CHG 3316

TRANSPORT PHENOMENA

Assignment 2

1. Schematic Diagram:

 

      

Referring to the above figure, you are asked to:

 

1) Calculate the shear stress on each plate when lower plate velocity is 10 ft/min. in the positive x-direction and the upper plate velocity is 35 ft/min. in the negative x-direction. The plates are placed 2 inches apart and the fluid viscosity between the plates remains constant at 150 cp.

 

2) Calculate the fluid velocity at every 0.5 in. interval.

 

Solution:

 

It was clear from the question that the fluid had a constant viscosity. i.e Newtonian fluid, that is why we are going to use Eq (2-8) from the text book. (See also Figure 2-9 to see how viscosities are changing as well for different fluids). Eq (2-8) is a first order ordinary differential equation. To solve it, first you have to separate the variables.

 

After variables separation, and by re-writing the general equation for the shear stress, we have:

                 (1 - 1)

Where  is constant.

 

 

 

Given that the plates are 2 inches a part, this means that y1=0.0 and y2=2. Also, Vo=10 ft/min and V1= - 35 ft/min (remember that the two plates are moving in opposite directions). By doing the integration and substitution in Eq (1) above:

This gives                                  (1-2)

Substituting,

 

Substituting in eq. (2) gives

Now we have the shear stress value.

 

To work for number 2) and find the velocities at  in., 1 in, and 1 in., distance from the lower plate, we will use Eq(2), since the shear stress and the viscosity remain unchanged and constant, then:

 

or

                                    (1-3)

 

 

 

For any value of y:

                          (1-4)

As a result, the following are equal:

or       or   

Also    and  

In the same way,

So at  in.,

At   in. , and at  in.

Results:

X value (inches)

Velocity (ft/min)

0.0

10

0.5

-1.25

1.0

-12.5

1.5

-23.75

2.0

-35

 

Comments: It can be clearly seen from the velocities in the table that the fluid is moving toward the negative x-direction following the resultant velocity vector.

 

 

2. Oil is flowing down (in the z-direction) a vertical wall as a film 1.7 mm thick. The oil density is 820  and the viscosity is 0.20 Pa s.

a) Draw a schematic diagram of the system, choose your shell and show it in the diagram, and do a shell momentum balance. Make sure you show your axes, momentum in and out directions, clearly in the diagram.

b) Develop an expression for the velocity  as a function of x,, the thickness of the layer of liquid and relevant fluid properties.

c) What is the maximum velocity, ?

d) Derive an expression for the average velocity, , relative to the maximum velocity.

 

Solution:    (a) Control volume for falling film at steady state:

 

 

b) In order to develop an expression for the velocity  as a function of x,, the thickness of the layer of liquid and relevant fluid properties, we have to perform momentum balance at the Cartesian coordinates at steady state:

 

    

 momentum-flux profile is linear (figure 2) & Max. value is at the wall.

For a Newtonian fluid using Newton’s law of viscosity.

   ®      and  

Boundary condition to evaluate the integration constant:

 at   

So, vz becomes as follows:

                                                    (3- 1)

(c) Max. Velocity occurs at is at  as can be seen from the velocity profile shown in Fig 2. Using equation 3-1 and setting x=0 results in the maximum velocity as shown:

 

         (3-2)

(d) To derive an expression for the average velocity,, we have to integrate over the area (along both the width and the thickness as well) as follows:

 

 

                                (3 - 3)

and by comparing 3-2 to 3-3 it can be seen that:

                                                      (3-4)

Due Friday, October 7, 2011 in the assignment box at 4:00 p.m.