I have a fluid mechanics question, i don't know which equation i should integrate and how?

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 )

Is...

$\displaystyle \displaystyle v_{m} = \frac{v_{0}}{R} \ \int _{0} ^{R} (1-\frac{r}{R})^{\frac{1}{7}}\ dr$ (1)

... that setting $\displaystyle x=\frac{r}{R}$ becomes...

$\displaystyle \displaystyle v_{m} = v_{0} \ \int_{0}^{1} (1-x)^{\frac{1}{7}}\ dx = \frac{7}{8}\ v_{0} \ |-(1-x)^{\frac{8}{7}}|_{0}^{1} = \frac{7}{8}\ v_{0} $ (2)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

Originally Posted by chisigma

Is...

$\displaystyle \displaystyle v_{m} = \frac{v_{0}}{R} \ \int _{0} ^{R} (1-\frac{r}{R})^{\frac{1}{7}}\ dr$ (1)

... that setting $\displaystyle x=\frac{r}{R}$ becomes...

$\displaystyle \displaystyle v_{m} = v_{0} \ \int_{0}^{1} (1-x)^{\frac{1}{7}}\ dx = \frac{7}{8}\ v_{0} \ |-(1-x)^{\frac{8}{7}}|_{0}^{1} = \frac{7}{8}\ v_{0} $ (2)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

but the result is (49/60)* Vmax, we should use hint, i think

Effectively I tried to symplify the problem integrating in the only $\displaystyle r$ variable ... integrating in the $\displaystyle r$ and $\displaystyle \theta$ variables it results...

$\displaystyle \displaystyle v_{m} = \frac{v_{0}}{\pi \ R^{2}}\ \int_{0}^{2 \pi}\ \int_{0}^{R} (1-\frac{r}{R})^{\frac{1}{7}}\ r \ dr \ d\theta$ (1)

... that setting again $\displaystyle x= \frac{r}{R}$ becomes...

$\displaystyle \displaystyle v_{m} = \frac{v_{0}}{\pi}\ \int_{0}^{2 \pi}\ \int_{0}^{1} x\ (1-x)^{\frac{1}{7}} \ dx \ d\theta= $

$\displaystyle \displaystyle = 2\ v_{0}\ \int_{0}^{1} x\ (1-x)^{\frac{1}{7}}\ dx = $

$\displaystyle \displaystyle = 2\ v_{0} \ \frac{7}{8}\ |x\ (1-x)^{\frac{8}{7}} - \frac{7}{15} \ (1-x)^{\frac{15}{7}}|_{0}^{1}= \frac{49}{60}$ (3)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$