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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 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.


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 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 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) / (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 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, entitled "circular mills area."


COPPER WIRE TABLE





























































































































































































B.& S.

Gauge
Feet per Lb. Area in

circular mills
(R) Ohms

per 1,000 feet
Feet

per Ohm
(R) Ohms

per pound
0000 1.561 211,600.0 .04904 20,392.90 .00007653
000 1.969 167,805.0 .06184 16,172.10 .00012169
00 2.482 133,079.0 .07797 12,825.40 .00019438
0 3.130 105,534.0 .09829 10,176.40 .00030734
1 3.947 83,694.0 .12398 8,066.00 .00048920
2 4.977 66,373.0 .15633 6,396.70 .00077784
3 6.276 52,634.0 .19714 5,072.50 .00123700
4 7.914 41,742.0 .24858 4,022.90 .00196660
5 9.980 33,102.0 .31346 3,190.20 .00312730
6 12.58 26,250.0 .39528 2,529.90 .00497280
7 15.87 20,816.0 .49845 2,006.20 .00790780
8 20.01 16,509.0 .62840 1,591.10 .01257190
9 25.23 13,094.0 .79242 1,262.00 .01998530
10 31.82 10,381.0 .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

















































































































B. & S.

Gauge No.
Weight 1,000 ft.

Weatherproof (Pounds)
Carrying capacity

Weatherproof (Amperes)
Carrying capacity

rubber cov. (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? Substituting these values in the above formula, the result is:


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, 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?


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 the voltage a further 7 volts by turning the rheostat handle another notch.


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. 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


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 would be prohibitive. If, however, he installed a 220-volt dynamo—at no increase in cost—then he would have to transmit only a half of 26 amperes, or 13 amperes, and he could allow 22 volts loss, counting 10 per cent. In this case, the problem would work out as follows:


(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. A safer rule is not to exceed 220 volts, for while 550 volts is not necessarily deadly, it is dangerous. When one goes into higher voltages, it is necessary to change the type of dynamo to alternating current, so that the current can be transformed to safe voltages at the point where it is used. Since only the occasional farm plant requires a high-tension system, the details of such a plant will not be gone into here.


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 voltage remains constant at the switchboard, from no load to full load, allowing for a slight curve which need not be taken into account.


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 one volt. If he drew 26 amperes, the voltage would fall, at the house, 26 volts. If his switchboard voltage was set at say 120, the voltage at his house, at 26 amperes of load, would fall to 94 volts, which would cause his lights to dim considerably. It would be a very unsatisfactory transmission line, with a flat-compounded dynamo.


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 remove the top or "series" winding from the field coils. Count the number of turns of this wire to each spool. Then procure some identical wire in town and begin experimenting. Say you found four turns of field wire to each spool. Now wind on five, or six, being careful to wind it in the same direction as the coils you removed and connect it in the same way. If this additional number of turns does not raise the voltage enough, in actual practice, when the dynamo is running from no load to full load, add another turn or two. With patience, the task can be done by any careful mechanic. The danger is in not winding the coils the same way as before, and getting the connections wrong. To prevent this mistake, make a chart of the "series" coils as you take them off.


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 be wound on each spool in place of the old "series" coil. They could tell you exactly, since they have mathematical diagrams of each machine they make.


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.







Next: Wiring The House

Previous: What Size Plant To Install



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