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Question 8.8: The velocity distribution for a fully developed laminar flow...

The velocity distribution for a fully developed laminar flow in a pipe is given by

v_{z}=-\frac{R^{2}}{4 \mu} \cdot \frac{\partial p}{\partial z}\left[1-(r / R)^{2}\right]

Determine the radial distance from the pipe axis at which the velocity equals the average velocity

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For a fully developed laminar flow in a pipe, we can write

 v_{z} =-\frac{R^{2}}{4 \mu} \frac{\partial p}{\partial z}\left[1-\left(\frac{r}{R}\right)^{2}\right]  

 

\bar{v}_{z} =\frac{Q}{A}=\frac{1}{\pi R^{2}} \int_{0}^{R}\left\{-\frac{R^{2}}{4 \mu} \frac{\partial p}{\partial z}\left[1-\left(\frac{r}{R}\right)^{2}\right]\right\} 2 \pi r \mathrm{~d} r  

 

=-\frac{R^{2}}{8 \mu} \frac{\partial p}{\partial z} 

Now, for v_{z}=\bar{v}_{z}   we have,

\frac{R^{2}}{4 \mu} \frac{\partial p}{\partial z}\left[1-\left(\frac{r}{R}\right)^{2}\right]=-\frac{R^{2}}{8 \mu} \frac{\partial p}{\partial z}

Or             1-\left(\frac{r}{R}\right)^{2}=\frac{1}{2}

Or             \left(\frac{r}{R}\right)^{2}=\frac{1}{2}               or     r=\frac{R}{\sqrt{2}}=0.707 \mathrm{R}

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