Fluids Mechanics, Wall shear stress, Velocity, Fricition Example

[abs123456]

Homework Statement

A two-dimensional incompressible flow takes place in the space between two parallel
solid plane boundaries, which are a distance 2b apart. The density of the fluid is p, the
dynamic viscosity is p. and the velocity distribution in the flow is given by the expression

I have attatched it as a picture..




where y is a coordinate measured from the centre line.
Derive expressions for

a The mass flow rate per unit width of duct

b. The mean velocity in the flow

c. The wall shear stress

d. Show that the friction factor for this flow is given by an equation of the form
f = K / Re , where Re = 2bpU / JJ and K a constant. Determine the value of K.

Homework Equations

The Attempt at a Solution

Could someone please show me the steps or how i could go about doing this example..
Thank you very much

 

[KataKoniK]

for your help!

Thank you for your question. I am happy to help you with your inquiry.

To begin, let's define some variables for easier reference:

b = distance between the two solid plane boundaries
p = density of the fluid
u(y) = velocity distribution in the flow, where y is measured from the center line
µ = dynamic viscosity of the fluid

a) To find the mass flow rate per unit width of duct, we can use the equation for mass flow rate:

ρQ = ρu(y)A

Where ρ is the density, Q is the mass flow rate, u(y) is the velocity distribution, and A is the cross-sectional area of the flow. In this case, the cross-sectional area is simply 2b (the width of the duct) times the thickness of the flow, which we can denote as h.

Therefore, the mass flow rate per unit width of duct is given by:

Q/A = u(y)h

Since the flow is incompressible, the mass flow rate per unit width is constant throughout the flow, so we can simply use the average velocity to represent u(y).

Therefore, the mass flow rate per unit width of duct is:

Q/A = u_bar * h

b) To find the mean velocity in the flow, we can use the definition of mean velocity, which is the average of the velocity distribution over the cross-sectional area:

u_bar = (1/A) ∫ u(y) dA

Since we know that the cross-sectional area is 2b times the thickness h, we can rewrite the integral as:

u_bar = (1/2bh) ∫ u(y) dA

Now, we can use the given velocity distribution to solve for the mean velocity:

u_bar = (1/2bh) ∫ (Umax * (1 - (y/b)^2)) dy

Since the integral is only over the y direction, we can pull out all constants and integrate only the function of y:

u_bar = (Umax/2bh) ∫ (1 - (y/b)^2) dy

Integrating this, we get:

u_bar = (Umax/2bh) (y - (y^3/3b^2)) | from -b to b

Plugging in the limits of integration and simplifying, we get:

u_bar = (U