Supply Systems and Conductor Economics

Set 2 — Supply Systems and Conductor Economics

Twenty worked problems on the question that decided the shape of every power system built since 1890: for a given power delivered over a given distance at a given loss, how much conductor does each possible arrangement require? The answer turns out to depend on the square of the voltage and the square of the power factor, and on very little else. Everything that follows — the choice of three phases over two, the choice of 400 kV over 132, the whole economics of the transmission network — is a consequence of that single scaling law.

Textbook Chapters 1–2 · 20 solved · 12 practice · 3 challenge · 12 MCQs

i Method Recap

Fix what is held constant before comparing anything. Every comparison in this set holds four things fixed — power delivered, distance, power loss, and either the voltage between conductors or the maximum voltage to earth. Change which voltage is held fixed and every ratio in the table changes. Most wrong answers here are right answers to a different question.
The master relation. For \(n\) conductors each of area \(a\) carrying \(I\), the loss is \(W = nI^2\rho l/a\), so \(a = nI^2\rho l/W\) and the volume of metal is \(nal = n^2I^2\rho l^2/W\). Everything else is arithmetic on this one line.
Volume varies as \(1/V^2\). Double the voltage and the current halves, the required area quarters, and three quarters of the metal disappears. This is the entire argument for high-voltage transmission, and Problem 1 is nothing more than this observation applied twice.
Volume varies as \(1/\cos^2\phi\). A poor power factor costs conductor material by exactly the same square law, which is why power-factor correction and voltage selection are two forms of the same economy.
On equal line voltage, three-phase three-wire needs three quarters of the metal of single-phase two-wire. The current falls by \(\sqrt3\) so the area falls by 3, but there are three conductors instead of two: \(3 \times \tfrac13 \div 2 = \tfrac34\). Problems 5 and 20 are the same calculation on different numbers.
Kelvin's law. Annual charges are \(P_1 + P_2a\) for the conductor and \(P_3/a\) for the energy lost in it. The total is least when \(P_2a = P_3/a\), that is \(a = \sqrt{P_3/P_2}\) — the annual cost of losses equals the annual charge on the part of the capital cost that varies with area.
A uniformly distributed load is not a load at the far end. Fed at one end the drop is \(\tfrac12 i l^2 r\) — half what a lumped load would give. Fed at both ends the maximum drop, at the centre, is \(\tfrac18 i l^2 r\) — a quarter of that again.

VideoWalkthrough

Problem 1Warm-upVoltage and Copper

What is the percentage saving in feeder copper if the line voltage in a two-wire d.c. system is raised from 200 V to 400 V, for the same power transmitted over the same distance with the same power loss?

Solution

The same power is delivered in both cases, so the current falls in inverse proportion to the voltage:

\[ P = V_1I_1 = 200I_1, \qquad P = V_2I_2 = 400I_2 \quad\Rightarrow\quad I_2 = \tfrac12 I_1 \]

The power loss is also the same. With two conductors in each case:

\[ 2I_1^{2}R_1 = 2I_2^{2}R_2 \quad\Rightarrow\quad \frac{R_2}{R_1} = \left(\frac{I_1}{I_2}\right)^{2} = 4 \]

Resistance is inversely proportional to area, so the new conductor may be four times thinner:

\[ \frac{a_2}{a_1} = \frac{R_1}{R_2} = \frac14 \]

Both lines are the same length, so the volume of copper is in the same ratio as the area: \(v_2/v_1 = 0.25\).

Hence the saving:

\[ \%\text{ saving} = \frac{v_1 - v_2}{v_1}\times100 = (1 - 0.25)\times100 = 75\% \]

Conductor volume varies as \(1/V^2\), and this single fact built the transmission network. Doubling the voltage removes three quarters of the metal; raising 11 kV to 400 kV removes 99.92% of it. What limits the voltage is not the conductor but everything around it — insulator strings, tower clearances, corona loss and switchgear, all of which grow with voltage while the copper shrinks. The economic voltage of Problem 14 is where those two opposing costs cross.

Answer\(75\%\) saving in copper

Problem 2Exam levelSystem Conversion

A d.c. two-wire system is to be converted into a three-phase, three-wire a.c. system by adding a third conductor of the same cross-section as the two existing ones. Calculate the percentage additional load that can then be supplied, if the voltage between wires and the percentage line loss both remain unchanged. Assume a balanced load at unity power factor.

Solution

Let \(R\) be the resistance of one conductor — the same in both cases, since the third conductor is identical to the other two and none is replaced.

Percentage power loss is what is held constant, not the absolute loss. For the d.c. two-wire system:

\[ \%\text{P.L.} = \frac{2I_1^{2}R}{VI_1}\times100 \]

For the three-phase three-wire system with the same voltage \(V\) between wires and unity power factor:

\[ \%\text{P.L.} = \frac{3I_2^{2}R}{\sqrt3\,VI_2}\times100 \]

Equating the two and cancelling \(R\), \(V\) and the factor of 100:

\[ \frac{2I_1}{1} = \frac{3I_2}{\sqrt3} = \sqrt3\,I_2 \quad\Rightarrow\quad I_2 = \frac{2}{\sqrt3}I_1 \]

The power ratio follows:

\[ \frac{P_2}{P_1} = \frac{\sqrt3\,VI_2}{VI_1} = \frac{\sqrt3 \times \frac{2}{\sqrt3}I_1}{I_1} = 2 \]

So the three-phase system carries twice the load:

\[ \text{Additional load} = 100\% \]

Adding 50% more metal bought 100% more capacity. That disproportion is the commercial case for three phase in one line. Note carefully that the constraint was percentage loss, not absolute loss — holding the absolute loss constant instead gives a different and smaller answer, and reading which one the question intends is half the work in every problem of this type.

Answer\(100\%\) additional load

Problem 3Exam levelSystem Conversion

A d.c. three-wire system is to be converted into a three-phase, four-wire system by adding a fourth wire equal in cross-section to each outer of the d.c. system. If the percentage power loss and the voltage at the consumer's terminals are to be the same in both cases, find the extra power at unity power factor that the a.c. system can supply. Assume balanced loads.

Solution

In both systems the load is balanced, so the neutral carries no current and contributes no loss. The comparison therefore involves only the outers in the d.c. case and the three phases in the a.c. case.

Let the consumer's terminal voltage be \(V\) — measured to neutral in both systems, since that is what a consumer's appliance sees. The d.c. three-wire system delivers across two outers at \(\pm V\):

\[ P_1 = 2VI_1, \qquad \%\text{P.L.} = \frac{2I_1^{2}R}{2VI_1}\times100 \]

The three-phase four-wire system delivers three phases each at \(V\) to neutral:

\[ P_2 = 3VI_2, \qquad \%\text{P.L.} = \frac{3I_2^{2}R}{3VI_2}\times100 \]

Equating the percentage losses:

\[ \frac{2I_1^{2}R}{2VI_1} = \frac{3I_2^{2}R}{3VI_2} \quad\Rightarrow\quad I_1 = I_2 \]

The currents come out equal — a pleasant simplification that occurs because in both systems every conductor carries the full load current of its own circuit.

Hence the power ratio:

\[ \frac{P_2}{P_1} = \frac{3VI_2}{2VI_1} = \frac{3}{2} = 1.5 \]

\[ \text{Extra power} = 50\% \]

Compare this with Problem 2 and the pattern becomes visible. There the d.c. system had two conductors and gained one; here it had three and gained one. The gain is essentially the ratio of load-carrying conductors — from 2 to 3 gives a doubling once the \(\sqrt3\) is accounted for, from 2 outers to 3 phases gives 1.5. The neutral in a balanced four-wire system is doing no work at all, which is why it is habitually installed at half the cross-section of the phases.

