Question

A solution of density 1100 kgm^-3 and viscosity 2×10^-3 Pa s is flowing under gravity at a rate of 0.24 kg/s through a bed of catalyst particles. The bed diameter is 0.2 m and the depth is 0.5 m. The particles are cylindrical, with a diameter of 1mm and length of 2 mm. They are loosely packed to give a voidage of 0.3. Calculate the depth of liquid above the top of the bed.

Answer 1

Determine the Reynolds number

Re = superficial velocity x density x diameter / viscosity (1 - voidage)

Density = 1100 kg/m3

Viscosity = 2*10^-3 Pa-s

Voidage = 0.3

Superficial velocity = mass flow / (density x area)

Area = (3.14/4)*(0.2*0.2)m2

= 0.0314 m2

Superficial velocity = 0.24/(1100*0.0314) = 6.94*10^-3 m/s

Surface volume ratio of particles

= surface area of 1 cylindrical particle / volume of 1 particle

= (2.5*3.14)/(3.14/2) = 5

For sphere, Surface volume ratio = 6/diameter

For same Surface volume ratio

5 = 6/diameter

Diameter of sphere = 1.20 mm

Re = (6.94*10^-3 * 1100 * 1.20) / (2*10^-3*(1-0.30))

= 6.50

Re < 10

The flow is laminar

From Ergun equation

??p = 13120 Pa

Head loss = 13120 Pa /(1100 kg/m3*9.81 m/s2)

= 1.216 m

Depth of liquid = head loss - height of bed

= 1.216 - 0.50

= 0.7160 m