[Olga from SJ]

I have a 10,000 gal water take. (I know the capacity does not matter). The outlet is a 4 inch pipe 9 feet below the tank water surface. The pipe then travels at a slope x1 to a valve where it splits the pipe to two pipes 2.25" diameter. This distance is approximately 80 feet and the drop is approximately 25 ft. One of the 2.25" pipes feeds my home. At the house level which is another 25 feet from the valve and another 6 foot drop has a pressure of 25psi. It seems to me that the water pressure should be higher but using the 0.433 pounds per foot of head rule, can I count this as 0.433x(9+25+6)=21.65 psi? This would explain the 25 psi pretty good, I am estimating the elevations.

So the question is really do I use the change in pipe elevations as the head in addition to the water head in the tank?

Olga

[ebaines]

You use the total drop in the calculation of head, or 9 + 25 + 6 = 40 feet. Then multiply by 0.433 to get the expected pressure, which is 0.433 x 40 = 17.3 psi. I think you may have used a head of 50 feet instead of 40 in your calculation. In practice however losses due to friction in the pipes and valves should result in a measured pressure that is less than that. As to why you are actually measuring 25psi - perhaps the tank at the top of the hill is pressurized (water tanks often are)?