Question

A cylindrical tank with inner diameter 2.0 m is designed to operate as a fluidized bed reactor for a heterogeneous catalytic reaction. The tank is loaded with a random packing of cubic particles to a height of 4.0 m with voidage 0.4 before the fluidization process. The density and length of each side of the cubic particles are 2600 kg m-3 and 1.0 cm respectively. During operation, a liquid reactant with density 1000 kg m-3 and viscosity 1.0 x 10-3 Pa·s is passed uniformly upwards through the bed to fluidize the particles. Determine the inlet superficial velocity U that needs to be applied to achieve a bed height of 6.0 m during operation of the fluidized bed reactor.

Answer 1

For 4 m Height and 2m diameter, Cylinder Volume = 3.14/4*D^2*H = 3.14/4*2^2*4 = 12.56 m3

Voidage =.4, Therefore, Total Cubes volume = 0.6*12.56 = 7.53 m3

length of cubic particle = 1 cm

Spherical Diameter, Ds = 6*Volume/Surface area

For cube, Ds = 6*1/6/1 = 1 cm

For 6 meter, 2 m diameter, Total Volume = 3.14/4*2^2*6 = 18.84 m3

Therefore, for 6 m Height, 2m Diametetr cylinder filled with 7.53 m3 cubes, Voidage, E = (18.84-7.53)/18.84 = 0.6

Density of particle, Pp = 2600 kg/m3

Density of water, Pf = 1000 kg/m3

g = 10 m2/s

Therfore, Delta P = (1-0.6)*(2600-1000)*6*10 = 38400 Pa

_{GRp =} Dp*Vs*Pf/((1-E)*u)

viscosity of fluid, u = 10^-3 pas

Vs = Fluid superficial velocity

Using Ergun equation,

}

By putting all the values,

38400/6*(.01/100/Vs^2)*(.6^3/(1-0.6) = 150/(1000*Vs*.01/(1-0.6)/10^-3)+1.75

Therefore, Superficial velocity, Vs =0.13m/s