This problem is related to the previous one, where we discussed a setup with a fluid flow between two horizontal plates. The solution method is very similar, with a difference that here we select a cylindrical element and consider forces acting on it.
Problem statement
A liquid is pumped from one container to another through a long pipe of radius $R$ and length $l$. Determine the flow velocity in the pipe as a function of distance to its axis (i.e. to the center of the pipe). Given that the pressure drop between the two ends of the pipe is $\Delta P$ and the dynamic viscosity of the liquid is $\eta$.
Problem statement is taken from Savchenko et al (1981) book and translated by the post author.
Exploiting the symmetry around the axis of the pipe we select a coaxial cylinder of radius $r$ in the flow and investigate forces acting on it.
Figure 1: Flow in pipe of length $l$ and radius $R$.
If we denote the viscosity stress along the cylinder's sides and parallel to its axis as $\tau_x(r)$, then we can write the balance of the viscous and pressure gradient forces acting on it, along the $x$-axis, as:
\begin{align*} & \tau_x(r) \cdot 2\pi r l + \Delta P \cdot \pi r^2 = 0 & \Rightarrow \\ & \tau_x(r) = -\frac{\Delta P r}{2 l} & \end{align*}
Now we can use the expression linking viscous stress and flow velocity gradient through the dynamic viscosity to determine the flow velocity as follows:
\begin{align*} & \tau_x(r) = -\frac{\Delta P r}{2 l} = \eta \frac{dv}{dr} & \Rightarrow \\ & \frac{dv}{dr} = -\frac{\Delta P r}{2 l \eta} & \Rightarrow \\ & \int\limits_{R}^{r} \frac{dv}{dr}\,dr = -\int\limits_{R}^{r}\frac{\Delta P r}{2 l \eta}\,dr & \Rightarrow \\ & v(r) - v(R) = -\frac{\Delta P \left( r^{2} - R^{2} \right) }{4 l \eta} \\ \end{align*}
Now using the non-slip condition at the pipe walls (i.e. $v(R) = 0$), we get the final expression for the flow velocity at a distance $r$ from the pipe axis: \begin{align*} \boxed{ v(r) = -\frac{\Delta P \left( r^{2} - R^{2} \right) }{4 l \eta} }. \end{align*}
