Sandra Enn Bahinting Homework 2

1. Consider the problem of coquette flow between parallel plates for a non-newtonian fluid.

Assuming constant pressure and temperature, solve for the velocity distribution u(y) between

the plates for a)n1. Plot the velocity profile. Compare with the Newtonian solution

Solution:

NON-NEWTONIAN

Conservation of momentum at x-direction:

Assumptions are follows:

Flow is infinite in the x-z plane and steady: Steady-state:

No flow in z direction: Constant Velocity:

Boundary conditions:

Simplifying the momentum equation:

Where

Substituting the said equation to the simplified momentum equation we have;

Boundary conditions:

Since pressure is constant,

Therefore,

NEWTONIAN

Conservation of momentum at x-direction:

Assumptions are follows:

Flow is infinite in the x-z plane and steady: Steady-state:

No flow in z direction: Constant Velocity:

Boundary conditions:

Simplifying the momentum equation:

Where

Substituting the said equation to the simplified momentum equation we have;

; n=1

Boundary conditions:

Since pressure is constant,

Therefore,

This shows that given the same the set of assumptions and conditions the Non-newtonian and

Newtonian have the same the velocity distribution. Furthermore, the plot for the velocity profile

would be same.

2. Stokes drag on a sphere

a. Obtain BSL Equation (2.6-9) from BSL Equation (2.6-7)

b. Obtain BSL Equation (2.6-12) from BSL Equation (2.6-10)

Solution:

a.

Eqn. 2.6-7

where and the normal stress is zero at r=R

-1

-0.5

0

0.5

1

1.5

0 0.2 0.4 0.6 0.8 1 1.2

Series1

Y h

Linear velocity profile u(y)

Let Let

=

Eqn. 2.6-9

b.

The shear stress distribution on the sphere surface, =

Eqn. 2.6-10

Eqn. 2.6-12

3. A sphere of specific gravity 7.8 is dropped into an oil of specific gravity of 0.88 and viscosity

= 0.15 Pa-sec. Estimate the terminal velocity of the sphere if its diameter is

a. 0.1 mm b. 1 mm c. 10 mm

Which of these is creeping motion?

Solution:

Equation for terminal velocity is given in equation 2. 6-17

Substituting the values for each variable in the equation above, we have;

a. 0.1 mm

Solving for the Reynolds Number, we obtained;

For b and c, same equations are used for terminal velocity and Reynolds number.

b. 1 mm

c. 10 mm

Therefore, it can be said that the 3 spheres having different diameters are the creeping

motion. Since these obey the condition for a creeping motion which is to have a Reynolds

number that is less than 0.1.