Pressure Drop for Flow of Air Through Packed Bed

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Calculate the pressure drop of air flowing at 30oC 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 (ρ) = 0.001156 gm/cc = 1.156 kg/m3

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

Bed porosity (ε) = 0.38

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

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

Dia of particles (Dp)= 1.25 cm = 0.0125 m

Sphericity (Φs) = 1 (sphere)

Formulae:

NRePM = DpVoρ/(μ (1 - ε) )

For laminar flow (i.e. NRePM < 10) pressure drop is given by Blake-Kozeny equation.

\(\displaystyle \frac{\Delta p \phi_s^2D_p^2\varepsilon^3}{LV_o\mu(1-\varepsilon)^2} = 150 \)

For turbulent flow (i.e. NRePM > 1000) pressure drop is given by Burke-Plummer equation.

\(\displaystyle \frac{\Delta p \phi_sD_p\varepsilon^3}{L\rho V_o^2(1-\varepsilon)} = 1.75 \)

For intermediate flows pressure drop is given by Ergun equation

\(\displaystyle \frac{\Delta p \phi_sD_p\varepsilon^3}{L\rho V_o^2(1-\varepsilon)} = \frac{150\mu(1-\varepsilon)}{\phi_sD_pV_o\rho} + 1.75 \)

Superficial velocity Vo = Volumetric flow rate/ cross-sectional area of bed

Calculations:

Volumetric flow rate = mass flow rate / density = 1 / 1.156 = 0.865 m3/sec

Superficial velocity Vo = 0.865 / ( (π/4)D2) = 0.865 / ( (π/4) 1.252 ) = 0.705 m/sec

NRePM = 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.

\(\displaystyle \frac{\Delta p \phi_sD_p\varepsilon^3}{L\rho V_o^2(1-\varepsilon)} = \frac{150\mu(1-\varepsilon)}{\phi_sD_pV_o\rho} + 1.75 \)

i.e. Δp x 0.0125 x 0.383 / ( 2.5 x 1.156 x 0.7052 x ( 1 - 0.38 ) ) = 150 / 903 + 1.75

Δp x 7.702 x 10-4= 1.92

Δp = 1.92 / 7.702 x 10-4 = 2492.92 N/m2


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Last Modified on: 04-Feb-2022

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