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/m2Last Modified on: 01-May-2024
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