Show compressed air notes for prospectors BY ROBERTS ROBERT S LEWIS the use of the gasoline engine as a source of power makes it possible for the prospector to em employ ploy compressed air for both drilling and pumping when he is sinking an exploratory shaft where a pump must be used it is commonly of the direct acting sinking pump type for relatively shallow shafts say up to feet in depth a pump having a capacity of about thirty gallons per minute and weighing about pounds is often quite satisfactory but for deeper shafts and for larger quantities of water a hevier sinking pump should be used these larger pumps range in capacity from eighty five to gallons per minute and vary in weight from 1600 to nearly pounds the piston speed may be from ninety to feet per minute and the length of stroke may be from six to sixteen inches most alost prospectors take it for granted that the calculating of the size of the pump needed or the cubic feet of tree free air required to operate the pump is beyond them so they will purchase whatever pump is recommended om mended by the salesman of a mining machinery house however calculations of this kind are not difficult and require only the use of simple arithmetic as the following discussion will show where drills are to be operated by the air from the compressor the pressure commonly used is pounds gauge the drills are often of the hammer type such drills work better at ninety or pounds pressure than at a lower now a pump has large clearance spaces to prevent the piston from striking the cylinder head and the air is used expansively non i e at full pressure throughout the stroke it is not well un understood der stood that as a result of these conditions the air will be used more efficiently in the pump if its pressure is low than if it is high thus if air is used at pounds gauge the highest theoretical efficiency that can be secured from the air is about 42 per cent but if the pressure is only fifty pounds gauge the possible efficiency is close to 52 per cent or 10 per cent higher of course the actual efficiency of a steam pump operated by compressed air is low say about 20 or 25 per cent but nevertheless it is 19 of advantage to use air at a lower pressure than that used at the drills this necessitates installing a reducing valve in the line supplying the pump it is better to do this than to throttle the air by a valve at the pump the best plan would be to let the air pass through the reducing valve into a small receiver where it would have time to reach atmospheric temperature before going to the professor of mining university of utah salt lake city pump this would lessen the danger of the exhaust ports becoming choked with ice since the temperature of the air as it expands to atmospheric pressure may otherwise be well below freezing the calculation of the size of pump required and the cubic feet of aree air needed for its operation will now be taken up in detail water pressure at the pump the pressure in pounds per square inch at the pump caused by the column of water in the discharge pipe may be taken at 12 the head in feet thus if the water is to be lifted feet the pressure is 50 2 pounds per square inch the actual weight of a column of water I 1 inch square and I 1 foot high is pounds so the pressure for a column feet high would be pounds per square inch but an allowance should be made for frictional resistance consequently the 50 pounds is a safe figure to use in case the stroke of the pump is greater than 7 inches it is advisable to add 13 of itself to the head before computing the pressure for the above lift and a pump with a stroke over 7 inches in length the head would be 10 0 1333 and the pressure would be 3 taken as 1333 3 pounds per square 2 inch such an allowance is to cover frictional resistance of the pump which increases with the size air pressure at pump if the air and water cylinders were the same size the air pressure necessary would be that just found for the water but the air cylinders are practically always made larger than the water cylinders since the total pressure in either cylinder is the area of the piston times the pressure per square inch the larger the air cylinder the less is the air pressure required to move the piston against the the resistance of the water the air pressure required is therefore proportional to the ratio of the areas of the water and air cylinders if the air cylinder is four times the area of the water cylinder the required air pressure is only one fourth of the water pressure in the above problem it if the cylinder ratio is 4 i the air pressure needs to be fief 6 only pounds per 4 square inch if the water cylinder has one half the area of the air cylinder the required air pressure is RG G G pounds per square 2 inch atmospheric and absolute air pressure the atmospheric pressure at sea level is usually taken at pounds per square inch now atmospheric pressure decreases as the altitude increased inc approximate atmospheric pressures for different altitudes are given in table 1 TABLE 1 atmospheric atmospheric altitude pressure altitude pres pressure 8 suri urt sea level 1475 lbs ibs lost 1087 1000 f ft t 1420 1046 lots 2000 1367 1007 1316 1267 1 g 9 6 1173 1130 air under more than atmospheric pressure has its pressure shown by a pressure pres sura baugue