Question

A fluid of density $\rho$ and coefficient of viscosity $\eta$ is flowing in a tube of length $l$ and radius $R$. Flow of the fluid is steady. Velocity $v$ of the flow is given by $v = v_0 (1 - \frac{r^2}{R^2})$, where $r$ is distance of flowing fluid from axis.

The difference of pressure at ends of the tube is

Answer 1

$\Delta P = \frac{4\eta\,v_0\,l}{R^2}$

Answer 2

We start with the known result for viscous flow in a tube. For steady, incompressible, laminar flow the Navier–Stokes equation in a tube gives the velocity profile as

$v(r) = -\frac{1}{4\eta}\frac{dp}{dx}\;(R^2 - r^2)$

Comparing with the given profile

$v(r) = v_0\left(1-\frac{r^2}{R^2}\right) = v_0 \frac{R^2 - r^2}{R^2}$,

we equate the two coefficients (the factor multiplying $(R^2 - r^2)$):

$\frac{-1}{4\eta}\frac{dp}{dx} = \frac{v_0}{R^2}$.

Solving for $\frac{dp}{dx}$ we have:

$\frac{dp}{dx} = -\frac{4\eta\,v_0}{R^2}$.

Over a tube of length $l$ the pressure difference is

$\Delta P = P_1 - P_2 = -\frac{dp}{dx}\,l = \frac{4\eta\,v_0\,l}{R^2}$.

Thus, the difference of pressure at the ends of the tube is

$\Delta P = \frac{4\eta\,v_0\,l}{R^2}$.

Core Explanation:

1. Compare the given velocity profile with the standard parabolic profile.
2. Equate coefficients to get $\frac{dp}{dx} = -\frac{4\eta\,v_0}{R^2}$.
3. Multiply by the tube length $l$ to obtain $\Delta P = \frac{4\eta\,v_0\,l}{R^2}$.