Answer\(50\%\) extra power

Problem 4Warm-upSystem Conversion

A single-phase a.c. system supplies a load of 200 kW. If this system is converted to a three-phase, three-wire a.c. system by running a third similar conductor, calculate the three-phase load that can now be supplied if the voltage between conductors is unchanged. Assume the power factor and the transmission efficiency are the same in both cases.

Solution

Equal transmission efficiency with equal power delivered means equal percentage power loss, so this is Problem 2 with a power factor carried through. Let \(R\) be the resistance per conductor and \(\cos\phi\) the power factor.

Single-phase two-wire:

\[ P_1 = VI_1\cos\phi, \qquad \%\text{P.L.} = \frac{2I_1^{2}R}{VI_1\cos\phi}\times100 \]

Three-phase three-wire, same voltage between conductors:

\[ P_2 = \sqrt3\,VI_2\cos\phi, \qquad \%\text{P.L.} = \frac{3I_2^{2}R}{\sqrt3\,VI_2\cos\phi}\times100 \]

Equating — and noting that \(\cos\phi\) cancels, so the answer is independent of it:

\[ 2I_1 = \sqrt3\,I_2 \quad\Rightarrow\quad I_2 = \frac{2}{\sqrt3}I_1, \qquad \frac{P_2}{P_1} = 2 \]

Therefore:

\[ P_2 = 2 \times 200 = 400\ \text{kW} \]

The power factor cancelled, and that is worth noticing. It appears once in the numerator of each percentage-loss expression and once in each power expression, so it leaves the ratio untouched. Power factor matters enormously for the absolute quantity of conductor — Problem 9 shows it entering as \(1/\cos^2\phi\) — but not at all for a comparison between two systems operating at the same power factor.

Answer\(P_2 = 400\) kW — a 100% increase

Problem 5Exam levelVolume of Conductor

A 50 km transmission line supplies a load of 5 MVA at 0.8 power factor lagging at 33 kV. The transmission efficiency is 90%. Calculate the volume of aluminium conductor required when

a single-phase, two-wire system is used;
a three-phase, three-wire system is used.

Take the resistivity of aluminium as \(2.85\times10^{-8}\ \Omega\text{m}\).

Solution

Power delivered and power lost:

\[ P = 5 \times 0.8 = 4\ \text{MW}, \qquad W = 10\%\text{ of }P = 4\times10^{5}\ \text{W} \]

An efficiency of 90% means the loss is 10% of the power transmitted. Taking 10% of the sending-end power instead is a common and avoidable slip.

aSingle-phase, two-wire. The line current follows from the apparent power:

\[ I_1 = \frac{5\times10^{6}}{33\times10^{3}} = 151.5\ \text{A} \]

The loss occurs in two conductors, so:

\[ W = 2I_1^{2}\left(\frac{\rho l}{a_1}\right) \quad\Rightarrow\quad a_1 = \frac{2I_1^{2}\rho l}{W} \]

\[ a_1 = \frac{2(151.5)^{2}(2.85\times10^{-8})(50\times10^{3})}{4\times10^{5}} = 1.635\times10^{-4}\ \text{m}^{2} \]

\[ \text{Volume} = 2a_1l = 2(1.635\times10^{-4})(50\times10^{3}) = 16.35\ \text{m}^{3} \]

bThree-phase, three-wire at the same line voltage:

\[ I_2 = \frac{5\times10^{6}}{\sqrt3(33\times10^{3})} = 87.5\ \text{A} \]

Now three conductors share the loss:

\[ a_2 = \frac{3I_2^{2}\rho l}{W} = \frac{3(87.5)^{2}(2.85\times10^{-8})(50\times10^{3})}{4\times10^{5}} = 0.818\times10^{-4}\ \text{m}^{2} \]

\[ \text{Volume} = 3a_2l = 3(0.818\times10^{-4})(50\times10^{3}) = 12.27\ \text{m}^{3} \]

The ratio is worth extracting, because it is universal:

\[ \frac{12.27}{16.35} = 0.75 \]

Three quarters — always, on equal line voltage. The general proof takes one line: \(I_2 = I_1/\sqrt3\), so \(a_2 = 3I_2^2\rho l/W = I_1^2\rho l/W = a_1/2\), and the volume ratio is \(3a_2/2a_1 = 3/4\). No numbers were needed. Whenever a problem gives resistivity, length and rating in this form, computing the ratio first and one absolute volume second is faster and far less error-prone than computing both from scratch — Problem 20 is the same result on different data.

Answer(a) \(16.35\ \text{m}^3\) · (b) \(12.27\ \text{m}^3\) — a 25% saving

Problem 6Challenge-liteVoltage Drop Constraint

A sub-station supplies power at 11 kV, 0.8 power factor lagging to a consumer through a single-phase transmission line of total resistance (go and return) 0.15 \(\Omega\). The voltage drop in the line is 15%. If the same power is to be supplied to the same consumer by a two-wire d.c. system over a new line of total resistance 0.05 \(\Omega\), and the allowable voltage drop is 25%, calculate the d.c. supply voltage.

Solution

Single-phase system. The drop is a stated percentage of the supply voltage, which fixes the current:

\[ \text{Drop} = \frac{15}{100}\times11\,000 = 1650\ \text{V} = I_1(0.15) \quad\Rightarrow\quad I_1 = 11\,000\ \text{A} \]

The power to be supplied to the consumer:

\[ P = VI_1\cos\phi = (11\,000)(11\,000)(0.8) = 9.68\times10^{7}\ \text{W} \]

D.C. system. The same power must be delivered, but the supply voltage \(V\) is now the unknown, so the current depends on it:

\[ I_2 = \frac{9.68\times10^{7}}{V} \]

The drop across the new line:

\[ \text{Drop} = I_2R_2 = \frac{9.68\times10^{7}}{V}\times 0.05 = \frac{4.84\times10^{6}}{V} \]

Setting this equal to the allowable 25% of \(V\) gives a quadratic in disguise:

\[ \frac{4.84\times10^{6}}{V} = 0.25V \quad\Rightarrow\quad V^{2} = 1.936\times10^{7} \quad\Rightarrow\quad V = 4400\ \text{V} \]

A percentage drop constraint is quadratic, not linear. The drop rises as \(1/V\) while the allowance rises as \(V\), so the two meet at a single voltage rather than admitting a whole range. This is why "keep the drop below 5%" behaves so differently from "keep the drop below 200 V" as a design rule — the first tightens automatically as the voltage falls, which is precisely the behaviour a distribution engineer wants.

AnswerD.C. supply voltage \(= 4400\) V

Problem 7Challenge-liteEqual Voltage to Earth

Compare the volume of conductor material required by a three-phase, three-wire a.c. system with that of a two-wire d.c. system having one conductor earthed, on the basis of equal power delivered, equal distance, equal loss and equal maximum voltage to earth.

Solution

Let the maximum voltage to earth be \(V_m\) in both systems, the power \(P\), the length \(l\), the loss \(W\) and the resistivity \(\rho\).

