The document provides an overview of fluidized bed reactors (FBR), including their characteristics, operational mechanics, and applications within chemical engineering. It discusses key concepts such as mass transfer between phases, reaction behavior in fluidized beds, and the balance of materials in different phases including bubbles and emulsions. Additionally, the document highlights advantages and disadvantages of FBRs, and presents the mathematical models that describe their performance and behavior.

1. Fluidized Bed Reactor
– An Overview
Submitted by :
Antarim Dutta Reg No :
2016CL04
Discipline : Chemical Engg.
Department
Course : Master of Technology
MNNIT, Allahabad
1

2. Contents
1. Introduction
2. The Mechanics of Fluidized Beds
2.1. Pressure Versus Gas Velocity Curve
2.2. Description of the Phenomena
2.3. The Minimum Fluidization Velocity
2.4. Maximum Fluidization
2.5. Descriptive Behavior of a Fluidized Bed - The Model of Kunii And
Levenspiel
2.5. Bubble Velocity and Cloud Size
2.6. Fraction of Bed in Bubble Phase
3. Mass Transfer in Fluidized Beds
3.1. Gas – Solid Mass Transfer
3.2. Mass Transfer Between the Fluidized-Bed Phases
4. Reaction Behaviour in a Fluidized Bed
5. Mole Balance on the Bubble, the Cloud, and the Emulsion
5.1. Balance on Bubble Phase
5.2. Balance on Cloud Phase
5.3. Balance on the Emulsion
5.4. Partitioning of the Catalyst
5.5. Solution to the Balance Equations for a First-Order Reaction
6. Advantages & Disadvantages
7. Current Applications of FBR
8. References 2

3. Introduction
O The catalytic reactor (which is in common use) is
analogous to the CSTR in that in content, though
heterogeneous, are well mixed and this results in
an even temperature distribution throughout the
bed.
O It consists of a vertical cylindrical vessel containing
fine solid catalyst particles. The fluid stream
(usually a gas) is introduced through the bottom at
a rate such that catalyst particle are suspended in
the fluid stream without being carried out.
O With this reactor, it is possible to regenerate the
catalyst continuously without shutting down the
reactor. This reactor is particularly suitable when
the heat effects are very large or when frequently
catalyst regeneration is required.
3

4. Continues…
4
O Fluidization occurs when small
solid particles are suspended in an
upward flowing stream of fluid.
O The fluid velocity is sufficient to
suspend the particles, but it is not
to large enough to carry them out
of vessel.
O The solid particles swirl around the
bed rapidly, creating excellent
mixing among them.
O The material “fluidized” is almost
always a solid and the “fluidizing
medium” is either a liquid or gas.
O The characteristics and behavior
of a fluidized bed are strongly
dependent on both the solid and
liquid or gas properties.
Figure : From Kunii and
Levenspiel Fluidization
Engineering, Melbourne, FL
32901: Robert E. Krieger Pub. Co.
1969.

5. The Mechanics of Fluidized Bed
Description of the Phenomena
5
Figure : Various kinds of contacting of a batch of solids by
fluid.

6. Continues…
O At low velocity pressure drop resulting from the drag
follows Ergun equation given as,
𝒅𝑷
𝒅𝒛
=
− 𝑮
𝝆𝒈 𝒄 𝑫 𝒑
𝟏 − 𝝋
𝝋 𝟑
𝟏𝟓𝟎 𝟏− 𝝋 𝝁 𝒈
𝑫 𝒑
+ 𝟏. 𝟕𝟓𝑮 --------- (1)
O The mass of solids in bed is given by,
𝑾 𝒔 = 𝝆 𝒄 𝑨 𝒄 𝒉 𝒔 𝟏 − 𝜺 𝒔 = 𝝆 𝒄 𝑨 𝒄 𝒉 𝟏 − 𝜺 ---------- (2)
O After the drag exerted on the particles equals the net
gravitational force exerted on the particles, that is,
∆𝑷 = 𝒈 𝝆 𝒄 − 𝝆 𝒈 𝟏 − 𝜺 𝒉 ------------ (3)
The pressure drop will not increase with an increase in
velocity beyond this point.
6

7. Pressure Versus Gas Velocity Curve
7
Figure : From Kunii and Levenspiel, Fluidization Engineering
(Melbourne, FL: Robert E. Krieger, Publishing Co. 1977).

