chemistry

1. UNIT OPERATIONS I
Ms. M. T. Rapoo,
Teaching Instructor; Block 109/002
Department of Chemical, Materials & Metallurgical Engineering

2. WHAT IS A UNIT OPERATION?
It is a basic step in a process involving a physical
change
Classes of Unit Operations
• Fluid flow processes – filtration and solids fluidization, settling
& sedimentation, mixing and agitation.
• Heat transfer processes – evaporation, and heat exchange
• Mass transfer processes – gas absorption, distillation, extraction,
adsorption, drying
• Thermodynamic processes – gas liquefaction, refrigeration
• Mechanical processes – solids transportation, crushing,
pulverisation, screening, sieving, froth flotation

3. A granular material is converted from a static solid-like state to a dynamic
fluid-like state. This process occurs when a fluid (liquid or gas) is passed up
through the granular material.
FLUIDIZATION

4. • The basic components required for a fluidized bed are a container, a
gas distributor, solid powder, and gas.
• Introduce a fluid at the bottom, upward at low linear velocities, there
is no movement.
• As the flow increases, frictional drag (FD) on the particles = their
weight (FG) less buoyancy (FB) (force exerted by a fluid that
opposes an object’s weight).
• As particles become rearranged and offer less resistance to flow, the
bed begins to expand.
• Expansion continues as the fluid linear velocity is increased,
reaching the loosest form of packing
• Further increase in velocity, results in the separation of individual
particles
FLUIDIZATION

5. The granular material is converted from a static packed bed to a dynamic
fluid-like state. This process occurs when a fluid (liquid or gas) is passed up
through the granular material at a particular velocity.
FLUIDIZATION

6. • Applied where intimate contact is required between solid
particles and a gas stream (fluid phase).
• Developed mainly for the chemical and petroleum
industries for processes where high heat transfer
coefficients and degree of temperature uniformity is
required.
• E.g.:
– Drying of finely divided solids is carried out in a fluidized
system.
– Applied also in pyrolysis and gasification processes
– Removal of suspended dusts and mists from gases
Applications of Fluidized Bed Techniques

7. • Application in mineral processing and metallurgical
engineering in the recovery of metals from ores.
– A good example is the gas phase extraction of metals, giving
high yields of recovery. A volatile organic reagent (eg.
acetylacetate) passes through the feed material and reacts
selectively with material to be extracted. The product is a volatile
metal complex which is removed from the residue gas by carrier
gas through absorption.
• One of the most important properties of a fluidized bed is
its good heat transfer characteristics and its ability to
handle particles with a wide size distribution.
Applications of Fluidized Bed Techniques

8. Applications of Fluidized Bed Techniques
Type Example Reason for using
fluidized bed
Homogenous catalytic
gas-phase reactions
Ethylene hydrogenation Rapid heating of
entering gas, Uniform
controllable temperature.
Heterogeneous non-
catalytic reactions
Sulphide ore roasting,
combustion
Ease of solids handling,
Temperature uniformity,
Good heat transfer.
Heterogeneous catalytic
reactions
Hydrocarbon cracking,
Phthalic Anhydride,
Acrylonitrile
Ease of solids handling,
Temperature uniformity,
Good heat transfer.

9. Particulate fluidization
• When fluid velocity is increased
the bed continues to expand and
maintains its uniform character.
• Average bed density at a given
velocity is the same in all
sections of the bed, characterized
by complete homogeneity.
• With gases particulate
fluidization only happens at low
velocities.
Fluidization Types

10. Aggregate fluidization
• Is observed at higher gas velocities
• It is characterized by a clear two-
phase structure.
• At high velocities two separate
phases may form.
- Dense or Emulsion –
continuous phase
- Lean or Bubble – is the
discontinuous phase
Fluidization Types

11. The Froude number distinguishes the two types of fluidization.
Where
minimum fluidization velocity
d – diameter of particles
g – acceleration due to gravity
Froude Number<1 – Particulate fluidization
Froude Number >1 – Aggregate Fluidization
Froude number
gd
umf
2
2
mf
u

12. Effect of fluidized bed on pressure gradient
• As the superficial velocity
approaches umf the bed starts
to expand, at higher
velocities it becomes
fluidized.
• At higher velocities, voidage
is higher, pressure gradient
decreases because weight of
particles per unit bed is
smaller
• With even higher velocities
transport of the solid
particles occur and pressure
gradient increases.
If the pressure gradient (−∆P/l) is plotted against the superficial velocity (uc) using
logarithmic co-ordinates, a straight line of unit slope is obtained.