D.C. two-wire, one conductor earthed. The other conductor sits at \(V_m\), and this is a d.c. system so \(V_m\) is both the peak and the effective value:

\[ I = \frac{P}{V_m}, \qquad W = \frac{2I^{2}\rho l}{a} \quad\Rightarrow\quad a = \frac{2P^{2}\rho l}{V_m^{2}W} \]

\[ K \equiv \text{Volume} = 2al = \frac{4P^{2}\rho l^{2}}{V_m^{2}W} \]

Three-phase, three-wire. Here \(V_m\) is the peak value of the phase voltage, so the r.m.s. phase voltage is smaller by \(\sqrt2\) — the step that decides the whole comparison:

\[ V_{ph} = \frac{V_m}{\sqrt2}, \qquad P = 3V_{ph}I'\cos\phi \quad\Rightarrow\quad I' = \frac{\sqrt2\,P}{3V_m\cos\phi} \]

Three conductors now carry the loss:

\[ W = \frac{3I'^{2}\rho l}{a'} \quad\Rightarrow\quad a' = \frac{3\rho l}{W}\cdot\frac{2P^{2}}{9V_m^{2}\cos^{2}\phi} = \frac{2P^{2}\rho l}{3V_m^{2}\cos^{2}\phi\,W} \]

\[ \text{Volume} = 3a'l = \frac{2P^{2}\rho l^{2}}{V_m^{2}\cos^{2}\phi\,W} \]

Taking the ratio:

\[ \frac{\text{Volume}_{3\phi}}{K} = \frac{2}{4\cos^{2}\phi} = \frac{0.5}{\cos^{2}\phi} \]

At unity power factor the three-phase system needs half the metal of the d.c. two-wire system; at 0.8 lagging it needs \(0.5/0.64 = 0.781\) of it — still less, but the margin has narrowed sharply.

Notice that this answer disagrees with Problem 5, and both are right. Problem 5 held the voltage between conductors equal and got 0.75; this problem holds the maximum voltage to earth equal and gets 0.5 at unity power factor. Neither basis is more correct — the first is the natural one when comparing systems on the same busbar, the second when the limiting factor is insulation to ground, as it is on an overhead line where the towers and insulator strings are sized by the phase-to-earth stress. Always read which basis the question intends before writing a single line.

AnswerRatio \(= 0.5/\cos^2\phi\) — half the d.c. volume at unity power factor

Problem 8Warm-upEfficiency

A three-phase line 20 km long delivers 5 MW at 0.85 power factor lagging at a receiving-end voltage of 33 kV. The conductor resistance is 0.4 \(\Omega\) per kilometre per phase. Find the line current, the total line loss, the sending-end power and the transmission efficiency.

Solution

Line current from the receiving-end conditions:

\[ I = \frac{P}{\sqrt3\,V_R\cos\phi} = \frac{5\times10^{6}}{\sqrt3(33\,000)(0.85)} = 102.9\ \text{A} \]

Resistance of one conductor over the full length:

\[ R = 0.4 \times 20 = 8\ \Omega\ \text{per phase} \]

Total loss in three conductors:

\[ W = 3I^{2}R = 3(102.9)^{2}(8) = 254.2\ \text{kW} \]

Sending-end power and efficiency:

\[ P_S = 5000 + 254.2 = 5254.2\ \text{kW} \]

\[ \eta = \frac{P_R}{P_S} = \frac{5000}{5254.2} = 95.16\% \]

A useful cross-check: the loss as a fraction of the delivered power is \(254.2/5000 = 5.08\%\), and \(1/1.0508 = 95.17\%\), agreeing to rounding.

Efficiency and percentage loss are close but not the same number, and the difference grows. A 5.08% loss gives 95.16% efficiency, not 94.92% — the loss is reckoned against the delivered power while the efficiency is reckoned against the sending-end power. At the 10% loss of Problem 5 the two differ by nearly a full point, and at the loss levels of a distribution feeder the distinction has to be made explicitly or the conductor comes out the wrong size.

Answer\(I = 102.9\) A, \(W = 254.2\) kW, \(P_S = 5254\) kW, \(\eta = 95.16\%\)

Problem 9Exam levelPower Factor and Metal

A three-phase line delivers a fixed power at a fixed voltage over a fixed distance with a fixed loss. Show that the volume of conductor material required varies as \(1/\cos^2\phi\), and hence find the percentage saving in conductor material if the power factor is improved from 0.7 to 0.9 lagging.

Solution

With \(P\), \(V\), \(l\) and \(W\) all fixed, the current is set entirely by the power factor:

\[ I = \frac{P}{\sqrt3\,V\cos\phi} \quad\Rightarrow\quad I \propto \frac{1}{\cos\phi} \]

The area needed to hold the loss constant follows the square of the current:

\[ a = \frac{3I^{2}\rho l}{W} \propto I^{2} \propto \frac{1}{\cos^{2}\phi} \]

Volume is \(3al\) with \(l\) fixed, so it carries the same dependence:

\[ \text{Volume} \propto \frac{1}{\cos^{2}\phi} \]

Applying it to the improvement from 0.7 to 0.9:

\[ \frac{v_2}{v_1} = \left(\frac{\cos\phi_1}{\cos\phi_2}\right)^{2} = \left(\frac{0.7}{0.9}\right)^{2} = 0.605 \]

\[ \%\text{ saving} = (1 - 0.605)\times100 = 39.5\% \]

Power factor and voltage are the same lever pulled in two directions. Volume goes as \(1/(V\cos\phi)^2\), so the product \(V\cos\phi\) is what actually matters and it makes no difference to the conductor whether an improvement comes from raising one or the other. A capacitor bank that lifts the power factor from 0.7 to 0.9 has the same effect on the metal as raising the voltage by 29% — and is very much cheaper, which is why correction is almost always tried first.

AnswerVolume \(\propto 1/\cos^2\phi\); saving \(= 39.5\%\)

Problem 10Challenge-liteKelvin's Law

State Kelvin's law for the most economical cross-section of a conductor, derive it, and explain in physical terms why the optimum occurs where it does.

Solution

The total annual cost of a line has two parts that move in opposite directions as the conductor is made thicker. Write the annual charge on capital as

\[ C_1 = P_1 + P_2a \]

\(P_1\) covers everything independent of conductor size — towers, insulators, erection, right of way — and \(P_2a\) the part proportional to the cross-section, namely the metal itself.

The annual cost of the energy lost varies inversely with the area, since \(R \propto 1/a\):

\[ C_2 = \frac{P_3}{a} \]

The total is

\[ C = P_1 + P_2a + \frac{P_3}{a} \]

Differentiating and setting to zero:

\[ \frac{dC}{da} = P_2 - \frac{P_3}{a^{2}} = 0 \quad\Rightarrow\quad a = \sqrt{\frac{P_3}{P_2}} \]

\[ \frac{d^{2}C}{da^{2}} = \frac{2P_3}{a^{3}} > 0 \quad\Rightarrow\quad \text{a minimum} \]

Substituting back reveals the statement of the law in its memorable form:

\[ P_2a = \frac{P_3}{a} \]

The most economical cross-section is that for which the annual cost of energy lost equals the annual charge on the part of the capital cost that varies with the cross-section. Note the qualification: \(P_1\) plays no part in the optimum at all.

Why the optimum sits there, physically: adding one more square millimetre of copper costs a fixed amount every year, and saves an amount that shrinks as the conductor grows, because the loss falls as \(1/a\) and its derivative as \(1/a^2\). The optimum is simply where the next millimetre stops paying for itself.

Kelvin's law gives the right answer to the wrong question more often than it is admitted. The cost of metal is rarely exactly proportional to area — bulk buying, standard sizes and handling costs all break the assumption — and the optimum it finds is usually too large to be mechanically or thermally sensible. In practice the law fixes the region and the final choice is the nearest standard size that also satisfies the current-carrying capacity, the short-circuit rating and the voltage-drop limit. Its enduring value is the insight, not the number: a well-designed line loses money in copper and money in heat at about the same rate.