8. The Minimum Fluidization Velocity
O The Ergun equation in (1) can be written as,
∆𝑷
𝒉
= 𝝆 𝒈 𝑼 𝟐 𝟏𝟓𝟎 𝟏 − 𝜺
𝑹𝒆 𝒅 𝝋
+
𝟕
𝟒
(𝟏 − 𝜺)
𝝋𝒅 𝒑 𝜺 𝟑 ----------- (4)
O At the point of minimum fluidization, the weight of the bed
just equals the pressure drop across the bed
𝑾 𝒔 = ∆𝑷𝑨 𝒄----------- (5)
𝒈 𝟏 − 𝜺 𝝆 𝒄 − 𝝆 𝒈 𝒉𝑨 𝒄 = 𝝆 𝒈 𝑼 𝟐 𝟏𝟓𝟎 𝟏 − 𝜺
𝑹𝒆 𝒅 𝝋
+
𝟕
𝟒
(𝟏 − 𝜺)
𝝋𝒅 𝒑 𝜺 𝟑 𝑨 𝒄 𝒉 ----------
(6)
O The minimum fluidization is given by
𝒖 𝒎𝒇 =
𝝋𝒅 𝒑
𝟐
𝟏𝟓𝟎𝝁
𝒈(𝝆 𝒄 − 𝝆 𝒈)
(𝜺 𝒎𝒇) 𝟐
𝟏 −𝜺 𝒎𝒇
-------------- (7)
Note : For Re < 10, 𝐸𝑞 𝑛
(7) can be solved. Where, Re =
𝜌 𝑔 𝑑 𝑝 𝑈
𝜇
; Reynolds
number less than 10 is usual in which fine particles are fluidized by gas. 8

9. O Introduction of two parameters are there.
O First one is 𝜑, the “sphericity” which is the measure of a particle’s non-
ideality in both shape and roughness. And Calculated as,
𝝋 =
𝑨 𝒔
𝑨 𝒑
=
𝝅
𝟏
𝟑 𝟔𝑽 𝒑
𝟐
𝟑
𝑨 𝒑
---------- (8)
O The second parameter is the void fraction at the time of minimum
fluidization 𝜺 𝒎𝒇.
𝜺 𝒎𝒇 = 𝟎. 𝟓𝟖𝟔𝝋−𝟎.𝟕𝟐 𝝁 𝟐
𝝆 𝒈 𝜼𝒅 𝒑
𝟑
𝟎.𝟎𝟐𝟗
𝝆 𝒈
𝝆 𝒄
𝟎.𝟎𝟐𝟏
-------- (9)
Another Correlation commonly used is that of Wen and Yu
Type equation here.
𝜺 𝒎𝒇 = 𝟎. 𝟎𝟕𝟏 𝝋
𝟏
𝟑------- (10) or, 𝜺 𝒎𝒇 =
𝟎.𝟎𝟗𝟏(𝟏 −𝜺 𝒎𝒇)
𝝋 𝟐 --------- (11)
O If the distribution of sizes of the particles covers too large a range, the
equation will not apply because smaller particles can fill the interstices
between larger particles. Then 𝒅 𝒑 is calculated as,
𝒅 𝒑 =
𝟏
𝒇 𝒊
𝒅 𝒑 𝒊
------- (12) ; 𝑓𝑖 is the fraction of particles with diameter 𝑑 𝑝 𝑖
9

10. Maximum Fluidization
O If the gas velocity is increased to a sufficiently high value, however, the
drag on an individual particle will surpass the gravitational force on the
particle, and the particle will be entrained in a gas and carried out of the
bed. The point at which the drag on an individual particle is about to
exceed the gravitational force exerted on it is called the maximum
fluidization velocity.
O Maximum Velocity through the bed 𝑢 𝑡 is given for fine particles, the
Reynolds number will be small, and the two relationships presented by
Kunii and Levenspiel are,
𝒖 𝒕 = 𝜼𝒅 𝒑
𝟐
𝟏𝟖𝝁 for Re < 0.4 ---------- (13)
𝒖 𝒕 = 𝟏. 𝟕𝟖 × 𝟏𝟎−𝟐
𝜼 𝟐
𝝆 𝒈 𝝁
𝟏 𝟑
𝒅 𝒑 for 0.4 < Re < 500 ------- (14)
O The entering superficial velocity, 𝑢0, must be above the the minimum
fluidization velocity but below the slugging 𝑢 𝑚𝑠 and terminal,
𝑢 𝑡, velocities.
Therefore, both 𝒖 𝒎𝒇 < 𝒖 𝟎 < 𝒖 𝒕 and 𝒖 𝒎𝒇 < 𝒖 𝟎 < 𝒖 𝒎𝒔 these conditions
must be satisfied for proper bed operation.
10

11. Descriptive Behavior of a Fluidized Bed – The Model of Kunii
and Levenspiel
O Early investigators saw that the fluidized bed had to be treated as a two-
phase system – an emulsion phase and a bubble phase (often called the
dense and lean phases). The bubbles contain very small amounts of
solids. They are not spherical; rather they have an approximately
hemispherical top and a pushed-in bottom. Each bubble of gas has a
wake that contains a significant amount of solids.
O These characteristics are illustrated in Figure, which were obtained from
x-rays of the wake and emulsion, the darkened portion being the bubble
phase.
11
Figure : Schematic of
bubble, cloud, wake and
emulsion.

