Question

 For laminar flow in a pipe of radius 0.9 m, the point of maximum instability would occur at a distance of _______ from the center of pipe cross section.
1. 0.60 m
2. 0.15 m
3. 0.30 m
4. Cannot say

Answer 1

Correct Answer - Option 3 : 0.30 m

Concept:

The parameter for instability since, the velocity distribution in pipe is

\(u = {V_{max}}\left( {1 - \frac{{{r^2}}}{{{R^2}}}} \right)\)

\(\frac{{du}}{{dr}} = - \frac{{du}}{{dy}}\)

\(\frac{{du}}{{dy}} = \frac{{2{V_{max}}r}}{{{R^2}}}\)

\(\frac{{du}}{{dy}} = \frac{{2{V_{max}}\left( {R - y} \right)}}{{{R^2}}}\)

\(u = 2{y^2}\frac{{\rho {V_{max}}\left( {R - y} \right)r}}{{{R^2}}}\)

Differentiating and equating to zero

\(\frac{{du}}{{dy}} = \frac{{2\rho {V_{max}}r}}{{{R^2}}}\frac{d}{{dy}}\;\left( {R{y^2} - {y^3}} \right)\)

0 = 2Ry - 3y2

0 = y × (2R-3y)

\(\therefore y = \frac{{2R}}{3}\)

Where y is distance from wall.

Here, R = 0.9 m

so y = 0.9 × 2/3 = 0.6 m

But it is asked from centre. So Distance = 0.6/2 = 0.3 m