Answer\(a = \sqrt{P_3/P_2}\); annual loss cost \(=\) annual charge on the variable capital cost

Problem 11Exam levelKelvin's Law

A two-wire feeder 1 km long carries a constant current of 100 A throughout the year. The cost of the installed line is \(\text{₹}(3000 + 40\,000a)\) per km, where \(a\) is the cross-section in cm². Interest and depreciation total 10% per annum and energy costs 40 paise per kWh. Taking \(\rho = 1.78\times10^{-8}\ \Omega\text{m}\), find the most economical cross-section.

Solution

Annual charge on capital. Ten per cent of the installed cost:

\[ C_1 = 0.10(3000 + 40\,000a) = 300 + 4000a \]

So \(P_1 = 300\) and \(P_2 = 4000\) in the notation of Problem 10.

Resistance of the loop. With \(a\) in cm², that is \(a\times10^{-4}\ \text{m}^2\), and the go-and-return path is two conductors:

\[ R = \frac{2\rho l}{a\times10^{-4}} = \frac{2(1.78\times10^{-8})(1000)}{a\times10^{-4}} = \frac{0.356}{a}\ \Omega \]

Annual cost of energy lost. The current is constant all year, so the loss is constant:

\[ \text{Loss} = I^{2}R = (100)^{2}\frac{0.356}{a} = \frac{3560}{a}\ \text{W} = \frac{3.56}{a}\ \text{kW} \]

\[ \text{Energy} = \frac{3.56}{a}\times 8760 = \frac{31\,186}{a}\ \text{kWh}, \qquad C_2 = 0.40\times\frac{31\,186}{a} = \frac{12\,474}{a} \]

So \(P_3 = 12\,474\).

Applying Kelvin's law:

\[ a = \sqrt{\frac{P_3}{P_2}} = \sqrt{\frac{12\,474}{4000}} = \sqrt{3.119} = 1.77\ \text{cm}^{2} \]

Verifying that the two costs balance at this section, which is what the law asserts:

\[ P_2a = 4000(1.77) = \text{₹}7064, \qquad \frac{P_3}{a} = \frac{12\,474}{1.77} = \text{₹}7063\ \checkmark \]

\[ \text{Total annual cost} = 300 + 7064 + 7063 = \text{₹}14\,427 \]

The optimum is remarkably flat, which is what makes the law usable. Try \(a = 1.5\) cm²: the total becomes \(300 + 6000 + 8316 = \text{₹}14\,616\), only 1.3% worse. At \(a = 2.0\) it is \(300 + 8000 + 6237 = \text{₹}14\,537\), 0.8% worse. A designer can therefore round freely to the nearest standard size — 1.77 cm² is not a size anyone manufactures — and lose almost nothing. Reporting the optimum to three decimal places, as the arithmetic invites, misrepresents how sharply it is really known.

Answer\(a = 1.77\ \text{cm}^2\), total annual cost \(\text{₹}14\,427\)

Problem 12Exam levelEarthing Arrangement

Compare the volume of conductor material required by a two-wire d.c. system with the midpoint earthed against the same system with one conductor earthed, on the basis of equal power, distance, loss and maximum voltage to earth.

Solution

From Problem 7, the two-wire system with one conductor earthed gives

\[ K = \frac{4P^{2}\rho l^{2}}{V_m^{2}W} \]

Midpoint earthed. Now neither conductor is at earth potential; one sits at \(+V_m\) and the other at \(-V_m\). The voltage between them is therefore \(2V_m\) while the maximum stress to earth is unchanged:

\[ I = \frac{P}{2V_m} \]

Two conductors carry the loss:

\[ a = \frac{2I^{2}\rho l}{W} = \frac{2\rho l}{W}\cdot\frac{P^{2}}{4V_m^{2}} = \frac{P^{2}\rho l}{2V_m^{2}W} \]

\[ \text{Volume} = 2al = \frac{P^{2}\rho l^{2}}{V_m^{2}W} \]

The ratio:

\[ \frac{\text{Volume}}{K} = \frac{1}{4} = 0.25 \]

Three quarters of the metal saved, purely by moving the earth connection.

Nothing was added and nothing was rebuilt — only the reference point moved. Earthing the midpoint doubles the useful voltage for the same insulation stress, and since volume goes as \(1/V^2\) that is a factor of four. The same reasoning explains why HVDC links are almost always bipolar rather than monopolar with an earth return, and why a 132 kV three-phase line is described by a number that is a line voltage while its insulators are designed to \(132/\sqrt3 = 76\) kV. Insulation responds to the stress to earth; economics responds to the voltage between conductors; the earthing arrangement is the only thing standing between them.

AnswerMidpoint earthed needs \(0.25K\) — a 75% saving

Problem 13Warm-upThree-Wire vs Four-Wire

A three-phase, four-wire system carries a neutral of half the cross-section of each phase conductor. On the equal-maximum-voltage-to-earth basis, compare its conductor volume with that of the three-phase, three-wire system of Problem 7, and state by what percentage the neutral increases the requirement.

Solution

With a balanced load the neutral carries no current, so it contributes no loss and the phase conductor area is exactly that found in Problem 7:

\[ a' = \frac{2P^{2}\rho l}{3V_m^{2}\cos^{2}\phi\,W} \]

The metal, however, must still be paid for. Three phases at \(a'\) plus a neutral at \(a'/2\):

\[ \text{Volume} = \left(3 + \tfrac12\right)a'l = 3.5\,a'l \]

Against the three-wire volume of \(3a'l\):

\[ \frac{3.5}{3} = 1.167 \]

\[ \frac{\text{Volume}_{4\text{-wire}}}{K} = 1.167 \times \frac{0.5}{\cos^{2}\phi} = \frac{0.583}{\cos^{2}\phi} \]

So the neutral costs 16.7% more conductor material and, under balanced conditions, delivers nothing whatever in return.

The neutral is bought for the unbalanced case, not the balanced one. Transmission systems are balanced by construction and therefore use three wires; distribution systems serve single-phase consumers who cannot be relied upon to balance, so they use four. The half-section neutral is the compromise: large enough for the imbalance that actually occurs, small enough that the 16.7% penalty is not paid twice. Where third-harmonic currents are present the compromise fails, because those add in the neutral instead of cancelling, and a full-section or oversized neutral becomes necessary.

Answer\(0.583/\cos^2\phi\) against \(0.5/\cos^2\phi\) — 16.7% more

Problem 14Exam levelEconomic Voltage

Estimate the most economical transmission voltage for a three-phase line 50 km long delivering 5000 kW, using the empirical relation

\[ V = 5.5\sqrt{\frac{L}{1.6} + \frac{P}{150}} \]

where \(V\) is in kV, \(L\) in km and \(P\) in kW. Comment on what standard voltage should actually be adopted.

Solution

Substituting the data:

\[ V = 5.5\sqrt{\frac{50}{1.6} + \frac{5000}{150}} = 5.5\sqrt{31.25 + 33.33} \]

Evaluating:

\[ V = 5.5\sqrt{64.58} = 5.5(8.036) = 44.2\ \text{kV} \]

No such standard voltage exists. The candidates on either side are 33 kV and 66 kV, and since the formula gives a lower bound below which the conductor cost dominates, the sensible choice is

\[ V = 66\ \text{kV} \]

The two contributions are worth reading separately: distance supplies 31.25 and power supplies 33.33, so at this duty the two are almost equally responsible. A line twice as long at the same power would give \(5.5\sqrt{62.5 + 33.33} = 53.9\) kV — still 66 kV, but now clearly so.

