Question

A cylindrical tube of length L and radius of cross-section R carries a steadily flowing liquid of density $\rho$ and viscosity $\eta$. The profile of the velocity of flow is given by $v=v_0(1-\frac{r^2}{R^2})$, where $r$ is the radial distance of the flowing liquid from the axis. Then, Which of the following statement(s) is (are) correct?

Answer 1

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

Answer 2

For steady, laminar, incompressible flow in a tube, the velocity profile is given by

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

The given profile is

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

Equate the coefficients of $(R^2 - r^2)$:

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

Thus,

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

For a tube of length $L$, the pressure difference is obtained by multiplying the pressure gradient by $L$:

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