Question

The velocity distribution in a laminar flow adjacent to a solid wall is given by u = 3.0 sin (5πy). The viscosity of the fluid is 5 poise. What is the shear stress at a section (i) y = 0.05 m; (ii) y = 0.12 m?
1. 16.7 N / m² and 7.3 N / m²
2. 33.4 N / m² and Zero
3. 16.7 N / m² and 12.3 N / m²
4. 16.7 N / m² and Zero

Answer 1

Correct Answer - Option 1 : 16.7 N / m² and 7.3 N / m²

Concept:

The shear stress distribution over a solid wall assuming the fluid is Newtonian is given as

τ = μ du/dy

where

μ is the dynamic viscosity

du/dy is the velocity gradient

Calculation:

Given,

μ = 5 Poise = 5 × 10^-1 N/m²

u = 3 sin (5 πy)

\(\frac{{du}}{{dy}} = 3 \times 5\pi \cos \left( {5\pi y} \right)\)

\({\left. {\frac{{du}}{{dy}}} \right|_{y = 0.05}} = 15\pi \cos \left( {5\pi \times .05} \right) = \frac{{15\pi }}{{\sqrt 2 }} = 33.31\)

\({\left. {\frac{{du}}{{dy}}} \right|_{y = 0.12\;m}} = 15\pi \cos \left( {5\pi \times 0.12} \right) = - 14.56\;\)

Now,

\({\tau _{y = 0.05}} = \mu {\left. {\frac{{du}}{{dy}}} \right|_{y = 0.05}} = 5 \times {10^{ - 1}} \times 33.31 = 16.66\;N/{m^2}\)

\({\tau _{y = 0.12}} = \;\mu {\left. {\frac{{du}}{{dy}}} \right|_{y = 0.12}} = 5 \times {10^{ - 1}} \times \left( { - 14.56} \right) = - 7.28\;N/{m^2}\)