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A smooth pipe line of 100 mm diameter carries 2.27 m3 per minute of water at 200C with kinematic viscosity of 0.0098 stokes.

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

1 Answer
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144views

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$

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

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