Calculate waterflow on inclined plane

[Stormer]

How can i calculate the waterflow to maintain X level flowing down a flat inclined plane?

For example say you have a waterslide that is 1 meter wide with a incline of 30 degrees and want a sheet of water 10 cm deep flowing down the slide. How many m3/h of waterflow do i need to supply on the top of the slide to maintain that level of water flowing down the waterslide?

 

[Baluncore]

How many m3/h of waterflow do i need to supply on the top of the slide to maintain that level of water flowing down the waterslide?

Your question assumes a stable, steady-state solution exists. Unfortunately, that is not the case.

The water depth on the slide will become unstable, because a slightly deeper flow, will travel faster, forming and supporting, discrete waves of water, that tumble down the wet surface.

 

[jrmichler]

Start by searching open channel flow, which discusses that type of flow. Unfortunately, the research behind open channel flow is based on studying rivers and culverts, few of which are similar to your case of smooth bottom and very steep angle. The MIT hit looks like a good place to start: https://ocw.mit.edu/courses/12-090-...2006/355d3c6b2ddfb45627c9b1fa7cd4463d_ch5.pdf. Then search Manning equation. Just keep in mind that the Manning equation is for flows with lesser slopes and rougher bottoms than your case. But it might give you a rough idea of the flow.

Are you sure that you want a water depth of 10 cm with a 30 degree incline? That is a LOT of flow. I would think that a waterslide would work very well with only a few millimeters depth.

 

[Baluncore]

Key terms are: Sheet flow, Kapitza_instability, and roll waves.
https://en.wikipedia.org/wiki/Kapitza_instability

https://mirjamglessmer.com/2019/01/...ated-friendlywaves-ive-gotten-over-the-years/

https://ponce.sdsu.edu/the_control_of_roll_waves.html





There are approximations for sheet flow on steep inclined plates. Those approximations will NOT hold for 100 mm thick flows on 30° plates.
"Consider a liquid (of density ρ) in laminar flow down an inclined flat plate of length L and width W. The fluid flows as a falling film with negligible rippling under the influence of gravity. End effects may be neglected because L and W are large compared to the film thickness δ."
https://www.syvum.com/cgi/online/serve.cgi/eng/fluid/fluid804.html

 

[Stormer]

Are you sure that you want a water depth of 10 cm with a 30 degree incline? That is a LOT of flow. I would think that a waterslide would work very well with only a few millimeters depth.

My application is not really a waterslide. That was just used as a example. My real application is in surfing and the depth is to accommodate some fins on the board so they dont just slide on the bottom plate, but actually function like they do in the open water.

 

[Stormer]

Keywords are: Kapitza_instability, sheet flow, and roll waves.

So why don't we see the roll waves in artificial wave surfing facility's?
As you can see from the inlet on the top all the way down to the bottom of the valley of the wave the water is almost completely flat, and it is a pretty thick watersheet over the floor of the artificial wave:

And you can see the same on standing wave river surfing:

Even dam overflow's have a pretty flat sheet of water over the concrete floor:

 

[Baluncore]

So why don't we see the roll waves in artificial wave surfing facility's?

They are there, but they have been minimised because otherwise they would hit the surfers and make it difficult to ride. The paper by Ponce V M and Guzmán B C shows how to minimise roll waves. Watch the figure 18 video to see what happens given a sufficiently long run.

As you can see from the inlet on the top all the way down to the bottom of the valley of the wave the water is almost completely flat, and it is a pretty thick watersheet over the floor of the artificial wave:

If the water was flat, you would see a reflection in it, but it is covered with small waves that have not yet, because of the short slope, had time to self organised into roll waves.

Even dam overflow's have a pretty flat sheet of water over the concrete floor:

Look at the foot of the dam wall, where there are patterns on the flat concrete, made by roll waves falling down the steep face of the dam wall.

