The voltage on a transmission line may be anything up to at least 60,000, and the weight of conductors varies inversely with the square of the figures selected, the power, length and loss being constant. Whatever the total line pressure, the weight of conductors varies inversely with the percentage of loss therein.

The case of maximum loss and minimum weight of conductors is that in which all of the transmitted energy is expended in heating the line wires. Such a case would never occur in practice, because the object of power transmission is to perform some useful work.

Minimum loss is theoretically zero, and the corresponding weight of conductors is infinite, but these conditions obviously cannot be attained in practice. Between these extremes of minimum and of infinite weights of conductors comes every practical transmission with a line loss greater than zero and less than 100 per cent.

To determine the weight and allowable cost of conductors, the cost of the energy that will be annually lost in them enters as one of the factors. At this point the distinction between the percentage of power lost at maximum load and the percentage of total energy lost should come into view.

Line loss ordinarily refers to the percentage of total power consumed in the conductors at maximum load. This percentage would correspond with that of total energy lost if the line current and voltage were constant during all periods of operation, but this is far from the case.

A system of transmission may operate with either constant volts or constant amperes on the line conductors, but in a practical case constancy of both these factors is seldom or never to be had. This is because the product of the line volts and amperes represents accurately in a continuous-current system, and approximately in an alternating-current system, the amount of power transmitted. In an actual transmission system, the load—that is, the demand for power—is subject to more or less variation at different times of the day, and the line volts or amperes, or both, must vary with it.

If the transmission system is devoted to the operation of one or more factories the required power may not vary more than twenty-five per cent during the hours of daily use; but if a system of general electrical supply is to be operated, the maximum load will usually be somewhere between twice and four times as great as the average load for each twenty-four hours. Such fluctuating loads imply corresponding changes in the volts or amperes of the transmission line.

A number of rather long transmissions is carried out in Europe with continuous, constant current, and in such systems the line voltage varies directly with the load. As the loss of power in an electrical conductor depends entirely on its ohms of resistance, which are constant at any given temperature, and on the amperes of current passing through it, the line loss in a constant-current system does not change during the period of operation, no matter how great may be its changes of load. For this reason the percentage of power loss in the line at maximum load is usually smaller than the percentage of energy loss for an entire day.

If, for example, the constant-current transmission line is designed to convert into heat 5 per cent of the maximum amount of energy that will be delivered to it per second—that is, to lose 5 per cent of its power at maximum load—then, when the power which the line receives drops to one-half of its maximum, the percentage of loss will rise to 10, because 0.05 ÷ 0.5 = 0.1. So again, when the power sent through the line falls to one-quarter of the full amount, the line loss will rise to 0.05 ÷ 0.25 = 0.2, or 20 per cent.

From these facts it is clear that a fair all-day efficiency for a constant-current transmission line can be obtained only in conjunction with a high efficiency at maximum load, if widely varying loads are to be operated. It does not necessarily follow from these facts as to losses in constant-current lines that such losses should always be small at maximum loads, for if a large loss may be permitted at full load a still greater percentage of loss at partial loads may not imply bad engineering.

In a large percentage of electric water-power plants some water goes over the dam during those hours of the day when loads are light, the storage capacity above the dam not being sufficient to hold all of the surplus water during most seasons of the year. If, therefore, the line loss in a constant-current transmission, where all of the daily flow of water cannot be used, is not great enough to reduce the maximum load that would otherwise be carried, then the fact that the percentage of line loss at small loads is still larger is not very important.

Obviously, it makes little difference whether water goes over a dam or through wheels to make up for a loss in the line. In a case where all the water can be stored during small loads and used during heavy loads, it is clearly desirable to keep the loss in a constant-current line down to a rather low figure, say not more than five per cent, at maximum load.

Much the greater number of electrical transmissions are carried out with nearly constant line voltage, mostly alternating, and the line current in such cases varies directly with the power transmitted, except as to certain results of inductance on alternating lines. As line resistance is constant, save for slight variations due to temperature, the rate of energy loss on a constant-pressure line varies with the square of the number of amperes flowing, and the percentage of loss with any load varies directly as the number of amperes.

These relations between line losses and the amperes carried follow from the law that the power, or rate of work, is represented by the product of the number of volts by the number of amperes, and the law that the power actually lost in the line is represented by the product of the number of ohms of line resistance and the square of the number of amperes flowing in it. In each of these cases the power delivered to the line is, of course, measured in watts, each of which is ¹/₇₄₆ of a horse-power.

