Kinetics and Reactor Design CHPE303

Kinetics and Reactor Design

CHPE303

Dr. YASIR ALI

Department of Chemical & Petrochemical

Engineering

Contact Details:

Office: 5D-44

Lectures: 3 hours per week

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

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)

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

Assessments

• Final exam 40%

• Mid semester test 40%

• Quiz 10%

• Assignments 10%

--------------------------------------------------------------------------

Total 100%

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

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

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 +

Classificationof

Reactions

HomogenousHeterogeneous

ReversibleIrreversible

ElementaryNon-

elementary

ExothermicEndothermic

CatalyticNoncatalytic

Single Multiple

Isothermal Nonisothermal

Constant DensityVariable density

ChemicalBiochemical

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

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

• 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

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

Rate of reaction :

Reaction Rate

A product

Rate of reaction (– rA)

– rA = dCA

dt

This definition for a constant-volume batch reactor only.

aA + bB pP + qQ

Reaction Rate

A B

Reactant A is consumed while the concentration of product B increases:

Rate of reaction = – d[A] = d[B]

dt dt

Reaction Rate

Rate of Reaction

A + B C

reactants A and B are consumed while the concentration of product C increases.

Rate of Reaction

A + 2B 3C

reactants A and B are consumed while the concentration of product C increases.

Rate of Reaction

Example:

Calculate the reaction rate of each compound in the following reaction:

Solution:

Rate of Reaction

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

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

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

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

The units of the specific reaction rate, kA, vary with the order of

the reaction.

First-order:

A B

Second -order:

A + B C

ASSUME : CA = CB

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

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 =

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.

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.

Non-Elementary reaction

Non-Elementary reactions involve more than one step

Elementary reactions involve only one step

CW

General Mole Balance

V Volume Systemon Balance Mole General

Note: Component A can be either reactant or product

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.



GjFjo

Fj

System volume


ΔV1

ΔV2

ΔV3

By dividing up the system volume

Subvolume V1, is: ΔGj1 = rj1 ×ΔV1



Gj = rj ×V

The equation will be:

The General Mole

Balance Equation

The basic equation for chemical

reaction engineering


dt

dN dV r - FF A

AAA0

onAccumulati Generation Out -In


Note: Component A can be either reactant or product

Batch Reactor Mole Balance

• The batch reactor is assumed well stirred:

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



0

0

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:

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

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

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:

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:

(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.



well mixed:

A + B C

Solution:

Example 2:

(c) If the temperature is 127°C, determine the initial and the final pressure

assuming the reaction goes to completion.



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:

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:

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

Reactor types

2. Continuous Flow Reactors

Continuous Stirred Tank Reactor (CSTR)

Packed-Bed Reactor

Tubular Reactor

Continuous Stirred Tank Reactor (CSTR)

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

:



remember

• 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:

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

?×

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?

Plug Flow Reactor

Plug Flow Reactor Mole Balance

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

(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


then, CAo = FAo /ʋo = (5 mol/hr)/(10 dm3/hr) = 0.5 mol/dm3

(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)


then, CAo = FAo /ʋo =(5 mol/hr)/(10 dm3/hr) = 0.5 mol/dm3

CA = 0.01CAo






(b) For PFR: rA dV = FA – FAₒ = CAʋo – CAoʋo = ʋo (CA – CAo) = ʋo dCA



rA dV = ʋo dCA

(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:(c)

Volume of PFR = 66,000 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)



(b) – rA = kCA with k = 0.0001 s –1 (c) – rA = kC 2A with k = 3 dm3/mol.h

(0.5)




(c) 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)



(b) – rA = kCA with k = 0.0001 s –1 (c) – rA = kC 2A with k = 3 dm3/mol.h



Packed Bed Flow Reactor Mole Balance

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,

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

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,

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

Straight Though Transport Reactor

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Kinetics and Reactor Design CHPE303

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