its own circuit, and the most onerous sensitivity check — R3 seeing the \(2800\) A feeder-end fault as a multiple of \(2800/600 = 4.67\) — is comfortably above unity. Both constraints hold, so no plug needs raising.

R1, the most downstream relay, evaluated at F1, the fault at the far end of its own section (\(2800\) A):

Nothing constrains R1 from below, so it takes the smallest useful TMS the relay offers. Every other time in the study is built on this one.

R2, graded against R1 at the common fault F1. This is the step students get wrong: the comparison is made at the fault both relays see, not at each relay's own section end.

R2 sees half the multiple R1 does, because its CT and pickup are both twice as large. That is grading falling out of the characteristic itself: the upstream relay is automatically slower even before any TMS is chosen.

R2's own duty, at F2, the fault at bus R (\(4500\) A):

R3, graded against R2 at the common fault F2:

The settings are fixed. The complete table evaluates every relay at every fault it can see:

Bold entries are primary clearance; every other entry is graded backup. R3 at F1 takes \(1.119\) s, which is \(0.506\) s above R2 — far more than the minimum, as always happens when a relay backs up a fault two sections away.

Now read the last two rows, because they contain the result the study exists to expose:

The most severe fault in the entire system, at the bus where the incoming transformer and the switchgear are, is cleared two and a half times more slowly than the mildest one. That is the structural weakness of graded overcurrent, and Problems 10 and 11 are its two standard remedies.

The setting rule. The element must never see a fault beyond bus Q, or it would trip R3 instantaneously for a fault that belongs to R2 and destroy the grading just computed. The \(1.3\) factor covers DC offset in the measured quantity and the difference between the fault study and the network on the day:

In relay terms:

High-set elements are always quoted as a multiple of rated current for this reason; a setting of \(1667\%\) would be true and unreadable. Numerical relays accept up to about \(30 I_n\), and thermal withstand of the input circuit, not the algorithm, is what limits it.

How far it reaches requires the fault-current profile along P–Q, which the two given fault levels determine. Working per phase at \(33\) kV:

The source is \(1.67\) times stiffer than the line section it feeds, which is typical of a distribution feeder and is exactly the situation that limits a high-set element.

Solve for the fraction \(x\) of the section at which the current falls to the setting:

The instantaneous element covers the first 33% of section P–Q. Beyond that point the graded curve of Problem 9 does the work, untouched.

What it buys. For the \(12\) kA fault at bus P — the fault the whole study identified as the badly protected one:

A twenty-eight-fold reduction in fault energy at the busbar, for the cost of one setting. Nothing else in a grading study offers that.

When it is worth fitting. The reach depends entirely on the ratio of section impedance to source impedance. Repeating the calculation for a longer section:

The result inverts the intuition. It is not the long feeder that suits a high-set element but the one whose fault level falls sharply from one end to the other relative to the source — and on a short feeder off a strong source, where the current at both ends is almost the same, the setting rule leaves no reach at all.

R1 on very inverse at F1. The characteristic is \(t = \mathrm{TMS}\times13.5/(M-1)\):

A very-inverse relay needs a TMS three times larger to give the same time at a high multiple — the curve sits much lower, so it must be scaled up further. TMS values are not comparable across characteristics.

R2, graded at F1:

R2 at F2, then R3 graded there:

The comparison at bus P, which is what the regrading was for:

The severe fault is now cleared in a quarter of a second instead of well over half. The fault energy at the busbar falls by \(56\%\), and the clearance is now inside the critical clearing times of Chapter 29 rather than outside them.

What it costs, which is at the other end of the curve:

Very inverse trades backup speed at low multiples for primary speed at high ones. That is the correct trade here — the slow entry is a backup time on a fault two sections away, which nobody is relying on for stability — but it is a trade and not a free improvement.

The condition for the trade to pay. The steeper curve extracts more time separation from the same current ratio, so it pays whenever fault current changes strongly along the feeder. Here the ratio between bus P and the feeder end is \(12{,}000/2800 = 4.3\), which is large; on a feeder fed from a very stiff source, where that ratio might be \(1.2\), the very-inverse relays would need almost the same TMS as the standard-inverse ones and nothing would be gained.