12. Assumptions in The Kunii- Levenspiel Model
O The bubbles are all of one size.
O The solids in the emulsion phase flow smoothly downward, essentially in
plug flow.
O The emulsion phase exists at minimum fluidizing conditions. The gas
occupies the same void fraction in this phase as it had in the entire bed at
the minimum fluidization point. In addition, because the solids are flowing
downward, the minimum fluidizing velocity refers to the gas velocity relative
to the moving solids, that is,
𝒖 𝒆 =
𝒖 𝒎𝒇
𝜺 𝒎𝒇
− 𝒖 𝒔 ------ (15)
The velocity of the moving solids, 𝒖 𝒔, is positive in the downward direction here,
as in most of the fluidization literature. The velocity of the gas in the emulsion,
𝒖 𝒆, is taken as a positive in the upward direction, but note that it can be
negative under some conditions.
O In the wakes, the concentration of solids is equal to the concentration of
solids in the emulsion phase, and therefore the gaseous void fraction in the
wake is also the same as in the emulsion phase. Because the emulsion
phase is at the minimum fluidizing condition, the void fraction in the wake is
equal to 𝜺 𝒎𝒇.
12

13. Bubble Velocity and Cloud Size
O For single bubble, 𝒖 𝒃𝒓 = (𝟎. 𝟕𝟏) 𝒈𝒅 𝒑
𝟏
𝟐
-------- (16)
O Velocities of bubble rise are given by,
𝒖 𝒃 = 𝒖 𝒃𝒓 + (𝒖 𝟎 − 𝒖 𝒎𝒇) --------- (17)
𝒖 𝒃 = 𝒖 𝟎 − 𝒖 𝒎𝒇 + (𝟎. 𝟕𝟏) 𝒈𝒅 𝒑
𝟏
𝟐
---------- (18)
O The best relationship between bubble diameter and height in the
column at this writing seems to be that of Mori and Wen, who correlated
the data of studies covering bed diameters of 7 to 130 cm, minimum
fluidization velocities of 0.5 to 20 cm/s, and solid particle sizes of 0.006
to 0.045 cm. Their principal equation was
𝒅 𝒃𝒎− 𝒅 𝒃
𝒅 𝒃𝒎− 𝒅 𝒃𝒐
= 𝒆−𝟎.𝟑𝒉 𝑫 𝒕 -------- (19)
O The maximum bubble diameter, 𝒅 𝒃𝒎 has been observed to follow the
relationship
𝒅 𝒃𝒎 = 𝟎. 𝟔𝟓𝟐 𝑨 𝒄(𝒖 𝟎 − 𝒖 𝒎𝒇)
𝟎.𝟒
--------- (20) for all beds.
13

14. O While the initial bubble diameter depends upon the type of
distributor plate. For porous plates, the relationship
𝒅 𝒃𝒐 = 𝟎. 𝟎𝟎𝟑𝟕𝟔 𝒖 𝟎 − 𝒖 𝒎𝒇
𝟐
, 𝒄𝒎 -------- (21)
is observed, and for the perforated plates, the relationship
𝒅 𝒃𝒐 = 𝟎. 𝟑𝟒𝟕 𝑨 𝒄(𝒖 𝟎 − 𝒖 𝒎𝒇)/𝒏 𝒅
𝟎.𝟒
-------- (22)
is observed.
O Werther developed the following correlation based on a
statistical coalescence model:
𝒅 𝒃
𝒄𝒎
= 𝟎. 𝟖𝟓𝟑
𝟑
𝟏 + 𝟎. 𝟐𝟕𝟐
𝒖 𝟎 − 𝒖 𝒎𝒔
𝒄𝒎/𝒔
𝟏 − 𝟎. 𝟎𝟔𝟖𝟒
𝒉
𝒄𝒎
𝟏.𝟐𝟏
-------- (23)
14

15. Fraction of Bed in the Bubble Phase
15
O 𝛿 = fraction of total bed
occupied by the part of the
bubbles that does not include
the wake.
O 𝛼 = volume of wake per
volume of bubble.
O 𝛼𝛿 = bed fraction in the
wakes.
O (1 - 𝛼 - 𝛼𝛿) = bed fraction in
the emulsion phase (which
includes the clouds).