Applying these laws, it appears that if the loss of a certain constant-pressure transmission line is 10 per cent of the power delivered to it at full load, then, when the power, and consequently the amperes, on the line is reduced one-half, the watts lost in the line as heat will be (¹/₂)² = ¹/₄ of the watts lost at full load, because the number of amperes flowing has been divided by 2.

But the amount of power delivered to the line at full load having been reduced by 50 per cent, while the power lost on the line dropped to one-fourth of 10 per cent, or to 2.5 per cent of the full line load, it follows that the power lost on the line at half-load is represented by 0.025 ÷ 0.5 = 0.05, or 5 per cent of the power then delivered to it.

This rise in the efficiency of a constant-pressure transmission line as the power delivered to it decreases, together with the fact that maximum loads on such lines continue during hardly more than one to two hours daily, tends to raise the allowable percentage of line loss at maximum loads.

This is so because a loss of fifteen per cent at maximum load may easily drop to an average loss of somewhere between five and ten per cent for the entire amount of energy delivered to a line during each day under ordinary conditions in electrical supply. In the practical design of transmission lines, therefore, the sizes of conductors are influenced by the relation of the largest load to be operated to the greatest amount of power available for its operation, and by questions of regulation, as well as by considerations of all-day efficiency.

If the maximum load that must be carried by a transmission system during a single hour per day requires nearly as much power as can be delivered to the line conductors, either because of lack of water storage or of water itself, even if it is stored, it may be desirable to design these conductors for a small loss at maximum load, rather than to install a steam plant.

So again, as the fluctuation in voltage at the delivery end of a transmission line between no load and full load will amount to the entire drop of volts in the line at full load, if the pressure at the generating end is constant, the requirements of pressure regulation on distribution circuits limit the drop of pressure in the transmission conductors. For good lighting service with incandescent lamps at about 110 volts, the usual pressure, it is necessary that variations be held within one volt either way of the pressure of the lamps—that is, between 109 and 111 volts.

Every long-transmission system for general electrical supply necessarily includes one or more sub-stations where the distribution lines join the transmission circuits, and where the voltage for lighting service is regulated. As the limits of voltage variations on lighting circuits are so narrow, it is necessary to keep the changes of pressure on the transmission lines themselves within moderate limits, or such as can be compensated for at sub-stations.

This is particularly true in cases where energy transmitted over a single circuit is distributed for both incandescent lamps and large electric motors, because the starting and operation of such motors causes large fluctuations of amperes and terminal voltage on the transmission circuits. To hold such fluctuations within limits which a sub-station can readily compensate for, it is necessary that the loss in the transmission line be moderate, say often within ten per cent of the total voltage delivered to it at maximum load.

Capacity and cost of equipment at generating stations go up with the percentage of line loss, and thus serve to limit its economical amount. For every horse-power delivered to a transmission line at a water-power station there must be somewhat more than one horse-power of capacity in water-wheels, at least one horse-power in generators, and frequently a further capacity of one horse-power in step-up transformers. Every additional horse-power lost in the line at maximum load, if the generating plant is to be worked up to its full capacity, implies an addition of somewhat more than one horse-power capacity in water-wheels, one horse-power in generators, and one horse-power in transformers.

Since the cost of a generating station is thus increased as the maximum line loss is raised, a point may be reached where any further saving in the cost of the line is more than offset by the corresponding addition to the cost of the station and of its operation. Just where this point, as indicated by a percentage of line loss, is to be found depends on the factors of each case, important among which is the length of the transmission line.

Much effort has been made to fix some exact relation for maximum economy between the first cost of conductors for a transmission line and the amount of energy annually lost as heat therein. The best-known statement applying to this case is that of Lord Kelvin, made in a paper read before the British Association in 1881. According to the rule there laid down, the most economical size for the conductors of a transmission line is that for which the annual interest on first cost equals the cost of the energy annually wasted in them.

If transmission systems were designed for the sole purpose of wasting energy in their line conductors this rule would exactly apply, for it simply shows how the cost of energy wasted, plus the interest on the cost of the conductor in which it is wasted, may be brought to a minimum. As a matter of fact, transmission systems are primarily intended to deliver energy rather than to waste it; but of the proportions of the entire energy to be delivered and wasted (which is exactly what we want to know), the rule of Kelvin takes no account.

According to his rule, the cheaper the cost of power where it is developed, the less should be paid for conductors to bring it to market. The obvious truth is that the less the cost of power development at a particular point, the more may be invested in a line to bring it to market. If power cost nothing whatever at its source it would not be worth while to build any transmission line at all if this rule is correct.

A modification of Lord Kelvin’s rule has been proposed by which it is said that the interest on the cost of the conductors and the annual value of the energy lost in them should be equal, value here meaning what the energy can be sold for. This rule would make an investment in line conductors too large.