The formula is a summary of the two costs that fight each other. Conductor cost falls as \(1/V^2\) from Problem 1; insulation, towers, switchgear and corona losses all rise with \(V\). The economic voltage is where the sum turns, and the empirical constants encode decades of accumulated cost data rather than any derivation. Because standard voltages come in a coarse ladder — 11, 33, 66, 132, 220, 400 — the formula rarely decides anything on its own; it decides which rung of the ladder to stand on, and the answer is normally the next one up.

Answer\(V = 44.2\) kV computed; adopt the standard \(66\) kV

Problem 15Exam levelDistributor Drop

A two-wire distributor 500 m long is fed at one end and carries a uniformly distributed load of 0.4 A per metre. The resistance of the go-and-return conductors is \(2\times10^{-4}\ \Omega\) per metre. Find the total current, the voltage drop at the far end, and compare with the drop that the same total current would produce if lumped at the far end.

Solution

Total current drawn:

\[ I = il = 0.4 \times 500 = 200\ \text{A} \]

Consider an element of length \(dx\) at a distance \(x\) from the feeding point. The current flowing through it is only what the remaining \((l-x)\) metres will draw:

\[ i(x) = i(l - x) \]

The drop across that element is \(i(l-x)\,r\,dx\). Integrating over the whole length:

\[ \Delta V = \int_0^{l} i(l-x)r\,dx = ir\left[lx - \frac{x^{2}}{2}\right]_0^{l} = \frac{1}{2}il^{2}r \]

Substituting:

\[ \Delta V = \tfrac12(0.4)(500)^{2}(2\times10^{-4}) = \tfrac12(0.4)(250\,000)(2\times10^{-4}) = 10\ \text{V} \]

If the same 200 A were lumped at the far end, the whole current would traverse the whole resistance:

\[ R_{\text{total}} = (2\times10^{-4})(500) = 0.1\ \Omega, \qquad \Delta V = 200(0.1) = 20\ \text{V} \]

So the distributed load gives exactly half the drop:

\[ \frac{10}{20} = \frac12 \]

A uniformly distributed load behaves as though the whole of it were concentrated at the midpoint. That is the practical content of the factor of one half, and it is worth carrying as a shortcut: replace the distribution by a lumped load at \(l/2\) and the drop comes out right without any integration. The same substitution does not work for the power loss, which involves \(I^2\) and integrates to \(\tfrac13 i^2l^3r\) — one third, not one half. Confusing the two factors is the standard error in this topic.

Answer\(I = 200\) A, \(\Delta V = 10\) V — half the lumped-load drop of 20 V

Problem 16Exam levelDistributor Drop

The distributor of Problem 15 is now fed at both ends at the same voltage. Find the position and magnitude of the maximum voltage drop, and compare with the single-end case.

Solution

With equal voltages at both ends and a symmetric load, the current supplied by each end is equal and the point of minimum potential lies at the centre. Each half of the distributor behaves as a 250 m distributor fed at one end.

Applying the result of Problem 15 to a half-length \(l/2\):

\[ \Delta V = \frac{1}{2}i\left(\frac{l}{2}\right)^{2}r = \frac{il^{2}r}{8} \]

Substituting:

\[ \Delta V = \frac{(0.4)(500)^{2}(2\times10^{-4})}{8} = \frac{20}{8} = 2.5\ \text{V} \]

Comparing with the 10 V of the single-end case:

\[ \frac{2.5}{10} = \frac14 \]

Each end now supplies only 100 A, and each half is only 250 m long — a factor of two in current and a factor of four in the \(l^2\) term, of which half is given back by the doubling of the load per unit length seen by each feed.

The point of minimum potential is at the centre, 250 m from either end, where the currents from the two feeds meet and cancel.

A second feed costs one extra cable and buys a factor of four. This is why ring mains dominate urban distribution: the drop collapses, and the arrangement also survives the loss of either feed, since the remaining one can still supply the whole distributor — at the 10 V drop of Problem 15 rather than 2.5 V, but supply it nonetheless. Redundancy and voltage quality arrive in the same cable, which is unusually good value.

AnswerMaximum drop \(= 2.5\) V at the centre — one quarter of the single-end value

Problem 17Warm-upCost of Losses

A three-phase feeder of resistance 0.2 \(\Omega\) per phase carries 150 A for 3000 hours and 60 A for 5000 hours in a year, and is off for the remainder. Energy costs \(\text{₹}5\) per kWh. Find the annual cost of the \(I^2R\) losses, and the loss load factor.

Solution

Loss at the higher loading:

\[ W_1 = 3I_1^{2}R = 3(150)^{2}(0.2) = 13\,500\ \text{W} = 13.5\ \text{kW} \]

\[ E_1 = 13.5 \times 3000 = 40\,500\ \text{kWh} \]

Loss at the lower loading:

\[ W_2 = 3(60)^{2}(0.2) = 2160\ \text{W} = 2.16\ \text{kW} \]

\[ E_2 = 2.16 \times 5000 = 10\,800\ \text{kWh} \]

Total annual energy lost and its cost:

\[ E = 40\,500 + 10\,800 = 51\,300\ \text{kWh}, \qquad \text{Cost} = 5 \times 51\,300 = \text{₹}256\,500 \]

The loss load factor is the average loss divided by the peak loss, taken over the whole year:

\[ W_{\text{avg}} = \frac{51\,300}{8760} = 5.86\ \text{kW}, \qquad \text{LLF} = \frac{5.86}{13.5} = 0.434 \]

The feeder is at peak for a third of the year yet loses at 43% of the peak rate on average. The gap is the whole reason loss load factor exists as a separate quantity from load factor: losses go as the square of the current, so they concentrate at the peak far more sharply than the energy does. A feeder loaded to its peak for only a few hundred hours a year can still spend most of its annual losses in those hours — which is why loss reduction schemes are almost always aimed at the peak rather than at the average.

Answer\(51\,300\) kWh costing \(\text{₹}256\,500\); loss load factor \(= 0.434\)

Problem 18Exam levelSingle-Phase Systems

On the equal-maximum-voltage-to-earth basis, a single-phase two-wire system with one conductor earthed requires \(2/\cos^2\phi\) times the volume of the d.c. two-wire reference, and a single-phase three-wire system with a half-section neutral requires \(0.625/\cos^2\phi\). Derive the second of these and find the percentage saving of the three-wire arrangement over the two-wire.

Solution

In the single-phase three-wire system the two outers sit at \(+V_m\) and \(-V_m\) peak with the neutral earthed between them, so each outer has an r.m.s. voltage to neutral of \(V_m/\sqrt2\).

With a balanced load the two halves each take \(P/2\), and the total is

\[ P = 2\left(\frac{V_m}{\sqrt2}\right)I\cos\phi = \sqrt2\,V_mI\cos\phi \quad\Rightarrow\quad I = \frac{P}{\sqrt2\,V_m\cos\phi} \]

The neutral carries nothing under balance, so only the two outers dissipate:

\[ W = \frac{2I^{2}\rho l}{a} \quad\Rightarrow\quad a = \frac{2\rho l}{W}\cdot\frac{P^{2}}{2V_m^{2}\cos^{2}\phi} = \frac{P^{2}\rho l}{V_m^{2}\cos^{2}\phi\,W} \]

Volume, counting the half-section neutral:

\[ \text{Volume} = \left(2 + \tfrac12\right)al = \frac{2.5\,P^{2}\rho l^{2}}{V_m^{2}\cos^{2}\phi\,W} \]

\[ \frac{\text{Volume}}{K} = \frac{2.5}{4\cos^{2}\phi} = \frac{0.625}{\cos^{2}\phi}\ \checkmark \]

The saving over the two-wire arrangement:

\[ 1 - \frac{0.625}{2} = 1 - 0.3125 = 68.75\% \]

Adding a neutral and splitting the load saved more than two thirds of the metal. The mechanism is exactly that of Problem 12 — the useful voltage doubled while the stress to earth did not — and the half-section neutral costs only a quarter of the gain back. This is the arrangement used for domestic supply in North America, where 240 V appears across the outers for heavy appliances and 120 V from either outer to neutral for everything else, from a distribution transformer whose secondary is a single winding with a centre tap.

