Water Pressure Explanation

Hi Smitty.

Where will this be used?

Is it for a sort of pelton wheel design or an obstruction in a channel or a water wheel design?

If it is a channel what is the clearance to the sides?

I don't quite understand the 16" - 18". What is the radius of the bend? it seems to be going through 90 deg.

Does the sharp end face the flow or is the intention to vary the face?

Your problem seems similar to the force acting on an elbow in a pipe with fluid flowing in it. If the elbow bends through an angle of θ, then the force acting on the elbow is 2F*sin(θ/2) where F is p*A, p being the pressure and A being the cross sectional area of the pipe. The resultant force is just the vector sum of the forces each side of the elbow.

The force required to anchor the elbow is the same magnitude as above but opposite in direction.

If I understand the OP correctly, he (I presume Smitty is a guy) wants to drag a cylindrical arc of sheet metal through the water. I'm assuming, as did Hendrik, a 90° arc.

Here is a top view of my interpretation:

The red force arrows are just slightly educated guesses of magnitude and direction.

There will be forces on both surfaces of the cylindrical arc, and I'm sure the calculations will be complex, especially at high speeds, where cavitation would complicate matters significantly.

He is talking about sheet metal. A piece of sheet metal that large is going to have to be quite strong to keep any semblance of its original shape at any significant velocity.

I interpreted the OP to be deflecting a stream of water in an open channel through some angle (could be 90 degrees). Smitty...we need clarification. Please describe your application in a little more detail.

Here is a neat little calculator for determining the force on an elbow in a pipe with various pressures and flow velocities. You could enter the calculator with a pipe diameter such that the area is equivalent to your channel, i.e. d = 2*(H*W/π)^0.5. In this case the diameter would be about 24.7".

F -> Force

ρ -> density of fluid

Q -> flow rate

V -> velocity of fluid

A -> projected Area

H -> Height

W -> Width

θ -> angle of attack (angle is 0° from the perpendicular surface)

F = ρQVcos(θ), Q = AV

F = ρAVVcos(θ), A = HWcos(θ)

F = ρHW [Vcos(θ)]^2

I assume that is for a flat surface. Now integrate that over the curve...

Your question doesn't really have a single answer. The force is caused by hydrodynamic lift which builds steadily with flow velocity until fluid shear develops at which point lift falls off noticeably and then starts to build again slowly. The angle of attack has a huge effect on the formation of turbulence which is basically the same as shear. There are hydrodynamic programs that can run a set of computations for you and they will give at data set that should be about what you actually see as a result but, they are really more an aide in design than an expression of an exact outcome.

If is a project that needs to be done perfectly the first time your in trouble. If you can make a estimate, do and experiment or two and get some data points and then use your data points to correct your estimate you should end up with a good estimate for further work. Hope this helps.

Mr. Gee

PS. If you would like more detailed information check out Wikipedia under the topic computational fluid dynamics or google this or the short form which is CFD.

OK.

I am no where near smart enough to work these calculations.

I was looking for a short, easy formula but i can see there is none. You all are way past my intellect on this.

Can I just ask this :

If i am looking to make a rudder of sorts to keep a gate open when a current runs into a pond from a tidal source.

I can just work on different sizes until i get the right one, but..............

If I want to maximize the force against this curved piece of metal, would it be better to go Higher or Wider ?

Thank you. This may in fact be the easiest.

Smitty