How would I calculate water pressure in a vertical column?

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SUMMARY

This discussion centers on calculating water pressure in a vertical column, specifically using a 17-foot PVC tube with a 4-inch diameter filled with water. The key formula for static pressure is derived from the hydrostatic pressure equation: Pressure = height * density * g, where water density is approximately 62.4 pounds per cubic foot. Participants emphasize that pressure depends solely on the height of the water column and not on the diameter of the tube. The discussion also touches on practical applications for designing water systems, including calculating static pressure at various heights in a tank.

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  • Understanding of hydrostatic pressure principles
  • Familiarity with basic physics equations (e.g., F = m*g)
  • Knowledge of units of measurement (imperial and metric)
  • Basic concepts of fluid dynamics
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  • Learn about the implications of pipe diameter on flow rate and pressure
  • Explore online calculators for fluid pressure and flow rate calculations
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Engineers, physics students, and anyone involved in designing water systems or studying fluid dynamics will benefit from this discussion.

milmar
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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?
 
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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).
 
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okay, so where do i go next?
 
Have you gone anywhere as yet?
Where's your equation derived on basis of what I told you?
 
What's the weight of water in the tube, and what is the area it is pressing on at the end of the tube?
 
Cant we use the formula 'hpg' to calculate the pressure? Because what I've learned 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 don't make fun of me and do correct me.
 
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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.
 
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.
 
You know that pressure is "force acting on a unit of area":
Pressure = \frac{F}{area}

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:
F = m*g

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

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

If you combine this expression of the force with the definition of pressure you get:
Pressure =\frac{height*area*density*g}{area} \Leftrightarrow Pressure =height*density*g
 
  • #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.
 
  • #11
gerben,

You make barometrics bearable. Let milmar mind the millibars!
 
  • #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?. . .)
 
  • #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...
 
  • #14
Spamlessjack-
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>

Toymkr
 
  • #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.
 
  • #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.

Lin
 
  • #17
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).
 
  • #18
SpamlessJack - 1 U.S. gallon = 231 cubic inches
1 cubic foot = 1728 cubic inches
 
  • #19
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)?
 
  • #20
sophiecentaur said:
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)?

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.
 
  • #21
To be honest, we still drink Pints of beer. This is probably because it is just bigger than a half litre, which always looks a bit stingey in a glass.
Our system is still not sorted out. We buy motor fuel in litres but our road distances are still in miles. You have to make a real effort to assess your car fuel consumption either in mpg or g/km. I think that's a nasty government trick!

But I am really amazed just how many posts pose questions in 'old money'.
 
  • #22
sophiecentaur said:
To be honest, we still drink Pints of beer. This is probably because it is just bigger than a half litre, which always looks a bit stingey in a glass.
Our system is still not sorted out. We buy motor fuel in litres but our road distances are still in miles. You have to make a real effort to assess your car fuel consumption either in mpg or g/km. I think that's a nasty government trick!

But I am really amazed just how many posts pose questions in 'old money'.

The people asking these questions obviously don't work physics math problems on a regular basis. Therefore they don't know that the first step would be to convert to metric.

As for pints of beer, the same thing that causes that caused the persistence in Gallons in america. It so happens that in such a plainly open market driven society, the markets description of things is what is seen as standard. If 3 cents bought you a pound of fish, "pound of fish" sort of became and easily identifiable brand name. If you switch that to 1/2 kg of fish, it loses this edge in the market and its why the change is so slow coming in such an overtly capitalistic society.
 
  • #23
But weren't they taught some Science at School and wasn't it delivered with SI units?
I guess the EU helped to get metric units accepted more in the UK. That Napoleon fellow certainly left his mark.

And "A pint of water weighs a pound and a quarter" comes to mind from primary school. owzatt?
 
  • #24
The metric system becomes the first step in physics problems, but is not necessary in most other way science fields. Still, all lab equipment is marked in metric. The conversions are taught, and the metric system is learned sometime around the 5th grade. "katy has diahreea but don't call mom." Kilo, hecto, deca, base, deci, centi, milli. Lol
 
  • #25
1 joule = 1 watt = 1 Newton metre as opposed to BTU, hp and watt, none of which even vaguely resemble one another. And what about ,God forbid, the slug.
 
  • #26
McQueen said:
1 joule = 1 watt = 1 Newton metre as opposed to BTU, hp and watt, none of which even vaguely resemble one another. And what about ,God forbid, the slug.
Owch!
One Watt (power) is one Joule (energy) per second. Unit alert.
 
  • #27
R.P.Feynman used to teach his (American...) students with furlong-pound-fortnight unit system to show that the choice of the system is a pure convention.
With all due respect to British tradition (and the superiority of pint over 1/2l), I feel happy to be born on the Continent...

Or maybe it was: furlong-bushel-fortnight?
 
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  • #28
I feel happy to be both Old and Continent.

I bet he never used that system for any serious work. As one who had to calculate the cost of 2tons 6cwt and 3 qtrs of coal at 5/6 (that's five shillings and sixpence) a cwt when I was about ten years of age I know that a bad choice of units can be a serious handicap.
 
  • #29
sophiecentaur said:
car fuel consumption either in mpg or g/km. I think that's a nasty government trick!
Oh no! that's even worse! We, Continentals, express it in litres per 100 km! My small Peugeot burns about 7l/100km...
 
  • #30
Actually esterling system was not the worst one. I liked sixpence coins ;) Don't even thinking about times when a pint of decent ale was 2 shillings, rather than 2 pounds (I haven't been in London for over three years, I am afraid it could hit 3 pounds already...)
 

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