Answer\(0.625/\cos^2\phi\); saving of \(68.75\%\) over the two-wire system

Problem 19Challenge-liteLoss Under Unbalance

A three-phase, four-wire distributor supplies unity-power-factor loads of 40 A, 30 A and 20 A on its three phases. Each conductor, neutral included, has a resistance of 0.5 \(\Omega\). Find the neutral current and the total copper loss, and compare with a balanced load of 30 A per phase drawing the same total current.

Solution

At unity power factor each current is in phase with its own phase voltage, so the three currents lie at \(0^\circ\), \(-120^\circ\) and \(+120^\circ\):

\[ \mathbf{I}_N = 40\angle0^\circ + 30\angle-120^\circ + 20\angle120^\circ \]

Resolving:

\[ = 40 + (-15 - j25.98) + (-10 + j17.32) = 15 - j8.66 \]

\[ |\mathbf{I}_N| = \sqrt{225 + 75} = 17.32\ \text{A} \quad\text{at}\quad -30^\circ \]

Total loss, now including the neutral which is carrying real current:

\[ W = \left(40^{2} + 30^{2} + 20^{2} + 17.32^{2}\right)(0.5) = (1600 + 900 + 400 + 300)(0.5) = 1600\ \text{W} \]

The balanced comparison — 30 A on each phase is the same total of 90 A, and the neutral is dead:

\[ W_{\text{bal}} = 3(30)^{2}(0.5) = 1350\ \text{W} \]

The penalty for unbalance:

\[ \frac{1600 - 1350}{1350}\times100 = 18.5\% \]

The same total current, spread unevenly, costs a fifth more in losses. Two separate effects combine: the \(I^2\) law penalises the heavily loaded phase more than the lightly loaded one relieves it, and the neutral — idle under balance — now dissipates 300 W of its own. Neither effect appears in an ammeter reading of the total current, which is identical in both cases. This is the quantitative case for the phase-balancing exercises that distribution utilities carry out on their low-voltage networks, and it is the same argument that Set 1 Problem 13 made in terms of voltage rather than loss.

Answer\(I_N = 17.32\angle-30^\circ\) A, \(W = 1600\) W against \(1350\) W balanced — 18.5% more

Problem 20Exam levelSystem Choice

A 20 km line is to deliver 2 MW at 0.9 power factor lagging with a line loss of 5% of the power delivered. Compare a single-phase two-wire system with a three-phase three-wire system, both at 11 kV between conductors, by finding the conductor cross-section and total volume of copper required for each. Take \(\rho = 1.78\times10^{-8}\ \Omega\text{m}\).

Solution

The permitted loss, common to both systems:

\[ W = 0.05 \times 2\times10^{6} = 1\times10^{5}\ \text{W} \]

aSingle-phase two-wire. Line current:

\[ I_1 = \frac{2\times10^{6}}{(11\,000)(0.9)} = 202.0\ \text{A} \]

\[ a_1 = \frac{2I_1^{2}\rho l}{W} = \frac{2(202.0)^{2}(1.78\times10^{-8})(20\times10^{3})}{1\times10^{5}} = 2.91\times10^{-4}\ \text{m}^{2} \]

\[ \text{Volume} = 2a_1l = 2(2.91\times10^{-4})(20\times10^{3}) = 11.62\ \text{m}^{3} \]

bThree-phase three-wire at the same 11 kV between conductors:

\[ I_2 = \frac{2\times10^{6}}{\sqrt3(11\,000)(0.9)} = 116.6\ \text{A} \]

\[ a_2 = \frac{3I_2^{2}\rho l}{W} = \frac{3(116.6)^{2}(1.78\times10^{-8})(20\times10^{3})}{1\times10^{5}} = 1.45\times10^{-4}\ \text{m}^{2} \]

\[ \text{Volume} = 3a_2l = 3(1.45\times10^{-4})(20\times10^{3}) = 8.72\ \text{m}^{3} \]

The ratio, as expected from Problem 5:

\[ \frac{8.72}{11.62} = 0.75 \]

Note also that each individual conductor is only half as thick — \(1.45\) against \(2.91\ \text{cm}^2\) — which matters for stringing, tower loading and handling quite apart from the cost of the metal.

Three thinner conductors beat two thicker ones, and the mechanical argument reinforces the economic one. The 25% saving in metal is the headline, but the halving of the individual conductor section reduces the tension each tower must withstand, the sag under ice loading, and the difficulty of jointing. None of that appears anywhere in the arithmetic, and all of it pushed the industry the same way. When a question asks you to "compare" two systems, the volume ratio is the answer expected; noticing what else changed is what separates an answer from an understanding.

Answer(a) \(2.91\ \text{cm}^2\), \(11.62\ \text{m}^3\) · (b) \(1.45\ \text{cm}^2\), \(8.72\ \text{m}^3\) — ratio 0.75

Practice

Practice Problems

Work each on paper before opening the answer. The answer is given so you can check yourself; the method is deliberately not.

P1. A d.c. two-wire line has its voltage raised from 250 V to 1000 V for the same power, distance and loss. Find the percentage saving in copper.

Show answer
Volume \(\propto 1/V^2\), so the ratio is \((250/1000)^2 = 1/16\) and the saving is \(\mathbf{93.75\%}\).
P2. A three-phase line delivers 4 MW at 0.8 p.f. over 30 km at 66 kV with a 4% loss. Find the line current and the conductor area, taking \(\rho = 2.85\times10^{-8}\ \Omega\text{m}\).

Show answer
\(I = 4\times10^6/(\sqrt3\cdot66\,000\cdot0.8) = 43.7\) A; \(W = 160\) kW; \(a = 3(43.7)^2(2.85\times10^{-8})(30\,000)/1.6\times10^5 = \mathbf{3.06\times10^{-5}\ \text{m}^2}\).
P3. On equal line voltage, what fraction of the single-phase two-wire conductor volume does a three-phase three-wire system need? Prove it without numbers.

Show answer
\(I_2 = I_1/\sqrt3\), so \(a_2 = 3I_2^2\rho l/W = a_1/2\), giving a volume ratio of \(3a_2/2a_1 = \mathbf{0.75}\).
P4. The power factor of a load falls from 0.95 to 0.75. By what percentage must the conductor volume increase for the same loss?

Show answer
\((0.95/0.75)^2 = 1.604\), an increase of \(\mathbf{60.4\%}\).
P5. A line loses 6% of the power it delivers. What is its transmission efficiency?

Show answer
\(\eta = 100/106 = \mathbf{94.34\%}\) — not 94%. The loss is reckoned on the delivered power, the efficiency on the sending-end power.
P6. For a feeder, \(P_2 = \text{₹}5000\) per cm² per annum and \(P_3 = \text{₹}20\,000\) cm² per annum. Find the most economical section and the minimum total variable cost.