The entire cost of production and transmission for the delivered energy should not be greater than the cost of a like amount of energy developed at the point where the delivery is made. In this entire cost of production and transmission, interest on the investment in line conductors is only one item.

It is perhaps impossible to state any exact rule for the most economical relation between the cost of conductors and the loss of energy therein that will apply to every transmission. A maximum limit to the weight of conductors may, however, be set for most cases. This limit should not allow the annual interest and depreciation charges on the investment in line conductors, plus all other costs of development and transmission, to raise the total cost of the transmitted energy above the cost of development for an equal amount of energy at the point where the transmitted energy is delivered.

While the maximum investment in transmission conductors may be properly limited in the way just stated, it by no means follows that this maximum limit should be reached in every case. In the varying requirements of actual cases, the problem may be to deliver a fixed amount of power at the least possible cost, or to deliver the largest possible amount of power at a cost per unit under that of development at the point of use. Frequently a transmission system has a possible capacity in excess of present requirements, and a line that would not be too heavy for future business might put an unreasonable burden of interest charges on present earnings.

The foregoing considerations apply to the design of conductors for a transmission line after the voltage at which it is to operate has been decided on. Quite a different set of facts should influence the selection of this voltage. A transmission that would be entirely impracticable with any percentage of line loss that might be selected, if carried out at some one voltage, might represent a paying business at some higher voltage and any one of several sizes of line conductors. The power that could be delivered by a line of practicable cost, operated at one voltage, might be too small for the purpose in hand, while the available power at a higher voltage might be ample.

If any given power is to be transmitted with a given percentage of maximum loss in line conductors, the weight of these conductors will increase as the square of their length, and decrease as the square of the full voltage of operation in every case.

Thus, if the length of this transmission is doubled, the weight of the conductors must be multiplied by four, the voltage remaining the same; but if the voltage is doubled and the line length remains unchanged, the weight of conductors must be divided by four. With the length of line and the voltage of transmission either lowered or raised together, the weight of the conductors remains fixed, for constant power and loss.

An illustration of this last rule may be drawn from the case of lines designed to transmit any given power a distance of ten miles at 10,000 volts, and a distance of fifty miles at 50,000 volts, in which the total weight of conductors would be the same for each line if the percentage of loss was constant.

This statement of the rule as to proportionate increase of voltage and distance presents the advantages of high voltages in their most favorable light. Though a uniform ratio between the voltage of operation and the length of line allows a constant weight of conductors to be employed for the transmission of a given power with unchanging efficiency of conductors, yet other considerations soon limit the advantage thus obtained.

Important among these considerations may be mentioned the mechanical strength of line conductors, difficulties of line insulation, losses between conductors through the air, limits of generator voltages, and the cost of transformers.

If the ten-mile transmission at 10,000 volts, above mentioned, requires a circuit of two No. 1/0 copper wires, the total weight of these wires will be represented by (5,500 × 10 × 2 × 320) ÷ 1,000 = 35,200 pounds, allowing 5,500 feet of wire per mile of single conductor to provide something for sag between poles, and 320 pounds being the weight of bare No. 1/0 copper wire per 1,000 feet.

When the length of line is raised to 50 miles, the two-wire circuit will contain 5,500 × 50 × 2 = 550,000 feet of single conductor, and since the voltage is raised to 50,000 at the same time, the total weight of conductors will be 35,200 pounds as before. The weight of single conductor per 1,000 feet is therefore only 64 pounds in the 50-mile line.

A No. 7 copper wire, B. & S. gauge, has a weight of 63 pounds per 1,000 feet, and is the nearest regular size to that required for the 50-mile line as just found. It would be poor policy to string a wire of this size for a transmission line, because it is so weak mechanically that breaks would probably be frequent in stormy weather. The element of unreliability introduced by the use of this small wire on a 50-mile line would cost far more in the end than a larger conductor.

As a rule, No. 4 B. & S. gauge wire is the smallest that should be used on a long transmission line in order to give fair mechanical strength, and this size has just twice the weight of a No. 7 wire of equal length. Here, then, is one of the practical limits to the advantages that may be gained by increasing the voltage with the length of line.

As line voltage goes up, the strain on line insulation increases rapidly, and the insulators for a circuit operated at 50,000 volts must be larger and of a much more expensive character than those for a 10,000-volt circuit. In this way a part of the saving in conductors effected by the use of very high voltages on long lines is offset by the increased cost of insulation.

