kinetics and reactor design chpe303
TRANSCRIPT
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Kinetics and Reactor Design
CHPE303
Dr. YASIR ALI
Department of Chemical & Petrochemical
Engineering
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Contact Details:
Office: 5D-44
Lectures: 3 hours per week
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Course Outcomes
1. define the meaning of chemical design
2. describe the meaning of rate of reaction and rate mathematical models
3. differentiate between different types of reactor system and their calculations
4. identify the effect of catalyst on rate of reaction and reactor design parameters
5. apply the fundamentals of reactor design on process selection based on selectivity and profitability
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Syllabus
• Introduction to chemical reaction engineering and mole balance (3 hrs)
• Expressing the design equation in terms of conversion, finding the size
of continuous reactor using Levinspiel diagram (6 hrs)
• Defining the meaning of reaction rate law, type of rate of reactions, and
relation to the stoichiometric equation (6 hrs)
• Design of isothermal reactors, study the effect of heat on the reaction
rate and reactor design (6 hrs)
• Calculating the reaction rate from experimental data. Find reaction
order and reaction constant from experimental data (6 hrs)
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Syllabus
• Design of reactors with multiple reactions. Finding the rate ofreaction for multi-component, study the selectivity of thereaction and their effect on the reactor design (6 hrs)
• Study the reaction mechanism, heterogeneous reactions andintroduction to the design of bioreactors (6 hrs)
• Study the catalysis, defining the catalyst, their role in thereaction (6 hrs)
Total: 45 hrs
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Assessments
• Final exam 40%
• Mid semester test 40%
• Quiz 10%
• Assignments 10%
--------------------------------------------------------------------------
Total 100%
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Objectives
After completing Chapter 1 of the text and associated the
reader will be able to:
• Define the rate of chemical reaction.
• Apply the mole balance equations to a batch reactor,
CFSTR, PFR, and PBR.
• Describe two industrial reaction engineering systems.
CFSTR = Continuous Flow Stirred Tank Reactor
PFR = Plug Flow Reactor
PBR = Packed Bed Reactor
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Chemical Identity
A chemical species is said to have reacted when it has lost its
chemical identity.
The identity of a chemical species is determined by the kind,
number, and configuration of that species’ atoms.
1. Decomposition CH3CH3 H2 + H2C=CH2
2. Combination N2 + O2 2NO
3. Isomerization C2H5CH=CH2CH2=C(CH3)2
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Rate of Reaction
When the chemical reaction take place?
Chemical reaction took place when a detectable number of
molecules of one or more species have lost their identity and
assumed a new form by a change in the kind or number of atoms in
the compound and/or by a change in structure or configuration of
these atoms. In this classical approach to chemical change, it is
assumed that the total mass is neither created nor destroyed when a
chemical reaction occurs.
Na OH + ClHNa OH ClH +
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Classificationof
Reactions
HomogenousHeterogeneous
ReversibleIrreversible
ElementaryNon-
elementary
ExothermicEndothermic
CatalyticNoncatalytic
Single Multiple
Isothermal Nonisothermal
Constant DensityVariable density
ChemicalBiochemical
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Type of Reactors
Homogeneous
Batch
Plug Flow
CSTR
Laminar flow
Recycle
Heterogeneous
Packed bed
Moving bed
Fluidized bed
Special
Slurry
Trickle bed
Bubble column
Ebullating Flow
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Reaction Rate
• The reaction rate is the rate at which a species looses its
chemical identity per unit volume.
• The rate of a reaction can be expressed as the rate of
disappearance of a reactant or as the rate of appearance of
a product.
Consider species A: A B
– rA = the rate of a disappearance of species A per unit volume
rB = the rate of formation of species B per unit volume
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• EXAMPLE: A B
o If B is being formed at 0.2 moles per decimeter cubed
per second, i.e,
o rB = 0.2 mole/dm3/s
Then A is disappearing at the same rate:
– rA= 0.2 mole/dm3/s
The rate of formation (generation of A) is
rA= – 0.2 mole/dm3/s
Reaction Rate
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Consider species j:
• rj is the rate of formation of species j per unit volume [e.g.
mol/dm3*s]
• rj is a function of concentration, temperature, pressure, and
the type of catalyst (if any)
• rj is independent of the type of reaction system (batch, plug
flow, etc.)
• rj is an algebraic equation, not a differential equation
• We use an algebraic equation to relate the rate of reaction,
-rj, to the concentration of reacting species and to the
temperature at which the reaction occurs
[e.g. – rj = k(T)Cj2].