Show answer
\(a = \sqrt{20\,000/5000} = \mathbf{2\ \text{cm}^2}\); each term is then \(\text{₹}10\,000\), total \(\mathbf{\text{₹}20\,000}\).
P7. A distributor 400 m long carries 0.5 A/m and is fed at one end; \(r = 3\times10^{-4}\ \Omega\)/m. Find the drop at the far end.

Show answer
\(\tfrac12 il^2r = \tfrac12(0.5)(400)^2(3\times10^{-4}) = \mathbf{12}\) V.
P8. The same distributor is fed at both ends at equal voltage. Find the maximum drop and where it occurs.

Show answer
\(il^2r/8 = \mathbf{3}\) V, at the centre, 200 m from either end — one quarter of the single-end value.
P9. Three-phase four-wire loads of 50 A, 50 A and 20 A at unity p.f. Find the neutral current.

Show answer
\(50 + 50\angle-120^\circ + 20\angle120^\circ = 15 - j25.98\), magnitude \(\mathbf{30}\) A at \(-60^\circ\).
P10. Using \(V = 5.5\sqrt{L/1.6 + P/150}\), estimate the economic voltage for 100 km and 20 MW, and name the standard voltage to adopt.

Show answer
\(5.5\sqrt{62.5 + 133.3} = 5.5(13.99) = \mathbf{77}\) kV; adopt \(\mathbf{132}\) kV, the next standard rung.
P11. A feeder carries 200 A for 2000 h and 80 A for 6000 h. Find its loss load factor.

Show answer
Losses \(\propto I^2\): \((40\,000\cdot2000 + 6400\cdot6000)/8760 = 13\,516\) against a peak of 40 000, so LLF \(= \mathbf{0.338}\).
P12. Why does a d.c. two-wire system with the midpoint earthed need only a quarter of the metal of one with a conductor earthed, when neither the power nor the loss changed?

Show answer
The useful voltage between conductors doubles for the same insulation stress to earth, and volume \(\propto 1/V^2\), giving \(\mathbf{1/4}\). Nothing was added — only the reference point moved.

Challenge

Challenge Problems

Each of these needs an idea rather than a formula. Decide what the governing principle is before opening the answer.

C1. The volume of conductor varies as \(1/V^2\), which appears to argue for transmitting at unlimited voltage. Identify every mechanism that opposes this, establish how each scales with \(V\), and explain why the economic optimum is a voltage rather than an infinity.

Show answer
Falling with \(V\): conductor volume, as \(1/V^2\). That is the only major cost that does.

Rising with \(V\):
— Insulation. Insulator string length rises roughly linearly with \(V\), and tower height and width with it, so tower cost rises faster than linearly once clearances start to dominate the structure.
— Corona. Loss depends on the surface gradient \(g \propto V/(r\ln(D/r))\). Above the critical disruptive voltage, Peek's law makes the loss rise roughly as \((V - V_0)^2\) — very steeply. Countering it needs larger or bundled conductors, which adds back metal that the \(1/V^2\) saving had just removed. This is the mechanism that bites first at EHV, and it is the subject of Set 8.
— Switchgear and terminal equipment. Circuit-breaker and transformer costs rise sharply and in steps, since each voltage class is a distinct product.
— Right of way. Wider corridors for clearance, and a cost that is often political rather than technical.

Why an optimum exists: total cost is \(A/V^2 + BV + C\) to leading order. Differentiating gives \(-2A/V^3 + B = 0\), so \(V_{\text{opt}} = (2A/B)^{1/3}\) — finite, and notably insensitive, since it depends on the cube root of the cost ratio. Halving the price of copper moves the economic voltage by only 21%, which is why the standard voltage ladder has been so stable for a century.

Note that \(A \propto P^2l^2\), so \(V_{\text{opt}} \propto (P^2l^2)^{1/3}\) — rising with both power and distance, exactly the behaviour the empirical formula of Problem 14 encodes.
C2. Kelvin's law says the most economical section makes the annual cost of losses equal the annual charge on the variable capital cost. Yet practising engineers rarely install the section it prescribes. Give at least four reasons, and explain what the law is nevertheless good for.

Show answer
1. The cost model is wrong in detail. \(P_2a\) assumes cost strictly proportional to area. Real cabling has fixed costs per metre — armouring, insulation, jointing, handling — that make the relation affine, not linear, and the optimum shifts accordingly.

2. Other constraints usually bind first. Current-carrying capacity, short-circuit withstand and the voltage-drop limit each set a minimum section. On short feeders the drop limit almost always dominates and Kelvin's answer is simply irrelevant.

3. The prescribed section is often mechanically absurd. When energy is dear relative to metal, the law can call for a conductor too heavy for the towers, or one whose weight demands stronger towers whose cost was never in the model.

4. Standard sizes exist. Conductors come in a discrete ladder, so the answer must be rounded regardless — and as Problem 11 showed, the optimum is so flat that rounding costs almost nothing.

5. The future is not the present. The law optimises for today's load and today's energy price over an asset lasting forty years. Loads grow; a section chosen for present load is undersized within a decade.

What it is good for: the insight that a well-designed line should lose comparable amounts of money to copper and to heat. That gives a designer an immediate check — if losses cost ten times the annual charge on the metal, the conductor is far too thin whatever else the calculation says. It also sets the region within which the other constraints are then applied.
C3. Problem 5 found a ratio of 0.75 for three-phase against single-phase, and Problem 7 found 0.5 for three-phase against d.c. Reconcile these with the fact that both compare a three-phase system with a two-wire one, and state precisely what determines which answer a given question wants.

Show answer
They compare against different two-wire systems on different bases, and both are correct.

Problem 5 compares three-phase a.c. with single-phase a.c. on equal voltage between conductors. Both are a.c., so peak-to-r.m.s. factors cancel and never appear. The result, 0.75, comes purely from \(3 \times \tfrac13 \div 2\).

Problem 7 compares three-phase a.c. with two-wire d.c. on equal maximum voltage to earth. Two separate changes occur: the reference is now peak-to-earth rather than between conductors, and one system is d.c. while the other is a.c., so the \(\sqrt2\) no longer cancels. Both changes favour the three-phase system, taking 0.75 down to 0.5.

Which basis a question wants is decided by what physically limits the design:
— Equal voltage between conductors when the systems share a busbar or a transformer secondary, so the terminal voltage is a given.
— Equal maximum voltage to earth when insulation to ground is the limit — overhead lines, where tower clearances and insulator strings are sized by phase-to-earth stress, and underground cables, where the dielectric sees the stress to the earthed sheath.

The second is the more common basis in transmission and the first in installation work. A question that says only "compare" without specifying is under-determined, and the honest response is to state the basis assumed before starting — which is what every worked comparison in this set does in its first line.