Forward fault. Substituting directly:

Ninety-one per cent of the available torque, with the operating point \(65^\circ\) clear of the boundary. The element is decisively in its operating region.

The same fault seen from behind. Reversing the direction of current flow through the relay reverses the current phasor, adding \(180^\circ\):

Equal and opposite. That symmetry is exact and does not depend on \(\tau\): whatever the MTA, a reverse fault is always exactly as far into the restraining region as the corresponding forward fault is into the operating one, because \(\cos(\phi+180^\circ) = -\cos\phi\).

The angular limits of the forward zone. Torque is positive over a half plane:

A width of \(180^\circ\), bounded by a straight line through the origin at right angles to the MTA. Any plausible fault angle for a network — anywhere from \(15^\circ\) on a resistive distribution feeder to \(85^\circ\) on a transmission line — falls inside it with room to spare.

What an exact MTA would buy:

Ten per cent of torque, which is worth nothing. The reason the MTA is not simply set to the network angle is that \(\theta\) is not a fixed number: it varies with fault type, with the arc resistance in the fault, with the proportion of cable to overhead line in the path, and with how much of the impedance is transformer. A setting of \(45^\circ\) keeps every one of those cases within \(0.7\) of maximum torque, and the boundary — not the maximum — is what actually matters.

Where the boundary does matter. The one case worth checking is a fault with substantial arc resistance on a distribution feeder, where \(\theta\) can fall to \(20^\circ\) or below:

Even a nearly resistive fault stays at \(91\%\). A directional element fails from loss of polarising voltage, not from being a few degrees off its maximum — which is what Problem 13 is about.

The connection, algebraically. With a balanced set \(V_a = V\angle0^\circ\), \(V_b = V\angle{-}120^\circ\), \(V_c = V\angle120^\circ\):

Quadrature with \(V_a\), and \(\sqrt3\) times larger. That \(90^\circ\) displacement names the connection; whether it is written as a lead or a lag depends only on whether \(V_{bc}\) or \(V_{cb}\) is taken, and the relay's characteristic angle is defined to absorb it either way. What matters physically is that the phase-A current element is polarised from the two phases the A-phase fault does not touch.

(i) A solid A-to-earth fault at the relay terminals. This is the case that decides the connection:

The phase voltage of the faulted phase is exactly the one quantity guaranteed to be destroyed by the fault the relay is trying to see. The two healthy phases are untouched, and their line voltage is at full rated secondary value.

(ii) Three-phase faults, where all three voltages collapse together. The relay-point phase voltage is the divider between source and line:

A three-phase fault collapses every voltage in proportion, so the \(90^\circ\) connection buys only the factor \(\sqrt3\) here — real, but a factor and not a rescue. At \(8\) km both quantities are usable; at \(0.5\) km neither is.

Where the boundary falls. An electromechanical directional element needs about \(1\)–\(2\) V of polarising voltage to develop reliable torque; a numerical relay needs a phasor estimate it can trust, which in practice means a few volts:

Inside about \(700\) m of the relay, a bolted three-phase fault leaves no usable polarising quantity of any kind — a genuine dead zone.

Memory polarising is the answer, and it is unavoidable for exactly this case. The relay stores the pre-fault positive-sequence voltage phasor and continues to rotate it at system frequency for a few cycles after the collapse:

The stored phasor has full pre-fault magnitude and the correct phase, so direction can be determined for a bolted three-phase fault at the relay terminals — something no live measurement can do. Its limitation is duration: after a few cycles the memory has drifted in frequency and must be abandoned, which is acceptable because a close-in three-phase fault is cleared in one or two cycles anyway.

Rated currents on both sides.

Their ratio is \(874.77/131.22 = 6.667 = 220/33\), as it must be. Checking that is the cheapest error trap in the whole calculation to close.

