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 a granular material is converted from a
static solid-like state to a dynamic fluid-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
̅̅̅̅𝟐
̅ 𝑶𝑴 (𝟏 − 𝜺𝑴 ) 𝟏. 𝟕𝟓𝝆𝑽
𝟏𝟓𝟎𝝁𝑽
𝑶𝑴 𝟏
+
= 𝒈(𝝆𝒑 − 𝝆)
𝜱𝒔 𝑫𝒑
𝜱𝟐𝑺 𝑫𝟐𝒑
𝜺𝟑𝑴
𝜺𝟑𝑴
̅ 𝑶𝑴 (𝟏 − 𝟎. 𝟒𝟓) 𝟏. 𝟕𝟓(𝟗𝟖𝟓)̅̅
𝟏𝟓𝟎(𝟎. 𝟔 𝐱 𝟏𝟎−𝟑 )𝑽
𝑽̅̅𝟐 𝑶𝑴
𝟏
+
𝟎. 𝟖𝟔𝟐 (𝟓 𝐱 𝟏𝟎−𝟑 )𝟐
𝟎. 𝟒𝟓𝟑
𝟎. 𝟖𝟔(𝟓 𝐱 𝟏𝟎−𝟑 ) 𝟎. 𝟒𝟓𝟑
= 𝟗. 𝟖𝟏(𝟏𝟕𝟔𝟎 − 𝟗𝟖𝟓)
̅̅̅̅̅
Solving for 𝑉
𝑂𝑀
̅ 𝑶𝑴 = 𝟎. 𝟎𝟑𝟖𝟒 𝒎/𝒔
𝑽
(ii)
𝑨=
Estimate the diameter of the vertical cylindrical vessel
𝒎̇
̅̅̅̅
𝝆𝑽𝑶𝑴
𝟒𝟓𝟎𝟎𝟎 𝒌𝒈
𝟏 𝒉𝒓
𝒎𝟑
𝒔
𝑨=
×
×
×
𝒉𝒓
𝟑𝟔𝟎𝟎𝒔 𝟗𝟖𝟓 𝒌𝒈 𝟎. 𝟎𝟑𝟖𝟒𝒎
𝑨 = 𝟎. 𝟑𝟑𝟎𝟓 𝒎𝟐 (Area of cylinder)
𝝅𝑫𝟐
𝑨=
𝟒
𝝅𝑫𝟐
𝟎. 𝟑𝟑𝟎𝟓 =
𝟒
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
∆𝒑
𝑳
= 𝟏𝟎𝟒𝟖. 𝟒𝟖 (𝐏𝐚/𝐦).
i)
Estimate the average size (Dp (m)) of the activated carbon granule packed in
the column.
𝒅𝒆 =
𝟒𝜺
𝑺𝑩
𝟓. 𝟓𝟔 × 𝟏𝟎−𝟒 =
𝟒(𝟎. 𝟐𝟓)
𝑺𝑩
𝑺𝑩 = 𝟏𝟕𝟗𝟖. 𝟓𝟔𝟏𝟐
𝑺𝑩 = 𝑺 (𝟏 − 𝜺)
𝟏𝟕𝟗𝟖. 𝟓𝟔𝟏𝟐 = 𝑺 (𝟎. 𝟕𝟓)
𝑺 = 𝟐𝟑𝟗𝟖. 𝟎𝟖𝟏𝟔
𝑺=
𝟔
𝑫𝑷
𝟐𝟑𝟗𝟖. 𝟎𝟖𝟏𝟔 =
𝟔
𝑫𝑷
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.
(𝟏 − 𝜺)𝟐 𝑺𝟐
∆𝒑
̅̅̅𝟎̅
=𝑲
𝝁𝑽
𝑳
𝜺𝟑
𝟏𝟎𝟒𝟖. 𝟒𝟖 = 𝟓
(𝟎. 𝟕𝟓)𝟐 𝟐𝟑𝟗𝟖. 𝟎𝟖𝟏𝟔𝟐
̅𝟎
𝟐. 𝟑𝟎𝟐𝟎 × 𝟏𝟎−𝟓 𝑽
𝟎. 𝟐𝟓𝟑
̅ 𝟎 = 𝟎. 𝟎𝟒𝟒 𝒎/𝒔
𝑽
𝑹𝒆 =
̅𝟎
𝝆𝑽
𝟎. 𝟖𝟕𝟔𝟖 (𝟎. 𝟎𝟒𝟒)
=
= 𝟎. 𝟗𝟑
𝑺(𝟏 − 𝜺)𝝁
𝟐𝟑𝟗𝟖. 𝟎𝟖𝟏𝟔 (𝟎. 𝟕𝟓)( 𝟐. 𝟑𝟎𝟐𝟎 × 𝟏𝟎−𝟓 )
Laminar flow because Re < 1