16. 16
O Letting 𝐴 𝑐 and 𝜌𝑐 represent the cross-sectional area of the bed and the
density of the solid particles, respectively, a material balance on the solids
gives
O A material balance on the gas flows gives
O The velocity of gas rise in the emulsion phase is
𝒖 𝒆 =
𝒖 𝒎𝒇
𝜺 𝒎𝒇
− 𝒖 𝒔
Solids flowing
downward in emulsion
= Solids flowing
upward in wakes
𝑨 𝒄 𝝆 𝒄(1 - 𝜶 - 𝜶𝜹)𝒖 𝒔 = 𝑨 𝒄 𝝆 𝒄 𝜶𝜹𝒖 𝒃
or, 𝒖 𝒔 =
𝜶𝜹𝒖 𝒃
(1 − 𝜶 − 𝜶𝜹)
-------- (24)
𝑨 𝒄 𝒖 𝟎 = 𝑨 𝒄 𝜹𝒖 𝒃 + 𝑨 𝒄 𝜺 𝒎𝒇 𝜶𝜹𝒖 𝒃 + 𝑨 𝒄 𝜺 𝒎𝒇(1
− 𝜶 − 𝜶𝜹)𝒖 𝒆
Total
gas
flow rate
= Gas flow
in
bubbles
+ Gas flow
in wakes
+ Gas flow in
emulsion

17. 17
O By combining the equations mentioned in the earlier slide, we
obtain an expression for the fraction 𝛿 of the bed occupied by the
bubbles
𝜹 =
𝒖 𝟎 − 𝒖 𝒎𝒇
𝒖 𝒃 − 𝒖 𝒎𝒇(𝟏+ 𝜶)
--------- (25)
O The wake parameter, α, is a function of the particle size. The value
of 𝛼 has been observed experimentally to vary between 0.25 and
1.0, with typical values close to 0.4. Kunii and Levenspiel assume
that the last equation can be simplified to
𝜹 =
𝒖 𝟎 − 𝒖 𝒎𝒇
𝒖 𝒃
--------- (26)
which is valid for 𝒖 𝒃 >> 𝒖 𝒎𝒇, e.g. 𝒖 𝒃 ≈
𝟓𝒖 𝒎𝒇
𝜺 𝒎𝒇

18. Mass Transfer In Fluidized Bed
18
O There are two types of mass transport important in fluidized-bed
operations.
O Transport between gas and solid.
O Transfer of materials between the bubbles and the clouds, and between the
clouds and the emulsion.
Figure : Transfer between bubble, cloud, and emulsion.

19. Gas – Solid Mass Transfer
19
O In the bubble phase of a fluidized bed, the solid particles are sufficiently
separated so that in effect there is mass transfer between a gas and
single particles. The most widely used correlation for this purpose is the
1938 equation of Fröessling (1938) for mass transfer to single spheres
given by
𝑺𝒉 = 𝟐. 𝟎 + (𝟎. 𝟔) 𝑹𝒆 𝟏 𝟐
𝑺𝒄 𝟏 𝟑
----------- (27)
O In the emulsion phase, the equation would be one that applied to fixed-
bed operation with a porosity in the bed equal to 𝜺 𝒎𝒇 and a velocity of
𝒖 𝒎𝒇. The equation recommended by Kunii and Levenspiel :
𝑺𝒉 = 𝟐. 𝟎 + (𝟏. 𝟓) 𝑺𝒄 𝟏 𝟑
(𝟏 − 𝜺) 𝑹𝒆 𝟏 𝟐
----------- (28)
For 5 < Re < 120, and 𝜺 < 0.84
O Mass transfer coefficients obtained from these relationships may then be
combined with mass transfer among the various phases in the fluidized
bed to yield the overall behavior with regard to the transport of mass.

20. Mass Transfer between the Fluidized-Bed Phases
20
O For the gas interchange between
the bubble and the cloud, Kunii
and Levenspiel defined the mass
transfer coefficient 𝐾𝑏𝑐 (𝑠𝑒𝑐)−1
in
the following manner :
𝑾 𝑨𝒃𝒄 = 𝑲 𝒃𝒄 ( 𝑪 𝑨𝒃 − 𝑪 𝑨𝒄) ------ (29)
O For the products, the rate of
transfer into the bubble from the
cloud is given by a similar
equation :
𝑾 𝑩𝒄𝒃 = 𝑲 𝒄𝒃 ( 𝑪 𝑩𝒄 − 𝑪 𝑩𝒃) ------ (30)
𝑾 𝑨𝒃𝒄 represents the number of
moles of A transferred from the
bubble to the cloud &
𝑾 𝑩𝒄𝒃 represents the number of
moles of B transferred from the
cloud to the bubble per unit time per
unit volume of bubble.
Figure : Sketch of flow pattern in a
fluidized bed for down flow of emulsion
gas, 𝑢 𝑒 𝑢0 < 0 or 𝑢0 𝑢 𝑚𝑓 > 6 𝑡𝑜 11.

