Water in freezing expands to the extent of 8 per cent. The specific heat of ice is one-half the specific heat of water. Ice 3 inches thick, will bear the passage of infantry; 5 inches thick, of cavalry and light guns. A cubic foot of fresh snow = 6 lbs. Snow has twelve times the bulk of water. A cubic foot of sea water = 64°10 lbs. Weight of sea water = 1027 the weight of fresh water. 35 cubic feet of sea water = 1 ton. I cubic yard of sea water weighs 15 cwt. I qr. 20 lbs. A column of water 1 inch diameter and 12 inches high = 341 lb. A column of water 1 inch square and 12 inches high ='434 lb. The capacity of a cylinder 1 inch diameter and 12 inches long = '034 gallon. The capacity of a cylinder 12 inches diameter and 12 inches long = 4.895 gallons. The capacity of a cylinder 1 inch diameter and 1 inch long 00283 gallon. The capacity of a 1-inch cube = '0036 gallon. The capacity of a 12-inch cube = 6'24 gallons. The capacity of a sphere 1 inch diameter 00188 gallon. The capacity of a sphere 12 inches diameter = 3.26 gallons. The cube of the diameter of a sphere in feet multiplied by 326 = gallons. Or the cube of the diameter of a sphere in inches multiplied by o0188 = gallons. A column of water produces approximately a pressure of half a lb. per square inch, for every foot in height. Pressure of Water.-The side of any vessel containing water sustains a pressure to the area of the side in feet multiplied by half the depth in feet, that product multiplied by 625 will give the pressure in lbs. on each side of the vessel. The pressure in lbs. on the bottom of a vessel is to the area of the bottom in feet multiplied by the depth of water in feet, that product multiplied by 625 will give the pressure in lbs. Contents of Cisterns.—To find the number of gallons contained in a cistern. Multiply the length, width, and depth together, all in feet. This will give the contents in cubic feet, which multiply by 6'24, and the product will be the number of gallons. If the dimensions are in inches use 003607 in place of 6.24. Two dimensions of a cistern being given to find the third, to contain at given number of gallons, multiply the required number of gallons by 16046 if the dimensions are in feet, or by 277 274 if the dimensions are in inches, and divide the result by the product of the two given dimensions. The quotient will be the third dimension required. To find the number of gallons contained in a cylinder, multiply the square of the diameter in feet by the length in feet of the cylinder, and multiply the product by 4'895; or multiply the square of the diameter in inches by the length in feet, and multiply the product by 034; or multiply the square of the diameter in inches by the length in inches, and multiply the product by '00283. The diameter of a cylinder being given, to find the length, multiply the number of gallons by 2043, and divide the product by the square of the diameter in feet, and the quotient is the length in feet. The length of a cylinder being given, to find the diameter, multiply the number of gallons by 2043, and divide the product by length in feet, and the square root of the quotient is the diameter in feet. If the dimensions are in inches, use 353 in place of '2043. PUMPS. Lifting Pumps.—When a pump lifts water it withdraws the pressure of the atmosphere from the surface of the water inside the suction-pipe, and the pressure of the atmosphere outside the suction-pipe forces up the water until the pressures inside and outside the suction-pipe become balanced. The distance the water is lifted is equal to the height of a column of water weighing 15 lbs. per square inch of area at its base, which is theoretically 34 feet; but, as it is impossible in practice to make perfect joints and prevent leakage of air, a perfect vacuum is never obtained, and 28 feet is the greatest distance above the level of the water from which a pump will lift water, although at that distance it will be liable to lose its water when the barometer is low. To prevent occasionally having a dry pump, the supply should never be drawn through a greater height than 25 feet; but, as the efficiency of a pump varies with the distance it lifts the water, the suction-pipe should be made as short as possible, and 15 feet is the maximum safe distance above the level of the water for a pump to work well and uniformly and draw its proper quantity of water at each stroke; but if the pump works quickly, better results will be obtained by making the distance 10 feet. The quicker the speed of the pump, the shorter should the suction-pipe be. Load on a Hand-worked Pump. In a common hand-pump, with lever-handle, the leverage is generally 6 to 1, and the resistance on the handle, exclusive of friction, is found by dividing the weight due to the column of water by 6-the leverage. Load on a Hand-Power Lift-Pump, with Crank and Well-Frame. -The radius of the winch-handle of a well-frame is generally 16 inches, and the leverage is found by dividing the radius of the winch-handle by the throw of the crank (or half-stroke of pump). Thus a pump, with 8-inch stroke, and with 16 inches radius of winch-handle, would have a leverage of 4 to 1, and the weight of the column of water it has to raise, divided by 4, will give the resistance to be overcome, exclusive of friction. Load on Hand-Power Geared Well-Frames. When gearing is applied to drive the crank of a well-frame, what is gained in power is lost in quantity in a given time. Thus, if a wheel and pinion of 2 to 1 are added to the above frame, only one-half the power will be required, but only one half the quantity of water will be raised at each turn of the handle. To find the resistance in working a geared well-frame, divide the radius of the winch-handle by the throw of the crank (or half-stroke of pump), and multiply the result by the proportion of the wheel and pinion, and with the product divide the weight of the column of water, which will give the resistance to be overcome, exclusive of friction. The power exerted by a man in turning the winch-handle of a pump may be reckoned at 20 lbs. In a single-barrel pump the whole lift comes at one half of the turn of the handle; but in a double-barrel pump it is distributed over the two halves of the turn; and in a treble-barrel pump the work is still more equalised. Therefore, it is easier to work a pump with two barrels than with one barrel, when the united capacity of the two barrels is the same as that of the single barrel. Suction and Delivery-Pipes of Pumps.