### Fluid Mechanics

# Pressure Drop for Flow of Air Through Packed Bed

Calculate the pressure drop of air flowing at 30^{o}C and 1 atm pressure through a bed of 1.25 cm diameter spheres, at a rate of 60 kg/min. The bed is 125 cm diameter and 250 cm height. The porosity of the bed is 0.38. The viscosity of air is 0.0182 cP and the density is 0.001156 gm/cc.

Data:

Mass flow rate of Air = 60 kg/min = 1 kg/sec

Density of Air (r) = 0.001156 gm/cc = 1.156 kg/m^{3}

Viscosity of Air (m) = 0.0182 cP = 0.0182 x 10^{-3} kg/(m.sec)

Bed porosity (e) = 0.38

Diameter of bed (D)= 125 cm = 1.25 m

Length of bed (L) = 250 cm = 2.5 m

Dia of particles (D_{p})= 1.25 cm = 0.0125 m

Sphericity (F _{s}) = 1 (sphere)

Formulae:

NRe_{PM }= D_{p}V_{o}r/(m (1 - e) )

For laminar flow (i.e. NRe_{PM} < 10) pressure drop is given by Blake-Kozeny equation.

For turbulent flow (i.e. NRe_{PM} > 1000) pressure drop is given by Burke-Plummer equation.

For intermediate flows pressure drop is given by Ergun equation

Superficial velocity V_{o} = Volumetric flow rate/ cross-sectional area of bed

Calculations:

Volumetric flow rate = mass flow rate / density = 1 / 1.156 = 0.865 m^{3}/sec

Superficial velocity V_{o} = 0.865 / ( (p/4) D^{2} ) = 0.865 / ( (p/4) 1.25^{2} ) = 0.705 m/sec

NRe_{PM }= 0.0125 x 0.705 x 1.156 / (0.0182 x 10^{-3} x ( 1- 0.38 ) ) = 903

We shall use Ergun equation to find the pressure drop.

i.e. Dp x 0.0125 x 0.38^{3} / ( 2.5 x 1.156 x 0.705^{2} x ( 1 - 0.38 ) ) = 150 / 903 + 1.75

Dp x 7.702 x 10^{-4} = 1.92

Dp = 1.92 / 7.702
x 10^{-4} = **2492.92
N/m ^{2}**