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Chemical Engineering

Department

CDB 2043 KINETICS AND REACTOR DESIGN

CHAPTER 8. STEADY STATE NONISOTHERMAL REACTOR

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At the end of this chapter, student should be able to:

1. Explain the importance of energy balance in designing

reactors

2. Apply the concept of energy balance to reactor design

for both adiabatic reaction and reaction involving heat

exchange

3. Evaluate the optimum conversion in reactors where

chemical equilibrium is the limitations.

2

LEARNING OUTCOME OF CHAPTER 8

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LECTURE 1WHY WE NEED THE ENERGY BALANCE EQUATION?

HOW TO DEVELOP THE ENERGY BALANCE EQUATION

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Why we need energy balance equation?

What do we have so far?

Mole balance

(design) equation

0dV

dX

PFor

:

A

A

F

r

FRExample

Rate law &

Arrhenius

equation

21

1

11exp

TTR

Ekk

kCrAA

Stoichiometry

)1(0

00

0

0

XCC

vCF

vCFvv

AA

AA

AA

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Why energy balance?

Mole balance(design) equation

0dV

dX

PFor

:

A

A

F

r

FR

Example

Rate law &

Arrhenius

equation

TTR

Ekk

kCrAA

11exp1

1

Stoichiometry

)1(0

00

0

0

XCCvCF

vCF

vv

AA

AA

AA

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Why energy balance?

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Combining these three equations gives:

Why energy balance?

XCF

TTR

Ek

dV

dXA

A

1

11exp

0

0

1

1

For non-isothermal operation: k depends on T (affects final

conversion, X)

T varies with V (reactor length)

Need either

X=f(T) or

T=f(V)

0dV

dX

A

A

F

r

AA C

TTR

Ekr

11exp

1

1

)1(0 XCC AA

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Set valuefor X

CalculateT

Calculatek

CalculaterA

Calculate(FA0/-rA)

8

General approach in solving non-isothermal

reactor problem

Use energy

balance to

relate X

and T

Use Levenspiel

plot techniqueto solve

problem

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The energy balance

General energy balance for open system

outou tinin

..

^

sysEFEFWQ

dt

Ed

Zero for closed system

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The energy balance

For n species, the equation becomes:

outi

n

1i

iini

n

1i

i

..

^

sysEFEFWQ

dt

Ed

Qs is how can we

make this equationUSEABLE @ USER

FRIENDLY?

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Steps to a more user-friendly equation

Evaluate the work term to get the equation in terms of enthalpies

ou ti

n

1i

iini

n

1i

i

..

^

sysEFEFWQ

dt

Ed

isys

Hfdt

Ed

^

Dissect the steady state molar flow rate terms to get the equation in terms of heats of

reaction

Rxsys

Hfdt

Ed

^

Dissect the enthalpies terms to get the equation in terms of heat capacity.

pisys

Cfdt

Ed

^

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Resulting equation based on energy balance* for steady stateflow system:

where:

The energy balance

*Full derivation: pg 474 - 486, Fogler

0)( 01

0

..

THXFTTCFWQ RXAA

n

i

iopiiAs

RpRRXRX TTCTHTH 0

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How do we apply the equation in designing

reactors?

0)( 01

0

..

THXFTTCFWQ RXAA

n

i

iopiiAs

dtr

dXN

V

r

XFV

F

r

A

A

A

A

A

A

0

0

0dV

dX

Solve using

Numerical technique

ODE solver

Graphical method

e.g: Levenspiel plot

method

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General Energy Balance Equation

Consider an adiabatic system with no work done.

)1(

)(

RpR

o

Rx

iopii

TTCTH

TTCX

0W,0Q s..

0)( 01

0

..

THXFTTCFWQ RXAA

n

i

iopiiAs

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Rearrangement of equation 1 gives:

Couple with mole balance equation:

we can develop the temperature, conversion and

concentration profiles along the reactor.

)2(

0

PPii

RPPiiRRX

CXC

TCXTCTHX

T

),(0 TXrdV

dXF

AA

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Since the second term in the denominator is usually zero, aplot of XEB vs T gives:

X

T

Use together with mole balanceequation in solving reactionengineering problem

XEB XMB

)( RoRxiopii

EBTH

TTCX

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CHECKPOINT!

For non-isothermal system, we need 2

equations to describe the conversion, X

XMB, which is based on Material

Balance

XEB, which is based on Energy

Balance

The final conversion is the value that satisfy

both these equations

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Example 8-2: Heat of reaction

Calculate the heat of reaction for the synthesis of ammonia from hydrogen

and nitrogen at 150oC in kcal/mol of N2 reacted and also in H2 reacted. The

reaction is given below:

N2 + 3H2 2NH3

Info:Heat of formation of NH3 at 25

oC = -11,020 cal/mol.N2 reacted

Heat of formation of elements at 25oC = 0

Heat capacities at 25oC

CP,NH3 = 8.92 cal/mol.K

CP,N2 = 6.984 cal/mol.K

CP,H2 = 6.992 cal/mol.K

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Heat of formation at reference temperature (25o

C):

Heat of reaction at any temperature:

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Solution to Example 8.2

)()()()()( 00000RARBRcRDRRx

THTHa

bTH

a

cTH

a

dTH

RPRRxRx

TTCTHTH )()( 0

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Using these equations, we will get:

