Question

The velocity distribution for laminar flow between parallel plates is given by:u/umax = 1- (2y/h)^2where h is the distance separating the plates and the origin is placed midway between the plates. Consider a flow of water at 15 C, with umax = 0.30m / sec and h = 0.50mm. Calculate the shear stress on the upper plate and give its direction.

Answer 1

Answer:

Explanation:

If the fluid is Newtonian, then the shear stress is directly proportional to rate of deformation.

${\tau _{xy}}\;\alpha \;\frac{{dy}}{{dx}}$

The constant of proportionality for the above equation is,

${\tau _{xy}}\; = \mu \frac{{dy}}{{dx}}$  

Here, $\mu$ is dynamic viscosity.

From the properties of water tables,

$\begin{array}{l}\\{\rm{At }}T = 15^\circ {\rm{C;}}\\\\\mu = 1.14 \times {10^{ - 3}}{\rm{ N}} \cdot {\rm{s/}}{{\rm{m}}^{\rm{2}}}\\\end{array}$

Sketch the velocity profile between the plates and show the direction of shear stress on the plate.

(see attachment)

The figure shows the velocity profile between the plates. The direction of shear flow of water is shown. The distance between the plates is found to be h.

Consider the equation for the velocity distribution for a laminar flow between parallel plates as follows:

$\begin{array}{l}\\\frac{u}{{{u_{\max }}}} = 1 - {\left( {\frac{{2y}}{h}} \right)^2}\\\\u = {u_{\max }}\left[ {1 - {{\left( {\frac{{2y}}{h}} \right)}^2}} \right]\\\end{array}$

Calculate the shear stress on the upper plate.

Using the boundary conditions on upper surface of plate, the shear flow on the upper plate will be as follows:

$\begin{array}{c}\\{\tau _w} = \mu {\left[ {\overrightarrow n \cdot \nabla u} \right]_{wall}}\\\\ = \mu {\left[ { < 0, - 1, < \frac{{\partial u}}{{\partial x}},\frac{{\partial u}}{{\partial y}}} \right]_{wall}}\\\\ = \mu {\left[ { - \frac{{\partial u}}{{\partial y}}} \right]_y}\\\end{array}$  

Substitute $\frac{h}{2}$ for $y$.

${\tau _w} = \mu {\left[ { - \frac{{\partial u}}{{\partial y}}} \right]_{y = \frac{h}{2}}}$  

Substitute ${u_{\max }}\left[ {1 - {{\left( {\frac{{2y}}{h}} \right)}^2}} \right]$ for $u$.

$\begin{array}{l}\\{\tau _{xy}} = - \mu \frac{d}{{dy}}{\left[ {{u_{\max }}\left( {1 - {{\left( {\frac{{2y}}{h}} \right)}^2}} \right)} \right]_{y = \frac{h}{2}}}\\\\{\tau _{xy}} = - \mu {u_{\max }}\frac{d}{{dy}}{\left[ {\left( {1 - {{\left( {\frac{{2y}}{h}} \right)}^2}} \right)} \right]_{y = \frac{h}{2}}}\\\end{array}$

Solving the equation further,

$\begin{array}{l}\\{\tau _{xy}} = - \mu {u_{\max }}{\left[ {\left( {0 - \frac{4}{{{h^2}}}2y} \right)} \right]_{y = \frac{h}{2}}}\\\\{\tau _{xy}} = \frac{{8\mu {u_{\max }}\left( {h/2} \right)}}{{{h^2}}}\\\\{\tau _{xy}} = \frac{{4\mu {u_{\max }}}}{h}\\\end{array}$

Substitute $1.14 \times {10^{ - 3}}{\rm{ N}} \cdot {\rm{s/}}{{\rm{m}}^{\rm{2}}}$ for $\mu , 0.30{\rm{ m/s}}$ for ${u_{\max }}$ and $0.5{\rm{ mm}}$ for $h$.

$\begin{array}{c}\\{\tau _{xy}} = \frac{{4\left( {1.14 \times {{10}^{ - 3}}} \right)\left( {0.3} \right)}}{{0.5 \times {{10}^{ - 3}}}}\\\\{\tau _{xy}} = 2.736\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}\left( \to \right)\\\end{array}$  

The direction indicates the flow of shear stress.

Therefore, the shear stress on the upper plate is $2.736\,{\rm{m/s}}\left( \to \right)$.