21. 21
O The mass transfer coefficient 𝑲 𝒃𝒄 can also be thought of as an exchange volume
q between the bubble and the cloud.
𝑾 𝑩𝒄𝒃 = 𝒒 𝒃 𝑪 𝑨𝒃 − 𝒒 𝒄 𝑪 𝑨𝒄 = 𝒒 𝟎 ( 𝑪 𝑨𝒃 − 𝑪 𝑨𝒄) ------ (31)
Where,
𝒒 𝒃 = Volume of gas flowing from the bubble to the cloud per unit time per unit
volume of bubble.
𝒒 𝒄 = Volume of gas flowing from the cloud to the bubble per unit time per unit
volume of bubble.
𝒒 𝟎 = Exchange volume between the bubble and cloud per unit time per unit volume
of bubble (i.e., 𝑲 𝒃𝒄 )
O Using Davidson’s expression for gas transfer between the bubble and the cloud,
and then basing it on the volume of the bubble, Kunii and Levenspiel obtained
this equation for evaluating 𝑲 𝒃𝒄 :
𝑲 𝒃𝒄 = 𝟒. 𝟓
𝒖 𝒎𝒇
𝒅 𝒃
+ 𝟓. 𝟖𝟓
𝑫 𝑨𝑩
𝟏 𝟐
𝒈 𝟏 𝟒
𝒅 𝒃
𝟓 𝟒 ----------- (32)
O Note, 𝑲 𝒃𝒄 = 𝑲 𝒄𝒆
O Similarly,
𝑾 𝑨𝒄𝒆 = 𝑲 𝒄𝒆 ( 𝑪 𝑨𝒄 − 𝑪 𝑨𝒆) −−−−−− (33)
𝑾 𝑩𝒄𝒆 = 𝑲 𝒄𝒆 ( 𝑪 𝑩𝒆 − 𝑪 𝑩𝒄) ------ (34)
O Using Higbie’s penetration theory and his analogy for mass transfer from a
bubble to a liquid, Kunii and Levenspiel developed an equation for evaluating
𝑲 𝒄𝒆:
𝑲 𝒄𝒆 = 𝟔. 𝟕𝟕
𝜺 𝒎𝒇 𝑫 𝑨𝑩 𝒖 𝒃
𝒅 𝒃
𝟑
𝟏 𝟐
---------- (35)

22. Reaction Behaviour in a Fluidized Bed
22
O To use the Kunii-Levenspiel model to predict reaction rates in a
fluidized-bed reactor, the reaction rate law for the heterogeneous
reaction per gram (or other fixed unit) of solid must be known. Then the
reaction rate in the bubble phase, the cloud, and the emulsion phase,
all per unit of bubble volume, can be calculated. Assuming that these
reaction rates are known, the overall reaction rate can be evaluated
using the mass transfer relationships presented in the preceding
section. All this is accomplished in the following fashion.
O We consider an nth order, constant-volume catalytic reaction.
O In the bubble phase, 𝒓 𝑨𝒃 = −𝒌 𝒃 𝑪 𝑨𝒃
𝒏
; in which the reaction rate is
defined per unit volume of bubble.
O In the cloud, 𝒓 𝑨𝒄 = −𝒌 𝒄 𝑪 𝑨𝒄
𝒏
O In the emulsion, 𝒓 𝑨𝒆 = −𝒌 𝒆 𝑪 𝑨𝒆
𝒏
Where 𝒌 𝒃, 𝒌 𝒆 and 𝒌 𝒄 are the specific reaction rates in the bubble, cloud
and emulsion respectively.

23. Mole Balance on the bubble, the Cloud, and the Emulsion
23
O Material balance will be written over an incremental height ∆𝑧 for
substance A in each of the three phases (bubble, cloud, and
emulsion)
Figure : Section
of a bubbling
fluidized bed

24. Balance on Bubble Phase
The amount of A entering at z in the bubble phase by flow,
A similar expression can be written for the amount of A leaving in
the bubble phase n flow at z + Δz.
𝒖 𝒃 𝑨 𝒄 𝑪 𝑨𝒃 𝜹 𝒛 − 𝒖 𝒃 𝑨 𝒄 𝑪 𝑨𝒃 𝜹 𝒛+ ∆𝒛 − 𝑲 𝒃𝒄 𝑪 𝑨𝒃 − 𝑪 𝑨𝒄 𝑨 𝒄∆𝒛𝜹 + 𝒌 𝒃 𝑪 𝑨𝒃
𝒏
𝑨 𝒄∆𝒛𝜹 = 𝟎
Dividing by 𝑨 𝒄∆𝒛𝜹 and taking limit as ∆𝒛 → 0.
A balance in bubble phase for steady state operation in section ∆𝒛,
𝒖 𝒃
𝒅𝑪 𝑨𝒃
𝒅𝒛
= − 𝒌 𝒃 𝑪 𝑨𝒃
𝒏
− 𝑲 𝒃𝒄(𝑪 𝑨𝒃 − 𝑪 𝑨𝒄) ----------- (36)
24
𝒖 𝒃 𝑨 𝒄 𝑪 𝑨𝒃 𝜹 = Molar flow rate of A
assuming the entire
bed is filled with
bubbles
Fraction of the
bed occupied
by bubbles
In by flow _ Out by
flow
_ Out by mass
transport
+ Generation = 𝐀𝐜𝐜𝐮 𝐧

