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.
[Illustration: 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.
[Illustration: 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, entitled "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?_ Substituting these
values in the above formula, the result is:
Answer: 600 × 22 × 45
------------- = 108,000 circular mills.
5.5
[Illustration: 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
--------------------------------- = Number of volts lost
circular mills
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.
[Illustration: 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_
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?
2640 × 22 × 26
Thus: -------------- = 151,000 circular mills.
10
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
-------------- = 34,320 circular mills,
22
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.