Question

If the point at which local velocity v is equal to mean velocity V is at a distance Kr from the centre of circular pipe having radius R, what will be the value of K?

1.

0.707

2.

0.5

3.

1.414

4. zero
5.

Answer 1

Correct Answer - Option 1 :

0.707

Concept:

The following formulas should be remembered for laminar flow in circular pipes:

1. Velocity distribution

\(u = \frac{1}{{4\mu }}\left( {\frac{{ - \delta P}}{{\delta x}}} \right)\left( {{R^2} - {r^2}} \right)\)

   Or

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

3. Relation between mean velocity and maximum velocity

Umax = 2 U̅

Calculation

Given:

local velocity, u = v 

mean velocity, U̅ = V 

Distance from the center of pipe = Kr  

If v = V Then K = ?

\(v = 2V\left( {1 - \frac{{{(Kr)^2}}}{{{R^2}}}} \right)\)

If v = V

\(\begin{array}{*{20}{l}} {\frac{1}{2} = \left( {1 - \frac{{{{(Kr)}^2}}}{{{R^2}}}} \right)}\\ {Kr = \frac{R}{{\sqrt 2 }} = 0.707R} \end{array}\)

Therefore, K = 0.707

Shear Stress Distribution

\(\tau = - \frac{r}{2}\frac{{\delta P}}{{\delta x}}\)