Let's say you have a cylinder vertically placed, and some fluid like water filling inside. To make it easier to visualize the situation, please look at the little figure at the upper right corner of the following web page (I hope the URL shows up in one line):

If you measure pressures due to the water at different depths (i.e., the depth, denoted h here, is the distance from the upper surface of the water) the following relationship applies:

P = ρ g h

where P is the pressure at depth h of the water, ρ the density of the water, and g is the gravitational acceleration constant. From this relation, you can see quntitatively that the pressure increases at greater depths (i.e., h is larger) within the cylinder.

In short, at greater depths within the water, there is more water above the level at height h which causes a greater downward pressure at depth h, due to gravity pulling the water down.

So "the static pressure rise per vertical foot" means exactly how much the pressure increases if you go one foot deeper in the depth h within the cylinder. I don't think I can say more than that given the problem statement.

Hope this helps!

Cheers! Taro

Taro and other Homework Specialists are ready to help you

The above answer gives a formula with incomplete
explanations of the formula in order to computate and use it.
eg.What is the density of water (in numerals),and what is the gravitational acceleration constant
as it applies to water (in numerals)? What units are "P" in, (psi)? Otherwise there is no way to work the formula. I only need a practical formula for use in piping estimation.

eg.A simple answer that would give me the information that I am looking for at this point, would be,"In a 'practical situation', the water pressure (psi) in a vertical pipe in a static state will increase by what psi per vertical foot of water column rise". This woud be at atmospheric pressure and measured at the bottom of the water column.

I will assume a practical situation, in which the density of water is estimated to be

ρ = 1000 kg / m^{3} = 62.43 lbm / ft^{3}.

(I used the values listed http://hypertextbook.com/facts/2007/AllenMa.shtml). The gravitational acceleration constant is

g = 9.8 m / s^{2} = 32.2 ft / s^{2}.

(http://en.wikipedia.org/wiki/Gravity).

We are interested in how much the pressure P changes if depth h is increased by 1 ft. Let's call the change in pressure between depths h_{1} and h_{2} in pressure ΔP = P_{1} - P_{2}. Writing out, we have

ΔP = P_{1} - P_{2} = ρ g h_{1} - ρ g h_{2} = ρ g (h_{1 }- h_{2} ) = ρ g Δh ,

where Δh = (h_{1 }- h_{2} ) = 1 ft, since we want the two depths to differ by 1 ft.

Then the rest is plugging in numbers. We try using both British and SI units (just for checking). Our goal is to express the pressure in units of psi for the final answer.

ΔP = ρ g Δh = (62.43 lbm / ft^{3}) (32.2 ft / s^{2} ) (1 ft) = 2010 lbm / ft / s^{2},

Since 1 psi = 1 lbf / in^{2}, we need to use the conversions

1 lbf = (1 lbm) g = (1 lbm) (32.2 ft / s^{2}) = 32.2 lbm ft / s^{2}

So that's my answer --- the pressure increases by 0.433 psi if the vertical depth changes by 1 ft.

Note: I see that you might be expecting a shorter solution via a formua which might exist specifically for piping estimation of which I may not be aware personally -- if you are not satisfied with my answer, you could wait for an answer from an expert in piping estimation, if s/he exists in the forum. Nonetheless, I belive the numerical answer should be the same. (It is sometimes hard to discern what kind of math is being assumed for this type of question; I'm sorry if I'm misunderstanding you.)

Thanks for the clarification. That was just the info. that I needed.
This was my first shot at using JustAnswer. I probably need to add a little more "clarity" to my questions in the future.