but the gauge shows only the pressure above that of the atmosphere where the absolute or actual pressure must ie be known it is found by adding the atmospheric pressure to the gauge pressure thus at sea level pounds gauge is pounds absolute at feet elevation pounds gauge is pounds absolute free air this term is used to denote air under conditions of atmospheric pressure and temperature pera ture since the atmospheric pressures at sea level and feet elevation are 14 and pounds respectively the figure is used instead of the more accurate value 1475 in order to simplify the calculations a cubic foot of air at sea level will hiegh more than a cubic foot of air at feet elevation to find the cubic feet of free air at any elevation that would be equivalent to a given volume A at sea level use the formula cu ft free aid at any altitude cu ft air at sea level x atmospheric pressure at altitude thus cubic feet at sea level 10 loof 1 cubic feet free air at feet elevation in other words a larger amount of free air is required at an altitude than is required at sea level to do the same amount of work volume of free air required by pump this is found by the formula V feet where 1 I bere iere V Vvo V lume of free air in cubic c per minute required by squat e aarel of air cylinder in sq inches PUMP of air in in r solute pressure in pounds per square inch feet ek pet sapiston Sp iston speed of pump in Pa atmospheric minute pressure at auf nuino let it be required to find the air press pres sare tt to required and the cubic feet of free air require operate a pump having a 3 inch water 01 inder a 6 inch air cylinder and a length leng stroke of 8 inches the head or lift li Is 1 1 j gal S lolis I 1 feet and the flow of water is sixty per minute the altitude is fe feet et the it j since the stroke is over 7 inches the aha water V head 0 feet and a 3 square pr pressure assure is 3 2 p 0 pounds per 2 pro p are inch the areas of the cy cylinders linders j to the squar squares es of the diameters it is therefore plain that the air cylinder has four timis times the area of the water cylinder the required air pressure is 4 go 40 pounds gauge gaub e at feet elevation the atmospheric pressure is pounds per square inch so the absolute air pressure is pounds per square inch to find the piston speed the gallons are reduced to cubic inches of water per minute A gallon contains cubic inches so 60 h cubic inches of water per minute that the pump must handle the diameter of the water cylinder is 3 inches and its area is square inches 1960 inches per minute that the piston must travel this equals feet per minute as the piston speed the diameter of the air cylinder is 6 inches and its area is square inches hence V 1654 cubic feet of free air per minute velocity of the water to prevent large friction losses the velocity e of the water in both the suction and discharge pipes should be kept low recommended values are a maximum of four feet per second in the suction pipe and six feet per second in the discharge values these values should not be exceeded the sizes of pipe required to give these velocities are given in bable table 2 TABLE NTO NO 2 gal al velocity minute per ier size of it f t size of ft pipe in per see sec pipe in per sec 15 1 20 39 1 SO 20 1 1 36 6 1 2 3 33 02 1 so 50 4 05 1 t 60 0 2 2 A 33 2 51 10 70 2 2 so 46 2 0 3 36 G ai po 3 M 3 2 ia 3 1 45 2 A ga 65 v la 2 3 ag 4 3 68 5 58 32 4 51 0 zoo O 6 4 4 GA 64 34 5 15 0 O 6 48 39 5 56 6 n 7 15 1 5 ga uw 1 6 42 6 56 size of compressor if the compressor is to supply air to both and drills the th drills the free air requirement of rills must be will known A hammer drill take air from 40 to 60 cubic feet of free der minute at go 90 can 1 I be pounds ovines gauge this as ex i a reduced to free air a at any elevation the a under r free ree air above thus if is IF feet for each drill we alave ul ave cubic feet of free air required ta it both are horii or the e pump and the drills 19 at sor the same time the compres the have a free air sun sum capacity equal to of ty by the th lne the air needed ceded by the drills and sump lessons lessors are rated on a basis of ps pis ton displacement in in cubic feet per minute this displacement is the volume swept through by the piston in one minute because of the effect of clearance and of the heating of the air as it enters the compressor the metal parts of which become heated after the compressor has been for some hours the actual amount of free air delivered as compressed air is not equal to the piston displacement it is always less the percentage of the piston displacement actually delivered as compressed air is called the volumetric efficiency of the compressor increase of altitude serves to lessen the volumetric efficiency of the compressor suppose that due to clearance and the altitude the volumetric efficiency of a compressor is only 80 per cent if the rated displacement of the compressor is cubic feet of free air per minute it will only deliver 80 per cent of or cubic feet per minute if cubic feet of free air was necessary a compressor with a volumetric efficiency of t 10 11 IT 1 I 1 1 1 IT a a s 3 d s 5 I 1 1 a 9 a 7 c 5 b a 1 a 7 aa a a A pa 1 1 1 1 A P 3 2 Is 1 1 1 Y 3 ui 1 va r si y y cam m ha UNO M I 1 pow awa i i n or rna 93 P MIX 00 raon y K 1 1 b P J t compression RATIO