111. Air Compressors.-An air compressor is like a double acting steam engine in appearance; but, instead of delivering up power, it requires power from some other source to run it. This power is stored in the air and later is recovered when the air is used. An air compressor takes air into the cylinder, raises its pressure by compressing it, and then forces it into the air line or the storage tank. In calculating the horse-power of a compressor, the same formula can be used as for a steam engine. The value of P to use is not the pressure to which the air is raised, but is the average or mean pressure during the stroke. It is usually somewhat less than half the final air pressure; for example, when an air compressor is delivering air at 80 lb. pressure, the mean pressure on the piston is about 33 lb. Most air compressors are double acting, though there are many small single acting ones. Example : A double acting 12 in. by 14 in. air compressor is running 150 R. P. M. It is supplying air at 100 lb. and the mean pressure in the cylinder is 37 lb. per square inch. Calculate the horse-power necessary to run it. In this case 12 appears in the denominator in order to reduce the 14 inches to feet. an It, 112. Brake Horse-power.-The Brake Horse-power of engine is the power actually available for outside use. therefore, is equal to the indicated horse-power minus the power lost in friction in the engine. Brake horse-power can be readily determined by putting a brake on the rim of the fly-wheel and thus absorbing and measuring the power actually delivered. Fig. 72 shows such a brake arranged for use. This form is known as the "Prony Brake." It consists of a steel or leather band carrying a number of wooden blocks. By tightening the bolt at A, the friction between the blocks and the rim of the wheel can be varied at will. The corresponding pull which this friction gives at a distance R ft. from the shaft is weighed by a platform scale or spring balance. From the scale reading must be deducted the weight due to the unbalanced weight of the brake arms, which can be determined by reading the scales when the brake is loose and the engine is not running. If an engine is capable of maintaining a certain net pressure W on the scale, and meanwhile maintains a speed of N revolutions per minute, we can readily see that this is equivalent to an effective belt pull of W pounds on a pulley of radius R running at N revolutions; or it can be considered as being equivalent to raising a weight equal to W by means of a rope wound around a pulley of radius R turning at N revolutions per minute. This weight would be lifted at the rate of 3.1416X2XRXN ft. per minute and the brake horse-power will be The brake and wheel rim will naturally get hot during a test, as all of the work done by the engine is transformed back into heat at the rubbing surfaces of the pulley rim and the brake. It is necessary to keep a stream of water playing on the rim to remove this heat and it is best to have special brake wheels for testing. These have thin rims and inwardly extending flanges on the rims so that a film of water can be maintained on the inner surface of the rim. Example: Suppose that, at the time of testing the 5X8 gas engine in article 110, we also determined the brake horse-power by means of a Prony brake having a radius of 3 ft. and that a net pressure of 22 lb. was exerted on the scales (the speed of the engine was 450 R. P. M.). Let us calculate the brake horse-power. 3.1416×2×3×450=8482 186,604÷33,000=5.65, Answer. Explanation: Our data is equivalent to that of hoisting a weight of 22 lb. by a rope winding upon a pulley of 3 ft. radius turning at 450 R. P. M. The 22 lb. weight would rise 8482 ft. per minute and the work done per minute would be 22 lb. X8482 ft. = 186604 footpounds per minute. Hence, the brake-horse power of the engine is 5.65. 113. Frictional Horse-power.-If this engine gave 7.13 indicated H. P. (I. H. P.), but the power available at the flywheel was only 5.65, it stands to reason that the difference, or 1.48 H. P., was lost between the cylinder and fly-wheel. The explanation is that this power is expended in simply overcoming the friction of the engine; and this horse-power is, therefore, called the Frictional Horse-power. At zero brake horsepower, the entire I. H. P. is used in overcoming friction. 114. Mechanical Efficiency.-The ratio of the Brake Horsepower to the Indicated Horse-power gives the mechanical efficiency, meaning the efficiency of the mechanism in transmitting the power through it from piston to fly-wheel. This is usually expressed in per cent. In the case of the engine of which we figured the I. H. P. and B. H. P., the mechanical efficiency was The mechanical efficiency of a gas engine is lower than that of a steam engine on account of the idle strokes which use up work in friction while no power is being generated, but at full load a well built gas engine should show over 80 per cent. mechanical efficiency. The mechanical efficiency of a steam engine should be above 90% at full load. PROBLEMS 191. The cage in a mine weighs 2200 lb. and the load hoisted is 3 tons, The hoisting speed is 20 ft. per second. Calculate horse-power necessary. allowing 25% additional for friction and rope losses. 192. A 10 in. by 12 in. air compressor runs 150 R. P. M. The M. E. P. is 30 lb. Calculate the horse-power required to run it. 193. A pump lifts 2000 gallons of water per minute into a tank 150 ft. above it. Find the horse-power of the pump. 194. Find the horse-power of a 10 in. by 12 in. steam engine running 250 R. P. M. with a M. E. P. of 60 lb. 195. What will be the horse-power of a single cylinder, four cycle gas engine with the following data: Size of cylinder, 12 in. by 16 in. Rev. per minute, 225 Mean effective pressure, 78 lb. per square inch? = Number of working strokes of the number of revolutions. 196. A body can do as much work in descending as is required to raise it. Knowing this fact, calculate the horse-power that could be developed by a water-power which discharges 800 cu. ft. of water per second from a height of 13.6 ft., assuming that 25% of the theoretical power is lost in the wheel and in friction. 197. What would be the brake horse-power of a steam engine which exerted a net pressure of 100 lb. on the scales, at a radius of 4 ft., when running at 250 R. P. M.? 198. How many foot-pounds of work per hour would be obtained from a 60 H. P. engine? 199. A centrifugal pump is designed to pump 3000 gallons of water per minute to a height of 70 ft. If the efficiency of the pump is 60%, what horse-power will be required to drive it? 200. The pump of problem 199 is to run 1500 R. P. M. and is to be beltdriven from a 48 in. pulley on a high speed automatic engine, running 275 R. P. M. What should be the diameter and width of face of the pulley on the pump, if the pulley is to be 1 in. wider than the belt? CHAPTER XVIII MECHANICS OF FLUIDS 115. Fluids. Nearly every shop of any size contains some devices which are operated by water or air pressure, so every upto-date mechanic should have a knowledge of how these machines work. A Fluid is any substance which has no particular form, but always shapes itself to the vessel which contains it. Water, oil, air, steam, gas-all are fluids. In some ways water, oil, and similar substances are different from the lighter substances—air, steam, etc. To separate these, we give the name of Liquids to such substances as water and oil; while air, steam, etc., are given the general name of Gases. In some respects liquids and gases are alike and in others they are different. The chief difference is that liquids have definite volumes; they cannot be compressed or expanded any visible amount, while gases can be readily compressed or expanded to almost any extent. For all practical purposes we can say that liquids cannot be compressed. The third form of matter-Solids-needs no explanation. The differences in these three forms can be stated as follows: A Solid has a definite shape and volume. A Liquid has no definite shape, but has a definite volume. A Gas has neither a definite shape nor volume. There are some substances that exist in states in between the solid and the liquid form. Among these are tar, glue, putty, gelatine, etc. 116. Specific Gravity.-By Specific Gravity of a substance we mean its relative weight as compared with the same volume of water. Thus we say that the specific gravity of cast iron is 7.21, meaning that cast iron is 7.21 times as heavy as water. A cubic foot of water weighs 62.4 lb. and a cubic foot of cast iron weighs about 450 lb. The quotient 450 -7.21 is the specific gravity of the iron. 62.4 In many hand books we find tables of specific gravities and, when we wish to get the actual weight per cubic inch or per cubic |