Question

The velocity distribution for flow over a flat plate is given \(u = \frac{3}{4}y - {y^2}\) in which u is the velocity in m/s at a distance y meter above the plate. Shear stress at y = 0.15 m is (Take μ for fluid = 8.5 poise)
1. 0.3825 Pa
2. 0.3815 Pa 
3. 0.3835 Pa
4. 0.3845 Pa

Answer 1

Correct Answer - Option 1 : 0.3825 Pa

Concept:

Shear stress in flow over a flat plate is given as:

\(τ=μ \frac{du}{dy}\)

where, μ = dynamic viscosity, y = distance in meter above the plate.

1 P = 0.1 Pa-sec

Calculation:

Given:

μ = 8.5 poise = 8.5 × 10-1 Pa-sec, y = 0.15 m, \(u\; = \frac{3}{4}\;y\;-\;{y^2}\)

\(\frac{du}{dy}=\frac{3}{4}-2y\) \(\Rightarrow{\left. {\frac{{du}}{{dy}}} \right|_{y = 0.15}} = \frac{3}{4} - 2 × 0.15 = 0.45\)

\(τ=μ{\left. {\frac{{du}}{{dy}}} \right|_{y = 0.15}}\)

∴ τ = 8.5 × 10-1 × 0.45 

τ = 0.3825 N/m2 or Pa