Another disadvantage that attends the operation of transmission lines at very high voltages is the continuous loss of energy by the silent passage of current through the air between wires of a circuit. This loss increases at a rapid rate after a pressure between 40,000 and 50,000 volts is reached with ordinary distances between the wires of each circuit. To keep losses of this sort within moderate limits, and also to lessen the probability of arcs on a circuit at very high voltage, the distance of eighteen inches or two feet between conductors that carry current at 10,000 volts should be increased to six feet or more on circuits that operate at 50,000 volts.

Such an increase in the distance between conductors makes the cost of poles and cross-arms greater, either by requiring them to be larger than would otherwise be necessary or by limiting the number of wires to two or three per pole and thus increasing the number of pole lines. These added expenses form another part of the penalty that must be paid for the use of very high voltages and the attendant saving in the cost of conductors.

Apparatus grows more expensive as the voltage at which it is to operate increases, because of the cost of insulating materials and the room which they take up, thereby adding to the size and weight of the iron parts.

Generators for alternating current can be had that develop as much as 13,500 volts, but such generators cost more than others of equal power that operate at between 2,000 and 2,500 volts. These latter voltages are as high as it is usually thought desirable to operate distribution circuits and service transformers in cities and towns, so that if more than 2,500 volts are employed on the transmission line, step-down transformers are required at a sub-station. For a transmission of more than ten miles the saving in line conductors by operation at 10,000 to 12,000 volts will usually more than offset the additional cost of generators designed for this pressure and of step-down transformers. If the voltage of transmission is to exceed that of distribution, it will generally be found desirable to carry the former voltage up to 10,000 or 12,000, at least.

As the cost of generators designed for the voltage last named is less than that of lower voltage generators plus transformers, step-up transformers should usually be omitted in systems where these pressures are not exceeded. For alternating pressures above 13,000 to 15,000 volts, step-up transformers must generally be employed. In order that the saving in the weight of line conductors may more than offset the additional cost of transformers when the voltage of transmission is carried above 15,000, this voltage should be pushed on up to as much as 25,000 in most cases.

Power transmission with continuous current has the advantage that the cost of generators remains nearly the same whatever the line voltage, and that no transformers are required. Such transmissions are common in Europe, but have hardly a footing as yet in the United States. The reason for the uniform cost of continuous-current generators is found in the fact that they are connected in series to give the desired line voltage, and the voltage of each machine is kept under 3,000 or 4,000. As a partial offset to the low cost of the continuous-current generators and to the absence of transformers, there is the necessity for motor-generators in a sub-station when current for lighting as well as power is to be distributed.

In spite of the various additions to the cost of transmission systems made necessary by the adoption of very high voltages, these additions are much more than offset by the saving in the cost of conductors on lines 30, 50, or 100 miles in length. In fact, it is only by means of voltages ranging from 25,000 to 50,000 that the greatest of these distances, and others up to more than 140 miles, have been successfully covered by transmission lines. Above 60,000 volts there has been but slight practical experience in the operation of transmission lines.

Calculations to determine the sizes of conductors for electric transmission lines are all based on the fundamental law discovered by Ohm, which is that the electric current flowing in a circuit at any instant equals the electric pressure that causes the current divided by the electric resistance of the circuit itself, or current = pressure ÷ resistance.

Substituting in this formula the units that have been selected because of their convenient sizes for practical use, it becomes, amperes = volts ÷ ohms, in which the ohm is simply the electrical resistance, taken as unity, of a certain standard copper bar with fixed dimensions.

The ampere is the unit flow of current that is maintained with the unit pressure of one volt between the terminals of a one-ohm conductor. When this formula is applied to the computation of transmission lines the volts represent the electrical pressure that is required to force the desired amperes of current through the ohms of resistance in any particular line, and these volts have no necessary or fixed relation to the total voltage at which the line may operate. Thus, if the total voltage of a transmission system is 10,000, it may be desirable to use 500, 1,000, or even 2,000 volts to force current through the line, so that one of these numbers will represent the actual drop or loss of volts in the line conductors when the number of amperes that represent full load is flowing. As it is a law of every electric circuit that the rate of transformation of electric energy to heat or work in each of its several parts is directly proportional to the drop of voltage therein, it follows that a drop of 500 or 1,000 or 2,000 volts in the conductors of a 10,000-volt transmission line at full load would correspond to a power loss of five to ten or twenty per cent respectively. Any other part of 10,000 volts might be selected in this case as the pressure to be lost in the line. Evidently no formula can give the number of volts that should be lost in line conductors at full load for a given power transmission, but this number must be decided on by consideration of the items of line efficiency, regulation, and the ratio of the available power to the required load.

