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472 notes on electrical terms.
Example.
An electrical motor is running on a 110 volt circuit and
using a current of 40 amperes. How many kilowatts ? How
many horse power ?
Solution
:
110 X 40
Power = ———— =4.4 kilowatts.
1000
110 X 40
Power = — = 6 horse power, very nearly.
Example.
A spool is wound with 3 pounds of No. 18 wire laid 4 layers,
250 turns to each layer. The voltage between its terminals is 10
volts. How many amperes are flowing through the spool ? How
many ampere-turns are there on the spool ? How much power
is consumed by the spool ? How many circular mils are there
in the wire for each ampere ?
Solution
:
In table No. 85 we find that 1 pound of No. 18 wire has a
resistance of 1.293 ohms; 3 pounds will, therefore, have a re-
sistance of 3X1.293 = 3.879 ohms and when the spool
warms up a little the resistance will practically be 4 ohms.
Then the current will be :
10
C = —— = 2yi amperes.
4
The spool has 4 X 250 = 1000 turns of wire.
It will, therefore, have 1000 X 2% = 2500 ampere-turns.
Power consumed will be :
W == 10 X 2 % = 25 watts.
In table No. 85 we find that the area of No. 18 wire is 1624
circular mils ; dividing 1624 by 2}£ we get 650 circular mils
per ampere.
It may be found in practice that a spool like this will heat
too much and, if so, the difficulty is overcome by winding more
turns of wire on the spool. This will not change the ampere-
turns and consequently not change the magnetic power in the
spool ; but it will increase the resistance, consequently decrease
the current flowing and thereby reduce the heat and also the
power consumed in the spool.
The temperature of any coil or spool depends greatly on the
area of the radiating surface, and may be determined by ex-
perience, but very seldom more than 1 watt of energy can be
allowed for each square inch of round radiating surface of the
spool. Frequently as much as 2 square inches of radiating sur-
face is required for each watt of energy lost in the spool.
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