## Q. 5.43

The velocity profile of an incompressible, fully developed laminar flow between two infinitely large stationary parallel plates is given by $u=\frac{1}{2 \mu} \frac{d p}{d x}\left(y^2-H^2\right), \text { where } \frac{d p}{d x}$ is a constant and 2H is the gap between the two plates. Compute (a) the wall shear stress, (b) the average velocity, (c) the stream function, (d) the vorticity and (e) the velocity potential.

## Verified Solution

Given data:

(a) Velocity profile is given as $u=\frac{1}{2 u} \frac{d p}{d x}\left(y^2-H^2\right)$

Differentiating with respect to y, we have

$\frac{d u}{d y}=\frac{1}{2 \mu} \frac{d p}{d x} 2 y=\frac{1}{\mu} \frac{d p}{d x} y$

The wall shear stress is found to be

$\tau_w=\left.\mu \frac{d u}{d y}\right|_{y=H}=\left.\mu \frac{1}{2 \mu} \frac{d p}{d x} 2 y\right|_{y=H}=\frac{d p}{d x} H$

(b) The average velocity is found to be

$u_{a v}=\frac{\int_{-H}^H u d y}{2 H}$

or          $u_{a v}=\frac{2}{2 H} \int_0^H u d y$

$=\frac{1}{H} \int_0^H \frac{1}{2 \mu} \frac{d p}{d x}\left(y^2-H^2\right) d y$

$=\frac{1}{2 \mu H} \frac{d p}{d x} \int_0^H\left(y^2-H^2\right) d y$

$=\frac{1}{2 \mu H} \frac{d p}{d x}\left[\frac{y^3}{3}-H^2 y\right]_0^H$

$=\frac{1}{2 \mu H} \frac{d p}{d x}\left[\frac{H^3}{3}-H^3\right]$

(c) From the definition of stream function ψ, we get

$u=\frac{\partial \psi}{\partial y}$

or                                      $\psi=\int u d y$

or                                      $\psi=\int \frac{1}{2 \mu} \frac{d p}{d x}\left(y^2-H^2\right) d y$

or                                      $\psi=\frac{1}{2 \mu} \frac{d p}{d x}\left(\frac{y^3}{3}-H^2 y\right)+f(x)$                (5.82)

Again,                          $v=\frac{\partial \psi}{\partial x}$

$\psi=-\int v d x$

or                                    $\psi=g(y)$              (5.83)

Comparing Eqs. (5.82) and (5.83), we have

$\psi=\frac{1}{2 \mu} \frac{d p}{d x}\left(\frac{y^3}{3}-H^2 y\right)$

Hence, the stream function for the flow is

$\psi=\frac{1}{2 \mu} \frac{d p}{d x}\left(\frac{y^3}{3}-H^2 y\right)$

(d) The vorticity is given by

$\Omega_{x y}=\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}=0-\frac{1}{\mu} \frac{d p}{d x} y=-\frac{1}{\mu} \frac{d p}{d x} y$

(e) Since the vorticity is not equal to zero, the flow is rotational. Therefore, the velocity potential does not exist.