25. Balance on Cloud Phase
O Similarly for Cloud phase,
𝒖 𝒃 𝜹
𝟑 𝒖 𝒎𝒇 𝜺 𝒎𝒇
𝒖 𝒃𝒓 − 𝒖 𝒎𝒇 𝜺 𝒎𝒇
+ 𝜶
𝒅𝑪 𝑨𝒄
𝒅𝒛
= 𝑲 𝒃𝒄 𝑪 𝑨𝒃 − 𝑪 𝑨𝒄 − 𝑲 𝒄𝒆 𝑪 𝑨𝒄 − 𝑪 𝑨𝒆 − 𝒌 𝒄 𝑪 𝑨
𝒏
---------------
(37)
Balance on the Emulsion
O Similarly for the emulsion phase,
𝒖 𝒆
𝟏 − 𝜹 − 𝜶𝜹
𝜹
𝒅𝑪 𝑨𝒆
𝒅𝒛
= 𝑲 𝒄𝒆 𝑪 𝑨𝒄 − 𝑪 𝑨𝒆 − 𝒌 𝒆 𝑪 𝑨𝒆
𝒏
----------- (38)
O The three material balances thus result in three coupled ordinary
differential equations, with one independent variable (z) and three
dependent variables (𝐶𝐴𝑏, 𝐶𝐴𝑐, 𝐶𝐴𝑒). These equations can be
solved numerically.
25

26. O The Kunii-Levenspiel model simplifies these still further, by assuming
that the derivative terms on the left-hand side of the material balances
on the cloud and emulsion are negligible in comparison with the terms
on the right-hand side. Using this assumption, and letting 𝑡 = 𝑧 𝑢 𝑏 (i.e.,
the time the bubble has spent in the bed), the three equations take the
form :
𝒅𝑪 𝑨𝒃
𝒅𝒛
= − 𝒌 𝒃 𝑪 𝑨𝒃
𝒏
− 𝑲 𝒃𝒄(𝑪 𝑨𝒃 − 𝑪 𝑨𝒄) ----------- (39)
𝑲 𝒃𝒄 𝑪 𝑨𝒃 − 𝑪 𝑨𝒄 = 𝑲 𝒄𝒆 𝑪 𝑨𝒄 − 𝑪 𝑨𝒆 + 𝒌 𝒄 𝑪 𝑨𝒄
𝒏
-------------- (40)
𝑲 𝒄𝒆 𝑪 𝑨𝒄 − 𝑪 𝑨𝒆 = 𝒌 𝒆 𝑪 𝑨𝒆
𝒏
--------------- (41)
26

27. Partitioning of the Catalyst
O To solve these equations, it is necessary to have values of 𝑘 𝑏, 𝑘 𝑐, and
𝑘 𝑒. Three new parameters are defined:
O 𝛾 𝑏 =
Volume of solid catalyst dispersed in bubbles
Volume of bubbles
O 𝛾𝑐 =
Volume of solid catalystin clouds and wakes
Volume of bubbles
O 𝛾𝑒 =
Volume of solid catalystin emulsion phase
Volume of bubbles
O First of all the specific reaction rate of solid catalyst, 𝑘 𝑐𝑎𝑡 must be known.
It is normally determined from laboratory experiments. The term 𝒌 𝒄𝒂𝒕 𝑪 𝑨
𝒏
is the g-moles reacted per volume of solid catalyst. Then
𝒌 𝒃 = 𝜸 𝒃 𝒌 𝒄𝒂𝒕 ; 𝒌 𝒄 = 𝜸 𝒄 𝒌 𝒄𝒂𝒕; 𝒌 𝒆 = 𝜸 𝒆 𝒌 𝒄𝒂𝒕 ------------- (42)
O The volume fraction of catalyst in the clouds and wakes is (𝟏 − 𝜺 𝒎𝒇).
The volume of cloud and wakes per volume of bubble is
𝑉𝑐
𝑉𝑏
=
𝟑 𝒖 𝒎𝒇 𝜺 𝒎𝒇
𝒖 𝒃 − 𝒖 𝒎𝒇 𝜺 𝒎𝒇
27

