CYLINDER. When an engine in good order is performing its regular work, the effective pressure may be taken at 8 lbs. on each square inch of the surface of the piston. To calculate the power of an Engine. RULE 1.-Multiply the area of cylinder by the effective pressure say = 8 lbs. the product is the weight the engine can raise. Multiply this weight by the number of feet the piston travels in one minute, which will give the momentum, or weight, the engine can lift 1 foot high per minute; divide this momentum by a horse power, as previously stated, and the quotient will be the number of horse power the engine is equal to do. RULE 2.-25 inches of the area of cylinder is equal to one horse power, the velocity of the engine being constantly 220 feet per minute. EXAMPLE I. What is the power of an engine, the cylinder being 42 inches diameter, and stroke 5 feet? 42 x .7854 × 10 x 210 44000 = 66.12 horse power. EXAMPLE II. What size of cylinder will a 60 horse power engine require, when the stroke is 6 feet? 44000 × 60 228 × 10 1158 inches area of cylinder. Note. To find the power to lift a weight at any velocity, multiply the weight in lbs. by the velocity in feet, and divide by the horse power; the quotient will be the number of horse power required. NOZZLES.-The diameter of the valves of nozzles ought to be fully one-fifth of the diameter of the cylinder. AIR-PUMP.-The solid contents of the air-pump is equal to the fourth of the solid contents of cylinder, or when the air-pump is half the length of the stroke of the engine, its area is equal to half the area of the cylinder. CONDENSER is generally equal in capacity to the airpump; but when convenient, it ought to be more; for when large, there is a greater space of vacuum, and the steam is sooner condensed. Many COLD WATER PUMP. The capacity of the cold water pump depends on the temperature of the water. engines return their water, which cannot be so cold as water newly drawn from a river, well, &c.; but when water is at the common temperature, each horse power requires nearly 7 gallons per minute. Taking this quantity as a standard, the size of the pump is easily found by the following rule, viz. - Multiply the number of horse power by 7 gallons, and divide by the number of strokes per minute; this will give the quantity of water to be raised each stroke of pump. Multiply this quantity by 231, (the number of cubic inches in a gallon,) and divide by the length of effective stroke of pump: the quotient will be the area. EXAMPLE. What diameter of pump is requisite for a 20 horse power steam engine having a 3 feet stroke, the effective stroke of pump to be 15 inches? 20×71=150 32 4.6875 × 231 15 =4.6875 gallons the pump lifts each stroke, =72.1875 inches area of pump. HOT WATER PUMP.-The quantity of water raised at each stroke ought to be equal in bulk to the 900th part of the capacity of the cylinder. PROPORTIONS.-The length of stroke being 1, the length of beam to centre will be 2, the length of crank .5, and the length of connecting rod 3. The following table shows the force which the connecting rod has to turn round the crank at different parts of the motion. .228 Col. B. Angle between the connecting Rod and Crank. .0 180° .0 .4 .271 .308 .342 .377 Col. C. Effective length of the Lever upon which the connecting Rod acts, the whole Crank being 1. .45 913 1. .441 .5 85 1. .473 .55 80 .986 .507 .6 75 .956 .538 Col. D. Decimal proportions of half a .65 69 .92 .572 .7 621 .88 .607 revolution of the Fly-wheel. .746 .68 Col. C. also shows the force which is .723 .66 .9 34 .546 .776 communicated to the Fly-wheel, expressed in decimals, the force of the Piston being 1. FLY WHEEL - Is used to regulate the motion of the engine, and to bring the crank past its centres. The rule for finding its weight is, Multiply the number of horses' rower of the engine by 2000, and divide by the square of the velocity of the circumference of the wheel per second: the quotient will be the weight in cwts. EXAMPLE. Require the weight of a fly-wheel proper for an engine af 20 horse power, 18 feet diameter, and making 22 revoLutions per minute? 18 feet diameter tions per minute === 56 feet circumference, × 22 revolu1232 feet, motion per minute ÷ 60 = 20 feet motion per second; then 2012-420 the divisor. 20 horse powerx 2000-40000 dividend. PARALLEL MOTION. -The radius and parallel bars are of the same dimensions; their length being generally 1.4 of the length of the beam between the two glands, or one half of the distance between the fulcrum and gland. Both pairs of straps are the same length between the centres, and which is generally three inches less than the half of the length of stroke. GOVERNOR OR DOUBLE PENDULUM.-If the revolutions be the same, whatever be the length of the arms, the balls will revolve in the same plane, and the distance of that plane from the point of suspension is equal to the length of a pendulum, the vibrations of which will be double the revolutions of the balls. For example: suppose the distance between the point of suspension and plane of revolution be 36 inches, the vibrations that a pendulum of-36 inches will make per minute is, 62 2 375 = 36 =62 vibrations, and =31 revolutions per minute the balls ought to make. WATER WHEEL. WATER. (Hydrostatics.) Hydrostatics is the science which treats of the pressure, or weight, and equilibrium of water and other fluids, especially those that are non-elastic. Note 1.-The pressure of water at any depth, is as its depth; for the pressure is as the weight, and the weight is as the height. Note 2.-The pressure of water on a surface any how immersed in it, either perpendicular, horizontal, or oblique, is equal to the weight of a column of water, the base being equal to the surface pressed, and the altitude equal to the depth of the centre of gravity, of the surface pressed, below the top or surface of the fluid. PROBLEM I. In a vessel filled with water, the sides of which are upright and parallel to each other, having the top of the same dimensions as the bottom, the pressure exerted against the bottom will be equal to the area of the bottom multiplied by the depth of water. EXAMPLE. A vessel 3 feet square and 7 feet deep is filled with water; what pressure does the bottom support? 32+7+1000 3937 lbs. Avoirdupois. PROBLEM II. A side of any vessel sustains a pressure equal to the area' of the side multiplied by half the depth, therefore the sides and bottom of a cubical vessel sustain a pressure equal to three times the weight of water in a vessel. EXAMPLE I. The gate of a sluice is 12 feet deep and 20 feet broad; what is the pressure of water against it? 20×12 × 6 × 1000 16 -90000 -40 tons nearly. From Note 2d. The pressure exerted upon the side of a vessel, of whatever shape it may be, is as the area of the side and centre of gravity below the surface of water. |