The delta connection multiplies by \(\sqrt3\). Three CT secondaries in delta deliver the difference of two phase currents, so what reaches the relay is

This is the step most often forgotten, and forgetting it produces a mismatch of \(73\%\) of rated current rather than a few per cent. The delta is there to cancel the \(30^\circ\) of the Yd11 vector group and to trap zero-sequence current from an external earth fault on the star winding; the \(\sqrt3\) is a side effect that must be paid for in the ratio.

The LV side, star-connected, passes the phase current straight through:

Take this as \(1\) per unit. Every quantity in Problems 15 and 16 is referred to it.

The ideal HV ratio is the one that makes the two equal:

Not a standard ratio, which is the normal situation — this is why interposing CTs, relay taps and software scaling constants all exist.

Test the available ratios, expressing the mismatch as a fraction of the mean of the two secondary currents, since the mean is exactly what feeds the bias circuit:

Adopt \(250/1\). The residual mismatch is \(3.85\%\) — small, permanent, and present at every load and every through fault.

Add the tap changer. The CTs were matched at the nominal tap; at the extremes of a \(\pm10\%\) range the true transformer ratio departs from the matched value by that much, and the relay sees the departure as difference current:

The two add in the worst case because nothing constrains their signs to oppose.

The minimum first bias slope. The relay must restrain when it sees this mismatch at any through current, and the errors of the CTs themselves must be allowed on top:

Which is why the standard first slope is \(20\)–\(30\%\) and not something smaller. The number is not a convention: it is the ratio mismatch, the tap range and the CT class added up, and a transformer with a \(\pm15\%\) tap range needs a correspondingly steeper first slope.

What a numerical relay does instead. With the delta and the ratio correction both done in software, the CT ratio can be chosen for the switchgear rather than the relay:

The mismatch is reduced to the accuracy of a floating-point constant, and the first slope then need only cover the tap changer and the CT class — which is how modern schemes reach \(20\%\) rather than \(30\%\).

Write the characteristic once and evaluate it thereafter:

Above \(I_b = 4\) the fixed part is \(0.20+0.25\times3 = 0.95\) pu, so \(I_{op} = 0.95+0.70(I_b-4)\) — the form used in every part below.

(i) Internal three-phase fault at the LV terminals. With no source on the LV side, all the fault current enters from HV and none leaves:

\(10.0 \gg 1.65\): the relay operates with a factor of \(6.1\) in hand. Note that \(I_{bias}\) is half the fault current, not the fault current — a single-end infeed puts the operating point on the line \(I_{diff} = 2I_{bias}\), which is where Problem 16 begins.

(ii) Through fault of 12 pu with a 7% CT error. The two terminal currents are \(12.00\) and \(0.93\times12 = 11.16\) pu:

\(0.84 \ll 6.26\): the relay restrains, with a factor of \(7.4\) of margin. This is the whole purpose of the second slope — the threshold has grown from \(1.65\) to \(6.26\) pu precisely because the through current, and therefore the CT error, has grown.

(iii) Add the tap changer. A \(+10\%\) tap and a \(7\%\) CT error both produce genuine difference current, and nothing forces their signs to oppose:

Still stable, by a factor of \(2.9\). The two worst error sources acting together at the worst through current is the condition a bias setting must survive, and it is the one to check.

The largest tolerable CT error. Let the error be a fraction \(e\) of the true current at \(12\) pu:

Check: \(I_{diff} = 4.85\), \(I_{bias} = 9.57\), threshold \(0.95+0.70(5.57) = 4.85\) — equal, as required. A CT would have to lose forty per cent of the current before the scheme maloperated, which is a deeply saturated CT and not a merely inaccurate one.

The operating point. Inrush enters one winding and leaves nowhere — it is magnetising current, not through current:

A factor of \(6.9\) inside the operating region, which is more than the internal fault of Problem 15 had at its own bias current. On magnitude alone the two events are indistinguishable, because on magnitude alone they are the same event.

Why no slope can help. Because the current enters one side and leaves nowhere, the two quantities are locked together:

The inrush locus is a straight line through the origin of slope \(200\%\). Any bias characteristic with a slope \(k \le 1\) lies below it everywhere, so the locus is inside the operating region at every point along its length.