28. O So the expression for 𝜸 𝒄 = (𝟏 − 𝜺 𝒎𝒇)
𝟑 𝒖 𝒎𝒇 𝜺 𝒎𝒇
𝒖 𝒃 − 𝒖 𝒎𝒇 𝜺 𝒎𝒇
+ 𝜶 ------ (43)
O It turns out that the value of 𝜶 is normally far from insignificant in this
expression for 𝜸 𝒄 and represents a weakness in the model because there
does not yet exist a reliable method for determining 𝜶.
O The volume fraction of the solids in the emulsion phase is again (𝟏 −
𝜺 𝒎𝒇). The volume of emulsion per volume of bubble is
𝑽 𝒆
𝑽 𝒃
=
𝟏 − 𝜹
𝜹
−
Volume of clouds and wakes
Volume of bubbles
And the expression for 𝜸 𝒆 is : 𝜸 𝒆 = 𝟏 − 𝜺 𝒎𝒇
𝟏 − 𝜹
𝜹
− 𝜸 𝒄 - 𝜸 𝒃 ---------- (44)
O Using all the above equation, the three balance equations become
𝒅𝑪 𝑨𝒃
𝒅𝒛
= −(𝜸 𝒃 𝒌 𝒄𝒂𝒕 𝑪 𝑨𝒃
𝒏
) − 𝑲 𝒃𝒄(𝑪 𝑨𝒃 − 𝑪 𝑨𝒄) ----------- (45)
𝑲 𝒃𝒄 𝑪 𝑨𝒃 − 𝑪 𝑨𝒄 = 𝑲 𝒄 𝑪 𝑨𝒄 − 𝑪 𝑨𝒆 + 𝜸 𝒄 𝒌 𝒄𝒂𝒕 𝑪 𝑨𝒄
𝒏
-------------- (46)
𝑲 𝒄𝒆 𝑪 𝑨𝒄 − 𝑪 𝑨𝒆 = 𝜸 𝒆 𝒌 𝒄𝒂𝒕 𝑪 𝑨𝒆
𝒏
--------------- (47)
NOTE : For reactors other than first order and zero order, these
equations must be solved numerically.
28

29. Solution to the Equations for a First-Order Reaction
O If the reaction is first order, then the 𝑪 𝑨𝒄 and 𝑪 𝑨𝒃𝒆 can be eliminated
using the two algebraic equations, and the differential equation can be
solved analytically for 𝑪 𝑨𝒃 as a function of t. An analogous situation
would exist if the reaction were zero. Except for these two situations,
solution to these two equations must be obtained numerically.
O To arrive at our fluidized-bed design equation for a first order reaction,
we simply express both the concentration of A in the emulsion, 𝑪 𝑨𝒆, and
the cloud, 𝑪 𝑨𝒄, in terms if the bubble concentration, 𝑪 𝑨𝒃. First, we use
the emulsion balance
𝑲 𝒄𝒆 𝑪 𝑨𝒄 − 𝑪 𝑨𝒆 = 𝜸 𝒆 𝒌 𝒄𝒂𝒕 𝑪 𝑨𝒆
𝒏
---------- (48)
to solve for 𝑪 𝑨𝒆 in terms of 𝑪 𝑨𝒄.
Rearranging equation (48), for a first-order reaction (n = 1), we obtain
𝑪 𝑨𝒆 =
𝑲 𝒄𝒆
𝜸 𝒆 𝒌 𝒄𝒂𝒕+ 𝑲 𝒄𝒆
𝑪 𝑨𝒄----------- (49)
We now use this equation to substitute for 𝐶𝐴𝑒 in the cloud balance
𝑲 𝒃𝒄 𝑪 𝑨𝒃 − 𝑪 𝑨𝒄 = 𝑪 𝑨𝒄 𝒌 𝒄𝒂𝒕 𝜸 𝒄 + 𝑲 𝒄𝒆 𝑪 𝑨𝒄 −
𝑲 𝒄𝒆 𝑪 𝑨𝒄
𝜸 𝒆 𝒌 𝒄𝒂𝒕+ 𝑲 𝒄𝒆
29

30. O Solving for 𝑪 𝑨𝒄 in terms of 𝑪 𝑨𝒃
𝑪 𝑨𝒄 =
𝑲 𝒃𝒄
𝜸 𝒄 𝒌 𝒄𝒂𝒕+
𝑲 𝒄𝒆 𝜸 𝒆 𝒌 𝒄𝒂𝒕
𝜸 𝒄 𝒌 𝒄𝒂𝒕+𝑲 𝒄𝒆
+𝑲 𝒃𝒄
𝑪 𝑨𝒃 ----------- (50)
We now substitute for 𝑪 𝑨𝒄 in the bubble balance
𝒅𝑪 𝑨𝒃
𝒅𝒛
= 𝜸 𝒃 𝒌 𝒄𝒂𝒕 𝑪 𝑨𝒃 + 𝑪 𝑨𝒃 −
𝑲 𝒃𝒄 𝑪 𝑨𝒃
𝜸 𝒄 𝒌 𝒄𝒂𝒕+
𝑲 𝒄𝒆 𝜸 𝒆 𝒌 𝒄𝒂𝒕
𝜸 𝒄 𝒌 𝒄𝒂𝒕+𝑲 𝒄𝒆
+𝑲 𝒃𝒄
------- (51)
Rearranging,
𝒅𝑪 𝑨𝒃
𝒅𝒛
= 𝒌 𝒄𝒂𝒕 𝑪 𝑨𝒃 𝜸 𝒃 +
𝜸 𝒆 𝜸 𝒄 𝒌 𝒄𝒂𝒕 𝑲 𝒃𝒄 + 𝜸 𝒄 𝑲 𝒄𝒆 𝑲 𝒃𝒄 + 𝜸 𝒆 𝑲 𝒄𝒆 𝑲 𝒃𝒄
𝜸 𝒆 𝜸 𝒄 𝒌 𝒄𝒂𝒕
𝟐
+ 𝜸 𝒄 𝒌 𝒄𝒂𝒕 𝑲 𝒄𝒆 + 𝜸 𝒆 𝒌 𝒄𝒂𝒕 𝑲 𝒃𝒄 + 𝑲 𝒄𝒆 𝑲 𝒃𝒄 + 𝜸 𝒆 𝒌 𝒄𝒂𝒕 𝑲 𝒄𝒆
After some further arrangements,
−
𝒅𝑪 𝑨𝒃
𝒅𝒛
= 𝒌 𝒄𝒂𝒕 𝑪 𝑨𝒃 𝜸 𝒃 +
𝟏
𝒌 𝒄𝒂𝒕
𝑲 𝒃𝒄
+
𝟏
𝜸 𝒄+
𝟏
𝟏
𝜸 𝒆
+
𝒌 𝒄𝒂𝒕
𝑲 𝒄𝒆
----------- (52)
30

