Question

Solid particles of density 2500 kgm-3 were packed into a fluidized bed reactor of 2 m in height and 0.04 m2 in cross-sectional area. The mass and height of the packed bed are 45 kg and 1 m, respectively. The surface-volume mean diameter of the particles is 1 mm.

(a) Determine the porosity of the packed bed.

(b) Determine the pressure drop across the bed when a liquid of density 800 kgm-3 and viscosity of 0.002 Pa s flows through the bed at a flow rate of 1.44 m3h-1.

(c) What is the pressure drop across the bed at the point of fluidization if the porosity is 0.6?

Thanks in advance

Answer 1

(a) The porosity of the bed is the ratio of the free space to that of the solid:

= Area* height = 0.04*1 = 0.04 m³

= Weight of solid/density of solid = 45/2500 = 0.018 m³

Thus, 0.022/0.04 = 0.55

(b) For the pressure drop, we will use the Ergun equation:

Now from the information provided, we know that:

V₀ = Flow rate/Area of cross section = (1.44/0.04) m³/ hr = 36 m/hr = 36/3600 m/s = 0.01 m/s

L = 1 m

We can now plug in these values into our equation and we will get:

Thus, ( 3651.39 + 3.79) Pa = 3655.18 Pa

(c) This new porosity has been mentioned in the question and thus, we have to use this for this part. Now, if the porosity was 0.6, the height of the bed would be higher as there would be more space between the solid particles

But,

V_solid does not change and is still 0.018 m³

Thus, for

Thus, L= V/Area of cross section = 0.045/0.04 = 1.125 m

Let us now plug in these values into the ergun equation: