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 should be set on the fence line; and in crossing public highways, the ordinance of your own town must guide you. Some towns prescribe a height of 19 feet above the road, others 27 feet, some 30. Direct current, such as is advised for farm installations, under ordinary circumstances, does not affect telephone wires, and therefore transmission lines may be strung on telephone poles. Poles are set at an average distance of 8 rods; they are set inclined outward on corners. Sometimes it is necessary to brace them with guy wires or wooden braces. Glass insulators are used to fasten the wires to the cross-arms of the poles, and the tie-wires used for this purpose must be the same size as the main wire and carry the same insulation.
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 pipes. It is subject to resistance, which cuts down the amount of current (in watts) delivered.
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 amperes carried is 10, and the ohms resistance of the line is 5, then the loss in watts to convey that current would be (10 × 10) × 5, or 500 watts, nearly a horsepower.
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.