Question

A packed bed of particles of density 2500 kg/m3 occupies a depth of 0.5 m in a cylindrical vessel of inside diameter 0.1 m. The mass of solids in the bed is 3 kg. The mean diameter of the particles is 300 μm and the sphericity is 0.67. Water (density 1000 kg and viscosity 0.001 Pa s) flows upwards through the bed.

Determine the pressure drop across the bed when being fluidized with water superficial velocity being higher than the minimum fluidisation flow velocity.

Answer 1

Density of particles, rhop = 2500 kg/m3

Density of water, rhof = 1000 kg/m3

Depth of bed, L = 0.5 m

Cylindrical inside vessel, d = 0.1 m

Mean diameter of particle, dp = 300 um = 3*10^-4 m

Sphericity of particles , ¤ = 0.67

Mass of solid in the bed, mp = 3 kg

Volume of solid in the bed Vp = mp/rhop = 3kg/(2500kg/m3) = 1.2*10^-3 m3

Volume of bed without solid, Vb = π*d^2*L = 3.14*0.1^2*0.5 = 0.0157 m3

Voidage in the bed e = 1- Vb/Vp = 1- 0.0012/0.0157= 0.92

For minimum fluidization velocity,

vm = gdp^2(rhop - rhof) /18mu = 9.8*(3*10^-4)^2(2500-1000)/(18*0.001) = 0.0735 m/s

Finding Reynolds number,

Re = rho*vm*dp/mu = 1000*0.0735*3*10^-4/0.001 = 22

Hence flow is laminar,

ΔP/L = 150mu*vm*(1-e)^2/e^3(¤*dp)^2

ΔP/0.5 = 150*0.001*0.0735*(1-0.92)^2/0.92^3*(0.67*3*10^-4)^2

ΔP = 1121 Pa

Pressure drop in packed bed when flow with  minimum fluidization velocity = 1121 Pa