13. • A linear relationship
occurs up to point A where
expansion of the bed
starts.
• The pressure drop reaches
a maximum at point B.
• It then decreases to point
C
• Beyond point C, the
pressure drop is
independent of the
superficial velocity
Effect of fluidized bed on pressure drop
When pressure drop across the whole bed is plotted against velocity using
logarithmic coordinates.

14. Pressure drop
• When the gas velocity is high enough that the frictional drag force
on the particles equals the weight of the particles (m×g) the bed
becomes fluidized;
• -∆P is equal the bed weight per unit cross-sectional area.
• Thus, in a bed of unit cross-sectional area, length l, and porosity
e, the pressure drop across the bed caused by the layout weight of
particles is given by:
Where
l – is length of bed
e – porosity of the bed (fluid volume fraction of the bed)
- density of solid
- density of fluid
g – acceleration due to gravity
p


−∆𝑃 = 1 − 𝑒 𝜌𝑝 − 𝜌 𝑙𝑔 (1)

15. Pressure drop
• Equation (1) applies from the expansion of the bed until
transport/flight of solids occur.
• In a streamline flow in a fixed bed of spherical particles, the
following relation between fluid velocity uc and pressure
drop applies (Carman-Kozeny equation)
𝑢𝑐 = 0.0055
𝑒3
1 − 𝑒 2
−∆𝑃𝑑2
𝜇𝑙
(2)
• For a fluidized bed, the weight of particles is
counterbalanced by frictional drag, substitute (1) into (2)
𝑢𝑐 = 0.0055
𝑒3
(1 − 𝑒)
𝑑2
𝜌𝑠 − 𝜌 𝑔
𝜇
3

16. Pressure drop
Increasing superficial velocity will lead to the point of
incipient fluidization, where particles are just supported in
the fluid, eqn. (4) applies with voidage at being
𝑢𝑚𝑓 = 0.0055
𝑒𝑚𝑓
3
1 − 𝑒𝑚𝑓
𝜌𝑠 − 𝜌 𝑔𝑑2
𝜇
(4)
The value of emf depends on shape, size distribution and
surface properties of particles.
A typical value is
Then eqn. 4 becomes
mf
u mf
e
4
.
0

mf
e
   
5
00059
.
0 2
)
4
.
0
(


 g
d
u s
e
mf
mf




17. Minimum fluidizing velocity
• Equations (4) and (5) apply only at streamline flow and
they are therefore restricted to very fine particles only.
• For large particles, where streamline flow cannot be
achieved at incipient fluidization, the Ergun equation is
applied, equation (6).
−∆𝑃 = 1 − 𝑒 𝜌𝑝 − 𝜌 𝑙𝑔 (1)

18. Substituting for e=emf and uc=umf at incipient fluidization
and -∆P from equation (1)
Minimum fluidizing velocity
   
7
1
75
.
1
)
1
(
150
)
)(
1
(
2
3
2
3
2
d
u
e
e
d
u
e
e
g
e
mf
mf
mf
mf
mf
mf
s
mf










Multiplying eqn. (7) by yields
)
1
(
2
3
mf
e
d



 
8
75
.
1
)
1
(
150
)
(
2
3
3
2
2
3



























 d
u
e
d
u
e
e
gd mf
mf
mf
mf
mf
p
Ga
gd
p


2
3
)
(




Ga is the Galileo number
mf
mf d
u
Re


 Remf is the Reynolds number at the
minimum fluidizing velocity

19. Equation (8) can be written as
Minimum fluidizing velocity
 
9
Re
75
.
1
Re
)
1
(
150 2
3
3 mf
mf
mf
mf
mf
e
e
e
Ga 


Equation 9 is a quadratic equation!!!
For a typical value of emf = 0.4, eqn. (9) becomes
)
11
(
]
1
)
10
53
.
5
1
(
[
7
.
25
Re 5



