Solving Turbulent Flow Velocity Distribution in a Pipe

[mathfied]

Hi Guys.
I have an absurd problem and no idea to solve it. Please could you help me?

$$\frac{u}{\hat u} = \left [ \frac{y}{R} \right ]^{1/n} = \left [1- \frac{r}{R} \right ]^{1/n}$$

Suppose the turbulent flow velocity distribution in a pipe of Radius R can be described by the equation above (Power Law relation).

y= distance from wall
r= radial distance from the axis
$$\hat u$$ = velocity on the axis

If $$\bar u$$ is the space mean average velocity in the pipe, show that:

$$\frac{\bar u}{\hat u} = \frac{2n^2}{(n+1)(2n+1)}$$

Any solutions please? I know I haven't posted my solutions but that's simply because I have no idea on this question.

 

[ideasrule]

Typically, "average velocity" is defined so that average velocity * area = volume flow, because that's the most useful definition of the "average". So the first step is to find the volume flow across this pipe, using integration.

 

[kerimek]

Hello, as a scientist, I would be happy to help you with this problem. Turbulent flow velocity distribution in a pipe is a complex problem, but it can be solved using various methods. One approach is to use the power law relation provided, which describes the relationship between the velocity and distance from the wall in a turbulent flow.

To solve for the space mean average velocity, we can integrate the given equation from y = 0 to R, which represents the entire pipe. This will give us the following equation:

\bar u = \frac{1}{R} \int_{0}^{R}\hat u \left [ \frac{y}{R} \right ]^{1/n} dy

Next, we can substitute the given values for y and R, and use the power law relation to simplify the equation. This will give us:

\bar u = \frac{1}{R} \int_{0}^{R}\hat u \left [1- \frac{r}{R} \right ]^{1/n} dy

Using the substitution r = y, we can rewrite this as:

\bar u = \frac{1}{R} \int_{0}^{R}\hat u \left [1- \frac{y}{R} \right ]^{1/n} dy

Now, we can use the substitution u = 1- (y/R) and du = -dy/R to transform the integral into a form that can be easily solved using the power rule. This will give us:

\bar u = \frac{1}{R} \int_{1}^{0}\hat u (-R) \left [u \right ]^{1/n} du

Simplifying this further, we get:

\bar u = -\frac{\hat u}{R} \int_{1}^{0}u^{1/n} du

Using the power rule to solve the integral, we get:

\bar u = -\frac{\hat u}{R} \left [ \frac{nu^{1/n+1}}{n+1} \right ]_{1}^{0}

Simplifying this, we get:

\bar u = \frac{n\hat u}{(n+1)R}

Finally, to find the ratio of the space mean average velocity to the velocity on the axis, we can divide this equation by \hat u