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Solution to Example 8.2

reactedNmol

kcalKH

Rx

2

004.22298

KreactedNmol

calCp

2

12.10

2

3.23Nmol

kcalH

Rx

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Example 8-3: Liquid phase isomerisation of normal

butane

HRx = -6900 J/mol butane

Activation Energy = 65.7 kJ/molKc = 3.03 at 60

oC

CA0 = 9.3 kmol/dm3

CPn-B = 141 J/mol.KCPi-B = 141 J/mol.K

CPi-P = 161 J/mol.K

Handout

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END OF LECTURE

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LECTURE 2STEADY STATE TUBULAR REACTOR WITH HEAT EXCHANGE

CASE 1: CONSTANT TEMPERATURE HEAT EXCHANGE

CASE 2: VARIABLE TEMPERATURE HEAT EXCHANGE

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Steady state tubular reactor with heat

exchange (Case 1)

d b l h h

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From:

Deriving the energy balance equation General energy balance equation, with no work done:

Steady state tubular reactor with heat

exchange (Case 1)

TTVUaQ

where

HFHFQ

a

vviivii

.

.

:

0

0

..

vviiviiS HFHFWQ

S d b l h h

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Deriving the energy balance equation After further substitution of the terms, the energy balance

equation for tubular reactor with heat exchange is:

Steady state tubular reactor with heat

exchange (Case 1)

Pii

aRxA

CF

TTUaHr

dV

dT

Heat generated Heat removed

Full derivation: Fogler, pg 496

VALID FOR CONSTANT

TEMPERATURE HEAT EXCHANGE

MEDIUM ONLY!!

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From Example 8.3, the vapour pressure at the exit to theadiabatic reactor at 360K is 1.5MPa for isobutene, which is

greater than the rupture pressure of the glass vessel used. Given

the information below, will the reactor temperature rise above

325K?

Reactor volume (10 banks): 6 m3 (volume of each reactor)

Ua: 5000 kJ/h.m3.K

Ta = 37oC

All other information : refer to Example 8.3

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Example 8.4: Butane Isomerisation

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WHAT IF THETEMPERATURE OF THE

COOLING / HEATINGMEDIUM IS NOT A

CONSTANT?

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S d b l i h h

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Energy balance on coolant fluid

Steady state tubular reactor with heat

exchange (Case 2)

St d t t t b l t ith h t

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Energy balance on coolant fluid For co-current case:

For counter-current case:

Steady state tubular reactor with heat

exchange (Case 2)

Pcc

aa

Cm

TTUa

dV

dT

.

Pcc

aa

Cm

TTUa

dV

dT

.

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One of the steps in the production of acetic anhydride is the vapourphase cracking of acetone to ketene and methane. The reaction is first

order with respect to acetone and the reaction rate is given below (T in

K).

The feed to the reactor consist og 7850 kg/hr acetone. The reactor

consist of a bank of 1000 tubes of 1 inch diameter. The inlet

temperature and pressure are the same of both cases at 1035 K and 162

kPa. Plot the conversion and temperature profile along the length of thereactor for the following cases:

Example 8-5: Production of acetic anhydride

Tk

CHCOCHCOCHCH

222,3434.34ln

4233

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1. The reactor is operated adiabatically

2. The reactor is surrounded by a heat exchanger where the heat

transfer coefficient is 110 J/m2.s.K and the temperature of the

heating medium is constant at 1150 K.

3. Same as part (2) but reactor is subjected to heating medium with

changing temperature

Example 8-5: Production of acetic anhydride

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END OF LECTURE 2

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LECTURE 3EQUILIBRIUM CONVERSION

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Equilibrium conversion

Will consider the following:A. How adiabatic temperature is related to equilibrium

conversion for the following cases:

1. Exothermic reaction

2. Endothermic reaction

B. How to find optimum feed temperature

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Exothermic reaction

Equilibrium

TO TO1

TO1>TO

T

X

Xe

temperature

RX

opiiEB

H

TTCX

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How do we increase Xe for exothermic reaction?

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Endothermic reaction

X

T

Xe

temperature

TO

RX

opiiEB

H

TTCX

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How do we increase Xe for endothermic reaction?

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Consider reversible & exothermic reaction under adiabaticoperation

B. Optimum feed temperature

Analysis on Xe vs. T(from energybalance) gives thefollowing profile:

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Analysis on conversion vs. reactor length gives the following profile:

B. Optimum feed temperature

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B. Optimum feed temperature

Therefore, if we analyse the achievable conversion in a reactor for a

given entering temperature, we will find:

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END OF LECTURE

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STEP 1 Evaluate the work term to get the equation in terms

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STEP 1: Evaluate the work term to get the equation in terms

of enthalpies

STEP 2: Dissect the steady state molar flow rate terms to

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STEP 2: Dissect the steady state molar flow rate terms to

get the equation in terms of heats of reaction

STEP 2: Dissect the steady state molar flow rate terms to

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STEP 2: Dissect the steady state molar flow rate terms to

get the equation in terms of heats of reaction

STEP 3: Dissect the enthalpies terms to get the equation in

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STEP 3: Dissect the enthalpies terms to get the equation in

terms of heat capacity.

STEP 3: Dissect the enthalpies terms to get the equation in

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STEP 3: Dissect the enthalpies terms to get the equation in

terms of heat capacity.

STEP 3: Dissect the enthalpies terms to get the equation in

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STEP 3: Dissect the enthalpies terms to get the equation in

terms of heat capacity.

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