WATER SUPPLY, PLUMBING, ETC., FOR COUNTRY HOMES. 29 



Example 2. It is desired to find the necessary fall from the spring 

 to the ram in order to supply the storage tank with 2 gallons per 

 minute, when the drive water supply in the spring is 10 gallons i)er 

 minute and the height to which the water is to be pumped is 40 feet. 



Substituting in the equation H = 



?t X 2 _ 2 X 40 



= 8 feet. Add one-third 



Q 10 



to allow for friction, making the total fall necessary 10.66 feet. 



Example 3. It is desired to find how much water will be delivered 

 into the storage tank if the drive water supply is 6 gallons per minute, 

 the fall is 10 feet, and the height to which the water is to be pumped 



buDstitutmg m the equation q^ = 



is 40 feet. 



1.5 



li 40 



gallons per minute. Deduct one-third of this result to allow for 

 friction, making the quantity delivered per minute 1 gallon. 



Example 4. It is desired to find how high 1 gallon per minute 

 can be pumped if the drive water supply is 4 gallons per minute 



QxH 



and the fall is 15 feet 

 4X15 



Substituting in the equation it 



60 feet. Deduct one-third to allow for friction, maldng the 



result 40 feet. 



The above computations are only approximate, but should give a 

 good general idea of the operation of a ram. 



The following table gives commercial estimates of the quantities of 

 water dehvered in 24 hours under certain conditions: 



Capacity of hydraulic rams. 



Power 



head 



in 



feet. 



Pumping head in feet — 



4 



10 



15 



20 



30 



140 



50 



60 



70 



80 



90 



100 



120 



140 



160 



ISO 



200 



2 

 3 

 4 

 5 

 6 

 7 

 8 

 9 

 10 

 12 

 14 

 16 

 18 

 20 

 22 

 24 

 26 

 28 

 30 



540 



192 

 301 

 432 

 540 



128 

 192 

 256 

 345 

 432 

 505 



96 

 144 

 192 

 240 

 302 

 378 

 432 

 485 

 540 



64 

 96 

 128 

 160 

 192 

 235 

 270 

 300 

 360 

 430 

 505 



43 



72 



96 



120 



144 



168 



192 



216 



1252 



301 



353 



432 



48a 



540 



29 

 58 

 77 

 96 

 115 

 134 

 154 

 173 

 192 

 230 

 270 

 323 

 390 

 430 

 475 

 520 



24 

 43 

 64 

 80 

 96 

 112 

 128 

 144 

 160 

 192 

 224 

 257 

 303 

 336 

 370 

 405 

 470 

 505 

 540 





















37 



55 



69 



82 



96 



110 



124 



137 



165 



192 



220 



247 



288 



303 



346 



375 



430 



465 



27 



43 



60 



72 



84 



96 



108 



120 



144 



168 



192 



216 



240 



264 



288 



328 



354 



405 



24 

 38 

 53 

 64 

 75 

 86 

 96 

 107 

 128 

 150 

 171 

 192 

 214 

 235 

 256 

 278 

 300 

 336 















29 

 43 

 57 

 67 

 77 

 86 

 96 

 115 

 135 

 154 

 173 

 192 

 212 

 230 

 250 

 269 

 288 



24 

 30 

 43 

 50 

 64 

 72 

 80 

 96 

 112 

 128 

 144 

 160 

 176 

 192 

 208 

 224 

 240 











26 

 31 

 36 

 55 

 62 

 68 

 82 

 96 

 110 

 124 

 137 

 151 

 164 

 178 

 192 

 206 









27 

 31 

 43 

 54 

 60 

 72 

 84 

 96 

 108 

 120 

 132 

 144 

 156 

 168 

 180 



24 

 28 

 38 

 43 

 53 

 64 

 75 

 85 

 96 

 107 

 118 

 128 

 139 

 149 

 160 



'25" 



29 



39 



43 



57 



67 



77 



86 



96 

 105 

 115 

 125 

 134 

 144 







































































































































1 Multiply factor opposite "power head" and under "pumpuig head" by the number of gallons per 

 minute used by the engine and the result will be the number of gallons delivered per day. Example: With 

 a supply of 6 gallons per minute, 10 feet fall, 40 feet elevation, No. 10 or 15 engine will deliver 1,512 gallons 

 per day; 6X252=1,512. 



This table will give only approximate quantities since the results will vary with the length of delivery 

 pipe. Due_ consideration of pipe friction will give more correct results. 



The efficiency developed is governed by the ratio of fall to pumping head, being greatest for a ratio of 1 to 

 2^ or 1 to 3, and the ram will not usually work well when the ratio is over 1 to 25, friction in the delivery 

 pipe being duly considered. 



