Science > Physics > Calculating water pressure - struggling with the concept!
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Science > Physics |
| User: |
"grant" |
| Date: |
12 Apr 2005 03:18:35 AM |
| Object: |
Calculating water pressure - struggling with the concept! |
Perhaps someone can shed some light on some calculations I am
attempting. (In simple terms!!)
I am intending to pressure test an acrylic vessel, and I would like to
calculate the max static water pressure when filled (prior to
commencing testing)
The vessel is non-uniform, in that it is made up of chambers connected
by large diameter (120mm) tubes in all 3 planes.
I have done some research and found that Pressure = g (9.81) x water
depth. While I can apply this formula it seems to me to oversimplify
(someone please correct me if I'm off base here) as I was under the
impression the volume of water pressing down creates the pressure.
To give some more figures - The max water depth is 758mm and the
total volume of water is around 250L
When I use this formula I get:
P = 9.81 x 758
Pressure = 7436 N/m2 (This is equal to 1.08 PSI or 0.074 bar)
My feeling is that this seems awfully low, and I am having difficulty
with the idea that the depth is purely the factor that contributes to
the static pressure.
Considering the testing procedure:
If (P=g x depth) holds true would it be possible to close the vessel
and add a flexible pipe (say of just 10mm diameter) to the top and fill
vessel and tube with water to twice the normal working depth to double
the working pressure?? The equation suggests this, but my sense of
physical mechanics says this cannot be true...to double the pressure
there needs to be 500L present.
Am I mixing pressure and force applied by weight...?
I will welcome any suggestions to help clarify this. (On a simple level
please!)
Many Thanks
Grant
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| User: "JM Albuquerque" |
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| Title: Re: Calculating water pressure - struggling with the concept! |
12 Apr 2005 02:08:39 PM |
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"grant" <grant.linscott@gmail.com> escreveu na mensagem
news:1113293915.187800.281030@o13g2000cwo.googlegroups.com...
Perhaps someone can shed some light on some calculations I am
attempting. (In simple terms!!)
(snip)
To give some more figures - The max water depth is 758mm and the
total volume of water is around 250L
The volume doesn't matter.
The pressure evolution is linear from top to bottom,
starting at zero on water surface and will be maximum
at the maximum depth.
Pressure = density x g x depth
density = 1000 kg/m^3 (pure water at 4 degree Celsius)
g = 9.8 m/s^2
your maximum depth = 0.758 m
Hence the maximum hydrostatic pressure will be:
1000 x 9.8 x 0.758 = 7428 N/m^2 = 728 kg/m2^=
= 0.728 kg/cm^2 at the bottom.
Which means that for every square centimeter of the
side wall at the bottom the force will be 728 grams.
Now you must evaluate the area of the biggest wall
at the bottom and calculate the maximum force on that
wall:
Force (in kg) = pressure (in kg/cm^2) x area (in cm^2)
(don't mess with units - units must be coherent)
Force = 0.728 kg/cm^2 x area (in cm^2)
If you have an area of 300 cm^2 the force will be
about 22 kg of force.
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| User: "CWatters" |
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| Title: Re: Calculating water pressure - struggling with the concept! |
12 Apr 2005 11:11:44 AM |
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"grant" <grant.linscott@gmail.com> wrote in message
news:1113293915.187800.281030@o13g2000cwo.googlegroups.com...
<snip> would it be possible to close the vessel
and add a flexible pipe (say of just 10mm diameter) to the top and fill
vessel and tube with water to twice the normal working depth to double
the working pressure??
Yes. The pressure at a depth doesn't depend on the shape or size of the
vessel.
Otherwise a diver who trained 10 foot down in a swimming pool would get a
big surprise when he went to the same depth in the Atlantic!
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| User: "" |
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| Title: Re: Calculating water pressure - struggling with the concept! |
12 Apr 2005 02:15:48 PM |
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In article <1113293915.187800.281030@o13g2000cwo.googlegroups.com>, "grant" <grant.linscott@gmail.com> writes:
Perhaps someone can shed some light on some calculations I am
attempting. (In simple terms!!)
I am intending to pressure test an acrylic vessel, and I would like to
calculate the max static water pressure when filled (prior to
commencing testing)
The vessel is non-uniform, in that it is made up of chambers connected
by large diameter (120mm) tubes in all 3 planes.
I have done some research and found that Pressure = g (9.81) x water
depth.
to be exact, g*density*depth. If you work in MKS, water density is
1000 kg/m^3.
While I can apply this formula it seems to me to oversimplify
(someone please correct me if I'm off base here) as I was under the
impression the volume of water pressing down creates the pressure.
It is the depth.
To give some more figures - The max water depth is 758mm and the
total volume of water is around 250L
When I use this formula I get:
P = 9.81 x 758
Pressure = 7436 N/m2 (This is equal to 1.08 PSI or 0.074 bar)
You're lucky, you got the right result for the wrong reasons:-)
You're using mixed units in the calculation (always a bad idea)
should've used depth = 0.758 m. So that's a factor of a thousand off.
