Question

Turbulent flow takes place in a circular pipe having friction coefficient $0.08$. The maximum velocity in the pipe is $4m/sec$. The average velocity (in m/sec, correct upto two decimal place) is _____.

• A. 2.8

Answer 1

The correct option is A 2.8
$U_{max}=4m/sec,f=0.08$
$\frac{U_{max}-U_{avg}}{U_x}=3.75$
$U_x=\sqrt{\frac{{\tau }_0}{\rho }}\times U_{avg}=\sqrt{\frac{f}{8}}\times U_{avg}$ where ${\tau }_0=$ boundary shear stress
$=\sqrt{\frac{0.08}{8}}{U}_{avg}$
$=\frac{U_{avg}}{10}$
$\frac{10(4-U_{avg})}{U_{avg}}=3.75$
$4-U_{avg}=0.375U_{avg}$
$U_{avg}=\frac{4}{1.375}=2.91m/sec$
Alternatively,
$\frac{U_{max}}{U_{avg}}=1.33\sqrt{f}+1$
$U_{avg}=\frac{4}{1.33\sqrt{0.08}+1}=2.91m/sec$