HYDRAULIC MACHINERY. WATER has been employed as a motive power from the remotest ages, and has been applied in numberless ways to a variety of machines, some of which display much invention and ingenuity. Since the introduction of the steam-engine, both the science and construction of water machinery have been, to a considerable extent, neglected. In our mines, the application of water-power has been mostly confined to wheels in which little refinement of construction has been attempted. Water is a compound body, composed of eight parts by weight of oxygen and one of hydrogen. It expands in bulk, and decreases in density from a temperature of 39° Fahrenheit up to 212°, and below 39° it dilates and decreases in density until it reaches the freezing point. 1 Experiment has determined that water may be compressed about 46,500,000 th part per atmosphere; a quantity so small, that, practically speaking, water may be regarded as incompressible. The varying density and volume of water, although small, should be borne in mind by those entrusted with the construction of hydraulic apparatus. If it be allowed to freeze within cast-iron pipes, and no means be provided to allow free expansion, destruction of the arrangement is sure to follow; since a cubic inch of water exerts within the range of its expansion a force equal to 13 tons. Water is subject to the same law of gravitation as other heavy bodies; but this must in practice be considered as applicable to a dense column or sheet of this fluid only, and not to a divided jet, which is much affected by the resistance of air. In England and Ireland the average fall of rain is about three feet per annum, of which about two-thirds is evaporated; the remaining twelve inches ultimately finding its way to the sea. A cubic foot of water ... weighs 1000 oz., or 621⁄2 lbs. nearly. A cubic inch ⚫03617 A column of water, 12in. high, lin. sq. weighs 434 lbs. inch An imperial gallon A pound avoirdupois 49.1 A cubic foot (1728 inches). A cylindrical yard, 12in. diameter lin. The subject of effluent water has received considerable attention from engineers, and the following deductions have been the result of numerous experiments : That the quantities of fluid discharged in equal times from different-sized apertures, the altitude of the fluid in the reservoirs being the same, are to each other nearly as the area of the apertures. That the quantities discharged in equal times by the same orifice, under different heads, are nearly as the square roots of the corresponding heights of the water in the reservoir above the centre of the apertures. That the quantities discharged in equal times under different heights, are to each other in the compound ratio of the areas of the apertures, and of the square roots of the height nearly. That, on account of friction, the smaller orifices discharge proportionally less water than those which are larger, and of a similar figure, under the same head, and that, from circular apertures presenting less surface under the same area than other figures, they are most advantageous. That if a horizontal tube be of greater length than the extent of its diameter, the discharge of water is much increased. DEFINITIONS. 1.—A stream cut vertically and perpendicularly to the direction of the current, presents a transverse section. 2.-If a stream be supposed to flow in a new channel whose sides are vertical, and whose bottom is flat, with a breadth and sectional area equal to that of its real channel, we arrive at its mean hydraulic depth. 3.-The velocity of water in a river is most rapid in the middle of the upper surface of the stream, and gradually diminishes towards the bottom and sides of the channel. The mean velocity is assumed to be the central velocity of the transverse section, and the declivity is the rate of fall or descent in a given distance. 4. When water issues from a small orifice in the bottom or side of a vessel or reservoir, it acquires the velocity which a dense body would acquire by falling from the horizontal surface of the water. This is called its natural velocity. 5. The height due to the velocity of water issuing from a cistern or reservoir, is known as the head of water. 6. The head of water in water-wheels is the distance from the surface of the water to that point at which it strikes upon the wheel. RULES. 1.-The mean hydraulic depth is found by dividing the area of the transverse section by the breadth of the bottom of the new channel. 2.-The mean or central velocity may be found by multiplying the mean hydraulic depth by the declivity, both in feet, and extracting the square root of the product. The result, diminished byth part, is the mean velocity in miles per hour. 3.-The force of water acting directly against a plane surface is found by multiplying the area of the surface in feet by the square of the velocity in feet per second. This product, less th, will give the force required in pounds nearly. Thus for the following velocities in feet per second, the force in pounds on a square foot will be Velocities, 1, 2, 3, 4, 5, 6, 7, 8, 9. Forces,... 1, 4, 9, 15, 248, 35, 48, 62, 79. 4.-If the velocity be given in miles per hour, then multiply the area of the surface in feet by the square of the velocity, and double the product, increased byth, will be the force required in pounds, nearly. Thus, for the following velocities in miles per hour, the force in pounds on a square foot will be— 9. Velocities, 1, 2, 3, 4, 5, 6, 7, 8, Forces,... 