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How would I calculate water pressure in a vertical column?

  1. Nov 20, 2004 #1
    How would I calculate water pressure in a vertical column? I had about a 17 foot PVC tube with a 4" diameter filled with water. What equations could I use?
  2. jcsd
  3. Nov 20, 2004 #2


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    Well, the water pressure at a given level must give rise to a force on the fluid above the level which balances the weight of the fluid above the level (otherwise, that part of the fluid would fall down due to gravity).
    Last edited: Nov 20, 2004
  4. Nov 20, 2004 #3
    okay, so where do i go next?
  5. Nov 20, 2004 #4


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    Have you gone anywhere as yet?
    Where's your equation derived on basis of what I told you?
  6. Nov 20, 2004 #5
    What's the weight of water in the tube, and what is the area it is pressing on at the end of the tube?
  7. Nov 21, 2004 #6
    Cant we use the formula 'hpg' to calculate the pressure? Because what i've learnt so far (i've just started A'levels) pressure of a fluid depends on its depth not on the width/diameter of the container. I can be wrong. If it is the case please dont make fun of me and do correct me.
    Last edited: Nov 21, 2004
  8. Nov 21, 2004 #7
    You're right that the pressure depends only on the depth (height) of the liquid, its density, and the acceleration due to gravity.

    With a constant diameter tube of water though, it's easy to visualise why this is the case by calculating the total weight of water in the tube, and then considering what force is keeping the water where it is.
  9. Nov 24, 2004 #8
    The formula for psi of static pressure is the height in feet times approximately .43 (or divided by approximately 2.33). Multiply the height of the water above any point by the area of the tube at that point and you can determine the total static pounds of pressure acting on that area.

    Such as a point 10' down the tube would be about 4.3 psi x pi x 2"^2 = 54 lbs approximately.
  10. Nov 25, 2004 #9
    You know that pressure is "force acting on a unit of area":
    [tex] Pressure = \frac{F}{area}[/tex]

    A water column will apply a certain force on the water below due to the gravity acting on it. That force depends on the mass of the water column and on the acceleration due to gravity:
    [tex] F = m*g[/tex]

    because the mass depends on the volume and on the density of the water you get:
    [tex]F = volume*density*g [/tex]

    because the volume depends on the height and cross sectional area of the water column you get:
    [tex]F = height*area*density*g[/tex]

    If you combine this expression of the force with the definition of pressure you get:
    [tex]Pressure =\frac{height*area*density*g}{area} \Leftrightarrow Pressure =height*density*g [/tex]
  11. Nov 25, 2004 #10
    Water weighs 62.4 pounds per cubic foot here on earth. There are 1728 cubic inches in a cubic foot. If you stack all 1728 cubes on top of each other the bottom cube would have 62.4 psi. 1728 divided by 12 inches per foot, then divided by 62.4 is approx. 2.3, so every foot of water increases the pressure by about 2.3.
  12. Nov 25, 2004 #11

    You make barometrics bearable. Let milmar mind the millibars!
  13. Aug 24, 2008 #12
    Milmar -

    I am a novice, here, but I think there is a standing formula for what you're asking.
    Water pressure has a specific/general weight/foot-of-elevation (here on earth...) - there's a rule-of-thumb (I THINK!!!) for calculating this water pressure –
    The rest of you PLEASE correct me if I'm wrong.

    I am in the process of designing a water system for a remote location without a public or well supply (electricity/wind power limited or unavailable). I am going to collect rain water and pump it up to a water tower, so I can have a static pressure supply. I am trying to calculate the water pressure/foot of height. The constants are:

    25’ tower
    1000 gal tank – 64” radius x 80” height.

    What I need to know is:

    • When the tank is full (25’+6.6’=31.6’) how much static pressure will I have at the bottom?
    • When the tank is pert near empty (25’) how much static pressure will I have at the bottom?
    • Will it matter if I use a 2” pipe vs. a 4” pipe for the riser? (flow=yes, pressure=???)

