Question 8.6: The analysis of a fully developed laminar flow through a pip...
The analysis of a fully developed laminar flow through a pipe can alternatively be derived from control volume approach. Derive the expression v_{z}=\frac{R^{2}}{4 \mu}\left(-\frac{ d p}{ d z}\right)\left(1-\frac{r^{2}}{R^{2}}\right) starting from the control volume approach.
Let us have a look at Fig. 8.17. The fluid moves due to the pressure gradient which acts in the direction of the axis and in the sections perpendicular to it the pressure may be regarded as constant. Due to viscous friction, individual layers act on each other producing a shearing stress which is proportional to \frac{\partial v_{z}}{\partial r}.
In order to establish the condition of equilibrium, we consider a fluid cylinder of length \delta l and radius r. Now we can write
[p-(p+\delta p)] \pi r^{2}=-\tau 2 \pi r \delta l
or -\delta p \pi r^{2}=-\mu \frac{\partial v_{z}}{\partial r} 2 \pi r \delta l
or \frac{\partial v_{z}}{\partial r}=\frac{1}{2 \mu} \frac{ d p}{ d l} r=\frac{1}{2 \mu} \frac{ d p}{ d z} r
upon integration,
v_{z}=\frac{1}{4 \mu} \frac{ d p}{ d z} r^{2}+K
\text { at } r=R, v_{z}=0, \text { hence } K=-\left(\frac{1}{4 \mu} \frac{ d p}{ d z}\right) \cdot R^{2}
\text { So, } \quad v_{z}=\frac{R^{2}}{4 \mu}\left(-\frac{ d p}{ d z}\right)\left(1-\frac{r^{2}}{R^{2}}\right)