How Does Bulk Velocity Relate to Maximum Velocity in Turbulent Tube Flow?

[chronicals]

Homework Statement

For turbulent flow in a smooth, circular tube with a radius R, the velocity profile varies according to the following expression at a Reynolds number of about 10^5.

Vx= Vxmax * [(R-r)/R)]^(1/7)

where r is the radial distance from the center and Vmax the maximum velocity at the center. Derive equation relating the average velocity ( bulk velocity ) Vav to Vmax for an incompressible fluid.
( Hint: The integration can be simplified by substitution z for R-r )

Homework Equations

The Attempt at a Solution

i don't know which equation i should integrate and how?

 

[hikaru1221]

I guess the average speed is defined as $$<v>=\frac{Q}{A}$$ where Q is the volume of fluid passing through the cross section of area A in 1 second. How do you find Q?

 

[chronicals]

I guess the average speed is defined as $$<v>=\frac{Q}{A}$$ where Q is the volume of fluid passing through the cross section of area A in 1 second. How do you find Q?

i think working with volume rate is useless

 

[hikaru1221]

Okay, why and what do you think of the solution?

 

[paranormal]

Hello, as a scientist, I can provide a response to the given content on fluid mechanic integration.

Firstly, we need to understand the concept of bulk velocity and its relationship with maximum velocity. Bulk velocity, also known as average velocity, is the average velocity of the fluid flow over a given cross-sectional area. On the other hand, maximum velocity is the highest velocity at the center of the tube.

To derive an equation relating bulk velocity (Vav) to maximum velocity (Vmax), we need to integrate the given velocity profile equation over the cross-sectional area of the tube. Since the velocity profile equation is expressed in terms of r, we can simplify the integration by substituting z for R-r. This will make the integration easier and help us to obtain the desired equation.

Using the substitution z = R-r, we can rewrite the given velocity profile equation as:

Vx = Vmax * (z/R)^(1/7)

Now, we can integrate this equation over the cross-sectional area of the tube, which is given by A = πR^2.

Integrating both sides, we get:

∫Vx dA = ∫Vmax * (z/R)^(1/7) * dA

Since Vx is the velocity at any point in the tube, we can write it as Vx = Vav, where Vav is the average velocity over the cross-sectional area. Therefore, the left-hand side of the equation becomes:

∫Vx dA = Vav * ∫dA = Vav * A

Substituting the value of A, we get:

∫Vx dA = Vav * πR^2

On the right-hand side, we can take Vmax out of the integral as it is a constant. Also, we can substitute z = R-r in the integral and change the variable to dz to make the integration easier. Therefore, the right-hand side becomes:

∫Vmax * (z/R)^(1/7) * dA = Vmax * ∫(1-z/R)^(1/7) * dA = Vmax * ∫(1-z/R)^(1/7) * 2πz * dz

Integrating the right-hand side, we get:

∫Vmax * (z/R)^(1/7) * dA = Vmax * (2πR/8) * (