Reaction Rate
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Rate of reaction :
Reaction Rate
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A product
Rate of reaction (– rA)
– rA = dCA
dt
This definition for a constant-volume batch reactor only.
aA + bB pP + qQ
Reaction Rate
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A B
Reactant A is consumed while the concentration of product B increases:
Rate of reaction = – d[A] = d[B]
dt dt
Reaction Rate
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Rate of Reaction
A + B C
reactants A and B are consumed while the concentration of product C increases.
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Rate of Reaction
A + 2B 3C
reactants A and B are consumed while the concentration of product C increases.
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Rate of Reaction
Example:
Calculate the reaction rate of each compound in the following reaction:
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Solution:
Rate of Reaction
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Using the data presented in the below table, calculate the reaction rate of each
compound in the following reaction:
Time (s) Fe (M) S (M) FeS (M)0 8 7 01 6.1 6.77 1.35
1.2 5.6 5.74 1.91.5 5.1 4.61 2.851.9 4.4 4.48 3.82.2 3.5 4.35 4.752.6 2.6 4.22 5.72.9 1.7 4.09 6.653.4 0.8 3.96 7.64 0.6 3.83 7.8
4.4 0.6 3.75 0.6 3.57 7.6
8 Fe + S8 8 FeS
CW
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Rate Laws and Stoichiometry
Basic Definitions
Homogenous Reaction Heterogeneous Reaction Irreversible Reaction
involves only
one phaseinvolves more
than one phase
is happening in
the direction of
products
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Reaction order
The Overall reaction order = n = α + β
For example, in the gas-phase reaction:
The reaction is second-order with respect to 2 for NO.
First-order with respect to 1O2.
Overall is a third-order reaction = 2 (NO) + 1 (O2) = 3.
αA + βB cC + dD
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The units of the specific reaction rate, kA, vary with the order of the reaction.
Consider a reaction involving only one reactant, such as:
A Product
Zero-Order
Reaction order
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The units of the specific reaction rate, kA, vary with the order of
the reaction.
First-order:
A B
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Second -order:
A + B C
ASSUME : CA = CB
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Rate of Reaction
Example:
For the given reaction below:
1. State the rate law.
2. State the overall order of the reaction.
3. Find the rate, given k = 1.14 x 10-2 dm3/mol.s and [H2O] = 2.04M
2H2O O2 + 2H2
Solution:
1. .
2. First - Order.
3. Rate = k [H2O] = 1.14 x 10-2 × 2.04 = 2.33 x 10-2 s-1
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Reversible Reactions
2B D + H2
k1
k2
A reversible reaction is a chemical change in which the products can be
converted back to the original reactants under suitable conditions.
Concentration equilibrium constant =
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Elementary reaction
AA kCrodAarUnimolecul .Pr
Order & Molecularity for Elementary reactions
2.Pr2 AA kCrodArBimolecula BAA CkCrodBArBimolecula .Pr
A unimolecular reaction occurs when a molecule rearrangesitself to produce one or more products.
A bimolecular reaction involves the collision of two particles.
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Elementary reaction
Order & Molecularity for Elementary reactions
3.Pr3 AA kCrodAarTrimolecul
BAA CkCrodBAarTrimolecul 2.Pr2
CBAA CCkCrodCBAarTrimolecul .Pr
A trimolecular or termolecular reaction requires the collision ofthree particles at the same place and time.
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Non-Elementary reaction
Non-Elementary reactions involve more than one step
Elementary reactions involve only one step
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CW
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General Mole Balance
V Volume Systemon Balance Mole General
Note: Component A can be either reactant or product
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Rate of flow of j into the system [moles/time]
–Rate of flow of j out of the system
[moles/time]+
Rate of generation of j by chemical reaction withinthe system [moles/time]
=Rate of
Accumulation of j within the system [moles/time]
In – Out + Generation = Accumulation
Fj0 – Fj + Gj = dNj/dt
Gj = rj∙V
GjFjo
Fj
System volume
where Nj represents the number of moles of species j in the system at a time t.