Having decided on the maximum loss of volts in the line conductors, and knowing the full voltage of operation, the power and consequently the number of amperes delivered to the line at maximum load, the resistance of the conductors may then be calculated by the formula, amperes = volts ÷ ohms. Thus, if the proposition is to deliver 2,000,000 watts or 2,000 kilowatts to a two-wire transmission line with a voltage of 20,000, the amperes in each wire must be represented by 2,000,000 ÷ 20,000 = 100. With a drop of ten per cent or 2,000 volts in the two line conductors, their combined resistance must be found from 100 = 2,000 ÷ ohms, and the ohms are therefore twenty. If the combined length of the two conductors is 200,000 feet, corresponding to a transmission line of a little under twenty miles, the resistance of these conductors must be 20 ÷ 200 = 0.1 ohm per 1,000 feet. From a wire table it may be seen that a No. 1/0 wire of copper, B. & S. gauge, with a diameter of 0.3249 inch, has a resistance of 0.1001 ohm per 1,000 feet at the temperature of 90° Fahrenheit, a little less at lower temperatures, and is thus the required size. Obviously, the resistance of twenty ohms is entirely independent of the length of the line, all the other factors being constant, and wires of various sizes will be required for other distances of transmission.

It is often convenient to find the area of cross section for the desired transmission conductor instead of finding its resistance. This can be done by substituting in the formula, amperes = volts ÷ ohms, the expression for the number of ohms in any conductor, and then solving as before.

Electrical resistance in every conductor varies directly with its length, inversely with its area of cross section, and also has a constant factor that depends on the material of which the conductor is composed. This constant factor is always the same for any given material, as pure iron, copper, or aluminum, and is usually taken as the resistance in ohms of a round wire one foot long and 0.001 inch in diameter, of the material to be used for conductors. Such a wire is said to have an area in cross section of one circular mil, because the square of its diameter taken as unity is still unity, that is, 1 × 1 = 1. In like manner, for the convenient designation of wires by their areas of cross-section, each round wire of any size is said to have an area in circular mils equal to the square of its diameter measured in units of 0.001 inch each. Thus, a round wire of 0.1 inch diameter has an area of 100 × 100 = 10,000 circular mils, and a round wire one inch in diameter has an area of 1,000 × 1000 = 1,000,000 circular mils. The circular mils of a wire do not express its area of cross section in terms of square inches, but this is not necessary since the resistance of a wire of one circular mil is taken as unity. Obviously, the areas of all round wires are to each other as are their circular mils.

From the foregoing it may be seen that the resistance of any round conductor is represented by the formula, ohms = l × s ÷ circular mils, in which l represents the length of the conductor in feet, s is the resistance in ohms of a wire of the same material as the conductor but with an area of one circular mil and a length of one foot, and the circular mils are those of the required conductor. Substituting the quantity, l × s ÷ circular mils, for ohms in the formula, amperes = volts ÷ ohms, the equation, amperes = volts ÷ (l × s ÷ circular mils), is obtained, and this reduces to circular mils = amperes × l × s ÷ volts. For any proposed transmission all of the quantities in this formula are known, except the desired circular mils of the line conductors. The constant quantity s is about 10.8 for copper, but is conveniently used as eleven in calculation, and this allows a trifle for the effects of impurities that may exist in the line wire.

The case above mentioned, where 2,000 kilowatts were to be delivered to a transmission line at 20,000 volts, and a loss of 2,000 volts at full load was allowed in the line conductors, may now be solved by the formula for circular mils. Taking the resistance of a round copper wire 0.001 inch in diameter and one foot long as eleven ohms, and substituting the 100 amperes, 2,000 volts, and 200,000 feet of the present case in the formula, gives circular mils = (100 × 200,000 × 11) ÷ 2,000 = 110,000. The square root of this 110,000 will give the diameter of a copper wire that will exactly meet the conditions of the case, or the more practical course of consulting a table of standard sizes of wire will show that a No. 1-0 B. & S. gauge, with a diameter of 0.3249 inch, has a cross section of 105,500 circular mils, or about five per cent less than the calculated number, and is the size nearest to that wanted. As this No. 1-0 wire will give a line loss at full load of about 10.5 per cent, or only one-half of one per cent more than the loss at first selected, it should be adopted for the line in this case.

The formula just made use of is perfectly general in its application, and may be applied to the calculation of lines of aluminum or iron or any other metal just as well as to lines of copper. In order to use the formula for any desired metal, it is necessary that the resistance in ohms of a round wire of that metal one foot long and 0.001 inch in diameter be known and substituted for s in the formula. This resistance of a wire one foot long and 0.001 inch in diameter is called the specific resistance of the substance of which the wire is composed. For pure aluminum this specific resistance is about 17.7, for soft iron about sixty, and for hard steel about eighty ohms. The use of these values for s in the formula will therefore give the areas in circular mils for wires of these three substances, respectively, for any proposed transmission line. In the same way the specific resistance of any other metal or alloy, when known, may be applied in the formula.

