Question 8.6: The analysis of a fully developed laminar flow through a pip...

Let us have a look at Fig. 8.15. The fluid moves due to the axial pressure gradient while the pressure across a section 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

Or     -\mathrm{d} 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{\mathrm{d} p}{\mathrm{~d} l} r=\frac{1}{2 \mu} \frac{\mathrm{d} p}{\mathrm{~d} z} r

Upon integration,
v_{z}=\frac{1}{4 \mu} \frac{\mathrm{d} p}{\mathrm{~d} z} r^{2}+K

At     r=R, v_{z}=0   hence K=-\left(\frac{1}{4 \mu} \frac{\mathrm{d} p}{\mathrm{~d} z}\right) \cdot R^{2}

So,    v_{z}=\frac{R^{2}}{4 \mu}\left(-\frac{\mathrm{d} p}{\mathrm{~d} z}\right)\left(1-\frac{r^{2}}{R^{2}}\right)