Atmospheric Pressure in Feet of Head
In last weeks blog we touched briefly on the subject of atmospheric pressure. I wanted to write today's blog to ensure you understand exactly what atmospheric pressure is and why it affects you and your pumps.
As illustrated below, the weight of the air in a 1 inch square column extending to the edge of outer space would be 14.7 lbs. This is, of course, how standard atmospheric pressure at sea level is established as 14.7 pound per square inch ( 14.7 Psi). When dealing with pumps it is very useful to be able to relate the measure of atmospheric pressure at 14.7 pound per square inch into a water pressure. More specifically to the height of a column of water that would exert that same 14.7 psi.
The simplest way to determine this is to start by finding the weight of non aerated water. A quick search on the internet would provide a number of 62.427 pounds per cubic foot.
As a cubic foot is 12 inches by 12 inches by 12 inches, a single cubic foot contains 1728 cubic inches.
The weight of one cubic inch of water would then be the weight of one cubic foot of water divided by 1728 cubic inches per cubic foot, or 62.427 lbs divided by 1728 equalling 0.0361lbs per cubic inch. The next step would be to determine how many cubic inches of water it would take to equal the atmospheric weight 14.7 lbs. Calculating 14.7 lbs divided by 0.0361 lbs gives us an answer of 407. Or in other words, 407 cubic inches of water weighs the same as the one inch square column of our atmosphere that we discussed earlier.
If we then took all 407 cubic inches and stacked them on top of each other to form a vertical column 407 inches tall (33.9 feet), the weight of the water in that column would be the same weight as the air in our atmospheric column. With both columns weighing the same, they would both exert the same pressure at the bottom of the column.
Based on this equivalency, standard atmospheric pressure can be expressed as 33.9 ft of water column, or as pump people refer to it, 33.9 ft of head.
The relationship between Psi and feet of water head is one of the most basic principals in the pump industry. Having spent the previous few paragraphs discussing one specific example of this relationship, it would seem appropriate at this time to use this example to confirm a well known constant. If 33.9 ft of water head is equivalent to 14.7 Psi then 1 psi will be equal to 33.9 divided by 14.7 or 2.31.
In my next blog I will discuss the relationship between atmospheric pressure and the inlet pressure required by a centrifugal pump.
Until next time,