## Textbooks & Solution Manuals

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

## WriteWise AI Model by Holooly Genius

Your Ultimate AI Essay Writer & Assistant.

## Holooly Arabia

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

## Holooly Help Desk

Need Help? We got you covered.

Products

## Textbooks & Solution Manuals

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

## WriteWise AI Model by Holooly Genius

Your Ultimate AI Essay Writer & Assistant.

## Holooly Arabia

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

## Holooly Help Desk

Need Help? We got you covered.

## Q. 5.P.1

It is required to transport sand of particle size 1.25 mm and density 2600 kg/m³ at the rate of 1 kg/s through a horizontal pipe 200 m long. Estimate the air flowrate required, the pipe diameter, and the pressure drop in the pipe-line.

## Verified Solution

For conventional pneumatic transport in pipelines, a solids-gas mass ratio of about 5 is employed. Mass flow of air $=(1 / 5)=0.20$ kg/s
and, taking the density of air as 1.0 kg/m3, volumetric flowrate of air $=(1.0 \times 0.20)$

$=\underline{\underline{0.20 m ^3 / s }}$

In order to avoid excessive pressure drops, an air velocity of 30 m/s seems reasonable.
Ignoring the volume occupied by the sand (which is about 0.2% of that occupied by the air), the cross-sectional area of pipe required $=(0.20 / 30)=0.0067 m ^2$, equivalent to a pipe diameter of $\sqrt{ }(4 \times 0.0067 / \pi)=0.092$ m or 92 mm.
Thus a pipe diameter of $\underline{\underline{101.6 mm }} (100 mm )$ would be specified.
From Table 5.3 for sand of particle size 1.25 mm and density 2600 kg/m³, the freefalling velocity is:

$u_0=4.7 m / s$

In equation 5.37, $\left(u_G-u_s\right)=4.7 /[0.468+7.25 \sqrt{ }(4.7 / 2600)]=6.05$ m/s
The cross-sectional area of a 101.6 mm i.d. pipe $=\left(\pi \times 0.1016^2 / 4\right)=0.0081$ m².

∴                      air velocity, $u_G=(0.20 / 0.0081)=24.7$ m/s

and:                                      $u_s=(24.7-6.05)=18.65$ m/s

Taking the viscosity and density of air as $1.7 \times 10^{-5}$ N s/m² and 1.0 kg/m3 respectively, the Reynolds number for the air flow alone is:

$\operatorname{Re}=(0.102 \times 24.7 \times 1.0) /\left(1.7 \times 10^{-5}\right)=148,000$

and from Fig. 3.7, the friction factor $\phi=0.002$.

$-\Delta P_{\text {air }}=4 \phi(l / d) \rho u^2$             (equation 3.18)

$=\left(4 \times 0.002(200 / 0.102) \times 1.0 \times 24.7^2\right)=9570 N / m ^2$ or $9.57 kN / m ^2$

assuming isothermal conditions and incompressible flow.

$\left(-\Delta P_x /-\Delta P_{\text {air }}\right)\left(u_s^2 / F\right)=\left(2805 / u_0\right)$              (equation 5.38)

∴       $-\Delta P_x=\left(2805\left(-\Delta P_{\text {air }}\right) F\right) /\left(u_0 u_s^2\right)=(2805 \times 9.57 \times 1.0) /\left(4.7 \times 18.65^2\right)$

$=\underline{\underline{16.4 kN / m ^2}}$