General Mole Balance
V Volume Systemon Balance Mole General
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GjFjo
Fj
System volume
V Volume Systemon Balance Mole General
ΔV1
ΔV2
ΔV3
By dividing up the system volume
Subvolume V1, is: ΔGj1 = rj1 ×ΔV1
Subvolume V2, is: ΔGj2 = rj2 ×ΔV2
Subvolume V3, is: ΔGj3 = rj3 ×ΔV3
Gj = rj ×V
The equation will be:
The General Mole
Balance Equation
The basic equation for chemical
reaction engineering
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General Mole Balance
dt
dN dV r - FF A
AAA0
onAccumulati Generation Out -In
V Volume Systemon Balance Mole General
Note: Component A can be either reactant or product
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Batch Reactor Mole Balance
• The batch reactor is assumed well stirred:
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Batch Reactor Mole Balance
Vrdt
dN
VrdVr
Assumption
F F
dt
dN dV r - FF
AA
AA
AA
AAAA
reactor mixed Well:
0 :outflowor inflow No
onAccumulati Generation Out -In
V Volume Systemon Balance Mole General
0
0
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Examples: Batch Reactor Times
A B
Calculate the time to reduce the number of moles by a factor of 10 (NA = NA0/10 ) in a batch reactor for the above reaction with
-rA = kCA, when k = 0.046 min-1
Example 1:
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Solution
Mole balance: In - Out + Generation = Accumulation
A0A
A0A
N 0.1N t;t
NN 0;t
:law rate
00
conditionsBoundary
kNdt
dN
kNVr
V
NkkCr
kCr
dt
dNVr
AA
AA
AAA
AA
AA
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minutes50)10ln(min046.0
1
10
ln1
0
0
0
00
tt
NN
N
N
kkN
dNt
kNVV
NkVkCVr
Vr
dN
Vr
dNt
AA
A
A
NA
NA A
A
AA
AA
NA
NA A
A
NA
NA A
A
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Example 2: Calculate the required time to reduce the number of moles
of A to 1 % of its initial value in a constant-volume batch reactor.
Take: k = 0.23 min–1.
Mole balance Constant-volume & batch reactor:
Rate law – first order:
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A + B C
The gas-phase reaction is carried out isothermally in a 20 dm3 constant-volume
batch reactor. 20 moles of pure A is initially placed in the reactor. The reactor is
well mixed:
(a) If the reaction is first order: –rA = kCA with k = 0.865 min-1, calculate the time
necessary to reduce the number of moles of A in the reactor to 0.2 mol. (Note:
NA = CAV).
Solution: (a)
Example 2:
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(b) If the reaction is second order. –rA = kCA2 with k = 2 dm3/mol.min. Calculate the time
necessary to consume 19 mol of A.
The gas-phase reaction is carried out isothermally in a 20 dm3 constant-volume
batch reactor. 20 moles of pure A is initially placed in the reactor. The reactor is
well mixed:
A + B C
Solution:
Example 2:
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(c) If the temperature is 127°C, determine the initial and the final pressure
assuming the reaction goes to completion.
The gas-phase reaction is carried out isothermally in a 20 dm3 constant-volume
batch reactor. 20 moles of pure A is initially placed in the reactor. The reactor is
well mixed:
A + B C
Solution:
Temperature = 127 °C = 400 K (isothermal), volume = 20 dm3, initial mole = 20 mol,
Final total mole = 20 + 20 = 40 mol
Example 2:
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Class work
(a) If the reaction is zero order: –rA = k with k = 0.005 mol/dm3.s, calculate the time (in
min) required to reduce the number of moles of A in the reactor to 12.5 mol.
(b) If the reaction is first order: –rA = kCA with k = 0.95 min-1. calculate the time
required to reduce the number of moles of A in the reactor to 0.25 mol.
(c) If the reaction is second order. –rA = kCA2 with k = 2 dm3/mol.min. Calculate the
time necessary to consume 19.0 mol of A.
(d) If the temperature is 127 °C, determine the initial and the final pressure assuming
the reaction goes to completion.
A + B C
The gas-phase reaction is carried out isothermally in a 1.8 m3 constant-volume
batch reactor. A 25 moles of pure A is initially placed in the reactor. The reactor
is well mixed:
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Advantages & Disadvantages of batch
1. High operating cost.
2. Product quality more variable than with continuous operation.
1. High conversion per unit volume for one pass
2. Flexibility of operation-same reactor can produce one product
one time and a different product the next
3. Easy to clean
Advantages Disadvantages
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Reactor types
2. Continuous Flow Reactors
Continuous Stirred Tank Reactor (CSTR)
Packed-Bed Reactor
Tubular Reactor
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Continuous Stirred Tank Reactor (CSTR)
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CSTR Mole Balance
A
AA
AAA
AA
AAAA
r
F F
Vr F F
VrdVr
sAssumption
dt
dN dV r - FF
0
0
A
0
V
0
thereforemixed, Well
0dt
dN thereforeState,Steady
:
onAccumulati Generation Out -In
V Volume Systemon Balance Mole General
remember
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• The irreversible liquid phase
second order reaction is carried
out in a CSTR.
