The velocity profile for laminar flow of water between two fixed parallel plates is given by $u=0.01[1-1000y^2]$ where u is in $m/s$ and is in $N-s/m^2$. The viscosity of water can be assumed to be $10^{-3}N-s/m^2$ and the gap between the two plates is 2 cm. Calculate the shear stress on each plate.

Data:- $u=0.01[1-1000y^2]$; where $u\rightarrow m/s, y\rightarrow m$

$u=10^{-3}N-s/m^2$; Gap $t=2 cm = 2\times 10^{-2}m$

To Find:- $T_1=?$ $T_2=?$

Solution:

Plate (1)

Let us consider two layers of water one is fixed with plate (1) and other moves with flow

$\therefore dy=\frac{t}{2}=\frac{2\times 10^{-2}}{2}=10^{-2}m$

We have,

$u=0.01[1-1000y^2]$

$=0.01[1-1000(10^{-2})^2]$..........$[\because y=10^{-2}m]$

$=-0.09m/s$

Now, By using Newtons law of viscosity,

$T_1=u\cdot \frac{du}{dy}$

But $du=u-0=u=-0.09m/s$

$T_1=10^{-3}\times \frac{-0.09}{10^{-2}}$

$=-9\times 10^{-3}N/m^2$

Plate (2)

Shear stress (T) will be same for plate (2) as distance and viscosity is same (Refer shear stress distribution)

$\therefore T_2=-9\times 10^{-3}N/m^2$