 
Ga
mf
Solving Remf
 
10
Re
3
.
27
Re
1406 2
mf
mf
Ga 

a
c
b

20. Similarly, for emf = 0.45
Minimum fluidizing velocity
)
12
(
]
1
)
10
39
.
9
1
(
[
6
.
23
Re 5



 
Ga
mf
Therefore by definition, if , then the minimum
fluidization velocity is obtained from:
)
13
(
Remf
mf
d
u



mf
mf d
u
Re




21. A bed consists of uniform spherical particles of diameter
3mm and density 4200kg/m3, what will be the minimum
fluidizing velocity in a liquid of viscosity 3 x10-3 Pasm-2 and
density of 1100kg/m3, Assume emf = 0.4
= 1.0035 x 10
5
Example 1
2
3
)
(



 gd
Ga
p 

2
3
3
3
)
10
3
(
81
.
9
)
1100
4200
(
1100
)
10
3
(




x
x
Ga

22. Assuming emf = 0.4 from equation (11)
Example 1
78
.
8115
]
1
)
10
0035
.
1
)(
10
53
.
5
1
(
[
7
.
25
Re 5
5





 
mf
s
m
x
x
u
d
mf
mf /
377
.
7
1100
10
3
)
10
3
(
78
.
8115
Re 3
3



 




23. In the case of non-spherical particles, to use the above
equations, we account for non sphericity as follows, from
eqn. (8)
Where
- mean linear dimension of the particles
- particle shape factor
When either and / or are unknown, the following
approximations by Wen and Yu (1966) can be used
Minimum fluidizing velocity
 
14
75
.
1
1
1
150
)
(
2
2
2
2
3
2
3
2
3
























 











 p
mf
mf
mf
p
mf
mf
p d
u
e
u
d
e
e
gd
p
d

mf
e 
14
1
3

mf
e
 )
15
(
11
1
3
2


mf
mf
e
e

and

24. The void fraction e is defined as
The specific area of the particle is given by
Shape factors
v
a
)
17
(
p
p
v
V
S
a 
Where
surface area of the particle
volume of the particle
p
S
p
V
 
)
16
(
solids
voids
bed
of
volume
total
bed
in
voids
of
volume
e



25. Shape factor
The sphericity shape factor ϕ of a particle is the ratio of the
surface area of the sphere having the same volume as the
particle to the actual surface area of the particle.
The surface area of a sphere,
And the volume;
2
p
p D
S 

6
3
p
p
D
V



26. Shape factors
For any particle the shape factor is given by
Where
actual surface of the particle
is the equivalent diameter of the sphere having the same
volume as the particle.
Therefore the specific area av of an irregular particle is:

p
p
S
D
2

 
p
S
p
D
)
20
(
6
6
3
2
p
p
p
p
p
v
D
D
D
V
S
a








































27. Shape factor
Therefore for irregular shapes
(20)
For a sphere ϕ = 1
For a Cylinder where diameter = length, ϕ= 0.874
For a cube ϕ = 0.806
p
p
p
v
D
V
S
a

6



28. Bed height & Porosity Relationship
The porosity of a packed bed is given by
Where
density of the solid particle
bulk density of the bed
The relationship between the height of packed bed and
fluidised bed is then
Height of bed at minimum fluidising conditions
p
b
p
e


 

b

p

p
f
f
p
e
e
L
L



1
1
p
mf
mf
p
e
e
L
L



1
1

29. Heat Transfer
Fluidised beds have extremely good heat transfer properties
Nu is a dimensionless parameter used in calculations of heat transfer
between a moving fluid and a solid body.
Where for the particle =
for the particle =
- the superficial velocity
K
hd

p
c d
u
c
u
'
Rec
'
u
N Pr for the particle =
𝑐𝑝𝜇
𝐾

30. Heat Transfer
Equation (29) is valued for
(Reynolds Number)
(Prantl Number)
(porosity Number)
For gas – solid systems
3
'
1
10
Re
10 


