velocity through a pipe similarly situated, of equal length, having a diameter of inch only, a head of 15 feet is required. To increase the velocity through the last mentioned pipe to 2 feet a second, requires a head 412 feet; to 3 feet, a head of 10; to 4 feet, a head of 1712, &c. From the foregoing, the following, it is believed, reliable rules, are deduced. To find the velocity of water passing through a straight horizontal pipe of any length and diameter, the head, or height of the fluid above the centre of the orifice, being known. RULE. Multiply the head, in feet, by 2500, and divide the product by the length of the pipe, in feet, multiplied by 13.9, divided by the interior diameter of the pipe in inches; the square root of the quotient will be the velocity in feet per second. EXAMPLE. The head is 6 feet, the length of the pipe 1340 feet, and its diameter 5 inches; required the velocity of the water passing through it. 2500 X 6 = 15000 ÷ (1340X13.9) second. = 4.03 =2 feet per Ans. To find the head necessary to produce a required velocity through a pipe of given length and diameter. RULE. Multiply the square of the required velocity, in feet, per second, by the length of the pipe multiplied by the quotient obtained by dividing 13.9 by the diameter of the pipe in inches, and divide the product thus obtained by 2500; the quotient will be the head in feet. EXAMPLE. The length of a pipe lying horizontal and straight is 1340 feet, and its diameter is 5 inches; what head is necessary to cause the water to flow through it at the rate of 2 feet a second? 22 X 1340 X 13:9 ÷ 25006 feet. Ans. To find the quantity of water flowing through a pipe of any length and diameter. RULE.- Multiply the velocity in feet per second by the area of the discharging orifice, in feet, and the product is the quantity in cubic feet discharged per second. EXAMPLE. The velocity is 2 feet a second, and the diameter of the pipe 5 inches; what quantity of water is discharged in each second of time? 5 ÷ 12.4166, and .41662 X .7854 X 2 = = .273 cubic foot. Ans. MISCELLANEOUS PROBLEMS. To find the specific gravity of a body heavier than water. RULE.Weigh the body in water and out of water, and divide the weight out of water by the difference of the two weights. EXAMPLE. A piece of metal weighs 10 lbs. in atmosphere, and but 8 in water; required its specific gravity. 10- 8.25= 1.75, and 101.755.714. Ans. To find the specific gravity of a body lighter than water. RULE. Weigh the body in air; then connect it with a piece of metal whose weight, both in and out of water, is known, and of sufficient weight that the two will sink in water; and find their combined weight in water; then divide the weight of the body in air by the weight of the two substances in air, less the sum of the difference of the weight of the metal in air and water and the combined weight of the two substances in water, and the quotient will be the specific gravity sought. EXAMPLE. The combined weight, in water, of a piece of wood, and piece of metal, is 4 lbs. ; the wood weighs in atmosphere 10 lbs. ; and the metal in atmosphere 12, and in water 11 lbs., required the specific gravity of the wood. 10 ÷ (10+12 1211+4) = .588. Ans. To find the specific gravity of a fluid. RULE. Multiply the known specific gravity of a body by the difference of its weight in and out of the fluid, and divide the product by its weight out of the fluid; the quotient will be the specific gravity of the fluid in which the body is weighed. EXAMPLE.The specific gravity of a brass ball is 8.6; its weight in atmosphere is 8 oz., and in a certain fluid 7 oz.; required the specific gravity of the fluid. 8 - 7.25 = .75, and 8.6 X .75 = 6.45, and 6.45 ÷ 8 = .806. Ans. To find the proportion of one to the other of two simples forming a compound, or the extent to which a metal is debased, (the metal and the alloy used being known.) The Rule strictly bears upon that of Alligation Alternate, which see. EXAMPLE.The specific gravity of gold is 19.258, and that of copper, 8.788; an article composed of the two metals, has a specific gravity of 18; in what proportion are the metals mixed? 