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
u=-\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.
For a fully-developed laminar flow in a pipe, we can write
u=-\frac{R^{2}}{4 \mu} \frac{\partial p}{\partial z}\left[1-\left(\frac{r}{R}\right)^{2}\right]
V_{ av }=\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 d r
=-\frac{R^{2}}{8 \mu} \frac{\partial p}{\partial z}
\text { Now, for } u=V_{ av } \text { 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} \quad \text { or } \quad r=\frac{R}{\sqrt{2}}=0.707 R
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