-The suction-pipe of a pump should always be larger than the delivery-pipe, because the friction. has to be overcome in the suction-pipe by the pressure of the atmosphere only; but in the delivery-pipe the friction is overcome by the power of the pump. The suction and delivery pipes should never be less than one-half the diameter of the pump-barrel. A good proportion for the suction-pipe is two-thirds the diameter of barrel. In quick-working pumps it is sometimes necessary to make it as large as the barrel. A long suction-pipe should fall evenly along its length towards the well, as, if any portion of it is higher than the pump-end, a trap will be formed in which air will accumulate, and from which it cannot easily be drawn away. A long suction-pipe should have a retaining or foot-valve placed near the water to prevent it losing its water, and to obviate the charging of the suction-pipe at each stroke. Pumps for Hot Water.-A pump will not lift hot water efficiently, because the steam destroys the vacuum; therefore the pump should be placed at the same level as the supply tank, so that the water may flow into the barrel by its own gravity. The valves of hot-water pumps should be made one-half larger in diameter than the ram, in order to obtain a large escape for the water with a small lift of the valve. Force-Pump. The barrel of a force-pump should be as close to the ram as possible, otherwise air will accumulate and impair the working of the pump. The diameter of the valves should never be less than three fourths the diameter of the ram; but it is preferable to make them of the same diameter as the ram; they should be placed as near to the ram as convenient, and they should only lift sufficiently to deliver their full capacity of water. An air-vessel should be placed on the suction side of a pump, the air in which becomes compressed, and its elastic force causes the water to flow uniformly into the barrel, and ensures the barrel being properly and continuously filled at each stroke. The neck of the air-vessel should be long and narrow, to prevent the action of the pump disturbing its water-level. An air-vessel also greatly reduces the percussion and wear and tear of the valves. Calculations for Pumps.-In addition to the weight of the water, allowance must be made for the friction of the pump and the friction of the water in the pipes, and also for the weight of the valves and for the resistance caused by the water passing through the valves, and likewise for the "slip," or water lost by the pump, as all pumps throw considerably less water than their capacity. In the following rules allowance is made for these contingencies. Capacity of a Pump.-The capacity of a pump with piston or bucket is the product of the area of the barrel multiplied by the length of stroke, and the capacity of a pump with a ram is the product of the area of the end of the ram multiplied by the length of stroke. Gallons of water delivered per Stroke.-Multiply the square of the diameter in inches of the pump-bucket, or ram, by '034, and by the length of the stroke in feet, and the product will be the number of gallons which the pump will deliver per stroke, provided the barrel gets properly filled with water at each stroke. But as all pumps throw considerably less than their capacity, deduct one-third from the number of gallons thus obtained for leakage, or "slip," and the remainder will be the actual quantity of water delivered per stroke, provided the pump is in first-rate order. But if the pump is of second-rate quality, it will be necessary to deduct one-half instead of one-third for "slip." Actual Horse-power of Pumps.-Find the number of gallons per stroke by the above rule, and multiply it by 10 (the weight of a gallon of water), and by the number of strokes per minute. The product will be the weight of water lifted per minute, which multiply by the height in feet from the water to the point of delivery. The product will be the total work done per minute in foot-lbs. Divide by 21,780, then add 4th for the friction of the engine itself, and the sum will be the actual horse-power of the engine required to drive that pump. Nominal Horse-power of a Pump.-Find the total work done per minute, as in the last rule, and divide it by 32,670, then add th for the friction of the engine itself. The product will be the nominal horse-power of the engine required to drive that pump. In calculating the horse-power of deep-well pumps, the weight of the spears and spear-plates, rods, bucket, &c., must be added to the total work per minute before dividing by the above given divisor. The effective work done by a pump is equal to the product of the weight. of the water by the height it is raised, and the efficiency of that pump is the ratio of the effective work to the total work expended in driving it. In ordinary pumps the efficiency is about 66 per cent. Diameter of Pump.-To find the diameter of a pump, multiply 034 by the length of stroke in feet, then multiply by the number of strokes per minute, and divide the number of gallons to be delivered per minute by the said product. The square root of the quotient will be the diameter of the pump in inches; but as all pumps throw considerably less water than their capacity, add a third to the area of the pump, to allow for leakage or "slip;" this allowance for "slip" only applies to pumps in first-rate order; if the pump is of second-rate quality, it will be necessary to add one-half instead of one-third for slip. The length of stroke of pump, is found thus-Divide 277 27 (the number of cubic inches in one gallon) by the area of the pump-barrel, which will give the length of stroke for each gallon. Multiply this by the number of gallons per stroke, and the product will be the length of stroke in inches of that pump. The velocity of the water in feet per minute in a pump, is found by multiplying the length of stroke in feet by the number of strokes per minute. Centrifugal Pumps are particularly adapted for irrigation, drainage, and sewage works. The maximum distance they will effectively draw water is 25 feet, but 15 feet will give better results; and the maximum height from surface of the water to the point of delivery to which they will lift water effectively is 70 feet. High lifts require very high velocities and large pumps. TABLE 7.-HORSE POWER REQUIRED FOR CENTRIFUGAL PUMPS. Quantity of Water in Horse-power each foot in Height which the water is lifted. 16 50 100 200 300 500 700 800 1000 1500 2000 2500 3000 3500 4200 012025 056 085 16 25 35 40 50 75 1 12 13 16 2° Hydraulic Ram.-This useful self-acting apparatus is used where |