1873 - Old Forge Coal Mines
1873 - Old Forge Coal Mines
1873 - Old Forge Coal Mines
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62 ANNUAL REPORT OF THE<br />
The work is done before the steam is cut off=2 feetX20 lbs., pressure=<br />
40 lbs. plus 36.61 lbs=76.61 lbs., minus work done against the piston by<br />
the vapor in the condenser=4 lbs.X5 feet—20 lbs.<br />
These vaWes give 36.61 plus 40 minus 20=56.61.<br />
When total work on 1 inch of piston=56.61 pounds.<br />
Example 3.—Two oscillating marine engines, (condensers,) whose cylinders<br />
are 42 inches in diameter, and steam admitted by the elastic force of<br />
20 lbs. per square inch ; the length of stroke being 3 feet and 1 inches,<br />
making 30 strokes per minute, and working with piston rods 5 inches in<br />
diameter. It is desired to know what these valves will work up to in effective<br />
horse-power.<br />
General Rule for Finding Power op<br />
Engines.<br />
Square the diameter of the piston in inches and parts, deduct half the<br />
circular area of the piston rod, multiply .7854, multiply length of stroke in<br />
feet and decimal parts, multiply the numoer of strokes per minute, multiply<br />
the number of pounds pressure per square inch of safet}^ valve, minus<br />
one-fifth (I) of the whole for modulus or friction, and divide by 33000, or<br />
equal horse-power.<br />
In the above example,<br />
42X42—12.5, half are of piston rodX. 7854X3.83X30X20<br />
_i=76.635 horse-power<br />
33000.<br />
of one engine ; hence 76.635+76.635=153.270<br />
horse-power, the same valves working 23^ strokes per minute, gives 120<br />
horse-power.<br />
Steam Ships.<br />
How to ascertain the speed of a steam ship with revolving paddles.<br />
Rale.—A steamer of 17 feet 6-inch paddle-wheel, the rims of the wheels<br />
being 4 inches broad, for which deduct twice 4 inches and 1<br />
inch clearance<br />
for working, brings the wheel to 16 feet 9 inches diameter, making 30 revolutions<br />
per minute. How many miles per hour ?<br />
Here— 16 feet 9 inchesX3.1416=52.6218, or 52 feet 7£ inches=l revolution.<br />
Then 52.6218X30 revolutions X 60 minutes-=-3 feet, or 1 yard-=-<br />
1,760 yards for miles=17.93 miles, or 18 miles nearly per hour ; and 60<br />
minutes-=-l 7.93=3.34, or 3 minutes and 20 seconds in completing 1 mile.<br />
This rule can be relied upon as being correct.<br />
Horse Power op Steamers, Paddle-Wheels.<br />
Suppose the four engines that propel the paddles of the Great Eastern<br />
to have steam of the elastic force of 20 pounds to the square inch admitted<br />
into the cylinders; the pistons moving through 140 feet of space in one<br />
minute, thus describing 10 revolutions per minute, or equal 14 feet each<br />
per minute ;<br />
the diameter of pistons being 74 inches each, the piston rods<br />
being 9 inches each in diameter. It is desired to know the horse-power<br />
that these values will give.<br />
Here: 74X74—\ area of 9 inches for piston rodsX7,854X10X10X20<br />
_l-5=9562G53,40800-r-33000=289.77=horse-power of these engines. This<br />
rule holds good for all such measurements.<br />
Pumping Engine Work Considered.<br />
Suppose an engine drawing water from a mine 100 fathoms deep, through<br />
the aid of three sets of pumps ; the bottom set lifts the water 50 fathoms,