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Chemical Reaction Engineering Lecture-4 Module 1: Mole Balances, Conversion & Reactor Sizing (Chapters 1 and 2, Fogler)

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Chemical Reaction Engineering

Lecture-4

Module 1: Mole Balances, Conversion & Reactor Sizing (Chapters 1 and 2, Fogler)

Topics to be covered in today’s lecture

• Conversion (X)

• GMBE in terms of conversion (X) for the following reactors – Batch Reactor

– Continuous Stirred Tank Reactor

– Plug Flow Reactor

• Compare Volume of CSTR and PFR

• Introduction to Levenspiel Plots

Chapter 2. Conversion and Reactor Sizing

Conversion (X)

• Conversion: quantification of how a reaction has progressed

fedAspeciesofMolesreactedAspeciesofMolesX A ""

""= (A: limiting reactant)

)1(0 XNN AA −⋅=

Batch Reactors

AO

AAO

NNNX −

=

Continuous (or Flow) Reactors

0

0

A

AA

FFFX −

=

)1(0 XFF AA −⋅=

can be omitted

• Maximum conversion for irreversible reactions: X = 1.0 • Maximum conversion for reversible reactions: X = Xe

? 0

A

AA

CCCXWhen −

=

aA + bB → cC + dD ; A + b/a B → cC + dD Limiting Reactant

Batch Reactor Design Equation

Vrdt

dNA

A ⋅=0

0

A

AA

NNNX −

= )1(000 XNXNNN AAAA −=−=

VrdtdXN AA ⋅−=0

For a constant-volume batch reactor VrdtdXN AA ⋅−=0 A

A rdt

dC−=

Therefore, a batch reactor has been widely used to investigate the rate law equation.

VrdtdXN AA ⋅−=0 ∫∫ =

⋅−=

X

AA

X

AA XCf

dXCVr

dXNt0 0

00

0 ),(

X

VrN

A

A

⋅−0

0 t

constant-volume reactor

NA = NA0 X

Design Equation in Differential Form

Design Equation in Integral Form

Design Equation for Flow Reactors

X = f (t) for Batch Reactor

X = f (V) for Flow Reactor FA = FA0 (1 – X) [moles/time]

FA = CA0 0υ

CA0 : morality for Liquid System CA0 = PA0/RT0 = yA0P0/RT0 for Gas System

),()()( 0

000

XCfXF

rXF

rXFV

A

A

exitA

A

A

A =−

=−

=

CSTR Design Equation

)(0

A

AA

rFFV

−−

=0

0

A

AA

FFFX −

= )1(000 XFXFFF AAAA −=−=

000 AA CF υ= υυ =0For incompressible fluid

)(0

A

A

rF−

X

CA0

CA

FA0

υ0 CA

FA

υ

X0

X

Area = Reactor volume

Design Equation for CSTR

∫∫ =−

=X

AA

X

AA XCf

dXFr

dXFV0 0

00

0 ),(AA r

dVdXF −=0

PFR Design Equation

AA r

dVdF

=0

0

A

AA

FFFX −

= )1(000 XFXFFF AAAA −=−=

000 AA CF υ= υυ =0For incompressible fluid

)(0

A

A

rF−

X

Area = Reactor volume

PFR

CA0

FA0

υ0

X0

CA

FA

υ

X

No radial gradients

Design Equation for PFR

∫ −=X

AA r

dXFW0

0 'AA r

dWdXF '0 −=

PBR Design Equation

000 AA CF υ= υυ =0For incompressible fluid

)'(0

A

A

rF−

X

Area = Catalysts weight

No radial gradients

Design Equation for PBR

Design Equation in terms of Conversion

REACTOR DIFFERENETIAL ALGEBRAIC INTEGRAL FORM FORM FORM

VrdtdXN AAO )(−= ∫ −=

X

AAO Vr

dXNt0

)( AAO rdVdXF −= ∫ −=

X

AAO r

dXFV0

CSTR

PFR

ExitA

AO

rXFV

)()(

−=

BATCH

Batch and Levenspiel Plots

][])(

[ 0 Xr

FVA

ACSTR ×

−=

Continuous Stirred Tank Reactor (CSTR)

)(0

A

A

rF−

X

∫=

= −=

xx

x A

APFR dX

rFV

0

0

X

)(0

A

A

rF−

Plug Flow Reactor (PFR)

Isothermal system

∫=

= ⋅−=

xx

x AABatch Vr

dXNt0

0 )(X

VrN

A

A

⋅− )(0

Batch Reactor

Class Problem # 1

The following reaction is to be carried out isothermally in a continuous flow reactor: A → B Compare the volumes of CSTR and PFR that are necessary to consume 90% of A (i.e. CA=0.1 CAO). The entering molar and volumetric flow rates are 5 mol/h and 10 L/h, respectively. The reaction rate for the reaction follows a first-order rate law:

(-rA) = kCA where, k=0.0001 s-1

[Assume the volumetric flow rate is constant.]

Solution to Class Problem #1

0

2

4

6

8

10

12

14

16

18

20

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Conversion (X)

F A0/(

-r A)

• Calculating Reaction rate in a CSTR Pure gases reactant A(CAo=100 millimol/liter) is fed at steady rate into a

mixed reactor (V=0.1 liter) where it dimerizes (2A→R). For different gas feed rates the following data are obtained

Find the expansion factor, conversion of each run

rate of equation for this reaction - rA = k CA

n

Run number 1 2 3 4

liter/hr 30.0 9.0 3.6 1.5

CA, out, millimil/liter 85.7 66.7 50 33.3 0υ

Class Problem # 2

Steady state Mixed Flow Reactor

Input =output + disappearance by reaction + accumulation

0

FA0= FA0(1-XA) + (-rA)V

Replacing in equation (1)

(1)

FA0XA= (-rA)V (2)

(3)

SOLUTION ( Ref. Chemical Rxn Eng, O. Levenspiel pp104-106 ) For this stoichiometry, 2A → R,

The expansion factor is

and the corresponding relation between concentration and conversion is

or

The conversion for each run is then calculated and tabulated in Column 4 of Table E1.

V=V0(1+εAXA) for linear expansion (4)

(5)

(6)

Given Calculated

Run CA,out XA

1 30.0 85.7 0.25

2 9.0 66.7 0.50 4500

3 3.6 50 0.667 2400

4 1.5 33.3 0.80 1200

TABLE E1

From the performance equation, Eq. (3), the rate of reaction for each run is given by

(7)

These values are tabulated in Column 5 of Table E1. Having paired values of rA and CA (see Table E1) we are ready to test various kinetic expressions.

Instead of separately testing for first order (plot rA versus CA), second order (plot rA versus CA

2), etc., let us test directly for nth-order kinetics. For this take logarithms of –rA = kCAn,

giving

This shows that nth-order kinetics will give a straight line on a log(-rA) versus logCA plot. As shown in Fig. E1, the four actual data are reasonably represented by a straight line of slope 2, so the rate equation for this dimerization is

(8)

(9)

FIGURE E1.