The foregoing calculations apply accurately to all two-wire circuits that carry continuous currents, whether these circuits operate with constant current, constant pressure, or with pressure and current both variable. Where circuits are to carry alternating currents, certain other factors may require consideration. Almost all transmissions with alternating currents are carried out with three-phase three-wire, or two-phase four-wire, or single-phase two-wire circuits. Of the entire number of such transmissions, those with the three-phase three-wire circuits are in the majority, next in point of number come the two-phase transmissions, and lastly a few transmissions are carried out with single-phase currents. The voltage of a continuous-current circuit, by which the power of the transmission is computed and on which the percentage of line loss is based, is the maximum voltage operating there; but this is not true for circuits carrying alternating currents. Both the volts and amperes in an alternating circuit are constantly varying between maximum values in opposite directions along the wires. It follows from this fact that both the volts and amperes drop to zero as often as they rise to a maximum. It is fully demonstrated in books on the theory of alternating currents, that with certain ideal constructions in alternating generators, and certain conditions in the circuits to which they are connected, the equivalent or, as they are called, the virtual values of the constantly changing volts and amperes in these circuits are 0.707 of their respective maximum values. Or, to state the reverse of this proposition, the maximum volts and amperes respectively in these circuits rise to 1.414 times their equivalent or virtual values. These relations between maximum and virtual volts and amperes are subject to some variations with actual circuits and generators, but the virtual values of these factors, as measured by suitable volt- and amperemeters, are important in the design of transmission circuits, rather than their maximum values. When the volts or amperes of an alternating circuit are mentioned, the virtual values of these factors are usually meant unless some other value is specified. Thus, as commonly stated, the voltage of a single-phase circuit is the number of virtual volts between its two conductors, the voltage of a two-phase circuit is the number of virtual volts between each pair of its four conductors, and the voltage of a three-phase circuit is the number of virtual volts between either two of its three conductors.

Several factors not present with continuous currents tend to affect the losses in conductors where alternating currents are flowing, and the importance of such effects will be noted later. In spite of such effects, the formula above discussed should be applied to the calculation of transmission lines for alternating currents, and then the proper corrections of the results, if any are necessary, should be made. With this proviso as to corrections, the virtual volts and amperes of circuits carrying alternating currents may be used in the formula in the same way as the actual volts and amperes of continuous current circuits. Thus, reverting to the above example, where 2,000 kilowatts was to be delivered at 20,000 volts to a transmission line in which the loss was to be 2,000 volts, the kilowatts should be taken as the actual rate of work represented by the alternating current, and the volts named as the virtual volts on the line. The virtual amperes will now be 100, as were the actual amperes of continuous current, and the size of line conductor for a single-phase alternating transmission will therefore be 1-0, the same as for the continuous-current line. If the transmission is to be carried out on the two-phase four-wire system, the virtual amperes in each of these wires will be fifty instead of 100, as the power will be divided equally between the two pairs of conductors, and each of these four wires should have a cross-section in circular mils just one-half as great as that of the No. 1-0 wire. The required wire will thus be a No. 3 B. & S. gauge, of 52,630 circular mils, this being the nearest standard size. In weight the two No. 1-0 wires and the four No. 3 wires are almost equal, and they should be exactly equal to give the same loss in the single-phase and the two-phase lines. For a three-phase circuit to make the transmission above considered, each of the three conductors should have an area just one-half as great as that of each of the two conductors for a single phase circuit, the loss remaining as before, and the nearest standard size of wire is again No. 3, as it was for the two-phase line. This is not a self-evident proposition, but the proof can be found in books devoted to the theory of the subject. From the foregoing it is evident that while the single-phase and two-phase lines require equal weights of conductors, all other factors being the same, the weight of conductors in the three-phase line is only seventy-five per cent of that in either of the other two. Neglecting the special factors that tend to raise the size and weight of alternating-current circuits, the single-phase and two-phase lines require the same weight of conductors as does a continuous-current transmission of equal power, voltage, and line loss. It should be noted that in each of these cases the factor l in the formula for circular mils denotes the entire length of the pair of conductors for a continuous-current line, or double the distance of the transmission with either of the alternating-current lines.