 

[jrmichler]

It looks to me like a better problem description is roughly:

1) Water flows over the top of a dam at a known depth of water over the top and dam width.
2) The water flows down a slope with a known angle and height.
3) The water flows up over a curved obstruction with known dimensions.
4) Estimate the required flow rate.
5) The static head and pipe friction loss to pump it up to the upstream side of the dam are easily calculated with standard methods.

If you can reduce the problem as stated above, the water flow over a dam is easily calculated using standard methods. Search terms flow over a weir will find good information online. The challenge is to get the water depth over a dam. You need to find a dam that seems to be flowing about the right amount of water, then poke a stick down onto the top of the dam.

I would not want to get caught in the turbulence downstream from the curved obstruction. Might get banged up a little.

 

[Lnewqban]

So why don't we see the roll waves in artificial wave surfing facilities?

Since several methods exist, it seems that the wave creation has been optimized by experimentation, even using pre-design computer modeling perhaps.

Please, see:
https://wavepoolmag.com/techniques-to-create-waves/

https://www.ansys.com/advantage-magazine/volume-xiv-issue-2-2020/simulation-drive-design-waves

https://airshaper.com/blog/big-wave-surfing-simulating-Ian-walsh

 

[Stormer]

The challenge is to get the water depth over a dam. You need to find a dam that seems to be flowing about the right amount of water, then poke a stick down onto the top of the dam.

Why would that not be easy to calculate? Using gravity, mass of water and friction that is all known values.

 

[Baluncore]

Given that the mass flow will be determined by a function of the depth, over the crest of the spillway, the flow depth will then be an inverse function of the water velocity.

A hydraulic jump will convert the kinetic energy into potential energy, maybe building a deep standing wave, with a tumbling top.

 

[tylerkelley1980]

When I calculated water flow on an inclined surface, I used the Manning equation for open channels—factoring slope, surface roughness, and hydraulic radius. It’s surprisingly accurate even for small-scale setups. Just make sure your measurements are consistent and account for flow width. For more complex geometries, I’ve also used simulation tools like HEC-RAS to visualize surface runoff behavior.

 

[Chestermiller]

For flow water down an incline, the Manning equation (in SI units) reads: $$v=\left(\frac{1}{n}\right)R^{2/3}S^{1/2}$$where v is the average flow velocity parallel to the incline, R is the hydraulic radius of the flow channel (R = h for an flat incline, with h representing the thickness of the fluid layer), and S is the slope of the incline ($S = \sin{\alpha}$), with $\alpha$ representing the sine of the angle of incline relative to the horizontal), and n is a roughness parameter for the surface, equal to 0.01 for a smooth surface.

This equation is inconsistent with a dimensional analysis of the system. From dimensional analysis, we obtain the following relationship in terms of two key dimensionless parameters: $$\frac{vd}{\nu}=F\left(\frac{h^3 g\sin{\alpha}}{\nu^2}\right)$$where $\nu$ is the kinematic viscosity of the fluid, and g is the acceleration of gravity (9.8 m/s^2). If the velocity is supposed to vary with the liquid thickness h to a power, this equation must be expressible as $$\frac{vd}{\nu}=k\left(\frac{h^3 g\sin{\alpha}}{\nu^2}\right)^m$$ where k and m are dimensionless constants. For v to vary with h to the 2/3 power, the exponent m must have the value 5/9, such that $$v=kg^{5/9}\nu^{-1/9}h^{2/3}\sin {\alpha}^{5/9}$$
So we see that, according to dimensional analysis, the exponent of the channel slope in the Manning equation should be 5/9 = 0.555, rather than 1/2 = 0.5.

It is also possible to perform a standard turbulent Flow frictional analysis for this system and show that the Fanning friction factor is expressible as $$f=\frac{0.0311}{Re^{0.2}}$$ where Re is the Reynolds number for the flow, given by $$Re=\frac{vh}{\nu}$$ This compares with $$f=\frac{0.0791}{Re^{0.25}}$$ for flow in a pipe.