84 BULLETIN 414, U. S. DEPARTMEITT OF AGEICULTIJEE. 



at the source to the level of the water in the elevated tank plus the 

 frictional loss in the pipes between the source and the tank as deter- 

 mined from Table 10; or if a hydropneumatic tank is to be used the 

 total lift will be found by adding the frictional loss to the sums of 

 the vertical distance from the surface of the water at the source to 

 the tank and of the height which is found by Table 9 to be equivalent 

 to the maximum pressure desired in the tank. Divide the result of 

 the last operation by 33,000 and the quotient wiU be the theoretical 

 horsepower required. But, as a pumping outfit usually is only about 

 50 per cent efficient, the theoretical horsepower derived by the fore- 

 going operation should be doubled to determine the horsepower 

 actually necessary. 



Example: It is desired to determine the horsepower necessary to 

 force water from a well 50 feet deep into the hydropneumatic tank, 

 48 inches in diameter and 14 feet long, selected for use in the camp of 

 40 convicts considered in the foregoing examples. The water is to 

 be pumped at the rate of 900 gallons per hour, and a 1-inch pipe is 

 to be used in the well. As a minimum pressure of 11 pounds is 

 needed, the tank when three-fourths full of water will be under an 

 internal pressure of 88 pounds. 



Solution: By Table 9, page 79, the maximum pressure of 88 pounds 

 is equivalent to a head of 204 feet of water. The vertical distance 

 from the surface of the water in the well to the surface of the water 

 in the tank is 50 feet. By Table 10, on page 80, the frictional loss in 

 50 feet of 1-inch pipe with water pumped at the rate of 900 gallons 

 per hour, or 15 gallons per minute, will be equivalent to a head of 8 

 feet. Therefore the total hft is equal to 204 + 50 + 8, or 262 feet. 

 The volume of water to be pumped per minute is 15 gallons, or 15 

 divided by 7.48 = 2 cubic feet. The weight of this volume of water is 

 2x62.5, or 125 pounds. Multiplying this weight by the total lift deter- 

 mined above, and dividing the result by 33,000, the power theoretically 



269 X 1*^5 

 necessary is found to be „„ „ " > or 1 horsepower, and allowing for 



50 per cent eflEiciency of the outfit the power actually necessary is 2 

 horsepower. 



Detailed information as to pumping mstallations may be obtamed 

 from pump manufacturers. In applying for such information it 

 is proper to advise the manufacturer fully regarding the following 

 points: 



(1) The source of the supply (whether well, cistern, lake, or spring) ; 

 (2) if a well, the inside diameter and total depth; (3) the distance 

 from the ground surface to the level of the water in the well; (4) 

 the flow of the well; (5) the number of gallons to be pumped per 

 hour; (6) the relative positions of the source anJ the pomt to 

 which the water is to be forced; (7) the position m which the pump 



