Water flow down a porous channel (Navier-Stokes/Fluid dynamics)

In summary: Your second boundary condition is u(H)=U. This is also correct, but you need to solve for H. You can do this by substituting y*=Hy* into the second equation and solving for H. You'll get a value that matches the given boundary condition.
  • #1
tome101
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Homework Statement


Water flows down a channel whose floor is porous, so that water seeps out of the bottom of the channel at a speed v, where v is constant and much less than the flow speed, U, along the surface of the channel. The seepage rate is slow so that H may be regarded as constant.

The x−component of the Navier-Stokes equation for such a system is

[itex]\nu\frac{d^{2}u}{dy^{2}}+v\frac{du}{dy}+G=0[/itex]

where G is a constant.

Verify (i.e. no need to derive) that the general solution of this equation is

[itex]u=A-\frac{G}{v}y+Be^{-vy/\nu}[/itex]

where A and B are integration constants.
State clearly what the boundary conditions are that determine A, B and G, and verify that the required solution is:

[itex]u*=\frac{1-Re^{-R}y*-e^{-Ry*}}{1-(R+1)e^{-R}}[/itex]

where u = Uu*, y = Hy* and R = vH/[itex]\nu[/itex].

Homework Equations


u(0)=0

u(H)=U


The Attempt at a Solution


Ok, I managed the first part and have verified that the 2nd equation is indeed a solution of the first. However, I am having trouble removing the integration constants and G and rearranging to the required equation.

I know that at the top of the channel, u=U so u(H)=U. I'm also assuming that since v is 'much less' than the flow speed we can assum that at the bottom of the path the horizontal flow speed is equivalent to 0 (although I am less sure about this).

So that gives boundary conditions of:
u(H)=0
u(0)=0

Using u(0)=0 I can see that A+B=0

I can get an equation for U by using u(H).

However, I can't seem to find any useful way of rearranging these equations to a) remove A B and G or b) look anything like the final equation... I am also not 100% certain my boundary conditions are corect. Attached is a diagam of the problem. Any help would be greatly appreciated.

Thanks
 

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  • #2
290 views and no-one has any thoughts on this? I'm still completely stumped! I've also just noticed i should say u(H)=U under the attempted solution
 
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  • #3
I'm not very familiar with fluid dynamics at all, but here are some thoughts:

Since you have three constants to eliminate, you'll need three equations. Your boundary conditions only give you two. You need to get one more constraint from somewhere.

The condition u(0)=0 is correct — you can see this by setting y*=0 in the provided solution — but your reason for it is wrong. It doesn't have to do with the seepage rate. It's the no-slip condition. The water doesn't slide along the channel floor.
 

1. How does the porosity of a channel affect the flow of water?

The porosity of a channel refers to the amount of open space or voids within the material of the channel. In fluid dynamics, this porosity affects the ability of water to flow through the channel. A higher porosity means there is more open space for the water to flow through, resulting in a faster flow rate. On the other hand, a lower porosity will cause the water to flow slower due to less available space for the water to move through.

2. What is the Navier-Stokes equation and how is it used to model water flow in a porous channel?

The Navier-Stokes equation is a fundamental equation in fluid dynamics that describes the motion of a fluid. It takes into account factors such as velocity, pressure, viscosity, and density to calculate the flow of a fluid. In the case of water flow in a porous channel, the Navier-Stokes equation is used to model the flow by considering the properties of the channel material and the forces acting on the water as it flows through the porous medium.

3. How does the viscosity of water affect its flow through a porous channel?

Viscosity refers to the resistance of a fluid to flow, and it plays a crucial role in the flow of water through a porous channel. A higher viscosity means that the water will flow slower as it encounters more resistance from the porous material. On the other hand, a lower viscosity will result in faster flow as the water can move more easily through the open spaces in the porous channel.

4. What factors can cause water flow through a porous channel to become turbulent?

There are several factors that can lead to turbulent flow in a porous channel. These include a high flow rate, a high porosity of the channel material, and a large pressure drop along the length of the channel. Additionally, the presence of obstacles or changes in the channel geometry can also cause turbulence. Turbulent flow is characterized by irregular and chaotic motion of the water, which can affect the overall efficiency of the flow.

5. How is the concept of Darcy's law used to analyze water flow through a porous channel?

Darcy's law is a fundamental principle in fluid dynamics that relates the flow rate of a fluid to the properties of the channel and the driving forces acting on the fluid. In the case of water flow through a porous channel, Darcy's law can be used to analyze and predict the flow rate based on factors such as the porosity of the channel, the hydraulic gradient, and the viscosity of the water. It is an essential tool in understanding and modeling water flow in porous media.

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