Question

For air flow over a flat plate, velocity (U) and boundary layer thickness ($\delta$) can be expressed respectively, as

$\frac{u}{u_{\infty }}=\frac{3}{2}\frac{y}{\delta }-\frac{1}{2}{\left(\frac{y}{\delta }\right)}^3;$ $\delta =\frac{4.64x}{\sqrt{R{e}_x}}$

If the free stream velocity is 2 m/s and air has kinematic viscosity of 1.5$\times {10}^{-5}{m}^2/s$ and density of 1.23 kg/$m^3$, the wall shear stress at x = 1 m, is

• A. $2.36\times {10}^{2}N/m^2$
• B. $43.6\times {10}^{-3}N/m^2$
• C. $4.36\times {10}^{-3}N/m^2$
• D. $2.18\times {10}^{-3}N/m^2$

Answer 1

The correct option is C $4.36\times {10}^{-3}N/m^2$

Reynold number,

$R{e}_x=\frac{u_{\infty }x}{v}=\frac{2\times 1}{1.5\times {10}^{-5}}=1.33\times {10}^5$

$\delta =\frac{4.64x}{\sqrt{R{e}_x}}=\frac{4.64\times 1}{\sqrt{1.33\times {10}^5}}=0.0127$

$Now\frac{du}{dy}=u_{\infty }[\frac{3}{2}.\frac{1}{\delta }-\frac{3}{2}\left(\frac{y^2}{{\delta }^3}\right)]$

$\frac{du}{dy}{|}_{y=0}=\frac{3u_{\infty }}{2\delta }$

Now, shear stress

${\tau }_0=\mu {\left(\frac{du}{dy}\right)}_{y=0}=\mu \times \frac{3u_{\infty }}{2\delta }$

$=\frac{3u_{\infty }\times \nu \times \rho }{2\delta }\because \mu =\nu \rho$

$=\frac{3\times 2\times 1.5\times {10}^{-5}\times 1.23}{2\times 0.0127}$

$=4.36\times {10}^{-3}N/m^2$

Points To Remember :
It is important to note that, shear stress for laminar
$\tau =\mu \frac{3}{2\delta }{u}_{\infty }i.e.\tau \propto \frac{1}{\sqrt{x}}$
So, as the distance from the leading edge (x) increases, shear stress on the plate surface decreases.