Question

The velocity profile in fully developed laminar flow in a pipe of diameter D is given by $u=u_0(1-4r^2/D^2)$, where r is the radial distance from the center. If the viscosity of the fluid is $\mu$, the pressure drop across a length L of the pipe is

• A. $\frac{\mu u_{0}L}{D^2}$
• B. $\frac{4\mu u_{0}L}{D^2}$
• C. $\frac{8\mu u_{0}L}{D^2}$
• D. $\frac{16\mu u_{0}L}{D^2}$

Answer 1

The correct option is D $\frac{16\mu u_{0}L}{D^2}$

Method I:
Given velocity profile,

$u=u_0(1-\frac{4r^2}{D^2})$

At r = 0, u = $u_{max}$
$\therefore u_{max}=u_0$


We know that the pressure drop across a length L of the pipe,

$p_1-p_2=\frac{32\mu \overline{u}L}{D^2}$

where $\overline{u}$ = Average velocity

$\overline{u}=\frac{u_{max}}{2}=\frac{u_0}{2}$

$\therefore p_1-p_2=\frac{32\mu u_{0}L}{2\times D^2}=\frac{16\mu u_{0}L}{D^2}$

Method II:
For the laminar flow through circular pipe,
Shear stress,

$\tau =-\frac{r}{2}\left(\frac{\partial p}{\partial x}\right)$

$\mu \frac{\partial u}{\partial r}=-\frac{r}{2}\frac{\Delta p}{L}$

$-\frac{r}{2}\frac{\Delta p}{L}=\mu \frac{\partial }{\partial r}\left(u_0(1-\frac{4r^2}{D^2})\right)$

$\because \frac{r}{2}\frac{\Delta p}{L}=\mu \left(\frac{8u_{0}r}{D^2}\right)$

$\frac{r}{2}\frac{\Delta p}{L}=\left(\frac{8\mu u_{0}r}{D^2}\right)$

Pressure drop,

$\Delta p=\frac{16\mu u_{0}L}{D^2}$