Question

With reference to a standard Cartesian (x, y) plane, the parabolic velocity distribution profile of fully developed laminar flow in x-direction between two parallel, stationary and identical plates that are seperated by distance h, is given by the expression

$u=-\frac{h^2}{8\mu }\frac{dp}{dx}[1-4{\left(\frac{y}{h}\right)}^2]$

In this equation, the y = 0 axis lies equidistant between the plates at a distance h/2 from the two plates, p is the pressure variable and $\mu$ is the dynamic viscosity term. The maximum and average velocities are, respectively

• A. $u_{max}=-\frac{h^2}{8\mu }\frac{dp}{dx}and{u}_{average}=\frac{2}{3}{u}_{max}$
• B. $u_{max}=\frac{h^2}{8\mu }\frac{dp}{dx}and{u}_{average}=\frac{2}{3}{u}_{max}$
• C. $u_{max}=-\frac{h^2}{8\mu }\frac{dp}{dx}and{u}_{average}=\frac{3}{8}{u}_{max}$
• D. $u_{max}=\frac{h^2}{8\mu }\frac{dp}{dx}and{u}_{average}=\frac{3}{8}{u}_{max}$

Answer 1

The correct option is A $u_{max}=-\frac{h^2}{8\mu }\frac{dp}{dx}and{u}_{average}=\frac{2}{3}{u}_{max}$


Velocity expression for a laminar flow between two parallel plates is,

$U=-\frac{h^2}{8\mu }\left(\frac{dp}{dx}\right)[1-4{\left(\frac{y}{h}\right)}^2]$

At y = 0,
$U=U_{max}$
$\Rightarrow U_{max}=-\frac{h^2}{8\mu }\left(\frac{dp}{dx}\right)$

Discharge, dQ = Area × Velocity

$\Rightarrow dQ=[-\frac{h^2}{8\mu }\left(\frac{dp}{dx}\right)(1-4{\left(\frac{y}{h}\right)}^2)](dy\times 1)$

$\Rightarrow Q=-\frac{h^2}{8\mu }\left(\frac{dp}{dx}\right)\underset{-h/2}{\overset{h/2}{\int }}(1-\frac{4y^2}{h^2})dy$

$=-\frac{h^2}{8\mu }\left(\frac{dp}{dx}\right){[y-\frac{4y^3}{3h^2}]}_{-h/2}^{h/2}$

$=-\frac{h^2}{8\mu }\left(\frac{dp}{dx}\right)[\{\frac{h}{2}-(-\frac{h}{2})\}-\left\{\frac{4}{3h^2}(\frac{h^3}{8}-(-\frac{h^3}{8}))\right\}]$

$=-\frac{h^3}{12\mu }\left(\frac{dp}{dx}\right)$

$\because Q=A{U}_{avg}$

$-\frac{h^3}{12\mu }\left(\frac{dp}{dx}\right)=(h\times 1)\times U_{avg}$

$\Rightarrow U_{avg}=-\frac{h^2}{12\mu }\left(\frac{dp}{dx}\right)$

$\frac{U_{avg}}{U_{max}}=\frac{-\frac{h^2}{12\mu }\left(\frac{dp}{dx}\right)}{-\frac{h^2}{8\mu }\left(\frac{dp}{dx}\right)}=\frac{8}{12}=\frac{2}{3}$

$\therefore U_{avg}=\frac{2}{3}{U}_{max}$