• The entering concentration of A,
CA0, is 2 mol/dm3 and the exit
concentration of A, CA is 0.1
mol/dm3. The entering and
exiting volumetric flow rate, v0,
is constant at 3 dm3/s.
• What is the corresponding
reactor volume?
Example 3:
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Solution
33
2
3
3
3
3
0
3
3
000
2
0
2
0
547120
75
2030
306
30103
623
:
:
dm..
dm.
dm
mol
smol
dm.
s
mol).(
V
s
molA.
dm
molA.
s
dmCvF
s
molA
dm
molA
s
dmCvF
kC
FF VCombine
kCr Rate Law
r
FF Vce Mole Balan
AA
AA
A
AA
AA
A
AA
?×
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The entering concentration, CA0 = 2 mol/dm3 of A was
substituted into the rate law instead of the exit concentration
CA = 0.1 mol/dm3. Because the CSTR is perfectly mixed, the
concentration inside the CSTR where the reaction is taking
place is the same as that in the exit.
The reactor volume is fairly large at about 5 thousand
gallons. (3.785 dm3 = 1 US gallon)
34
2
3
3
109.1
1.003.0
)3.06(
dmV
dm
mol
smol
dm
s
mol
V
WHAT IS WRONG?
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Plug Flow Reactor
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Plug Flow Reactor Mole Balance
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APPLICATION: (CSTR & PFR)
The reaction (A B) is to be carried out isothermally in a continuous-flow reactor. Calculate both
the CSTR and PFR (plug flow reactor) volumes necessary to consume 99% of A (CA = 0.01CAₒ)
when the entering flow rate is 5 mol/h & the entering volumetric flow rate is 10 dm3/h, assuming the
reaction rate (– rA) is:
(a) – rA = k, with k = 0.05 mol/h.dm3 (b) – rA = kCA with k = 0.0001 s –1
(c) – rA = kC2A with k = 3 dm3/mol.h
FA = CAʋ . For a constant volumetric flow rate ʋ = ʋo, FA = CAʋo = 0.01CAoʋo = 0.01FAo
then, CAo = FAo /ʋo =(5 mol/h)/(10 dm3/h) = 0.5 mol/dm3
Volume of CSTR = 99 dm3
For CSTR:(a)
FAo = CAoʋo
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(b) – rA = kCA with k = 0.0001 s–1 (c) – rA = kC 2A with k = 3 dm3/mol.h
Volume of PFR = 99 dm3
(a) For PFR:
The reaction (A B) is to be carried out isothermally in a continuous-flow reactor. Calculate both the
CSTR and PFR (plug flow reactor) volumes necessary to consume 99% of A (CA = 0.01CAₒ) when the
entering flow rate is 5 mol/h & the entering volumetric flow rate is 10 dm3/h, assuming the reaction rate (– rA)
is: (a) – rA = k, with k = 0.05 mol/h.dm3
rA dV = FA – FAₒ = dFA
FA = CAʋ . For a constant volumetric flow rate ʋ = ʋo, FA = CAʋo = 0.01CAoʋo = 0.01FAo
then, CAo = FAo /ʋo = (5 mol/hr)/(10 dm3/hr) = 0.5 mol/dm3
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(a) – rA = k, with k=0.05mol/h.dm3 (b) – rA = kCA with k = 0.0001 s–1 (c) – rA = kC 2A with k = 3 dm3/mol.h
For CSTR:(b)
FA = CAʋ . For a constant volumetric flow rate ʋ = ʋo, FA = CAʋo = 0.01CAoʋo = 0.01FAo
then, CAo = FAo /ʋo =(5 mol/hr)/(10 dm3/hr) = 0.5 mol/dm3
CA = 0.01CAo
The reaction (A B) is to be carried out isothermally in a continuous-flow reactor. Calculate both
the CSTR and PFR (plug flow reactor) volumes necessary to consume 99% of A (CA = 0.01CAₒ)
when the entering flow rate is 5 mol/h & the entering volumetric flow rate is 10 dm3/h, assuming the
reaction rate (– rA) is:
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Volume of PFR = 128 dm3
(b) For PFR: rA dV = FA – FAₒ = CAʋo – CAoʋo = ʋo (CA – CAo) = ʋo dCA
FA = CAʋ . For a constant volumetric flow rate ʋ = ʋo, FA = CAʋo = 0.01CAoʋo = 0.01FAo
then, CAo = FAo /ʋo =(5 mol/hr)/(10 dm3/hr) = 0.5 mol/dm3
rA dV = ʋo dCA
(a) – rA = k, with k=0.05mol/h.dm3