Self-Test

Multiple-Choice Questions

MCQ 1. For a fixed power, distance and loss, the volume of conductor material varies as:
(a) \(V\) (b) \(1/V\) (c) \(V^2\) (d) \(1/V^2\)

Show answer
(d). Current falls as \(1/V\) and the area needed falls as \(I^2\). This one relation drives the entire set.
MCQ 2. On equal line voltage, a three-phase three-wire system needs what fraction of the conductor volume of a single-phase two-wire system?
(a) 0.5 (b) 0.583 (c) 0.75 (d) 1.0

Show answer
(c) 0.75. Option (a) is the answer on the equal-maximum-voltage-to-earth basis against d.c. — a different question, as Challenge C3 sets out.
MCQ 3. Kelvin's law states that the most economical conductor section is that for which:
(a) the total capital cost is least (b) the annual loss cost equals the annual charge on the variable capital cost (c) the loss is least (d) the loss cost equals the total capital charge

Show answer
(b). Option (d) is the common misstatement — the fixed part \(P_1\) plays no role in the optimum at all.
MCQ 4. A uniformly loaded distributor fed at one end has a far-end drop of:
(a) \(il^2r\) (b) \(\tfrac12 il^2r\) (c) \(\tfrac13 il^2r\) (d) \(\tfrac18 il^2r\)

Show answer
(b). Option (c) is the power loss coefficient and (d) the drop when fed at both ends.
MCQ 5. Feeding a uniformly loaded distributor at both ends instead of one reduces the maximum drop by a factor of:
(a) 2 (b) 3 (c) 4 (d) 8

Show answer
(c) 4. From \(\tfrac12 il^2r\) to \(\tfrac18 il^2r\), with the maximum now at the centre.
MCQ 6. Improving the power factor from 0.8 to 1.0 reduces the required conductor volume by:
(a) 20% (b) 36% (c) 56% (d) 64%

Show answer
(b) 36%. The ratio is \((0.8/1.0)^2 = 0.64\), so 36% is saved.
MCQ 7. A d.c. two-wire system with the midpoint earthed needs, relative to one with a conductor earthed and equal maximum voltage to earth:
(a) the same (b) half (c) a quarter (d) twice

Show answer
(c) a quarter. The useful voltage doubles for the same insulation stress, and volume goes as \(1/V^2\).
MCQ 8. In a balanced three-phase four-wire system the neutral conductor:
(a) carries one third of the line current (b) carries no current (c) carries \(\sqrt3\) times the phase current (d) may be omitted without any consequence

Show answer
(b). Option (d) is tempting but wrong — it may be omitted only so long as the load stays balanced, which in distribution it does not.
MCQ 9. A line delivering 100 kW with a 5 kW loss has a transmission efficiency of:
(a) 95% (b) 95.24% (c) 94.76% (d) 105%

Show answer
(b) 95.24%, from \(100/105\). Option (a) is the trap of subtracting the percentage loss directly.
MCQ 10. The economic transmission voltage rises with:
(a) distance only (b) power only (c) both distance and power (d) neither

Show answer
(c). Both enter the empirical formula, and the derivation in Challenge C1 gives \(V_{\text{opt}} \propto (P^2l^2)^{1/3}\).
MCQ 11. Loss load factor is generally:
(a) equal to load factor (b) greater than load factor (c) less than load factor (d) unrelated to load factor

Show answer
(c) less. Losses go as \(I^2\), so they concentrate at the peak more sharply than energy does, pulling the average down relative to the peak.
MCQ 12. Three-phase loads of 40, 30 and 20 A at unity p.f. give a neutral current of about:
(a) 0 A (b) 10 A (c) 17.3 A (d) 90 A

Show answer
(c) 17.3 A. The three phasors must be added vectorially, not arithmetically — Problem 19.

Reference

Key Formulas

Quantity	Relation	Notes
Conductor area	\(a = nI^2\rho l/W\)	\(n\) = number of loaded conductors
Conductor volume	\(nal = n^2I^2\rho l^2/W\)	The master relation of this set
Voltage scaling	volume \(\propto 1/V^2\)	Double \(V\), save 75%
Power-factor scaling	volume \(\propto 1/\cos^2\phi\)	Combines as \(1/(V\cos\phi)^2\)
3-φ vs 1-φ, equal line V	\(0.75\)	Independent of power factor
Equal max voltage to earth, relative to d.c. 2-wire (one earthed) = 1
D.C. 2-wire, midpoint earthed	\(0.25\)	Best of all — nothing added
D.C. 3-wire	\(0.3125\)	Half-section neutral
1-φ 2-wire, one earthed	\(2/\cos^2\phi\)	Worst of all
1-φ 2-wire, midpoint earthed	\(0.5/\cos^2\phi\)
1-φ 3-wire	\(0.625/\cos^2\phi\)	Half-section neutral
2-φ 4-wire, midpoints earthed	\(0.5/\cos^2\phi\)
3-φ 3-wire	\(0.5/\cos^2\phi\)	The transmission standard
3-φ 4-wire	\(0.583/\cos^2\phi\)	Half-section neutral; +16.7%
Economics and distributors
Kelvin's law	\(a = \sqrt{P_3/P_2}\)	Loss cost \(=\) variable capital charge
Economic voltage	\(V = 5.5\sqrt{L/1.6 + P/150}\)	kV, km, kW — empirical
Transmission efficiency	\(\eta = P_R/(P_R + W)\)	Not \(1 - W/P_R\)
Loss load factor	\(W_{\text{avg}}/W_{\text{peak}}\)	Always below the load factor
Distributor, one end	drop \(= \tfrac12 il^2r\)	As if lumped at the midpoint
Distributor, one end	loss \(= \tfrac13 i^2l^3r\)	One third — not one half
Distributor, both ends	drop \(= \tfrac18 il^2r\)	Maximum at the centre
Neutral current	\(\mathbf{I}_N = \mathbf{I}_A+\mathbf{I}_B+\mathbf{I}_C\)	Phasor sum, never arithmetic

Diagnostics

Common Mistakes

Not identifying which voltage is held constant. Equal voltage between conductors gives 0.75 for three-phase against single-phase; equal maximum voltage to earth gives 0.5 against d.c. Both appear in this set and they are answers to different questions — Challenge C3.
Confusing percentage loss with absolute loss. Problem 2 holds the percentage constant and gets 100% extra load; holding the absolute loss constant gives a different and smaller figure.
Forgetting the \(\sqrt2\) when comparing a.c. with d.c. on a peak-voltage basis. It cancels between two a.c. systems and does not cancel between an a.c. and a d.c. one.
Taking efficiency as \(1 - \%\text{loss}\). A 6% loss gives 94.34% efficiency, not 94%. The two are reckoned against different quantities — Problem 8.
Taking 10% loss as 10% of the sending-end power. It is 10% of the power transmitted, as in Problem 5.
Omitting the neutral from the volume when it carries no current. It carries no current but it is still metal that must be bought — Problem 13.
Including the fixed cost \(P_1\) in Kelvin's law. Only the part of the capital cost proportional to area enters the optimum.
Using \(\tfrac12\) for the loss in a uniformly loaded distributor. The drop coefficient is \(\tfrac12\); the loss coefficient is \(\tfrac13\), because loss involves \(I^2\).
Adding unbalanced phase currents arithmetically to find the neutral current. 40, 30 and 20 A give 17.3 A, not 90 A and not 10 A — Problem 19.
Quoting Kelvin's optimum to three decimal places. The cost curve is nearly flat near the minimum, so the section is far less precisely determined than the arithmetic suggests — Problem 11.
Treating the empirical economic-voltage formula as exact. It selects a rung of the standard voltage ladder; it does not determine a voltage.
Assuming a uniformly distributed load behaves like a load at the far end. It behaves like one at the midpoint, giving half the drop — Problem 15.

Looking Ahead

Every comparison in this set treated the conductor as a pure resistance of area \(a\) and length \(l\), and that was enough to settle the architecture of the system: three phases, high voltage, corrected power factor. It is not enough to settle anything else.

Set 3 removes the arithmetic burden that these comparisons have exposed — the endless tracking of which quantity is line and which is phase, and which voltage level a given impedance belongs to — by normalising everything to a per-unit base. Sets 5 to 8 then return to the conductor itself and ask what else it does besides dissipate: the inductance set by its spacing, the capacitance set by its diameter, and the corona loss set by its surface gradient, which is the mechanism that finally halts the \(1/V^2\) argument of Problem 1.