TRANSMISSION LINES Copper wire—Setting of poles—Loss of power in transmission—Ohm's Law and examples of how it is used in figuring size of wire—Copper-wire tables—Examples of transmission lines—When to use high voltages—Over-compounding a dynamo to overcome transmission loss. Having determined on the location of the farm water-power electric plant, and its capacity, in terms of electricity, there remains the wiring, for the transmission line, and the house and barn. For transmission lines, copper wire covered with waterproof braid—the so-called weatherproof wire of the trade—is used. Under no circumstances should a wire smaller than No. 8, B. & S. gauge be used for this purpose, as it would not be strong enough mechanically. The poles should be of chestnut or cedar, 25 feet long, and set four feet in the ground. Where it is necessary to follow highways, they Size of Wire for Transmission To determine the size of the transmission wires will require knowledge of the strength of current (in amperes) to be carried, and the distance in feet. In transmission, the electric current is again analogous to water flowing in The loss in transmission is primarily measured in volts; and since the capacity of an electric current for work equals the volts multiplied by amperes, which gives watts, every volt lost reduces the working capacity of the current by so much. This loss is referred to by electrical engineers as the "C^2R loss," which is another way of saying that the loss is equal to the square of the current in amperes, multiplied by ohms resistance. Thus, if the The pressure of one volt (as we have seen in another chapter) is sufficient to force one ampere, through a resistance of one ohm. Such a current would have no capacity for work, since its pressure would be consumed in the mere act of transmission. If, however, the pressure were 110 volts, and the current one ampere, and the resistance one ohm, the effective pressure after transmission would be 110-1, or 109 volts. To force a 110-volt current of 50 amperes through the resistance of one ohm, would require the expenditure of 50 volts pressure. Its capacity for work, after transmission, would be 110-50, or 60 volts, × 50 amperes, or 3,000 watts. As this current consisted of 110 × 50, or 5,500 watts at the point of starting, the loss would be 2,500 watts, or about 45 per cent. It is bad engineering to allow more than 10 per cent loss in transmission. There are two ways of keeping this loss down. One is by increasing the size of the transmission wires, thus cutting down the resistance in ohms; the other way is by raising the voltage, thus cutting down the per cent loss. For instance, suppose the pressure was 1,100 volts, instead of 110 volts. Five amperes at 1,100 volts pressure, gives the same number of watts, power, as 50 amperes, at 110 volts pressure. Therefore it would be necessary to carry only 5 amperes, at this rate. The loss would be 5 volts, or less than ½ of 1 per cent, as compared with 45 per cent with 110 volts. Splicing transmission wire In large generating stations, where individual dynamos frequently generate as much as 20,000 horsepower, and the current must be transmitted over several hundred miles of territory, the voltage is frequently as high as 150,000, with the amperes reduced in proportion. Then the voltage is lowered to a suitable rate, and the amperage raised in proportion, by special machinery, at the point of use. It is the principle of the C^2R loss, which the The formula by which the size of transmission wire is determined, for any given distance, and a given number of amperes, is as follows: (Distance ft. one way × 22 × No. of amperes) / (Number of volts lost) = circular mills. In other words, multiply the distance in feet from mill to house by 22, and multiply this product by the number of amperes to be COPPER WIRE TABLE
CARRYING CAPACITY OF WIRES AND WEIGHT
Since two wires are required for electrical transmission, the above formula is made simple by counting the distance only one way, in feet, and doubling the resistance constant, 10.6, which, for convenience is taken as 22, instead of 21.2. Examples of Transmission Lines As an example, let us say that Farmer Jones has installed a water-power electric plant on his brook, 200 yards distant from his house. The generator is a 5 kilowatt machine, capable of producing 45 amperes at 110 volts pressure. He has a 3 horsepower motor, drawing 26 amperes at full load; he has 20 lights of varying capacities, requiring Answer: (600 × 22 × 45) / 5.5 = 108,000 circular mills. Transmission wire on glass insulator Referring to the table, No. 0 wire is 105,534 circular mills, and is near enough; so this wire would be used. It would require 1,200 feet, Farmer Jones says this is more money than he cares to spend for transmission. As a matter of fact, he says, he never uses his motor except in the daytime, when his lights are not burning; so the maximum load on his line at any one time would be 26 amperes, not 45. What size wire would he use in this instance? Substituting 26 for 45 in the equation, the result is 61,300 circular mills, which corresponds to No. 2 wire. It would cost $57.00. Now, if Farmer Jones, in an emergency, wished to use his motor at the same time he was using all his lights and his wife was ironing and making toast—in other words, if he wanted to use the 45 amperes capacity of his dynamo, how many volts would he lose? To get this answer, we change the formula about, until it reads as follows: (Distance in feet × 22 × amperes) / circular