21, 84, 189, 33-6, 52.5, 75-6, 102-9, 134-4, 170.1. 5.-The effective power of a stream, as available for driving machinery, is determined as follows:-Multiply the force due to the velocity and the area of the transverse section by the velocity per minute, and divide the product by the estimate of a horse-power. The quotient will be the effective power required. 6. The velocity of water issuing through a circular orifice at any given depth from the surface, after allowing for friction and the contraction of vein thereby produced, will be found by multiplying the square root of the height or depth to the centre of the orifice by 8.1, and the product will give the velocity in feet per second. Thus: suppose the depth from the head to the centre of the fluid to be 16 feet, required its velocity; 16 4x8132.4 feet velocity per second. In the case of rectangular apertures, the use of a co-efficient, or number expressing the proportion between the theoretical discharge of the water calculated as a falling body, and the actual discharge, becomes necessary. NOTCHED WEIRS, &c. 1.-When an aperture extends to the surface of the fluid. Multiply the area of the opening in feet, by the square root of its depth also in feet, and the product by 5.1; then will two-thirds of the last product equal the quantity discharged in cubic feet per second. Ex. 1.-Required the quantity of water in cubic feet, per second, discharged through an opening in the side of a dam or weir; the width or length of the opening being 7 feet, and depth 10 inches, or 833 of a foot. ✓ of depth, 83 ='912. Then 7.5 ft. width × 83 depth,=6.24975 area, × 912 = 5.709772 × 5.1=29.1198372; for x2÷3-19-4132 cubic feet. Ex. 2.-When the aperture is under a given head : Multiply the area of the aperture in feet by the square root of the depth also in feet, and by 51, the product is the quantity discharged in cubic feet per second. Ex-What quantity of water would be discharged through an opening in the side of a dam or weir, the width or length being 7 feet, and its depth 10 inches, if under a head of water 5 feet in height ? √52.236 × 5.1 for velocity = 11.4 feet of water per second, And 7.5 x 83 x 2.236 × 5.1 = 71.269 cubic feet discharged in the same time. To find the time a vessel will take in filling, when a known quantity of water is flowing in and out in a given time.-Divide the contents of the vessel in gallons, by the difference of the quantity going in and the quantity flowing out. The quotient is the time in hours and parts of an hour that the vessel will take in filling. Ex.-If 40 gallons per hour run in, and 30 gallons flow out of a vessel in the same time, such vessel holding 400 gallons, in what time will the vessel be filled? Diff. between 30 and 40 = 10; then 400 ÷ 10 = 40 hours. To find the time a cistern will take in emptying itself of water through an aperture of a given area.-Multiply the square root of the depth in feet, by the area of the falling surface in inches; divide the product by the area of the orifice multiplied by 37, and the quotient is the time required in seconds nearly. Ex-Required the time it will occupy to empty a cylinder of water 16 feet high, and 24 inches diameter, through a hole 3-inch diameter. √16 = 4 = square root of depth, 242 = 576 × 7854 = 452.39 = area of falling surface. 752 x 7854 =5625 × 7854 = 4417875; then 452.39 × 4 4417875 x 3.7 m. sec. = 1107·02 seconds 18 27 nearly. Approximate Rule for finding the number of imperial gallons of water contained in a square or rectangular cistern. Find the contents of the cistern in cubic feet, and multiply by 6.232, or the contents in cubic inches by 003607, and the product is the number of gallons nearly. Ex.-A cistern is 10 feet long, 5 feet wide, and 4 feet deep; required the number of imperial gallons it will contain? Cistern 10 feet x 55 x 4 = 220 cubic feet; Then 220 × 6.232 = 1371.040 gallons. Or thus:-Cistern, 120 in. × 66 in. x 48 in. x 003607 =1371-237, as before nearly. TABLE, SHOWING THE CAPACITY OF CIRCULAR CISTERNS IN GALLONS, FOR EACH 12 INCHES IN DEPTH. There are three varieties of water-wheels, named Overshot, Undershot, and Breast Wheels, according to the positions at which the water is received upon the circumference. The last description is distinguished as High or Low Breast, according as the water is received upon it either above or below the horizontal plane of its axis. Overshot Wheels.—Wheels that have the water applied within a few degrees of the summit are termed Overshot Wheels. It is considered by some engineers that the periphery of a water-wheel ought to move at a velocity equal to about twice the square root of the fall in feet per second, and that the number of buckets should equal 2.1 times the wheel's diameter in feet, or that the head of water should be sufficient to cause the velocity of its flow to be as 3 to 2 of that of the wheel. The experiments of Smeaton tended to prove that the best effect was obtained when the velocity of a wheel's circumference was a little more than 3 feet per second; and hence practical men have constituted a general rule to make the speed of overshot wheels at their circumference 3 feet per second, or 210 feet |