31. O The overall transport coefficient 𝐾 𝑅 for a first-order reaction.
𝑲 𝑹 = 𝜸 𝒃 +
𝟏
𝒌 𝒄𝒂𝒕
𝑲 𝒃𝒄
+
𝟏
𝜸 𝒄+
𝟏
𝟏
𝜸 𝒆
+
𝒌 𝒄𝒂𝒕
𝑲 𝒄𝒆
---------- (53)
−
𝒅𝑪 𝑨𝒃
𝒅𝒛
= 𝑲 𝑹 𝒌 𝒄𝒂𝒕 𝑪 𝑨𝒃 ------- (54)
Expressing 𝑪 𝑨𝒃 as a function of X, that is 𝑪 𝑨𝒃 = 𝑪 𝑨𝟎 𝟏 − 𝑿
We can substitute to obtain
𝒅𝑿
𝒅𝒕
= 𝑲 𝑹 𝒌 𝒄𝒂𝒕(𝟏 − 𝐗)
And integrating
𝒍𝒏
𝟏
𝟏 −𝑿
= 𝑲 𝑹 𝒌 𝒄𝒂𝒕 𝒕 -------------- (55)
31

32. Advantages & Disadvantages
32
ADVANTAGES
O Uniform Particle
Mixing
O Uniform Temperature
Gradients
O Ability to Operate
Reactor in Continuous
State
DISADVANTAGES
O Increased Reactor
Vessel Size
O Pumping
Requirements and
Pressure Drop
O Particle Entrainment
O Erosion of Internal
Components
O Pressure Loss
Scenarios
O Lack of Current
Understanding

33. Current Application of FBR
33
PETROLEUM SECTOR
O Gasolines
O Aviation Fuel
O Diesel Feedstocks
O Jet Fuel Feedstocks
O Propane
O Butane
O Propylene ; For
Liquified Petroleum
Gas (LPG) and
Butanes
O Butylene ; For
Liquified Petroleum
Gas (LPG) and
Butanes
O Isobutane
O Cracked Naptha
O Gasoline from
MethanolFuel
O Oils from
Polyethylene
PETROCHEMICAL
SECTOR
O Acetone Recovery
O Aniline
O Aniline from
Nitrobenzene
O Ethanol from
Butadiene
O Polyethylene
O Hydrogen from Steam
O Coal Gasification
O Styrenes from
Hydrocarbons
O Cracking of
Methylcyclohexane
O Maleic Anhydride
O Maleic Anhydride from
Benzene and
Butylenes
O Vinyl Chloride
O Vulcanization of
Rubber
OTHERS
 Fertilizers from Coal
 Oil Decontamination
of Sand
 Industrial and
Municipal Waste
Treatment
 Radioactive Waste
Solidification

34. References
1. nptel.ac.in/courses/103103026/module2/lec18/1.html
2. D. Kunii and O. Levenspiel, Fluidization Engineering (New York:
Wiley, 1968).
3. H. S. Fogler and L. F. Brown [Reactors, ACS Symposium
Series, vol.168, p. 31 1981, H. S. Fogler ed.]
4. T.E. Broadhurst and H.A. Becker, AIChE J., 21, 238 (1975).
5. J. F. Davidson and D. Harrison, Fluidized Particles (New York:
Cambridge University Press, 1963).
6. S. Mori and C. Y. Wen, AIChE J., 21, 109 (1975).
7. J. Werther, ACS Symposium Series., 72, D. Luss & V. W.
Weekman, eds. (1978).
8. https://en.wikipedia.org/wiki/Fluidized_bed_reactor
9. http://faculty.washington.edu/finlayso/Fluidized_Bed/FBR_Intro/
uses_scroll.htm
34

35. THE END!
THANK YOU VERY MUCH FOR YOUR
ATTENTION!
ANY QUERIES???
35