Question

The velocity profile of a fully developed laminar flow in a straight circular pipe, as shown in the figure,




given by the expression

$u\left(r\right)=\frac{R^2}{4\mu }\left(\frac{\partial p}{\partial x}\right)(1-\frac{r^2}{R^2})$

where $\frac{\partial p}{\partial x}$ is a constant
The average velocity of fluid in the pipe is

• A. $-\frac{R^2}{8\mu }\left(\frac{\partial p}{\partial x}\right)$
• B. $-\frac{R^2}{4\mu }\left(\frac{\partial p}{\partial x}\right)$
• C. $-\frac{R^2}{2\mu }\left(\frac{\partial p}{\partial x}\right)$
• D. $-\frac{R^2}{\mu }\left(\frac{\partial p}{\partial x}\right)$

Answer 1

The correct option is A $-\frac{R^2}{8\mu }\left(\frac{\partial p}{\partial x}\right)$


The velocity profile of a fully developed laminar flow in a straight circular pipe, as shown in figure, is given by the expression

$u\left(r\right)=-\frac{R^2}{4\mu }\left(\frac{dp}{dx}\right)[1-\frac{r^2}{R^2}]$

$=U_{max}[1-\frac{r^2}{R^2}]$

$where{U}_{max}=-\frac{R^2}{4\mu }\left(\frac{dp}{dx}\right)$

$where\frac{dp}{dx}isaconstant$
The average velocity of fluid in the pipe,
$V_{avg}=\frac{U_{max}}{2}=-\frac{R^2}{8\mu }\left(\frac{dp}{dx}\right)$

Alternatively:

Given velocity profile expression,

$u\left(r\right)=-\frac{R^2}{4\mu }\left(\frac{dp}{dx}\right)[1-\frac{r^2}{R^2}]$

Consider a small circular ring element of radius r and thickness dr as shown in figure,



The fluid flowing per second through small elementary ring,
dQ = Velocity at radius r $\times$ Area of a small ring element)
$=u\times 2\pi rdr$
$=\frac{-1}{4\mu }{R}^2\left(\frac{dp}{dx}\right)[1-\frac{r^2}{R^2}]\times 2\pi rdr$
$=\frac{-\pi }{2\mu }{R}^2\left(\frac{dp}{dx}\right)[r-\frac{r^3}{R^2}]dr$

Net discharge flow through pipe,

$Q=\underset{0}{\overset{R}{\int }}\frac{-\pi }{2\mu }{R}^2\left(\frac{dp}{dx}\right)[r-\frac{r^3}{R^2}]dr$
$=\frac{-\pi }{2\mu }{R}^2\left(\frac{dp}{dx}\right)\underset{0}{\overset{R}{\int }}[r-\frac{r^3}{R^2}]dr$
$=\frac{-\pi }{2\mu }{R}^2\left(\frac{dp}{dx}\right){[\frac{r^2}{2}-\frac{r^4}{4R^2}]}_0^R$
$=\frac{-\pi }{2\mu }{R}^2\left(\frac{dp}{dx}\right)[\frac{R^2}{2}-\frac{R^4}{4R^2}]$
$=\frac{-\pi }{2\mu }{R}^2\left(\frac{dp}{dx}\right)[\frac{R^2}{2}-\frac{R^2}{4}]$
$=\frac{-\pi }{2\mu }{R}^2\left(\frac{dp}{dx}\right)\times \frac{R^2}{4}=\frac{-\pi }{8\mu }{R}^4\left(\frac{dp}{dx}\right)$

$\therefore Averagevelocity,$

$\overline{u}=\frac{Q}{\pi R^2}=\frac{\frac{-\pi }{8\mu }{R}^4\left(\frac{dp}{dx}\right)}{\pi R^2}$

$=\frac{-1}{8\mu }\left(\frac{dp}{dx}\right)R^2$