Electric Transmission of Water Power

by Alton D. Adams

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.