Calculate the friction factor, maximum velocity as well as shear stress at the boundary.

(10 marks) May-2018

Subject Fluid Mechanics 2

Topic Turbulent Flow

Difficulty Medium

Given :

Dia. of pipe, D = 100 mm = 0.1m,

$\therefore$ Radius R = 0.05 m

Discharge, Q = 2.27 $m^3/min = \frac{2.27}{60} = 0.0378 m^3/s$

kinematic viscosity,

v = 0.0098 stokes = 0.0098 $cm^{2/s}$

$= 0.0098 \times 10^{-4} m^{2/s}$

Now, avg. velocity is $ū = \frac{Q}{A}$

i.e $ū = \frac{0.0378}{\pi \times 0.1^2} = 4.1817 \ m/s$

$\therefore$ Reynold number is given by $Re = \frac{ū \times D}{V} = \frac{4.817 \times 0.1}{0.0098 \times 10^{-4}}$

$= 4.9154 \times 10^5$

The flow is turbulent and Re is more than $10^5$

Hence for smooth pipe, the co-efficient of friction F is,

$\frac{1}{\sqrt{4f}} = 2Log^{10}$ $(Re \sqrt{4f}) - 0.8$

i.e. $\frac{1}{\sqrt{4f}} = 2Log^{10}$ $(4.915 \times 10^5 \times \sqrt{4f}) - 0.8$

$\therefore \frac{1}{\sqrt{4f}} - Log_{10}$ (4f) = 11.383 - 0.8 = 10.583 --------- (1)

1) Friction factor,

Now, friction factor (f*) = 4 x co-efficient of friction

= 4 x f

$\therefore$ Equation (1) becomes, $\frac{1}{\sqrt f*} - Log_{10}$ $f* = 10.583$

$\therefore$ on solving we get,

f* = 0.0132

2) Maximum velocity ( u max)

we know that, p* = 4f

$\therefore$ co-efficient of friction, $f = \frac{f*}{4} = \frac{0.013}{4}$

= 0.00325

Now, shear velocity (u*) in terms of f and ū is,

$u* = ū \sqrt \frac{f}{2} = 4.817 \times \sqrt \frac{0.00325}{2}$

u* = 0.194 m/s

For smooth pipe, the velocity at any point is,

$u = u* [ 5.75 Log_{18}$ $\frac{u* \times y}{v} + 5.55]$

For velocity to be maximum at centre of pipe, then y = R = 0.05 m (radius of pipe)

Hence,

$u max = u * [5.75 Log_{10}$ $\frac{u* \times R}{v} + 5.55]$

$= 0.194 \times [ 5.75 Log_{10}$ $\times \frac{0.194 \times 0.05}{0.0098 \times 10^{-4}} + 5.55]$

u max = 5.528 m/s

3) Shear stress at the boundry $(\tau_o)$

Now, $ x* = \sqrt\frac{\tau_o}{\rho }$

$\therefore \tau_o = u^2* \times \rho $

$ = 0.194^2 \times 1000$

$\tau_o = 37.63 \ N/m^2$