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}}