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A fluid near a solid wall has an approximated velocity profile given by \(u\left( y \right) = \;{U_\infty } \times \sin \left( {\frac{{\pi y}}{{2\delta }}} \right),\;0 \le y \le \delta \) . The walls shear stress is given by:


1. \({\tau _{wall}} = \;\frac{{\pi \mu {U_\infty }}}{{2\delta }}\)
2. \({\tau _{wall}} = \;\frac{{3\pi \mu {U_\infty }}}{{\delta }}\)
3. \({\tau _{wall}} = \;\frac{{2\pi \mu {U_\infty }}}{{\delta }}\)
4. \({\tau _{wall}} = \;\frac{{\pi \mu {U_\infty }}}{{\delta }}\)

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Correct Answer - Option 1 : \({\tau _{wall}} = \;\frac{{\pi \mu {U_\infty }}}{{2\delta }}\)

Concept:

According to Newton’s law of viscosity, shear stress is given by:

\(\tau = \mu \times \frac{{du}}{{dy}} = \;\mu \times \frac{{d\theta }}{{dy}}\;\)

τ = shear stress, μ = coefficient of viscosity or absolute viscosity (or dynamic viscosity)

\(\frac{{du}}{{dy}} = Velocity\;gradient\)

\(\frac{{d\theta }}{{dt}}\; = Rate\;of\;angular\;deformation\;or\;Rate\;of\;shear\;strain\)

Calculation:

Given:

Velocity profile 

\(u\left( y \right) = \;{U_\infty } \times \sin \left( {\frac{{\pi y}}{{2\delta }}} \right),\;0 \le y \le \delta \)

\(\tau = \mu \times \frac{{du}}{{dy}}\)

\(\tau = \mu \times \frac{{du}}{{dy}} = \mu \times \;{U_\infty } \times\frac{\partial }{{\partial y}}\left( {\sin \left( {\frac{{\pi y}}{{2\delta }}} \right)} \right)\)

\( = \mu \times {U_\infty } \times \cos \left( {\frac{{\pi y}}{{2\delta }}} \right) \times \frac{\pi }{{2\delta }}\)

As we have to find the shear stress at the wall,

y = 0

\({\tau _{wall}} = \mu \times {U_\infty } \times \frac{\pi }{{2\delta }} = \frac{{\pi \mu {U_\infty }}}{{2\delta }}\)

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