MOTION OF PARTICLE TROUGH FLUID : FLUIDIZATION
(a) Fluidization is one of the processes involving the movement of particles through fluid.
(i) Define ‘ Fluidization ‘
Fluidization is a process whereby agranular materialis converted from a static solid-like state to a dynamicfluid-like state.
(ii) Describe the purpose of fluidization
The most common reason for fluidizing a bed is to obtain vigorous agitation of the solids in contact with the fluid, leading to an enhanced transport mechanism (diffusion, convection, and mass/energy transfer).
(iii) Sketch the schematic diagram depicts the fluidization process occurred in the
granular bed column.
(b) Catalyst pellets 5 mm in diameter are to be fluidized with 45 000 kg/h of water with density 985 kg/m3 at 50 ºC in a vertical cylindrical vessel. The density of the catalyst particles is 1760 kg/m3 and their sphericity is 0.86.
Particle: Pellet
Dp = 5 x 10-3 m
ρp = 1760 kg/m3
Φ = 0.86
Take εM = 0.45
Fluid: Water
m = 45000 kg/h
ρ = 985 kg/m3
µ = 0.6 x 10-3 kg/m.s ( at T=50 ºC)
(i) Calculate the minimum velocity VOM of the water used to fluidized the catalyst pellets
150μVOMΦS2Dp2 1-εMεM3+1.75ρV2OMΦsDp 1εM3=gρp-ρ
150(0.6 x 10-3)VOM0.8625 x 10-3 2 1-0.450.453+1.75(985)V2OM0.865 x 10-3 10.453=9.811760-985
Solving for VOM VOM=0.0384 m/s
(ii) Estimate the diameter of the vertical cylindrical vessel
A=mρVOM
A=45000 kghr×1 hr3600s×m3985 kg×s0.0384m
A=0.3305 m2 (Area of cylinder)
A=πD24
0.3305=πD24
D=0.65 m
FLOW OF FLUID THROUGH GRANULAR BEDS
(a) The specific surface area of particle (S) in one of the parameters affecting the pressure drop of the fluid flows through a granular beds or porous medium. In considering the effect of surface area, there is a difference between the specific surface or particle (S) and the specific surface area of bed (SB). Explain why these two parameters cannot be assumed similar.
S and SB are not equal due to the voidage which is present when the particles are packed into a bed.
(b) Adsorption process has been conducted in a packed column whereas air was subjected to flow through beds of granular activated carbon at 130 º C under the atmospheric pressure. A pore space which approximately 25% of granular bed form a channel like-tube with diameter of 5.56 x10-4 (L). Based on Carman-Kozeny, the pressure drop per length of the column is given by 1048.48 (Pa/m).
Air properties at T= 130 º C, (interpolation from table appendix A.3.3)
ρ = 0.8768 kg/m3
µ = 2.3020 x 10-5 kg/m.s or Pa.s
ε = 0.25
de = 5.56 x10-4 m
∆pL=1048.48 (Pa/m).
i) Estimate the average size (Dp (m)) of the activated carbon granule packed in the column.
de=4εSB
5.56 × 10-4=4(0.25)SB
SB=1798.5612
SB=S (1-ε)
1798.5612=S (0.75)
S=2398.0816
S=6DP
2398.0816=6DP
Dp=2.502 ×10-3 m
(c) Calculate the Reynolds number of the gas flow through the column and state the types of flow occurred in the process.
∆pL=K1-ε2S2ε3μV0
1048.48=50.7522398.081620.253 2.3020 ×10-5 V0
V0=0.044 m/s
Re=ρV0S1-εμ= 0.8768 (0.044)2398.0816 (0.75)( 2.3020 ×10-5 )=0.93
Laminar flow because Re < 1