Water Tank Wall Fluid Pressure 4

[silverfai]

I have to build a water tank with the dimension of 20 ft by 10ft and it holds up to 5 feet of water. I am trying to optimize my design and things like that. I would like to see if there's any equations to help me figure out the water pressure acting on the walls of my tank. Thanks

 

[BigInch]

62.4 * Depth_feet / 144 = static H2O pressure in psi

Going the Big Inch!

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[25362]

dP/dh = [ρ] g, meaning the hydrostatic pressure P increases linearly with depth h.

Where [ρ] is density of water, and g is the acceleration of gravity.

Assuming the tank surface is under atmospheric pressure, and the total depth is H:

Pressure at the deepest end, Pa = 1,000 kg/m³ [×] 9.81 m/s² [×] H, m

 

[sailoday28]

Dependent upon vents, etc, you may want to consider design for full or partial vacuum.

regards

 

[25362]

If one needs to estimate the total forces F, acting on the tank vertical walls of width W, and depth H, and since pressure varies with depth, one has to integrate thus getting:

F = P_oWH + [½] [ρ] gWH²​

The units are:

F, N
P_o, if atmospheric, (101,325) Pa (= N/m²)
[ρ], for water, (1,000) kg/m³
W and H, m
g, (9.81) m/s²

 

[BigInch]

Seems like I can't concentrate today. I see I also forgot the weight of air.

Going the Big Inch!

http://virtualpipeline.spaces.msn.com

 

[silverfai]

It's my first time to post a topic. I greatly appreciate the instant help from everybody. I guess I have to dig in to do some more math now. Thank you very much!

 

[silverfai]

25362, So, the froce acting on the vertical wall is
F = PoWH + ½ ? gWH^2

It makes sense the forces acting on the wall is based upon the width and the depth of water. The deeper it gets. The pressure will be found. Then, the forces acting on the wall are higher.

But, did you integrate the total pressure equation in order to get the force equation?

I understand the second part of the equation where you integrate "? gWH" with respect to H to get the ½ ? gWH^2. I still cannot explain to myself that where is the first term (PoWH) comes from. Is it just the atmospheric pressure acting on the walls in general?

If you can explain a little bit more, that would be great!

 

[silverfai]

25326, nevermind, I got it!

 

[sailoday28]

Big Inch--If you add the weight of the air to your calculation, the result gives the absolute pressure.
This could be done by simply adding approximately 1 atmos of pressure to your calc. These resulting pressures are on the inside of the tank.

The net pressure across a piece of the wall or bottom of tank can then be obtained by subtracting the outside ambient pressure, which would probably be one atmosphere.

Regards

 

[quark]

The maximum pressure, for water, can be calculated as suggested by BigInch (in FPS units) or much simply p = 0.433Hs where p is pressure in psi, H is height of liquid column in ft, and s is specific gravity of tank contents.

There is an excellent paper by Kanti Mahajan, 'A method for designing rectangular storage tanks' appeared in March 77 issue of Chemical Engineering Journal. The calculations are based on theory of bending for thin plates. There are methods to design tanks with and without stiffeners. The procedures are easy to do.

Sometime back, a comember converted that paper into an excellent excel spreadsheet. Google for it if you are fortunate.

 

[25362]

Silverfai, sorry for the delay due to the large time difference.

For example, when designing swimming pools with vertical walls, forces are the result of integration. The area d A = Wd h, then the force is d F = Pd A = PWd h, and for P = P_o + [ρ] gh,

F = [∫] d F = [∫]₀^H (P_o + [ρ] gh)Wd h = P_oWH + [½][ρ] gWH²​

 

[quark]

P₀WH may be redundant in this paticular case as it acts on both sides of the tank wall.

 

[BigInch]

Sailoday, Thanks, I know. I was taking a ... about adding it, although you can if you want to, as long as you subtract it later. I believe it is both an internal and external pressure, as the OP still makes no reference to any internal pressure in a pressurized tank. I've probably designed over 250 open to atmos tanks, retention basins, cooling tower basins, swimming pools, pump pits, etc. and never included air before. Still see no reason to do so as long as the tank is open to atmosphere and there is no floating roof.

For a nonpressurized water tank, I'd still use the first formula <it now appears correct above, thank you Zapster>. No integration required!!!

Going the Big Inch!

http://virtualpipeline.spaces.msn.com

 

[25362]

Quark, you are right as long as the external pressure is equal to P_o. However, this is not always the situation, and there are cases where the external pressure on the walls, P [&ne;] P_o. These external pressures would have to be taken into account when estimating the net pressure and forces acting on the walls, as indicated by sailoday28.

 

[LSThill]

TEAM MENBER

What about Sloshing Pressure and Sloshing Force on the tank wall?

 

[IFRs]

For a tank this size, the minimum thickness allowed in the code (AWWA or API etc) will be much greater than your calculated thickness. Adding the weight of air is an unnecessry complication unless you are building a swiss watch. Sloshing is not usually accounted for and tanks don't usually fall over in an earthquake. I'd advise you to get the code and follow thier lead.

 

[quark]

25362,

I like even being devil's advocate with you. You come up with better knowledge whenever you are teased, yet never losing your temper.

Coming to the OP, I presumed, here, many things. In my experience, all water tanks are atmospheric. That is why I considered P₀ as redundant.

Secondly, integration yields net pressure and net force on the tank wall and this acts at about 2/3rd height from the top of the tank. This calculation is generally used when we have to design hinged (or other)closures.

This calculation is generally not useful for designing thickness of tanks, for the tank should be designed for the maximum pressure possible in the system i.e at the bottom most section of the tank. Vertical section may not fail if we design the tank as per the pressure calculated by your method but bottom horizontal section may fail.

PS: I know that you are perfectly aware of this from your first post on 8th September. I clarified it to clear the misunderstanding(if any) of future viewers of this thread.

 

[25362]

Quark, I agree, please note that:

1. my example referred to a swimming pool, not just a
simple prismatic opened water tank,
2. although we all assumed so, silverfai didn't say that it
would be an isolated metal-, concrete-, wood- or
otherwise a dike-protected tank,
3. no details were given on the filling and emptying
procedure and connections,
4. you are right, the total "net" force may be assumed to
be resisted by a single force applied at a point, named
center of pressure or centroid of the submerged
surface, about which there will be no moment due to the
hydrostatic force, and that these concepts are applied
when designing, for example, submerged gates or
openings.