To find the contents of a square tank: Multiply length, breadth and height in feet together; the result by 7.4; this will give the number of gallons in the tank. Or multiply the length, breadth and height in inches and the result by .004329.

To find the contents of a cylinder: Multiply the diameter in inches by itself, this by the height, and the result by .3400; this gives the number of gallons.

To find the circumference of a circle: Multiply the diameter by 3.14.

To find the area of a circle: Square the diameter and multiply by .7854.

To find the pressure per square inch exerted by a column of water, multiply the head in feet by 0.434.

The result will be the pressure in pounds.

To find the head in feet, the pressure being known, multiply the pressure per square inch by 2.31.

To find the contents of a barrel: To twice the square of the largest diameter, add the square of the smallest diameter and multiply this by the height, and the result by 2618. This will give the cubic inches in the barrel, and this divided by 231 will give the number of gallons.

To find the lateral pressure of water upon the side of a tank, multiply in inches the area of the submerged side by the pressure due to one-half the depth.

Example. - Suppose a tank to be 12 feet long and 12 feet deep. Find the pressure upon the side of tank, 144X144=20,736 inches area of side, 12X.43=5.16, pressure at bottom of tank; pressure at top, o; average, 2.6; therefore, 20,736X2.6=53,914, pressure on side of tank.

The rule of 6-8-10, to ascertain if an article is square: Measure off six inches on one side of the angle, and on the other measure 8 inches. If the article is perfectly square it will measure exactly 10 inches across from point to point. This rule holds good in feet.

Rule for finding the size of a pipe necessary to fill a number of smaller pipes: Suppose it is desired to fill from one pipe a 2-, 2 1/2- and 4-inch pipe. Draw a right angle, one arm 2 inches in length, the other 1 1/2 inches in length. From the extreme ends of the two arms draw a line. The length of this line in inches will give the size of pipe necessary to fill the two smaller pipes - about 3 1/4 inches. From one end of this last line, draw another line at right angles to it, 4 inches in length. Now, from the end of the 2-inch line to the end of the last line draw another line. Its length will represent the size of pipe necessary to fill a 2-, 2 1/2-and 4-inch pipe. This may be continued as long as desired.

Discharge of water: The amount of water discharged through a given orifice during a given length of time and under different heads, is as the square roots of the corresponding heights of the water in the reservoir above the surface of the orifice.

To find the number of gallons in a foot of pipe of any diameter: Multiply the square of the diameter of the pipe in inches by .0408.

To find the number of gallons of water that will drain from a roof: Multiply the number of square feet of roof by the average number of inches of rainfall per month, and the product by .623. The result gives the number of gallons which will drain from the roof in a month. When the roof is not flat or very nearly so its area should be considered as the area which it actually covers.

To find the power necessary to raise water to any given height: Multiply the number of cubic feet required per minute by the number of feet through which it is to be lifted. Then multiply this result by 6.23 and divide by 33,000, which will give the nominal horse power required. If the amount of water required per minute is in gallons, the multiplier will be 8.3 instead of 6.23.

To find the capacity of a cistern or well: Multiply the square of the diameter in inches by the decimal .7854. Multiply this result by the depth in inches, and divide this by 231. The final figure gives the contents in gallons.

Example.- Cistern, 10 ft. deep, 5 ft. in diameter. The square of 60, the diameter in inches, is 3,600.

3,600X7854=2,827.44.

2,827.44X120=3,392.93 cubic inches in cistern.

3,392.93/ 231 (cubic in. in gal.)=1,469 gallons.

To find the thickness of lead pipe necessary for a given head of water: Multiply the head in feet by the size of pipe required, expressed as a decimal, and divide the result by 750. The quotient represents the thickness required, in one-hundredths of an inch.

Example. - What thickness should half-inch pipe have for a head of 50 ft.? 50X.50=25. 25/ 750=.033 inch.

To find the diameter of pipe to discharge a given amount of water per minute, in cubic feet: Multiply the square of the quantity in cubic feet per minute by .96. The result equals the diameter of the pipe in inches.

To find the head which will produce a given velocity of water through a pipe of a given diameter and length: Multiply the square of the velocity, expressed in feet per second, by the length of pipe multiplied by the quotient obtained by dividing 13.9 by the diameter of the pipe in inches, and divide the result obtained by 2,500. The final amount will give the head in feet.

Example. - The horizontal length of pipe is 1,200 feet, and the diameter is 4 inches. What head must be secured to produce a flow of 3 feet per second?

3X3=9; 13.9/ 4=3.475. 9X1,200X3.475=37,530. 37,530/ 2,500=15 ft.

Doubling the diameter of a pipe increases its capacity four times. .

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To find the velocity of water flowing through a horizontal straight pipe of given length and diameter, the head of water above the center of the pipe being known: Multiply the head in feet by 2,500, and divide the result by the length of pipe in feet multiplied by 13.9, divided by the inner diameter of the pipe in inches. The square root of the quotient gives the velocity in feet per second.

Example. - The head is 6 feet, length 1,340 feet, and diameter 5 inches. What is the velocity of water passing through the pipe?

2,500X6=15,000. 1,340X13.9=18,626/ 5=3,725.2.

15,000/ 3,725.2=4.03.

The square root of 4.03=2 feet per second.

To find the weight of any length of lead pipe, when the diameter and thickness of the lead are known: Multiply the square of the outer diameter in inches, by the weight of 12 cylindrical inches, then multiply the square of the inner diameter in inches by the same amount, subtracting the product of the latter from that of the former. The remainder multiplied by the length gives the desired result.

Example. - Find the weight of 1,200 feet of lead pipe, the outer diameter being 7/8 inch, and the inner diameter 9/16 inch.

N. B. - The weight of 12 cylindrical inches, 1 foot long, 1 inch in diameter, is 3.8697 lbs.

7/8X7/8=49/64=.765625.

9/46X9/16=91/256=.316406.

.765625-.316406=449219X3.8697X1,200=2,086 lbs.

Tests for Pure Water.

Color. - Fill a long clean bottle of colorless glass with the water. Look through it at some blank object. It should look colorless and free from suspended matter. A muddy or turbid appearance indicates soluble organic matter or solid matter in suspension.

Odor. - Fill the bottle half full, cork it and leave it in a warm place for a few hours. If, when uncorked, it has a smell the least repulsive, it should be rejected for domestic use.

Taste. - If water at any time, even after heating, has a repulsive or disagreeable taste, it should be rejected. A simple, semi-chemical test is known as the "Heisch Test." Fill a clean pint bottle three-fourths full of water; add a half teaspoonful of clean granulated or crushed loaf sugar; stop the bottle with glass or a clean cork, and let it stand in the light, in a moderately warm room, for forty-eight hours. If the water becomes cloudy, or milky, it is unfit for domestic use.

To compute the amount of caulking lead used on any amount of work: Allow for each joint one pound for each inch in size.

Example. - How much caulking lead would be necessary on a job of 30 4-inch joints, and 20 2-inch joints?

For a 4-inch joint, 4 lbs. would be used. For a 2-inch joint, 2 lbs. would be used. Therefore, the total amount would be, viz.:

30X4=120 +

20X2= 40

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=160 lbs.

To find the diameter of a pump cylinder to move a given quantity of water per minute (100 feet of piston being the standard of speed): Divide the number of gallons by 4, then take the square root, and the product will be the diameter in inches of the pump cylinder.

To find the quantity of water elevated in one minute, running at 100 feet of piston speed per minute: Square the diameter of the cylinder in inches and multiply by 4.