mills = Number of volts lost Substituting values, we have, in this case, (600 × 22 × 45) / 66,373 (No. 2) = 9 volts, nearly, less than 10 per cent. This is a very efficient line, under the circumstances. Now if he is willing to lose 10 per cent on half-load, instead of full load, he can save still more money in line wire. In that case (as you can find by applying the formula again), he could use No. 5 wire, at a cost of $28.50. He would lose 11 volts pressure drawing 26 amperes; and he would lose 18 volts pressure drawing 45 amperes, if by any chance he wished to use full load. In actual practice, this dynamo would be regulated, by means of the field resistance, to register 110 plus 11 volts, or 121 volts at the switchboard to make up for the loss at half-load. At full load, his voltage at the end of the line would be 121 minus 18, or 103 volts; his motor would run a shade slower, at this voltage, and his lights would be slightly dimmer. He would probably not notice the difference. If he did, he could walk over to his generating station, and raise A barn-yard light Thousands of plants can be located within 100 feet of the house. If Farmer Jones could do this, he could use No. 8 wire, costing $2.62. The drop in pressure would be 5.99 volts at full load—so small it could be ignored entirely. In this case the voltmeter should be made to read 116 volts at the switchboard, by means of the rheostat. If, on the other hand, this plant were 1,000 feet away from the house and the loss 10 volts the size wire would be (1,000 × 22 × 45) / 10 = 99,000 circular mills; a No. 0 wire comes nearest to this figure, and its cost, for 2,000 feet, at 19 cents a pound, would be $137.94. Benefit of Higher Voltages If Farmer Jones' plant is a half of a mile away from the house, he faces a more serious proposition in the way of transmission. Say he wishes to transmit 26 amperes with a loss of 10 volts. What size wire will be necessary? Thus: (2640 × 22 × 26) / 10 = 151,000 circular mills. A No. 000 wire is nearest this size, and 5,280 feet of it would cost over $650.00. This cost (2640 × 22 × 13) / 22 = 34,320 circular mills, or approximately a No. 5 wire which, at 19 cents a pound, would cost $120.65. Install a 550-volt generator, instead of a 220-volt machine and the amperes necessary would be cut to 5.2, and the volts lost would be raised to 55. In this case a No. 12 wire would carry the current; but since it would not be strong enough for stringing on poles, a No. 8 wire would be used, costing about $63. It will be readily seen from these examples how voltage influences the efficiency of transmission. Current generated at a pressure in excess of 550 volts is not to be recommended for farm plants unless an expert is in charge. In transmitting the electric current over miles of territory, engineers are accustomed to figure 1,000 volts for each mile. Since this is a deadly pressure, it should not be handled by any one not an expert, which, in this case, the farmer is not. Over-Compounding the Generator One can absorb the loss in transmission frequently, by over-compounding the machine. In describing the compound machine, in Chapter Five, it is shown that the usual compound dynamo on the market is the so-called flat-compounded type. In such a dynamo, the Now, by adding a few more turns to the series wires on the field coils of such a dynamo, a machine is to be had which gradually raises its voltage as the load comes on in increasing volume. Thus, one could secure such a machine, which would begin generating at 110 volts, and would gradually rise to 150 at full load. Yet the voltage would remain constant at the point of use, the excess being absorbed in transmission. A machine of this type can be made to respond to any required rise in voltage. As an example of how to take advantage of this very valuable fact, let us take an instance: Say that Farmer Jones has a transmission line 1,000 feet long strung with No. 7 copper wire. This 2,000 feet of wire would introduce a resistance of one ohm in the circuit. That is, every ampere of current drawn at his house would cause the working voltage there to fall On the other hand, if his dynamo was over-compounded 25 per cent—that is, if it gained 28 volts from no load to full load, the system would be perfect. In this case, the dynamo would be operated at 110 volts pressure at the switchboard with no load. At full load the voltmeter would indicate 110 plus 26, or 136 volts. The one or two lights burned at the power plant would be subject to a severe strain; but the 50 or 100 lights burned at the house and barn would burn at constant voltage, which is very economical for lamps. The task of over-compounding a dynamo can be done by any trained electrician. The farmer himself, if he progresses far enough in his study of electricity, can do it. It is necessary To make the task of over-compounding your own dynamo even more simple, write to the manufacturers, giving style and factory number of your machine. Tell them how much voltage rise you wish to secure, and ask them how many turns of "series" wire should Avoid overloading an over-compounded machine. Since its voltage is raised automatically, its output in watts is increased a similar amount at the switchboard, and, for a given resistance, its output in amperes would be increased the same amount, as can be ascertained by applying Ohm's Law. Your ammeter is the best guide. Your machine is built to stand a certain number of amperes, and this should not be exceeded in general practice. |