Lecture 21 - University of Michigan

Chemical Reaction Engineering (CRE) is the field that studies the rates and mechanisms of

chemical reactions and the design of the reactors in which they take place.

Lecture 21

Web Lecture 21

Class Lecture 17 – Tuesday

Gas Phase Reactions

Trends and Optimums

2

3

User Friendly Equations relate T, X, or Fi

Review Last Lecture

1. Adiabatic CSTR, PFR, Batch, PBR achieve this:

0ˆ PS CW

Rx

Pi

EBH

TTCX i

0

ˆ

Rx

P

H

TTCX A

0

~

iPi

Rx

C

XHTT 0

4


2. CSTR with heat exchanger, UA(Ta-T) and a

large coolant flow rate:

Rx

Pia

A

EBH

TTCTTF

UA

Xi

0

0

~

TTa

Cm

5


3. PFR/PBR with heat exchange:

FA0

T0

CoolantTa

3A. In terms of conversion, X

XCCF

THrTTUa

dW

dT

pPiA

RxAa

B

i

~0

6


3B. In terms of molar flow rates, Fi

i

ij

Pi

RxAa

B

CF

THrTTUa

dW

dT

4. For multiple reactions

i

ij

Pi

Rxija

B

CF

HrTTUa

dV

dT

5. Coolant Balance

cPc

aA

Cm

TTUa

dV

dT

7

Reversible Reactions

endothermic

reaction

exothermic

reaction

KP

T

endothermic

reaction

exothermic

reaction

Xe

T

Heat Exchange

8

Example: Elementary liquid phase reaction carried out in a PFR

FA0

FI

Ta

cmHeat Exchange

Fluid

BA

The feed consists of both inerts I and Species A

with the ratio of inerts to the species A being 2 to 1.

Heat Exchange

9

a) Adiabatic. Plot X, Xe, T and the rate of

disappearance as a function of V up to V = 40 dm3.

b) Constant Ta. Plot X, Xe, T, Ta and rate of

disappearance of A when there is a heat loss to the

coolant and the coolant temperature is constant at

300 K for V = 40 dm3. How do these curves differ

from the adiabatic case.

Heat Exchange

10

c) Variable Ta Co-Current. Plot X, Xe, T, Ta and

rate of disappearance of A when there is a heat

loss to the coolant and the coolant temperature

varies along the length of the reactor for V = 40

dm3. The coolant enters at 300 K. How do these

curves differ from those in the adiabatic case

and part (a) and (b)?

d) Variable Ta Countercurrent. Plot X, Xe, T, Ta

and rate of disappearance of A when there is a

heat loss to the coolant and the coolant

temperature varies along the length of the

reactor for V = 20 dm3. The coolant enters at 300

K. How do these curves differ from those in the

adiabatic case and part (a) and (b)?

Heat Exchange

11

Example: PBR A ↔ B

5) Parameters

• For adiabatic:

• Constant Ta:

• Co-current: Equations as is

• Counter-current:

0dW

dTa

T)-T toT-T flip(or )1(dW

dTaa

0Ua


12

1) Mole Balances)1( Fr

dW

dX0AA

0A

A

0A

BA

b

F

r

F

r

dV

dX

VW


13

2) Rate Laws )2(

C

BAA

K

CCkr

)3( 11

exp1

1

TTR

Ekk

)4( 11

exp2

2

TTR

HKK Rx

CC


14

3) Stoichiometry

5 CA CA0 1 X y T0 T

6 CB CA0Xy T0 T

0

00

0

2

20

T

T

ydV

dy

VW

T

T

yT

T

F

F

ydW

dy

FF

b

T

T

TT


15

Parameters

bARx

CA

TCTH

KTREkF

, , , , ,

)15()7( , , , , , ,

002

2110

3) Stoichiometry:Gas Phase

v v0 1X P0

P

T

T0

5 CA FA 0 1 X v0 1X

P

P0

T0

TCA 0 1 X

1X yT0

T

6 CB CA 0X

1X yT0

T

7 dy

dW

2y

FT

FT 0

T

T0

2y1X

T

T0

16



Gas Phase Heat Effects

KC CBe

CAe

CA 0X eyT0 T

CA 0 1 X e yT0 T

8 Xe KC

1KC

17




18

Exothermic Case:

Xe

T

KC

T

KC

T T

Xe

~1Endothermic Case:




19

dT

dV

rA HRx Ua T Ta FiCPi

FiCPi FA0 iCPi CPX

Case 1: Adiabatic and ΔCP=0

T T0 HRx XiCPi

(16A)

Additional Parameters (17A) & (17B)

T0, iCPi CPA ICPI



Heat effects:

9 0

PiiA

a

b

RxA

CF

TTUa

Hr

dW

dT

20

Case 2: Heat Exchange – Constant Ta



)C17( TT 0V ,

Cm

TTUa

dV

dTaoa

P

aa

cool

Case 3. Variable Ta Co-Current

Case 4. Variable Ta Countercurrent

? 0

a

P

aa TVCm

TTUa

dV

dT

cool

Guess Ta at V = 0 to match Ta0 = Ta0 at exit, i.e., V = Vf

21



22

23

24

25

26

Endothermic

PFR

A

B

dX

dV

k 1 11

KC

X

0

, Xe KC

1 KC

XEB iCPi

TT0 HRx

CPA

ICPI T T0 HRx

T0

XXEB

Xe

T T0 HRx X

CPA ICPI

27

Conversion on temperature

Exothermic ΔH is negative

Adiabatic Equilibrium temperature (Tadia) and conversion (Xeadia)

PA

Rx0

C

XHTT

C

Ce

K1

KX

X

Xeadia

Tadia T28

Adiabatic Equilibrium

X2

FA0 FA1 FA2 FA3

T0X1 X3T0 T0

Q1 Q2

29

X

T

X3

X2

X1

T0

Xe

Rx

Pii

EBH

TTCX

0

30

31

T

X

Adiabatic

T and Xe

T0

exothermic

T

X

T0

endothermic

PIIPA

Rx

CC

XHTT

0

Trends:

Adiabatic

Gas Flow Heat Effects

32

Effects of Inerts in the Feed

33

k

1I

Endothermic

34

As inert flow increases the

conversion will increase.

However as inerts increase,

reactant concentration

decreases, slowing down the

reaction. Therefore there is an

optimal inert flow rate to

maximize X.

First Order Irreversible

Adiabatic:

35

As T0 decreases the conversion X will increase, however the

reaction will progress slower to equilibrium conversion and may not

make it in the volume of reactor that you have.

Therefore, for exothermic reactions there is an optimum inlet

temperature, where X reaches Xeq right at the end of V. However,

for endothermic reactions there is no temperature maximum and

the X will continue to increase as T increases.

X

T

Xe

T0

X

T

X

T


Adiabatic:

36

Effect of adding inerts

X

T

V1 V2X

TT0

I

0I

Xe

X

X T T0 CpA ICpI

HRx


Exothermic Adiabatic

37

As θI increase, T decrease and

dX

dV

k

0 H I

k

θI

38

39

KC

V

Xe

X

V

T

V

Xe

V

k

V

Xe

X

Frozen

V

OR

Endothermic

T

V

KC

V

Xe

V

Xe

X

V

k

V

Exothermic

40

Adiabatic

Heat Exchange

T

V

KC

V

Xe

V

Xe

X

V

Endothermic

T

V

KC

V

Xe

V

Xe

X

V

Exothermic

41

End of Web Lecture 21

End of Class Lecture 17

42

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Lecture 21 - University of Michigan