PATIO chart 1 80 per cent would have to have a displacement of 30 0 cubic feet per minute in so order to deliver cubic feet of free air as compressed air for this reason the best way to purchase a compressor when the required quantity of free air is known is to specify that the compressor is to deliver cubic feet of free air per minute at P pounds gauge pressure when operating at feet elevation the manufacturer will then see that the compressor is large enough to do the work required the theoretical work required to compress a given amount of free air at any altitude is given gien in chart no 1 to use the chart it is first necessary I 1 to determine the ratio of coni compression ratio of compression absolute final air pressure e absolute initial air pressure atmospheric pressure gauge pressure atmospheric pressure thus for a gauge pressure of pounds at 1000 feet elevation the ratio of compression is 1004 1139 gg now enter the chart at the bottorof bottom of this calculated ratio of compression and pass vertically upward to intersect curve d which is for single stage compression curve c is for two stage compression curve b is for three stage compression and curve a is for theoretically perfect or isothermal c compression om this chart is from the catalogue of the nordberg manufacturing company from this point of intersection pass horizontally to the left margin and read the horsepower required per pound of atmospheric pressure this value is 14 to get the horsepower required to compress cubic feet of free air this quantity must be multiplied by the atmospheric mo press pressure ure therefore the horsepower required to compress cubic feet of free air from atmospheric pressure kressu re of pounds to pounds gauge is 14 if the total air required is cubic feet per minute the total theoretical horsepower required is HP the mechanical efficiency of small compressors pres sors may range from 80 to 85 per cent assuming an efficiency of 85 per cent the actual power required to operate the compressor would be HP and an 85 engine of this power should be used to run the compressor loss of pressure in air line the friction loss in an air line depends upon the length of the line and the velocity with which the air flows through the pipe a as s the chief factors it is poor economy to duvan a certain size of pipe because the next larger size costs a little more the loss in power due to excessive friction in the smaller line may cost far more in the end than would yie he larger pipe many lines are put in simply because some one guesses that this is the right size to use it may happen that it is the right size but often it is too small and causes excessive friction loss or drop in pressure between the compressor and the drills table 3 has been calculated to show the size of pipe that should be used in order that the drop in pressure between the ends of the line should no not t be greater than five pounds the gauge pressure at a t the compressor has been taken at and pounds per square inch since the pressure most likely to be used will be one of these figures in many cases the drop in pressure will be less than five pounds but it if the next smaller commercial size or of pipe was selected the drop would be greater than five pounds TABLE 3 size of all air pipe to use so that the drop in pressure pre ssuie through the pipe will not be greater than 5 pounds cubic feet of free air per minute NII nute pressure at Corn compressor 90 pounds gauge length of pipe i 50 1000 in i in 11 in in ill in ill in in in in n ill ft 1 11 1 1 1 1 2 2 0 2 0 ft I 1 A 1 1 2 2 2 az 3 I 1 ft I 1 li A 1 2 av 2 4 2 21 1 4 10 2 I 1 3 2 2 1000 ft I 1 it 2 2 1 2 A 21 2 2 1 A 3 3 IA 1250 ft 1 2 2 21 0 J 0 1 12 4 1500 ft 1 aj 12 2 2 h 2 hii A 3 3 3 3 Vs 12 4 pressure at compressor pounds G gauge a ug e f t 1 11 1 I 1 2 2 0 2 0 12 3 ft I 1 1 1 2 2 i 2 A 2 ft 1 2 2 2 21 aa 1 2 I 1 2 A 12 1000 ft 1 2 2 2 2 12 3 3 9 3 1250 1 50 ft 1 1 vi 2 0 U 2 91 A 2 91 3 rl S 3 3 3 IA 3 iza 1500 ft 1 2 2 A 2 2 ai 2 3 3 3 3 12 4 pressure at compressor pounds gau gauge 9 e f ft t 1 1 i 1 1 1 2 2 12 I 1 ft 1 A 1 2 2 2 2 vz 12 ft I 1 1 1 2 2 21 2 21 2 2 21 lb 3 1 12 1000 ft 1 1 2 2 ai A 2 ak ia 2 3 3 3 16 1250 ft 1 2 2 2 ai A 2 i 3 3 12 31 1500 0 ft I 1 is il 2 2 ak 1 4 oi oi 3 la 12 4 1 in iii making calculations for this table the elevation has been assumed at feet for higher elevations the friction loss would be slightly less and for lower elevations the loss due to friction would be slightly greater however the difference in any case is small and the sizes of pipe required remain practically athe same as given in the table if a certain volume of free air is heated from 60 degrees F to degrees F the volume will increase in proportion to the increase in absolute temperature absolute temperature is the temperature by the thermometer plus degrees consequently the absolute final and initial temperatures would be degrees and degrees re ively the increase in volume is 78 79 per cent for an increase 0 df f 40 degrees in temperature A rough rule is that the increase in volume is 1 per cent for each five degrees rise in temperature since the real measure of the quantity of air used is its weight and a cubic foot of air at 60 degrees weighs more than a cubic foot of air at degrees F it is plainly of advantage to take |