Liquid flow over a conveyor belt

[piscosour00]

1. A conveyor belt with one end immersed in a glycerin tank is used to transport small amounts of glycerin to a second tank. The belt surface is completely smooth, flow is laminar, and the specific weight of the fluid is 60 lbf/ft^3 with the belt velocity equal to 1 ft/s. Determine the volume flow rate and the mass flow rate from the first tank to the second tank per unit foot of width. Take the viscosity to be 0.002 lbf*s/ft^2 and the thickness of liquid (or the height above the conveyor belt) to be 0.04 ft. Assume flow is one-dimensional.

Homework Equations

The velocity of the fluid is given by the vector

$$\vec { V } =\left\{ u,v,w \right\} \quad$$

Navier-Stokes equation for flow (without the gravitational term):
$$\rho \frac { D\vec { V } }{ Dt } =-\nabla P+\mu { \nabla }^{ 2 }\vec { V }$$

Navier-Stokes equation in the x-direction:

$$\rho ( \frac { \partial u }{ \partial t } +u\frac { \partial u }{ \partial x } +v\frac { \partial u }{ \partial y } +w\frac { \partial u }{ \partial z } ) = -\frac { \partial P }{ \partial x } + \mu ( \frac { { \partial }^{ 2 }u }{ { { \partial }x }^{ 2 } } + \frac { { \partial }^{ 2 }u }{ { { \partial }y }^{ 2 } } +\frac { { \partial }^{ 2 }u }{ { { \partial }z }^{ 2 } } )$$

There may be some other equations that I need to use but I don't know for sure.

The Attempt at a Solution

If any of these assumptions are wrong, let me know please. Since it is the conveyor belt that moves the liquid, and the liquid itself is not flowing independent from the belt, the whole left side of the equation becomes 0 since u=0, v=0, and w=0. There is no pressure gradient. In other practice problems in my class, all the laplacian terms have been ignored except for:

$$\frac { { \partial }^{ 2 }u }{ { \partial y }^{ 2 } }$$

Though I can't say for sure why (is it because the flow is 1-D and laminar ??). So we're left with:

$$0=\mu \frac { { \partial }^{ 2 }u }{ { \partial y }^{ 2 } }$$

My TA told me that since the viscosity is non-zero, I should work instead with:

$$0= \frac { { \partial }^{ 2 }u }{ { \partial y }^{ 2 } }$$

And solving that differential equation gives me and equation of the form:

$$u(y)=Ay+B$$

Since the velocity at u(y=0) must equal the velocity of the belt, the equation becomes:

$$u(y)=Ay+{ V }_{ belt }$$

After this I'm stuck, since I don't know what other condition to use to solve for the constant A, or I don't even know if I'm on the right track. Any help would greatly be appreciated.

 

[KataKoniK]

Thank you for your post. I can help you solve this problem.

First, let's clarify some assumptions that need to be made in order to solve this problem. As mentioned in the question, we will assume that the flow is laminar, one-dimensional, and the conveyor belt is completely smooth. We will also assume that the height of the liquid above the conveyor belt is constant and equal to 0.04 ft.

Now, let's start by looking at the continuity equation:

\frac { \partial \rho }{ \partial t } + \nabla \cdot (\rho \vec { V }) = 0

Since the flow is one-dimensional, we can simplify this equation to:

\frac { \partial \rho }{ \partial t } + \frac { \partial \rho u }{ \partial x } = 0

Since the density, \rho, is constant, this equation becomes:

\frac { \partial u }{ \partial x } = 0

This means that the velocity, u, is constant in the x-direction. Therefore, we can use the equation of motion in the x-direction:

\rho \frac { D u }{ D t } = - \frac { \partial P }{ \partial x } + \mu \frac { \partial ^2 u }{ \partial y^2 }

Since we already know that \frac { \partial u }{ \partial x } = 0, the first term on the left side of the equation becomes 0. Also, since the flow is laminar and one-dimensional, we can assume that the velocity in the y-direction, v, is also equal to 0. Therefore, we are left with:

0 = \frac { \partial P }{ \partial x } + \mu \frac { \partial ^2 u }{ \partial y^2 }

Now, let's look at the boundary conditions. At the surface of the conveyor belt, the velocity of the fluid must equal the velocity of the belt, which is 1 ft/s. This means that at y = 0, u = 1. At y = 0.04 ft, we know that the velocity must be 0 since the liquid is not flowing above the conveyor belt. Therefore, our boundary conditions are:

u(y = 0) = 1

u(y =