84 BULLETIN 414, U. S. DEPARTMENT OF AGRICULTURE. 
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 hydro pneumatic 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 will 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 lift 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 
2X-62.5, 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 9 5 
necessary is found to be „» ~* ? or 1 horsepower, and allowing for 
50 per cent efficiency of the outfit the power actually necessary is 2 
horsepower. 
Detailed information as to pumping installations may be obtained 
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 and the point to 
which the water is to be forced; (7) the position in which the pump 