Confirm it by decay. Inrush decays over several seconds; track the operating point down:

The relay does not fall out of the trip region until the inrush has decayed below the minimum pickup \(I_s\) itself — several seconds after energisation. Waiting it out is not an option, and raising \(I_s\) above the inrush would destroy the sensitivity to internal faults.

The discriminant is the waveform. Saturation clips the current for part of every cycle, so inrush is strongly distorted, while a genuine fault current is close to sinusoidal:

The conventional \(15\%\) sits nearer the inrush end of the gap deliberately: failing to block on inrush is a nuisance trip on every energisation, while failing to trip on a fault whose second harmonic happens to be high is prevented by a separate unrestrained high-set differential element set well above any possible inrush.

Two refinements that follow. Inrush is not equal on the three phases — the switching instant is favourable for one and unfavourable for another — so one phase may show only \(8\%\) second harmonic while the others show \(20\%\). Cross-blocking therefore blocks all three phases when any one detects second harmonic. And a transformer run at high volts per hertz saturates too and also draws excess magnetising current, but its distortion is fifth harmonic rather than second, so overexcitation is blocked by a separate element:

The unrestrained element is the security release valve: above a current no inrush and no through fault can reach, the relay trips without consulting any harmonic or any bias.

The line impedances, in primary ohms.

All at \(80^\circ\). The angle never enters the reach calculation — a mho characteristic is set by its diameter along the line angle, and the angle only matters when load encroachment is checked in Problem 19.

The conversion factor, computed once and applied at the end:

Do all the engineering in primary ohms — a reach is a distance along a line, and a distance is a thing you can check. Convert once, at the last step, only because that is the unit the relay's keypad accepts.

Zone 1 — \(80\%\) of the protected line, instantaneous:

The \(20\%\) shortfall is the price of never over-reaching. There is no time delay to correct an over-reach with, so the errors of the CT, the VT, the relay and the line impedance data must all fit inside that margin.

Zone 2 — the line plus half the shortest adjacent line:

Both bounds on zone 2, which is the part of the question that carries the marks. The lower bound covers what zone 1 gave up:

The upper bound stops it reaching past the zone 1 of the shortest line leaving bus B, which itself reaches \(80\%\) of that line:

Both hold, with \(4.8\) Ω of room below and \(8.6\) Ω above. If they had not — a very short adjacent line makes the window vanish — the remedy is not to violate one of them but to add a teleprotection channel, which turns zone 2 into an instantaneous trip without changing any reach.

Zone 3 — remote backup, the line plus \(120\%\) of the longest adjacent line:

The \(120\%\) here is the opposite of zone 1's \(80\%\), and for the opposite reason: a backup zone must certainly cover the whole of what it backs up, and the time delay makes over-reach harmless. Zone 3 is the setting most likely to encroach on load, which Problem 19 checks.

The complete schedule:

Enter the secondary column; keep the primary column on the drawing, because it is the one that can be checked against a map.

The apparent impedance with infeed. The relay at A measures its own voltage over its own current, but the drop along the faulted section beyond bus B is produced by the total current:

The infeed factor \(K\) multiplies only the section beyond the infeed point. Nothing multiplies \(Z_{AB}\), which is why zone 1 is untouched: there is no infeed inside the protected line.

The infeed factor:

The actual reach. Let the fault be \(d\) km along B–C, so \(Z_{Bx} = 0.32d\):

Zone 2 has shrunk from \(50\%\) of line B–C to \(18/90 = 20\%\). The extra current from B makes the fault look further away than it is, so infeed always causes under-reach and never over-reach — a reassuring direction of error, and the reason the effect is tolerated rather than eliminated.

The setting that would restore 45 km:

And why it must not be used. Test it in the condition the setting must also survive — the source at B switched out, so \(K = 1\):

The relay at A would trip in \(0.35\) s for faults two lines away — an unselective operation on a network that has already lost a source, which is exactly when it can least afford one. A reach must be set for the weakest infeed condition and the reduced coverage accepted.