Having found the circular mils of any desired conductor, its weight per 1,000 feet can be found readily in a wire table. In some cases it is desirable to calculate the weight of the conductors for a transmission line without finding the circular mils of each, and this can be done by a modification of the above formula. A copper wire of 1,000,000 circular mils weighs nearly 3.03 pounds per foot of its length, and the weight of any copper wire may therefore be found from the formula, pounds = (circular mils × 3.03 × l) ÷ 1,000,000, in which pounds indicates the total weight of the conductor, l, its total length, and the circular mils are those of its cross-section. This formula reduces to the form, circular mils = (1,000,000 × pounds) ÷ (3.03 × l) and if this value for circular mils is substituted in the formula above given for the cross-section of any wire, the result is (1,000,000 × pounds) ÷ (3.03 × l) = (l × amperes × 11) ÷ volts. Transposition of the factors in this last equation brings it to the form, pounds = (3.03 × l² × amperes × 11) ÷ (1,000,000 × volts), which is the general formula for the total weight of copper conductors when l, the length of one pair, the total amperes flowing, and the volts lost in the conductors are known for either a continuous-current, a single-phase, or a two-phase four-wire line.

If the value of l, 200,000, of amperes, 100, and of volts, 2,000, for the transmission above considered are substituted in the formula for total weight, just found, the result is pounds = (3.03 (200,000)² × 100 × 11) ÷ (1,000,000 × 2,000), which reduced to pounds = 66,660, the weight of copper wire necessary for the transmission with either continuous, single-phase or two-phase current. With three-phase current the weight of copper in the line for this transmission will be 75 per cent of the 66,660 pounds just found. One or more two-wire circuits may be employed for the continuous current or for the single-phase transmission, and if one such circuit is used the weight for each of the two wires is obviously 33,660 pounds. For a two-phase transmission two or more circuits of two wires each will be used, and in the case of two circuits, if all four of the wires are of equal cross section, as would usually be the case, the total weight of each is 16,830 pounds. If the transmission is made with one three-phase circuit, the weight of each of the three wires is 16,830 pounds, and their combined weight, 50,490 pounds of copper. In each of these transmission lines the length of a single conductor in one direction is 100,000 feet, or one-half of the length of the wires in a single two-wire circuit. For the two-wire line the calculated weight of each conductor amounts to 66,660 ÷ 200 = 333.3 pounds per 1,000 feet. For a two-phase four-wire line and also for a three-phase three-wire line, the weight of each conductor is 16,830 ÷ 100 = 168.3 pounds per 1,000 feet. On inspection of a table of weights for bare copper wires it may be seen that a No. 1-0 B. & S. gauge wire has a weight of 320 pounds per 1,000 feet, and being much the nearest size to the calculated weight of 333 pounds should be selected for the two-wire circuit. It may also be seen that a No. 3 wire, with a weight of 159 pounds per 1,000 feet, is the size that comes nearest to the calculated weight of 168 pounds, and should therefore be employed in the three-wire and the four-wire circuits, for two- and three-phase transmissions. Either a continuous-current, single-phase, two-phase, or three-phase transmission line may of course be split up into as many circuits as desired, and these circuits may or may not be designed to carry equal portions of the entire power. In either case the combined weights of the several circuits should equal those above found, the conditions as to power, loss, and length of line remaining constant. It will be noted that the formulÆ for the calculation of the circular mils and for the weight of the conductors in the transmission line lead to the selection of the same sizes of wires, as they obviously should do.

Several laws governing the relations of volts lost, length and weight of line conductors, may be readily deduced from the above formulÆ. Evidently the circular mils and weight of line conductors vary inversely with the number of volts lost in them when carrying a given current, so that doubling this number of volts reduces the circular mils and weight of conductors by one-half. If the length of the line changes, the circular mils of the required conductors change directly with it, but the weight of these conductors varies as the square of their length. Thus, if the length of the line conductors is doubled, the cross-section in circular mils of each conductor is also doubled, and each conductor is therefore four times as heavy as before for the same current and loss in volts. Should the length of the conductors and also the number of volts lost in them be varied at the same rate, the circular mils of each conductor remain constant, and its weight increases directly with the distance of transmission. Thus, with the same size of line wire, both the number of volts lost and the total weight are twice as great for a 100- as for a fifty-mile transmission. If the total weight of conductors is to be held constant, then the number of volts lost therein must vary as the square of their length, and their circular mils must vary inversely as the length. So that if the length of a transmission line is doubled, the circular mils for conductors of constant weight are divided by two, and the volts lost are four times as great as before. Each of these rules assumes that the watts and percentage of loss in the line are constant.