But you forgot the density which gives you a factor of a thousand in
the opposite direction, so all is well.
My feeling is that this seems awfully low,
No, it is fine.
and I am having difficulty with the idea that the depth is purely the
factor that contributes to the static pressure.
But it is.
Considering the testing procedure:
If (P=g x depth) holds true would it be possible to close the vessel
and add a flexible pipe (say of just 10mm diameter) to the top and fill
vessel and tube with water to twice the normal working depth to double
the working pressure??
Indeed.
The equation suggests this, but my sense of
physical mechanics says this cannot be true...to double the pressure
there needs to be 500L present.
No, just the height. There is a lovely experiment (which kids
nowadays won't get to see due to safety concerns, where you get a
glass vessel, filled with few liters of water, plugged solid with a
stopper through which there goes a tall and slim glass tube, vertical
and open on top. Then you start adding water to the glass tube and at
some point the vessel bursts from the pressure.
Am I mixing pressure and force applied by weight...?
Yep.
I will welcome any suggestions to help clarify this. (On a simple level
please!)
Consider a vessel with base area A and height h. The volume is V= A*h.
The weight of the liquid is W = V*rho*g = A*h*rho*g, where rho is the
density. This weight is distributed over an area A. So the pressure
is P = W/A = h*rho*g. Depending on height alone. That's very
simplisitic, of course, but more detailed treatment will give the
same.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
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| User: "John C. Polasek" |
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| Title: Re: Calculating water pressure - struggling with the concept! |
12 Apr 2005 01:17:59 PM |
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On 12 Apr 2005 01:18:35 -0700, "grant" <grant.linscott@gmail.com>
wrote:
Perhaps someone can shed some light on some calculations I am
attempting. (In simple terms!!)
I am intending to pressure test an acrylic vessel, and I would like to
calculate the max static water pressure when filled (prior to
commencing testing)
snip.
Considering the testing procedure:
If (P=g x depth) holds true would it be possible to close the vessel
and add a flexible pipe (say of just 10mm diameter) to the top and fill
vessel and tube with water to twice the normal working depth to double
the working pressure?? The equation suggests this, but my sense of
physical mechanics says this cannot be true...to double the pressure
there needs to be 500L present.
Am I mixing pressure and force applied by weight...?
I will welcome any suggestions to help clarify this. (On a simple level
please!)
Many Thanks
Grant
Maarten has given you sound advice but here's some simple information
in English units.
The water pressure increases at 0.44 psi per foot
(because a cubic foot weighs 64 lbs. and
divided by 144 sq. inches = .44 psi/foot).
If the vessel has height H, adding a pipe to 2H will double the
pressure.
John Polasek
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| User: "John C. Polasek" |
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| Title: Re: Calculating water pressure - struggling with the concept! |
13 Apr 2005 08:23:33 AM |
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On 12 Apr 2005 01:18:35 -0700, "grant" <grant.linscott@gmail.com>
wrote:
Perhaps someone can shed some light on some calculations I am
attempting. (In simple terms!!)
I am intending to pressure test an acrylic vessel, and I would like to
calculate the max static water pressure when filled (prior to
commencing testing)
The vessel is non-uniform, in that it is made up of chambers connected
by large diameter (120mm) tubes in all 3 planes.
I have done some research and found that Pressure = g (9.81) x water
depth. While I can apply this formula it seems to me to oversimplify
(someone please correct me if I'm off base here) as I was under the
impression the volume of water pressing down creates the pressure.
To give some more figures - The max water depth is 758mm and the
total volume of water is around 250L
When I use this formula I get:
P = 9.81 x 758
Pressure = 7436 N/m2 (This is equal to 1.08 PSI or 0.074 bar)
My feeling is that this seems awfully low, and I am having difficulty
with the idea that the depth is purely the factor that contributes to
the static pressure.
Considering the testing procedure:
If (P=g x depth) holds true would it be possible to close the vessel
and add a flexible pipe (say of just 10mm diameter) to the top and fill
vessel and tube with water to twice the normal working depth to double
the working pressure?? The equation suggests this, but my sense of
physical mechanics says this cannot be true...to double the pressure
there needs to be 500L present.
Am I mixing pressure and force applied by weight...?
I will welcome any suggestions to help clarify this. (On a simple level
please!)
Many Thanks
Grant
You said you were pressure testing the vessel and it should be
apparent by now that static hydraulic head is not practical. A 100
foot pipe would only give 44psi.
Here's my suggestion:
Fill the tank nearly to the brim with water, put a cap on it with a
Shrader valve. Use an electric tire inflator that goes to 100 or more
psi to get to the pressure you want.
By filling it with water, only the tiniests amount of air will need to
be compressed, and if it bursts, little harm done, compared to filling
the empty vessel with air, which if it exploded could make you part of
the wallpaper in your garage.
John Polasek
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| User: "grant" |
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| Title: Re: Calculating water pressure - struggling with the concept! |
12 Apr 2005 02:20:20 PM |
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Thank you for all your inputs and insights.
