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.

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Accepted Answer

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

[latex]=\underline{\underline{0.20 m ^3 / s }}[/latex]

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 [latex]=(0.20 / 30)=0.0067 m ^2[/latex], equivalent to a pipe diameter of [latex]\sqrt{ }(4 imes 0.0067 / \pi)=0.092[/latex] m or 92 mm.
Thus a pipe diameter of [latex]\underline{\underline{101.6 mm }} (100 mm )[/latex] would be specified.
From Table 5.3 for sand of particle size 1.25 mm and density 2600 kg/m³, the freefalling velocity is:

[latex]u_0=4.7 m / s[/latex]

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

∴ air velocity, [latex]u_G=(0.20 / 0.0081)=24.7[/latex] m/s

and: [latex]u_s=(24.7-6.05)=18.65[/latex] m/s

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

[latex]\operatorname{Re}=(0.102 imes 24.7 imes 1.0) /\left(1.7 imes 10^{-5} ight)=148,000[/latex]

and from Fig. 3.7, the friction factor [latex]\phi=0.002[/latex].

[latex]-\Delta P_{ ext {air }}=4 \phi(l / d) ho u^2[/latex] (equation 3.18)

[latex]=\left(4 imes 0.002(200 / 0.102) imes 1.0 imes 24.7^2 ight)=9570 N / m ^2[/latex] or [latex]9.57 kN / m ^2[/latex]

assuming isothermal conditions and incompressible flow.

[latex]\left(-\Delta P_x /-\Delta P_{ ext {air }} ight)\left(u_s^2 / F ight)=\left(2805 / u_0 ight)[/latex] (equation 5.38)

∴ [latex]-\Delta P_x=\left(2805\left(-\Delta P_{ ext {air }} ight) F ight) /\left(u_0 u_s^2 ight)=(2805 imes 9.57 imes 1.0) /\left(4.7 imes 18.65^2 ight)[/latex]

[latex]=\underline{\underline{16.4 kN / m ^2}}[/latex]

Question 5.P.1: It is required to transport sand of particle size 1.25 mm an...

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