What is done about the shortfall. The purpose of zone 2 is not really to cover half of the adjacent line — it is to cover the last \(20\%\) of the protected line that zone 1 gave up, and infeed does not touch that:

The genuine remedy for the last \(20\%\) of A–B is a teleprotection channel: the relay at B sees the fault in its own zone 1 and sends a permissive signal to A, which converts A's zone 2 into an instantaneous trip without changing any reach at all. On a \(220\) kV line that channel is standard equipment.

The apparent load impedance. A relay measures phase voltage over phase current, whatever produced them:

A large impedance at a small angle — the signature of load. A fault, by contrast, is a small impedance at a large angle. Those two facts are the whole basis on which a distance relay tells them apart.

The mho circle's reach in the load direction. A mho characteristic is a circle through the origin whose diameter lies along the line angle, and the chord along any other direction is shorter by the cosine of the angle between:

The circle's \(51.46\) Ω of reach along the line angle has become \(28.08\) Ω in the direction load actually lies. That shrinkage in the load direction is the whole reason the mho shape exists.

The margin:

Adequate, though not generous. A margin of \(1.5\) is usually taken as the minimum, and it is checked at exactly this condition — maximum load at depressed voltage — rather than at nominal voltage, because both effects push the load point inward at once.

The current at which zone 3 would operate. Invert the load impedance relation at the reach value:

Seventy per cent more load, at the same depressed voltage and the same power factor, and the backup zone of a healthy line trips it. Since a system disturbance both depresses voltage and redistributes power onto surviving lines, this is not a hypothetical combination — it is the standard mechanism by which a local fault becomes a regional blackout.

What a plain impedance characteristic would have done. That characteristic is a circle centred on the origin, so its reach is \(Z_3\) in every direction:

The same reach, the same line, the same load, and the relay trips a perfectly healthy circuit. It is also non-directional, so it would trip for faults behind the relay as well. Both defects come from the same source: a shape that responds to \(|Z|\) alone discards the angle, which is the only information distinguishing load from fault.

The quadrilateral, and why numerical relays use it. Resolving the load point into its components:

A quadrilateral with a reactive reach of \(51.46\) Ω and a resistive reach set to \(20\) Ω excludes the load by a factor of \(2.2\) in the \(R\) direction while covering more arc resistance for genuine faults than the mho circle does. Setting the two reaches independently is exactly what a rotating disc could not do and a processor can, and it is why every modern transmission distance relay offers the shape.

Clearance time is relay time plus breaker time, always — the fault current persists until the arc is finally extinguished, not until the trip contact closes:

The let-through energy, which is what actually damages plant:

A factor of \(11.4\) between the fastest and the slowest, for the same fault on the same plant. Time, not current, is the variable the protection engineer controls.

Convert to cable size, which turns the abstraction into a purchase order. The adiabatic equation for a cable's short-circuit withstand is \(I^2t = k^2S^2\):

If the LV connection is a cable, the difference between the fastest and the slowest protection is the difference between a \(25\) mm² and a \(70\) mm² conductor — nearly three times the copper, on every phase, for the whole run.

What the comparison decides — and what it does not. The cable must withstand the slowest protection that could clear the fault, not the fastest, because backup exists precisely for the occasions when the fastest does not operate:

Sizing on the differential's \(0.080\) s would leave a cable that survives every fault the differential clears and none that it does not, which converts one relay failure into a burnt cable. This is the arithmetic behind the rule that plant is rated on backup clearance time.

Where the differential's speed does pay. Not in the cable size, but in three other places:

The stability entry is usually decisive on a transmission transformer. Chapter 29's critical clearing times run to a few hundred milliseconds at best, so a \(0.911\) s clearance is not slow protection — it is a loss of synchronism.

The resulting scheme, which is what every transformer of this size actually has:

Four layers, three of which are expected never to operate. That is not redundancy for its own sake: each layer covers a failure mode the one above cannot, and the ordering is exactly the dependability-versus-security trade of Section 36-1 resolved in favour of dependability, as transmission practice requires.