I'll continue the examination with a simple pressure test. (Using the
flexable pipe)
Thanks again
Grant
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| User: "Uncle Al" |
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| Title: Re: Calculating water pressure - struggling with the concept! |
12 Apr 2005 03:35:54 PM |
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grant wrote:
Perhaps someone can shed some light on some calculations I am
attempting. (In simple terms!!)
I am intending to pressure test an acrylic vessel, and I would like to
calculate the max static water pressure when filled (prior to
commencing testing)
The vessel is non-uniform, in that it is made up of chambers connected
by large diameter (120mm) tubes in all 3 planes.
I have done some research and found that Pressure = g (9.81) x water
depth. While I can apply this formula it seems to me to oversimplify
(someone please correct me if I'm off base here) as I was under the
impression the volume of water pressing down creates the pressure.
Pressure, force/cm^2, is water depth in cm times the density of the
water g/cm^3 times local gravitational acceleration cm/sec^2.
Pressure is overhead weight/area. Per cm^2 keeps it easy. Convert to
whatever units you like at the end. As water is reasonly
incompressible, pressure at constant temperature and composition is
directly proportional to depth - and independent of everything else
including shape - unless you worry about kilometers deep. Then you
can correct for gee vs. radius, too.
If you have a meter cube of water and a cm-diameter tube that rises
another meter from it, at the bottom of it all you have two meters'
depth of static pressure. If the extension is shaped like a silly
straw and has all sorts of bulges and whatnot, and its continuous
contiguous water level is a meter above your cube, nothing changes in
the static case. This has obvious dynamic caveats.
[snip]
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/qz.pdf
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| User: "Randy Poe" |
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| Title: Re: Calculating water pressure - struggling with the concept! |
12 Apr 2005 12:27:31 PM |
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grant wrote:
If (P=g x depth) holds true would it be possible to close the vessel
and add a flexible pipe (say of just 10mm diameter) to the top and
fill
vessel and tube with water to twice the normal working depth to
double
the working pressure?? The equation suggests this, but my sense of
physical mechanics says this cannot be true
Yet it is true. You've just made a fairly close
description of a hydraulic system, such as car
brakes. You can exert a small force (but large
pressure) on a small piston under your foot, and
this translates to the same pressure (but much
larger force) on some large working cylinder
elsewhere in the system.
..to double the pressure there needs to be 500L present.
Nope. Look into hydraulics, perhaps that will help
your insight in the right direction.
- Randy
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| User: "Maarten van Reeuwijk" |
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| Title: Re: Calculating water pressure - struggling with the concept! |
12 Apr 2005 04:31:03 AM |
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grant wrote:
Perhaps someone can shed some light on some calculations I am
attempting. (In simple terms!!)
I am intending to pressure test an acrylic vessel, and I would like to
calculate the max static water pressure when filled (prior to
commencing testing)
The vessel is non-uniform, in that it is made up of chambers connected
by large diameter (120mm) tubes in all 3 planes.
I have done some research and found that Pressure = g (9.81) x water
depth. While I can apply this formula it seems to me to oversimplify
(someone please correct me if I'm off base here) as I was under the
impression the volume of water pressing down creates the pressure.
No, this is a pressure per unit density. The hydrostatic pressure is defined
as
p = rho * g * d (1)
You can see this from the units:
[p] = kg / m3 * m / s2 * m = kg /m s2 = (kg m / s2) / m2 = N/m2
Search google for 'hydrostatic pressure', for example:
http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Pressure/HydroStatic.html
To give some more figures - The max water depth is 758mm and the
total volume of water is around 250L
When I use this formula I get:
P = 9.81 x 758
Pressure = 7436 N/m2 (This is equal to 1.08 PSI or 0.074 bar)
Remarkably this answer is right, as a result of two errors canceling each
other. When using formula's, you have to apply consistent units. This means
that the depth = 0.758 m. However, as the density is 1000 kg / m3, formula
(1) results in exactly the same answer, namely p = 7.4 kN / m2 = 0.074 bar.
You can check that this is correct because 1 bar corresponds to 10 m of
water pressure, as every diver knows.
If (P=g x depth) holds true would it be possible to close the vessel
and add a flexible pipe (say of just 10mm diameter) to the top and fill
vessel and tube with water to twice the normal working depth to double
the working pressure?? The equation suggests this, but my sense of
physical mechanics says this cannot be true...to double the pressure
there needs to be 500L present.
Yep this is exactly how it works. This is why water towers (approximately 10
m high, corresponding to 1 bar of pressure) are used to make sure that when
you open the tap water flows out of it.
HTH, Maarten
--
===================================================================
Maarten van Reeuwijk Thermal and Fluids Sciences
Phd student dept. of Multiscale Physics
www.ws.tn.tudelft.nl Delft University of Technology
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| User: "AJW" |
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| Title: Re: Calculating water pressure - struggling with the concept! |
12 Apr 2005 08:37:54 AM |
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the fluid pressure at any depth has to support the weight of the fluid
above it, not to its sides. Your concern about the totalo volume would
be correct if the material was a solid -- the bottom, no matter what
its size, would have to support the entire weight.
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