c
14000
Pr
22 

9
.
0
4
.
0 
 e
  )
30
(
1
55
.
0
80
.
0
25
.
0
17
.
0
65
.
0

















 















t
c
p
t
t
t d
u
epc
e
d
d
l
d
k
hd

31. Example 2
Spherical catalyst pellets 3mm in diameter are to be fluidized with
nitrogen at 101.3kPa at 600C. The density of the catalyst particles are
980 kg/m3. The molecular weight of nitrogen is 29kg/kmol. If it is
assumed that the point of incipient fluidization is reached
at emf = 0.43, Calculate the minimum fluidization velocity in the
vessel. Nitrogen is considered here to be an ideal gas with µ=
0.0000207 Ns/m2.
Ergun Equation
R (gas constant) = 8.314 m3∙Pa/(mol.K), g =9.81m/s2

32. Example 2 Solution
The pressure drop across the bed caused by the layout weight of
particles is given by
Combining equation (1) and the Ergun equation
This equation can be simplified to
−∆𝑃 = 1 − 𝑒 𝜌𝑝 − 𝜌 𝑙𝑔 (1)
1.75𝑑𝜌𝑢𝑚𝑓
2
+ 150 1 − 𝑒𝑚𝑓 𝜇𝑢𝑚𝑓 − 𝜌𝑝 − 𝜌 𝑔𝑑2𝑒𝑚𝑓
3
= 0
   
7
1
75
.
1
)
1
(
150
)
)(
1
(
2
3
2
3
2
d
u
e
e
d
u
e
e
g
e
mf
mf
mf
mf
mf
mf
s
mf











33. Example 2 Solution
 
3
/
061092365
.
1
/
1000
60
273
314
.
8
29
101300
m
kg
kmol
mol
x
x
RT
PM r
a





      
      0
003
.
0
43
.
0
81
.
9
061092365
.
1
980
0000207
.
0
43
.
0
1
150
061092365
.
1
003
.
0
75
.
1
2
3
2




 mf
mf u
u
0
10
87183105
.
6
10
76985
.
1
2
0055707349
.
0 3
3
2


 

x
u
x
u mf
mf
0

mf
u Ignore negative root
ρnitrogen is unknown
1.75𝑑𝜌𝑢𝑚𝑓
2
+ 150 1 − 𝑒𝑚𝑓 𝜇𝑢𝑚𝑓 − 𝜌𝑝 − 𝜌 𝑔𝑑2𝑒𝑚𝑓
3
= 0

34. Example 3
Non spherical catalyst pellets, 4mm in diameter, shape factor of
1.1, are to be fluidized with air at 101.3kPa at 700C. The density of
the catalyst particles are 1100 kg/m3. Take the molecular weight of
air as 26.9 kg/kmol. If it is assumed that the point of incipient
fluidization is reached at , calculate the minimum
fluidizing velocity in the vessel. Air is considered here to be an
ideal gas with Ns/m2.
43
.
0

mf

0000207
.
0

air


35. Example 3 Solution
At minimum fluidization velocity
  
    )
7
(
1
75
.
1
1
150
1 3
2
3
2
2
2
d
e
u
e
e
d
u
e
g
e
mf
mf
mf
mf
mf
mf
s
mf













 shape factor = 1.1
43
.
0

mf
e
3
3
3
3
/
9338
.
0
/
83
.
933
15
.
343
/
314
.
8
/
3
.
26
10
3
.
101
m
kg
m
g
K
mol
K
Pa
m
mol
g
Pa
x
RT
PMr










36. Example 3 Solution
Substitute values into equation (7)
  
   
     
  
    
004
.
0
1
.
1
43
.
0
9338
.
0
43
.
0
1
75
.
1
43
.
0
004
.
0
1
.
1
10
07
.
2
43
.
0
1
150
81
.
9
9338
.
0
1100
43
.
0
1
3
3
2
2
5
2
mf
mf
u
u
x







0
61
.
2662
39
.
655
6
.
6145 2



 mf
mf u
u
    
s
m
x
umf /
401
.
1
61
.
2662
2
61
.
2662
6
.
6145
4
39
.
655
39
.
655
2






37. Thank you for your attention!