CHAPTER VII

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 circ.u.mstances 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 circ.u.mstances, 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. Gla.s.s 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 a.n.a.logous to water flowing in pipes. It is subject to resistance, which cuts down the amount of current (in watts) delivered.

[Ill.u.s.tration: Bringing wires into the house or barn]

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.

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 1/2 of 1 per cent, as compared with 45 per cent with 110 volts.

[Ill.u.s.tration: 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 farmer must apply in determining the size of wire he is to use in transmitting his current from the generator switchboard to his house or barn. The wire table on page 159, together with the formula to be used in connection with it, reduce the calculations necessary to simple arithmetic. In this table the resistance of the various sizes of wire is computed from the fact that a wire of pure copper 1 foot long, and 1/1000 inch in diameter (equal to one circular mill) offers a resistance of 10.6 ohms to the foot. The principle of the C^2R loss is founded on Ohm's Law, which is explained in Chapter V.

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 circular ------------------------------------------ = mills.

Number of volts lost

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 carried. Then divide the product by the _number of volts_ to be lost; and the result will be the diameter of the wire required _in circular mills_. By referring to the table above, the B. & S. gauge of the wire necessary for transmission, can be found from the nearest corresponding number under the second column, ent.i.tled "circular mills area."

COPPER WIRE TABLE

--------+----------+-----------+-----------+-----------+------------ | | _Area in | _(R) Ohms | | _B.& S. | _Feet | circular | per 1,000 | _Feet | _(R) Ohms Gauge_ | per Lb._ | mills_ | feet_ | per Ohm_ | per pound_ --------+----------+-----------+-----------+-----------+------------ 0000 | 1.561 | 211,600 | .04904 | 20,392.90 | .00007653 000 | 1.969 | 167,805 | .06184 | 16,172.10 | .00012169 00 | 2.482 | 133,079 | .07797 | 12,825.40 | .00019438 0 | 3.130 | 105,534 | .09829 | 10,176.40 | .00030734 1 | 3.947 | 83,694 | .12398 | 8,066.00 | .00048920 2 | 4.977 | 66,373 | .15633 | 6,396.70 | .00077784 3 | 6.276 | 52,634 | .19714 | 5,072.50 | .00123700 4 | 7.914 | 41,742 | .24858 | 4,022.90 | .00196660 5 | 9.980 | 33,102 | .31346 | 3,190.20 | .00312730 6 | 12.58 | 26,250 | .39528 | 2,529.90 | .00497280 7 | 15.87 | 20,816 | .49845 | 2,006.20 | .00790780 8 | 20.01 | 16,509 | .62840 | 1,591.10 | .01257190 9 | 25.23 | 13,094 | .79242 | 1,262.00 | .01998530 10 | 31.82 | 10,381 | .99948 | 1,000.50 | .03178460 11 | 40.12 | 8,234.0 | 1.26020 | 793.56 | .05054130 12 | 50.59 | 6,529.9 | 1.58900 | 629.32 | .08036410 13 | 63.79 | 5,178.4 | 2.00370 | 499.06 | .12778800 14 | 80.44 | 4,106.8 | 2.52660 | 395.79 | .20318000 15 | 101.4 | 3,256.7 | 3.18600 | 313.87 | .32307900 16 | 127.9 | 2,582.9 | 4.01760 | 248.90 | .51373700 17 | 161.3 | 2,048.2 | 5.06600 | 197.39 | .81683900 18 | 203.4 | 1,624.3 | 6.38800 | 156.54 | 1.29876400 --------+----------+-----------+-----------+-----------+------------

CARRYING CAPACITY OF WIRES AND WEIGHT

-----------+-------------------+--------------------+-------------------- | _Weight 1,000 ft. | _Carrying capacity | _Carrying capacity _B. & S. | Weatherproof | Weatherproof | rubber cov.

Gauge No._ | (Pounds)_ | (Amperes)_ | (Amperes)_ -----------+-------------------+--------------------+-------------------- 0000 | 800 | 312 | 175 000 | 666 | 262 | 145 00 | 500 | 220 | 120 0 | 363 | 185 | 100 1 | 313 | 156 | 95 2 | 250 | 131 | 70 3 | 200 | 110 | 60 4 | 144 | 92 | 50 5 | 125 | 77 | 45 6 | 105 | 65 | 35 7 | 87 | 55 | 30 8 | 69 | 46 | 25 10 | 50 | 32 | 20 12 | 31 | 23 | 15 14 | 22 | 16 | 10 16 | 14 | 8 | 5 18 | 11 | 5 | 3 -----------+-------------------+--------------------+--------------------

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 1,200 watts, or 10 amperes when all on; and his wife uses irons, toasters, etc., which amount to another 9 or 10 amperes--say 45 altogether. The chances are that he will never use all of the apparatus at one time; but for flexibility, and his own satisfaction in not having to stop to think if he is overloading his wires, he would like to be able to draw the full _45 amperes_ if he wishes to.

He is willing to allow _5 per cent loss_ in transmission. _What size wires will be necessary, and what will they cost?_ Subst.i.tuting these values in the above formula, the result is:

Answer: 600 22 45 ------------- = 108,000 circular mills.

5.5

[Ill.u.s.tration: Transmission wire on gla.s.s 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, which would weigh, by the second table, 435.6 pounds. At 19 cents a pound, it would cost $82.76.

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?_

Subst.i.tuting 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 --------------------------------- = Number of volts lost circular mills

Subst.i.tuting 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 circ.u.mstances. 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 the voltage a further 7 volts by turning the rheostat handle another notch.

[Ill.u.s.tration: 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 --------------- = 99,000 circular mills; 10

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. A No. 0000 wire, costing $294.00, would give a 5 per cent drop at full load. In this case, the cost of transmission can be reduced to a much lower figure, by allowing a bigger drop at half-load, with regulation at the switchboard. Thus, a No. 2 wire here, costing but $95, would be satisfactory in every way. The loss at half-load would be about 9 volts, and the rheostat would be set permanently for 119 or 120 volts.

A modern dynamo can be regulated in voltage by over 25 per cent in either direction, without harm, if care is taken not to overload it.

_Benefit of Higher Voltages_