The reaction (A B) is to be carried out isothermally in a continuous-flow reactor. Calculate both
the CSTR and PFR (plug flow reactor) volumes necessary to consume 99% of A (CA = 0.01CAₒ)
when the entering flow rate is 5 mol/h & the entering volumetric flow rate is 10 dm3/h, assuming the
reaction rate (– rA) is:
(b) – rA = kCA with k = 0.0001 s–1 (c) – rA = kC 2A with k = 3 dm3/mol.h
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For CSTR:(c)
Volume of PFR = 66,000 dm3
(a) – rA = k, with k=0.05mol/h.dm3
The reaction (A B) is to be carried out isothermally in a continuous-flow reactor. Calculate
both the CSTR and PFR (plug flow reactor) volumes necessary to consume 99% of A (CA = 0.01CAₒ)
when the entering flow rate is 5 mol/h & the entering volumetric flow rate is 10 dm3/h, assuming the
reaction rate (– rA) is:
(b) – rA = kCA with k = 0.0001 s –1 (c) – rA = kC 2A with k = 3 dm3/mol.h
(0.5)
FA = CAʋ . For a constant volumetric flow rate ʋ = ʋo, FA = CAʋo = 0.01CAoʋo = 0.01FAo
then, CAo = FAo /ʋo =(5 mol/hr)/(10 dm3/hr) = 0.5 mol/dm3
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Volume of PFR = 660 dm3
(c) For PFR:
(a) – rA = k, with k=0.05mol/h.dm3
The reaction (A B) is to be carried out isothermally in a continuous-flow reactor. Calculate
both the CSTR and PFR (plug flow reactor) volumes necessary to consume 99% of A (CA = 0.01CAₒ)
when the entering flow rate is 5 mol/h & the entering volumetric flow rate is 10 dm3/h, assuming the
reaction rate (– rA) is:
(b) – rA = kCA with k = 0.0001 s –1 (c) – rA = kC 2A with k = 3 dm3/mol.h
FA = CAʋ . For a constant volumetric flow rate ʋ = ʋo, FA = CAʋo = 0.01CAoʋo = 0.01FAo
then, CAo = FAo /ʋo =(5 mol/hr)/(10 dm3/hr) = 0.5 mol/dm3
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Packed Bed Flow Reactor Mole Balance
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Reactor types
Tubular Reactor
Also called Plug flow reactor, Tubular Reactor consists of a cylindrical pipe and is
normally operated at steady state, as is the CSTR.
For a tubular reactor operated at steady state, accumulation = 0.
In a spatially uniform sub-volume ∆V,
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Reactor types
For a tubular reactor operated at steady state, accumulation = 0.
In a spatially uniform sub-volume ∆V,
Then the equation will be,
resemble
Tubular Reactor
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Reactor types
The first-order reaction is carried out in a tubular reactor in which the volumetric flow
rate, ʋ, is constant. Derive an equation relating the reactor volume to the entering
and exiting concentrations of A, the rate constant k, and the volumetric flow rate ʋ.
Determine the reactor volume necessary to reduce the exiting concentration to 10%
of the entering concentration when the volumetric flow rate is 10 dm3/min and the
specific reaction rate, k, is 0.23 min –1.
A Bk
Tubular reactor mole balance (j=A):
A first-order reaction: – rA = kCA
Since the volumetric flow rate, ʋₒ is constant,
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Reactor Mole Balance Summary
Reactor Differential Algebraic Integral
Batch
CSTR
PFR
PBR
Vrdt
dNA
A
NA
NA A
A
Vr
dNt
0
A
AA
r
FFV
0
AA r
dV
dF
FA
FA A
A
r
dFV
0
AA r
dW
dF
FA
FA A
A
r
dFW
0
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Straight Though Transport Reactor
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Automotive Catalytic Converter
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Sasol Advanced Synthol (SAS) Reactor