18 19.258 X 8.788 = 18 = 11.055 8.78819.258 = 177.4, then Or, 18 11.055177.4 11.055: 18 = 1.056 copper, Ans. 1.056= 16.944 gold. Copper to gold as 1 to 16.04 -. To find the lifting power of a balloon. RULE. Multiply the capacity of the balloon, in feet, by the difference of weight between a cubic foot of atmosphere and a cubic foot of the gas used to inflate the balloon, and the product is the weight te balloon will raise. EXAMPLE. A balloon, whose diameter is 24 feet, is inflated with hydrogen; what weight will it raise? Specific gravity of air is 1, weight of a cubic foot 527.04 grains; specific gravity of hydrogen is .0689. 36.31 grains = weight of 1 cubic foot of hydrogen. 243 X.5236 7238.24 Then, 7238.24 X 490.73 To find the diameter of a balloon that shall be equal to the raising of a given weight. The weight to be raised is 507 507.4 X 7000÷490.73= 7238.24, and 7238.24.5236 = 13824 = = 24 feet. Ans. To find the thickness of a concave or hollow metallic ball or globe, that shall have a given buoyancy in a given liquid. EXAMPLE. - A concave globe is to be made of brass, specific gravity 8.6, and its diameter is to be 12 inches; what must be its thickness that it may sink exactly to its centre in pure water? Weight of a cubic inch of water .036169 lb.; of the brass .3112 lb. Then, 123.5236 × .036169 ÷ 2 16.3625 cubic inches of water to be displaced. 16.3625.3112 52.5787 cubic inches of metal in the ball. 122 X 3.1416 452.39 square inches of surface of the ball. And, 52.5787452.39.1162 + = inch thick, full. Ans. To cut a square sheet of copper, tin, etc., so as to form a vessel of the greatest cubical capacity the sheet admits of. RULE. From each corner of the sheet, at right angles to the side, cut part of the length of the side, and turn up the sides till the corners meet. Comparative Cohesive Force of Metals, Woods, and other substances, Wrought Iron (medium quality) being the unit of comparison, or 1; the cohesive force of which is 60000 lbs. per inch, cross section. The strength of white oak to cast iron, is as 2 to 9. 66 66 66 is as 1 to 13. To determine the weight, or force, in pounds, necessary to tear asunder a bar, rod, or piece of any of the above named substances, of any given cross section: RULE.-Multiply the comparative cohesive force of the substance, as given in the table, by the cohesive force per square inch, area of cross section (60000 lbs.) of wrought iron, which gives the cohesive force of 1 square inch area of cross section of the substance whose power is sought to be ascertained, and the product of 1 square inch thus found, multiplied by area of cross section, in inches, of the rod, piece, or bar itself, gives the cohesive force thereof. 66 Alloys having a tenacity greater than the sum of their constituents. Swedish copper 6 pts., Malacca tin 1; tenacity per sq. inch, 64000 lbs. Chili copper 6 pts., Malacca tin 1; Japan copper 5 pts., Banca tin 1; Anglesea copper 6 pts., Cornish tin 1; 66 66 60000" 66 57000" 66 66 41000" Common block-tin 4 pts., lead 1, zinc 1; tenacity per sq. in., 13000 lbs. Malacca tin 4 pts., regulus of antimony 1; " Block-tin 3 pts., lead 1 part; Block-tin 8 pts., zinc 1 part; Zinc 1 part, lead 1 part; 66 66 12000 " 10000" Alloys having a density greater than the mean of their constituents. GOLD with antimony, bismuth, cobalt, tin, or zinc. SILVER with antimony, bismuth, lead, tin, or zinc. COPPER with bismuth, palladium, tin, or zinc. PLATINUM With molybdinum. PALLADIUM with bismuth. Alloys having a density less than the mean of their constituents. GOLD with copper, iron, iridium, lead, nickel, or silver. SILVER with copper or lead. IRON with antimony, bismuth, or lead. TIN with antimony, lead, or palladium. NICKEL with arsenic. ZINC with antimony. LINEAR DILATION OF SOLIDS BY HEAT. Length which a bar heated to 212° has greater than when at the tem perature of 32°. NOTE. -To find the surface dilation of any particular article, double its linear dilation, and to find the dilation in volume, triple it. To find the elongation in linear inches per linear foot, of any particular article, multiply its respective linear dilation, as given in the TABLE. by 12. |