The above principles and formulÆ apply to the design of transmission lines for either continuous or alternating currents, but where the alternating current is employed certain additional factors should be considered. One of these factors is inductance, by which is meant the counter-electromotive force that is always present and opposed to the regular voltage in an alternating current circuit. One effect of inductance is to cut down the voltage at that end of the line where the power is delivered to a sub-station, just as is also done by the ohmic resistance of the line conductors. Between the loss of voltage due to line resistance and the loss due to inductance there is the very important difference that the former represents an actual conversion of electrical energy into heat, while the latter is simply the loss of pressure without any material decrease in the amount of energy. While the loss of energy in a transmission line depends directly on its resistance, the loss of pressure due to inductance depends on the diameter of conductors without regard to their resistance, on the length of the circuit, the distance between the conductors, and on the frequency or number of cycles per second through which the current passes. As a result of these facts, it is not desirable or even practicable to use inductance as a factor in the calculation of the resistance or weight of a transmission line. On transmission lines, as ordinarily constructed, the loss of voltage due to inductance generally ranges between 25 and 100 per cent of the number of volts lost at full load because of the resistance of the conductors. This loss through inductance may be lowered by reducing the diameter of individual wires, though the resistance of all the circuits concerned in the transmission remains the same, by bringing the wires nearer together and by adopting smaller frequencies. In practice the volts lost through inductance are compensated for by operating generators or transformers in the power-plant at a voltage that insures the delivery of energy in the receiving-station at the required pressure. Thus, in a certain case, it may be desirable to transmit energy with a maximum loss of ten per cent in the line at full load, due to the resistance of the conductors, when the effective voltage at the generator end of the line is 10,000, so that the pressure at the receiving-station will be 9,000 volts. If it appears that the loss of pressure due to inductance on this line will be 1,000 volts, then the generators should be operated at 11,000 volts, which will provide for the loss of 1,000 volts by inductance, leave an effective voltage of 10,000 on the line, and allow the delivery of energy at the sub-station with a pressure of 9,000 volts, when there is a ten-percent loss of power due to the line resistance.

Inductance not only sets up a counter-electromotive force in the line, which reduces the voltage delivered to it by generators or transformers, but also causes a larger current to flow in the line than is indicated by the division of the number of watts delivered to it by the virtual voltage of delivery. The amount of current increase depends on both the inductance of the line itself and also on the character of its connected apparatus. In a system with a mixed load of lamps and motors there is quite certain to be some inductance, but it is very hard to predetermine its exact amount. Experience with such systems shows, however, that the increase of line current due to inductance is often not above five and usually less than ten per cent of the current that would flow if there were no inductance. To provide for the flow of this additional current, due to inductance, without an increase of the loss in volts because of ohmic resistance, the cross section of the line conductors must be enlarged by a percentage equal to that of the additional current. This means that in an ordinary case of a transmission with either single, two, or three-phase alternating current, the circular mils of each line wire, as computed with the formulÆ above given, should be increased by five to ten per cent. Such increase in the cross section of wires of course carries with it a like rise in the total weight of the conductors for the transmission. If wire of the cross section computed with the formulÆ is employed for the alternating current transmission, inductance in an ordinary case will raise the assumed line loss of power by five to ten per cent of what it would be if no inductance existed. Thus, with conductors calculated by the formulÆ for a power loss of ten per cent at full load, inductance in an ordinary case would raise this loss to somewhere between 10.5 and eleven per cent. As a rule it may therefore be said that inductance will seldom increase the weight of line conductors, or the loss of power therein, by more than ten per cent.

When an alternating current flows along a conductor its density is not uniform in all parts of each cross section, but the current density is least at the centre of the conductor and increases toward the outside surface. This unequal distribution of the alternating current over each cross section of a conductor through which it is passing increases with the diameter or thickness of the conductor and with the frequency of the alternating current. By reason of this action the ohmic resistance of any conductor is somewhat greater for an alternating than for a continuous current, because the full cross section of the conductor cannot be utilized with the former current. Fortunately, the practical importance of this unequal distribution of alternating current over each cross section of its conductor is usually slight, so far as the sizes of wires for transmission lines are concerned, because the usual frequencies of current and diameters of conductors concerned are not great enough to give the effect mentioned a large numerical value. Thus, sixty cycles per second is the highest frequency commonly employed for the current on transmission lines. With a 4-0 wire, and the current frequency named, the increase in the ohmic resistance for alternating over that for continuous current does not reach one-half of one per cent.

Having calculated the circular mils of weight of a transmission line by the foregoing formulÆ, it appears that the only material increase of this weight required by the use of alternating current is that due to inductance. This increase cannot be calculated exactly beforehand because of the uncertain elements in future loads, but experience shows that it is seldom more than ten per cent of the calculated size or weight of conductors.