Question

A Pitot-static tube is mounted on a 2.5 cm pipe where oil (???? = 860 kg/m3, ???? = 0.0103 kg/m·s) is flowing. The Pitot tube is positioned such that the local velocity measured is similar to the mean velocity through the pipe cross section. The pressure difference is measured to be 95.8 Pa. Find the volumetric flow rate in m3/s.

Answer 1

Answer:

$\dot{V}=0.0733 \,m^3.s^{-1}$

Explanation:

Given:

density, $\rho=860\,kg.m^{-3}$

diameter of the pipe, $d=2.5\times 10^{-2}m$

pressure difference, $\Delta P=95.8\times 10^{5}\,Pa$

In case of  pitot tube, the velocity is given by:

$v=\sqrt{\frac{2.\Delta P}{\rho} }$

$v=\sqrt{\frac{2\times 95.8\times 10^{5}}{860} }$

$v=149.26\,m.s^{-1}$

Now we know that volumetric flow rate is given as:

$\dot{V}=a.v$

where :

a= cross sectional area of the pipe

v= velocity of flow

$\dot{V}=(\pi\times \frac{(2.5\times 10^{-2})^2}{4} ) \times 149.26$

$\dot{V}=0.0733 \,m^3.s^{-1}$