3°4

MODERN STEAM PRACTICE.

Area in sq. feet.

Diameter of the plunger, 23 inches = 2*8852Weight in Ibs. of

Plunger area. Lift in feet. 1 cubic ft. Ibs.

2*8852 x 285 x 62*5 = 51392

Now allowing -l-th to overcome the load on the air pump and thefriction of the engine, we have:

Load on the Length of Strohespiston in ibs. stroke. per minute.

61670 x io x io = 6167000

and

6167000

33000

= 186 horse-power nearly.

Approximate rille forpower of Cornish engine. —A simple rule usedby some engineers for calculating the quantity of water delivered froma given pump is as follows:—Let D =the diameter of the pump, then

—- represents the quantity of water in gallons delivered per i foot

stroke of pump nearly. Let S = the speed of the plunger of bücket

per minute, then S -~= the number of gallons delivered per minute.

Let L = the lift in feet, and the horse-power will be thus obtained:/ D 2 \

L 10 30/ . The following is an example:—Diameter of pump

33000

= 16, stroke of pump=7'5, number of strokes per minute = f$,lift =190 fathoms = 1140 feet.

Diameter of pump.16 X 16

Speed.

30

The work done =

- 8'5 x (7*5 x 7*5) = 478 gallons per minute.1140 x 10 x 478 _

33 000

• = 165 horse-power nearly.

This rule evidently allows for waste in the pump, but one-fifthmust be added to the sum for the friction of the engine.

To find the area of cylinder required to perform a given amountof work. —We may consider the mean pressure in the cylinder asfrom 14 to 15 Ibs. per square inch, and the velocity of the pistonfrom 80 to 85 feet per minute. It must be remembered that thepressure per square inch is derived from the actual water loaddivided by the area of the piston, and that one-fifth more powermust be allowed for friction. Thus the pressure multiplied by thevelocity equals so many foot pounds, which may be taken on anaverage as 1000. Therefore we divide the number of Ibs. of waterraised 1 foot high by 1000, and the quotient is the area of thecylinder in square inches. For example:—Suppose it be required