    Please help. . .

    (How can a cubic foot of water weigh 62.4Lb??? I thought water only weighed 8.4Lb/Gal – Is there 7.4Gal of water in a CF? – that’s the biggest CF/H2O I ever saw. . . what did I miss?. . .)
  14. Aug 24, 2008 #13
    Yes, spamlessjack, a cubic foot is bigger than you think (or that gallon jug is smaller than it looks). There are 7.4805 gallons (US) per cubic foot. I don't know that one offhand for imperial gallons...
  15. Aug 25, 2008 #14
    Don't let your calculations be distracted by the diameter of your pipe or tank. Weight of the column is not the same as pressure on a gauge.Don't feel bad when it seems to you that volume should matter, lots of other bright folk make the same error. The ooonly thing that affects head pressure of water is elevation at .43 psi per foot of height. ( 10 foot head = 4.3 psi, 100 foot head = 43 psi) Do a search for "calculators online" and you'll find that there are more than you can count for a plethora of different things. They're set up so you just plug in the values you want to convert, hit "calculate" and you get an answer...as if by magic.
    Be sure to design supports for your pipe and tank with consideration for total weight of contents plus pipe and tank plus a sizeable factor for wind pressure on the total structure.
    Milmar & gmax137 are right on the weight of water and gallons per cubic foot.

    As in all things......do the math. <grin>

  16. Nov 4, 2009 #15
    I was trying to find out why someone had tried equating .43 psi/ft to kPA in an owner's manual I was proofreading, did some searching to find some kind of context that would help me understand what they were trying to say, and was able to use the posts here to understand what they were trying to do. THANKS very much.
  17. Aug 31, 2011 #16
    Dear All,

    I have w=1meter, L=1meter and h=1.5meter. I will put inside full of tap water. And I will

    make 10mm hole at height 200mm from the bottom. How much pressure (kgf/cm2) of water

    will be comeout? Anyone can explain for me by equation?

    Thank you very much.

  18. Sep 1, 2011 #17


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    All that counts is the total head of water above the point of interest. You can have any shape or size of pipes above you, taking any route (Utubes or spirals or whatever). The hydrostatic pressure will be the same. For a large sealed tank with a long thin tube sticking up out of the top, the head will be governed by the level of water at the very top of the tube, for instance. Try it with a simple U tube of transparent plastic. Whatever the slope or shapes of the two sides of the tube, the two water surfaces will always be at the same height. This must mean that the pressure at the bottom, from both directions must be balanced - or water would flow until it was.

    The simple formula is P=ρgh (where ρ is density).
    As a rule of thumb, you can say that the pressure will increase by about one Atmosphere for every 10m (about 30ft ) of head.

    This applies strictly to just hydroSTATIC pressure. If water is flowing then pressures can be affected by constrictions / thin pipes (which is why you can get a low pressure jet out of your mains supply by turning the tap nearly off).
  19. Sep 1, 2011 #18


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    SpamlessJack - 1 U.S. gallon = 231 cubic inches
    1 cubic foot = 1728 cubic inches
  20. Sep 1, 2011 #19


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    Ye Gods.
    When I was at School, in the early 1950's, we had exercise books with tables on the back cover. They told you how many yards in a Rod, pole or perch, how many Fluid Oz in a Pint, Yards in a mile, square yards in an acre etc. etc.
    By the time I left School, in the early 60's, the UK had kicked all that stuff into touch and were using a decimal system in Science. How is it that the US still uses such a numbingly naff system? You'd have thought the example of Dollars and Cents would have given them the clue. Added to the fact that it is called the "Imperial System", I can't think how the Republicans or Democrats could stand using the very word "Imperial". How does the Tea Party stand on units (or would they recognise one if they trod on it)?
  21. Sep 1, 2011 #20
    We learn the metric system as well. The reason the Imperial System stays around is mostly due to the market always providing things in these quantities. Even here however, the first step in working a physics problem is to convert to metric, since the systems lends itself better to math.
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