To find the Diameter of a Cylinder to work a Pump of a given diameter for a given depth.

RULE.-Multiply the square of the pump's diameter in inches by of the depth of the pit in fathoms, and the square root of the product will be the cylinder's diameter in inches.

EXAMPLE.-Required the diameter of a cylinder to work a pump 12 inches diameter and 27 fathoms deep.

√122 × 9 = 36 inches diameter.

To find the Diameter of a Pump that a Cylinder of a given diameter can work at a given depth.

RULE.-Divide three times the square of the cylinder's diameter in inches by the depth of the pit in fathoms, and the square root of the quotient will be the pump's diameter in inches.

EXAMPLE.-What diameter of a pump will a 36 inch cylinder be capable of working 27 fathoms deep? 362 × 3 27

12 inches diameter.

To find the Depth from which a Pump of a given diameter will work by means of a Cylinder of a given diameter.

RULE. Divide three times the square of the cylinder's diameter in inches by the square of the pump's diameter, also in inches; and the quotient will be the depth of the pit in fathoms.

EXAMPLE.-Required the depth that a cylinder of 36 inches diameter will work a pump of 12 inches diameter.

APPROXIMATE RULES FOR CALCULATING LIQUIDS.

To find the number of Imperial Gallons contained in any square or rectangular Cistern.

RULE.-Multiply the content of the cistern in cubic feet by 6.232, or the content in cubic inches by .003607, and the product is the number of gallons nearly.

EXAMPLE 1.-A cistern that is 8 feet long, 4 feet wide, and 3 feet deep; required its content in imperial gallons.

And

8 x 4.5 x 3 = 108 cubic feet,

108 x 6.232 673.056 gallons nearly.

Or, 8 feet 96 inches; 4 feet 54 inches, and 3 feet 36 inches; then,

96 x 54 x 36 = 186624 cubic inches,

And 186624 × .003607

673.152 gallons nearly.

Any two Dimensions of a square or rectangular Cistern being given to find the third, that shall contain any number of Imperial Gallons required.

RULE. Divide the number of gallons that the cistern is required to contain, by the product of the two dimensions multiplied by either of the multipliers as above, according as the dimensions are given in feet or inches, and the quotient will be the third dimensions of the cistern nearly.

EXAMPLE.-Required the depth of a cistern to contain 800 Imperial gallons, the length being 6 feet, and width 42 feet.

6.5 x 4.75 x 6.232 = 192.413; and

800 192.413 4.16 feet deep nearly.

To find the Content of a Cylinder in Imperial Gallons.

RULE.-Multiply the square of the diameter in feet by the length of the cylinder, also in feet, and by 4.895;

Or, the square of the diameter in inches by the length in feet and by .034;

Or, the square of the diameter in inches, by the length also in inches, and by .002832, and the product will be the content in gallons nearly.

EXAMPLE.-How many imperial gallons is contained in a well 22 feet deep, and 3 feet diameter.

3.52 x 22.5 x 4.895 1349.18 gallons nearly.

Or, 3 feet 42 inches,

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And 422 x 22.5 x .034 = 1349.46 gallons nearly. Also, 224 feet 270 inches,

And 422 × 270 x .002832 = 1349.3 gallons nearly.

The Length of a Cylinder given, to find the Diameter, or the Diameter given, to find the Length, that shall contain any number of Imperial Gallons required.

RULE.-Divide the number of gallons that the cylinder is required to contain, by the length in feet multiplied by 4.895, and the square root of the quotient is the diameter in feet, and parts of a foot;

Or, divide the number of gallons by the square of the diameter in feet multiplied by 4.895, and the quotient is the length in feet and parts of a foot,-and,

If the dimensions are inches in place of feet, use 354 in place of 4.895.

L

EXAMPLE. What must be the diameter of a cylinder to contain 5 imperial gallons, when the length is 20 inches?

354 x 5 9.4 inches diameter nearly.

20

The cube of the diameter of a sphere in feet, multiplied by 3.263= imperial gallons;

Or, the cube of the diameter of a sphere in inches, multiplied by .001888 = imperial gallons.

NOTE. The weight of a cubic foot of water = 62.5 lbs. avoirdupois.

Weight of a cubic inch = .03617 lbs. avoirdupois. Weight of a column of water 12 inches high and 1 inch square .434 lbs. avoirdupois.

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Weight of a cylindrical foot of water = 49.1 lbs. avoirdupois.

Weight of a cylindrical inch = .02842 lbs. avoirdupois.

Weight of a column of water 12 inches high and 1 inch diameter = .341 lbs. avoirdupois.

Take for example a column of water 11 inches diameter and 15 feet high, required its weight.

112 x 15 = 1815 × .341 = 618.915 lbs. avoirdupois nearly.

OF STEAM AND THE STEAM ENGINE.

Steam is the visible moist vapour which arises from all bodies that contain juices easily expelled from them by heats not sufficient for their combustion.

But steam, as applicable at present to the steamengine, is highly rarified water, the particles of which are expanded by the absorption of caloric, or the matter of heat.

Water rises in vapour at all temperatures, but is confined to the surface of the fluid acted upon, until it has attained 212° Fahrenheit, called the boiling point; at that heat steam ascends through it, preventing its elevation to a higher temperature by carrying the heat off in a latent form

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The latent heat of steam at the common pressure of the atmosphere, according to very accurate experiments, is found to be 1000°; and we know that the sensible, or thermometric heat 212°.--Now 212° 32° = 180°, and 1000+ 180 1180°, therefore, steam at 212 is highly rarified water, containing 1180° of heat; hence, to find the latent heat of steam at any other temperature, subtract the sensible heat from 1180° and the difference plus 32° the latent heat.

EXAMPLE. Required the latent heat of steam whose sensible heat is 224°.

And 956 +32= 988° latent heat.

Again, by adding together the latent heat, and sensible heat of steam at any temperature, another constant number is obtained, thus 988° + 224° = 1212°; for, as the sensible heat increases, the latent heat diminishes; hence, to find the quantity of water necessary to reduce or condense steam to any temperature required,—

Subtract the required temperature of the water from 1212°, and divide the remainder by the required