Additional boundary conditions for inclined flow?

In summary: JainIn summary, the problem is to find the volumetric flow rate of inclined flow in a square channel, but the attempt at a solution is not set up correctly. The coordinates should be taken parallel and perpendicular to the inclined plane, and the problem implies that the flow is fully developed. The partial differential equation for the velocity vz should be a Poisson equation with boundary conditions of vz being zero at the walls of the square duct.
  • #1
Runner 1
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Homework Statement



I am solving an inclined flow problem, and am stuck. The problem is to find the volumetric flow rate of inclined flow in a square channel. Once I have the velocity profile, I can just integrate over that to get the flow rate.

2. The attempt at a solution

Letting the x-axis be along the direction of flow, I start with Navier-Stokes:

$$\rho v_x \frac{\partial v_x}{\partial x} + \rho v_y \frac{\partial v_x}{\partial y} = \rho g \sin \theta + \mu \left( \frac{\partial^2 v_x}{\partial x^2} + \frac{\partial^2 v_x}{\partial y^2} \right)$$

$$\rho v_x \frac{\partial v_y}{\partial x} + \rho v_y \frac{\partial v_y}{\partial y} = \rho g \cos \theta + \mu \left( \frac{\partial^2 v_y}{\partial x^2} + \frac{\partial^2 v_y}{\partial y^2} \right)$$

No pressure terms since it's a free fluid. I found in a textbook the solution for inclined plane flow (but my problem is inclined flow in a square channel). In that solution, [itex]\frac{\partial v_x}{\partial x}[/itex] and [itex]v_y[/itex] are assumed to be 0, which gives

$$-\frac{\rho}{\mu} g \sin \theta = \mu\frac{\partial^2 v_x}{\partial y^2}$$

which is easy enough to solve with the boundary conditions [itex]v_x\big|_{y = 0} = 0[/itex] and [itex]\frac{\partial v_x}{\partial y}\big|_{y=L} = 0[/itex]. However, I don't know why they just assume [itex]\frac{\partial v_x}{\partial x}[/itex] and [itex]v_y[/itex] are 0; where does that come from? The fluid is flowing under the influence of gravity so I would think [itex]v_x[/itex] is increasing along x (imagine if you roll a ball down an inclined plane), and I would also think a free flow would decrease in height (L) as it travels. I mean, if you continuously dump water at the top of the incline, you're going to have a thicker water "height" at the top I would think...

And even if I accept that those assumptions are true, I don't know how to extend the problem from a plane to a square channel. There's no z-flow anyway, so I don't need any [itex]v_z \big|_{z=0}=0[/itex] or [itex]\big|_{z=W}=0[/itex] boundary conditions anyway. So the solution would be the same...
 
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  • #2
Runner 1 said:

Homework Statement



I am solving an inclined flow problem, and am stuck. The problem is to find the volumetric flow rate of inclined flow in a square channel. Once I have the velocity profile, I can just integrate over that to get the flow rate.

2. The attempt at a solution

Letting the x-axis be along the direction of flow, I start with Navier-Stokes:

$$\rho v_x \frac{\partial v_x}{\partial x} + \rho v_y \frac{\partial v_x}{\partial y} = \rho g \sin \theta + \mu \left( \frac{\partial^2 v_x}{\partial x^2} + \frac{\partial^2 v_x}{\partial y^2} \right)$$

$$\rho v_x \frac{\partial v_y}{\partial x} + \rho v_y \frac{\partial v_y}{\partial y} = \rho g \cos \theta + \mu \left( \frac{\partial^2 v_y}{\partial x^2} + \frac{\partial^2 v_y}{\partial y^2} \right)$$

No pressure terms since it's a free fluid. I found in a textbook the solution for inclined plane flow (but my problem is inclined flow in a square channel). In that solution, [itex]\frac{\partial v_x}{\partial x}[/itex] and [itex]v_y[/itex] are assumed to be 0, which gives

$$-\frac{\rho}{\mu} g \sin \theta = \mu\frac{\partial^2 v_x}{\partial y^2}$$

which is easy enough to solve with the boundary conditions [itex]v_x\big|_{y = 0} = 0[/itex] and [itex]\frac{\partial v_x}{\partial y}\big|_{y=L} = 0[/itex]. However, I don't know why they just assume [itex]\frac{\partial v_x}{\partial x}[/itex] and [itex]v_y[/itex] are 0; where does that come from? The fluid is flowing under the influence of gravity so I would think [itex]v_x[/itex] is increasing along x (imagine if you roll a ball down an inclined plane), and I would also think a free flow would decrease in height (L) as it travels. I mean, if you continuously dump water at the top of the incline, you're going to have a thicker water "height" at the top I would think...

And even if I accept that those assumptions are true, I don't know how to extend the problem from a plane to a square channel. There's no z-flow anyway, so I don't need any [itex]v_z \big|_{z=0}=0[/itex] or [itex]\big|_{z=W}=0[/itex] boundary conditions anyway. So the solution would be the same...

This is not set up correctly. First of all, the coordinates should be taken parallel and perpendicular to the inclined plane. Secondly, the problem statement implies that the flow is fully developed (the length of the incline is very large compared to the side of the square channel). The coordinate along the incline should be z, and the two velocity components in the cross channel directions should be zero. Furthermore, the derivative on the down-channel velocity with respect to the down-channel coordinate z should be zero (again, fully developed). Write your partial differential equation for the velocity vz under these circumstances. You should end up with a Poisson equation in vz. Boundary conditions are that vz is zero at the walls of the square duct.

Chet
 

1. What is inclined flow and why is it important in boundary conditions?

Inclined flow refers to the movement of fluids on an inclined surface, such as a sloped channel or pipe. It is important in boundary conditions because it can significantly affect the behavior and characteristics of the fluid flow, such as velocity and pressure distribution.

2. What are the additional boundary conditions for inclined flow?

The additional boundary conditions for inclined flow include the inclination angle of the surface, the roughness of the surface, and the presence of any obstructions or objects that may affect the flow. Other factors such as fluid viscosity, density, and flow rate may also need to be considered.

3. How do these additional boundary conditions affect the flow behavior?

The inclination angle of the surface can affect the direction and magnitude of the flow, with steeper angles resulting in faster flow. The roughness of the surface can create turbulence and alter the flow patterns. Obstructions or objects can cause flow separation and pressure changes in the fluid.

4. How are these additional boundary conditions incorporated into mathematical models?

These additional boundary conditions are incorporated into mathematical models through the use of specific equations and parameters. For example, the inclination angle can be included in the Navier-Stokes equations, while the roughness of the surface can be accounted for through the use of friction coefficients.

5. What are the practical applications of considering additional boundary conditions in inclined flow?

Considering additional boundary conditions in inclined flow is crucial for accurate predictions and understanding of fluid behavior in various systems, such as pipelines, rivers, and channels. It also has practical applications in engineering, such as designing efficient irrigation systems and optimizing the performance of hydraulic structures.

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