Pressure exerted by water in a pond

[Mr. A]

I am trying to figure out how much pressure will water exert on the walls of the pond. The circular pond has a radius of 10 ft and depth of 2.5 ft.

Pressure = density (water) * depth
Pressure = 62.4 lb/ft3 * 2.5 ft = 156 lb/ft2 or 1.083 psi​

That will give me the maximum pressure which will be at the bottom of the pond. Correct? Or I need to add gravity to the equation?

I am basically trying to figure out what will be the maximum pressure applied by the water so that based on that I can choose the strength of the material for the pond.

Thanks for help.

 

[Mech_Engineer]

Pressure = density (water) * depth
Pressure = 62.4 lb/ft3 * 2.5 ft = 156 lb/ft2 or 1.083

Correct, the pressure at the bottom of the tank's wall will be 1.083 psi; however, don't forget that the pressure on the wall will vary linearly from 1.083 psi at the bottom to 0 psi at the top. Therefore, the total force on the wall will not simply be the wall's surface area multiplied by the pressure exerted on it...

 

[Mr. A]

Thanks Mech_Engineer.

In the file attached how do I find the resultant force F(r). I seem to have forgotten the formula. And, is that the maximum force that water will exert on the walls of the pond?

F (r) = 1/3 * height = 1/3 * 2.5 ft?

 

[Mech_Engineer]

You're on the right track, in that the resultant force when viewed in a 2-D cross-section will occur at the centroid of the triangle's area, so 1/3 of the tank's height, however its magnitude will be the average pressure exerted on the tank's wall multiplied by the wall's area. This force can be used for static reaction force calculations. Don't forget that your tank is a cylindrical cross section, which means its analysis will look rather similar to the analysis of a pressure vessel, other than the fact that the pressure is changing with height (and it will not have a top).

The area you will have to be most concerned with in terms of material strength will be where the wall is attached to the floor of the tank. Unless you're using a tapered wall, you can use the same thickness throughout as long as the tank can hold up to the forces at this critical sopt.

 

[JohnGardner]

I am trying to find a formula for a similar calculation.

The project I'm working on is digging out a 15 acre pond with an island. The dirt/land of the island will be resting on top of a concrete structure.

The concrete structure is a room that will be submerged under the water after the pond/lake is full and will have access from the surface above.

One wall of the structure will have an acrylic glass wall for viewing fish under the water.

I understand that pressure on that wall will exponentially increase with surface area.

I am trying to figure a calculation for measuring the size and thickness of the acrylic/plexi-glass needed, but am haveing difficulties can anyone assist?

 

[russ_watters]

You haven't given us enough information to figure it out, so you're going to need to do some work here yourself. What is the size, depth, and orientation of the window? Perhaps you can do the integration described above to calculate the force yourself...

 

[nvn]

JohnGardner: Assuming your flat acrylic plexiglas (also called PMMA) does not exceed 2.5 m high, 6.25 m long, and does not exceed a depth from the water surface to the top of the plexiglas during flooding conditions of 2.5 m, then a pane thickness of t = 140 mm should be adequate. If your pane height, length, and depth dimensions are different than assumed above, post the dimensions.

The above refers to Plexiglas G, Plexiglas GS, and Altuglas CN, not Plexiglas MC, Plexiglas XT, nor Altuglas EX.

 

[JohnGardner]

Fantastic! Thank you so much - that was in the vicinity that I was looking in - and special thanks for the clarification on Plexiglass types!

 

[Harold in CR]

Hi
Hope this post gets some attention.
I live in Costa Rica and am trying to figure out the best way to set up a small Hydro system.

I have a spring fed flow of water across our small farm. I have measured the flow at 5 Gallons at 14 seconds. I realize this is not a lot of water.

I have dug a small pond uphill from the water source. It will be 4' deep and have around 9000 gallons of storage, but, be fed with possible 2 waterwheels, each pushing 2 piston pumps to push the water up to the pond. I plan on using the overflow for the generation, downhill. Hopefully, I will have 1500 gallons in 24 hr period of time, OR MORE.??

First, the pond is right at 110' elevation, at 200 feet distance to the stream and waterwheel location. I have 2-4" PVC pipes set in concrete for the drain and overflow. I will cap the ends inside the pond, and let the overflow drop down a 4" pipe, into the "T" that attaches the drains to each overflow. That will drop the water to the generator site, just above and upstream of the waterwheel site. The spent water from the generator, can be piped right back downstream, to help the waterwheels, in case there is insufficient water available to run both. Figure to wear out the water, using it over and over and over. This is a second water supply, coming as a direct result of the first supply being piped to the waterwheels from the stream source to the waterwheels.

Today, I was digging around below the elevation of the waterwheels location, trying to see how deep the bedrock is, because, I found out that the small dam I installed upstream, to put the 3" pipes into, that will feed the waterwheels, has less flow than at the location where I was digging. SOOooooo, I am thinking about putting a Dam at THIS location, and have it hold back a small pond, that would be 18' wide, X 40'+ long, X nearly 36" deep, throughout the whole pond area. There is plenty of room to build a small cement block, "powerhouse" at the base of the Dam. IT will be cement blocks, poured full of concrete, and tied into the existing boulder type rocks that are set into a "concretion" type mix, that looks exactly like moist coarse sand mix concrete. I can drill holes into this concretion, and put pieces of rebar in the holes, that will allow me to tied the blocks into the concretion.

This whole dam will be 4' high by the thickness of the blocks, 4½", and span a distance of 4-5 FEET or so. Pretty small set up.

What I am needing to find out is, I can figure the amount of power I can produce at the Pond overflow system, but, is it possible, I could put a drain pipe low in the dam, and have sufficient pressure to get the same amount of power, without the Rube Goldberg piping-waterwheel system? Just picture a MINI-Hoover Dam using this SPRING fed stream.

I can supply all the photos needed, but, first, would like someone to try to figure out if there is sufficient FORCE of water pressure, to run a Pelton Wheel set up, at the discharge of the dam ?? AND how much power would be available from the info I have provided. Anyone? PLEASE

 

[russ_watters]

I'm not quite following: why can't you use the water wheels themselves to generate energy? What you are describing is an energy storage system, using pumping to move water uphill (spending energy), then getting some of the energy back when it falls.

 

[Harold in CR]

You are absolutely correct. There is not much power to be generated by the water wheels. There is less than 3' of head, from the dam upstream to the waterwheels pit.

I am copying another persons water wheel design. It is possible, that I will need to run the water back down, to the original waterwheel, to get sufficient flow to start pushing water into the pond uphill.

As long as I can do all this with no other power being used, I can just be patient to allow the buildup of sufficient water to fill the pond.

This is why I am having thoughts about the larger pond and the generator being powered from the bottom discharge. I am thinking there is not enough pressure to be had from the pond, with only 3' of head, but, there will be a lot of water stored, so, MAYBE that will add more down force to the drain ?

Sorry, physics is NOT my strong suit.