How JustAnswer Works:

  • Ask an Expert
    Experts are full of valuable knowledge and are ready to help with any question. Credentials confirmed by a Fortune 500 verification firm.
  • Get a Professional Answer
    Via email, text message, or notification as you wait on our site.
    Ask follow up questions if you need to.
  • 100% Satisfaction Guarantee
    Rate the answer you receive.

Ask Taro Your Own Question

Taro, Professional w/Adv. Degree (JD/MBA/PhD/Eng.)
Category: Homework
Satisfied Customers: 18
Experience:  Physics Ph.D.
Type Your Homework Question Here...
Taro is online now
A new question is answered every 9 seconds

what is the static pressure rise per vertical foot in a ...

This answer was rated:

what is the static pressure rise per vertical foot in a vertical cylinder, pipe or column
Hi Kerry.

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!

Taro and 3 other Homework Specialists are ready to help you
Customer: replied 8 years ago.
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.
Kerry -- Okay, thank you for the clarifications.

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

ρ = 1000 kg / m3 = 62.43 lbm / ft3.

(I used the values listed The gravitational acceleration constant is

g = 9.8 m / s2 = 32.2 ft / s2.


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 h1 and h2 in pressure ΔP = P1 - P2. Writing out, we have

ΔP = P1 - P2 = ρ g h1 - ρ g h2 = ρ g (h1 - h2 ) = ρ g Δh ,

where Δh = (h1 - h2 ) = 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 / ft3) (32.2 ft / s2 ) (1 ft) = 2010 lbm / ft / s2,

Since 1 psi = 1 lbf / in2, we need to use the conversions

1 lbf = (1 lbm) g = (1 lbm) (32.2 ft / s2) = 32.2 lbm ft / s2

and 1 ft = 12 in. Then step-by-step we have

ΔP = 2010 lbm / ft / s2 = (2010 lbm / ft / s2) (1 lbf / (32.2 lbm ft / s2) )
ΔP = 62.42 lbf / ft2 = (62.42 lbf / ft2) (1 ft / 12 in)2 = 0.433 lbf / in2 .

Hence ΔP = 0.4335 psi.

In SI units instead, 1 ft = 0.3048 m, and we have

ΔP = ρ g Δh = (1000 kg / m3) (9.8 m / s2) (0.3048 m) = 2987 kg / m / s2 = 2987 Pa,

where Pa is Pascal and the conversion factor is 1 psi = 6894.8 Pa ( So in psi, we have again

ΔP = 2987 Pa = (2987 Pa) (1 psi / (6894.8 Pa) ) = 0.433 psi.

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.)


Customer: replied 8 years ago.
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.


You need to spend $3 to view this post. Add Funds to your account and buy credits.

Related Homework Questions