lecture 21 (chapter 12)

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 (Chapter 12)

User Friendly Equations Relate T and X or Fi

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

0CW PS

Rx

0PiEB H

TTCX i

Rx

0P

HTTC~

X A

iPi

Rx0 C

XHTT

Last Lecture

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Web Lecture 21Class Lecture 19–Tuesday 3/15/2011Gas Phase ReactionsTrends and Optimums

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2. CSTR with heat exchanger, UA(Ta-T) and a large coolant flow rate:

Rx

0Pia0A

EB H

TTC~TTFUA

Xi

TTa

Cm


4

3. PFR/PBR with heat exchange:

FA0T0

CoolantTa


3A. In terms of conversion, X

XCC~F

THrTTUa

dWdT

pPi0A

RxAaB

i5


3B. In terms of molar flow rates, Fi

i

ij

Pi

RxAaB

CF

THrTTUa

dWdT

4. For multiple reactions

i

ij

Pi

RxijaB

CF

HrTTUa

dVdT

5. Coolant Balance

cPc

aA

CmTTUa

dVdT

6

Reversible Reactions

endothermic reaction

exothermic reaction

KP

T

endothermic reaction

exothermic reaction

Xe

T

7

Heat ExchangeExample: Elementary liquid phase reaction carried out in a PFR

FA0FI

Tacm Heat

Exchange Fluid

A B

The feed consists of both inerts I and Species A with the ratio of inerts to the species A being 2 to 1.8

Heat Exchangea) 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.

9

Heat Exchangec) 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 Counter 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 = 20 dm3. The coolant enters at 300 K. How do these curves differ from those in the adiabatic case and part (a) and (b)?

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Example: PBR A ↔ B

5) Parameters….

Reversible ReactionsGas Phase Heat Effects

• For adiabatic:• Constant Ta:

• Co-current: Equations as is • Counter-current:

0dWdTa

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

aa

0UA

11

)1( FrdWdX

0AAMole

Balance:

0A

A

0A

BA

b

Fr

Fr

dVdX

VW

12


)3( T1

T1

REexpkk

11

)4( T1

T1

RHexpKK

2

Rx2CC

)2( KCCkr

C

BAA

Rate Law:

13


Stoichiometry:

5 CA CA 0 1 X y T0 T

6 CB CA 0Xy T0 T

dydW

yFTFT0

TT0

2y

TT0

W V

dydV

b2y

TT0

14


b00A2Rx

2C110A

, ,T ,C ,T ,H

)15()7( ,K ,T ,R ,E ,k ,F

Parameters:

15



3) Stoich (gas):

v v0 1X P0

PTT0

5 CA FA 0 1 X v0 1X

PP0

T0

TCA 0 1 X

1X yT0

T

6 CB CA 0X1X y

T0

T

7 dydW

2y

FTFT 0

TT0

2y

1X TT0


16

KC CBeCAe

CA 0X eyT0 T

CA 0 1 Xe yT0 T

8 Xe KC

1KC


17




18

Exothermic Case: Xe

T

KC

T

KC

T T

Xe~1

Endothermic Case:

Example A ↔ BGas Phase Heat Effects

19

dTdV

rA HRx Ua T Ta

FiCPi

FiCPi FA 0 iCPi CPX Case 1: Adiabtic and ΔCP=0

T T0 HRx XiCPi

(16A)

Additional Parameters (17A) & (17B)

T0, iCPi CPA ICPI

Heat effects:

dTdW

ra HR

Ua

bT Ta

FA 0 iCPi 9

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Case 2: Heat Exchange – Constant Ta

Example A ↔ BGas Phase Heat Effects

)C17( TT 0V , Cm

TTUadVdT

aoaP

aa

cool

Case 3. Variable Ta Co-Current

Case 4. Variable Ta Counter Current Guess ?T 0V

CmTTUa

dVdT

aP

aa

cool

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

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Example A ↔ B

Endothermic

PFR

A B

dXdV

k 1 1

1KC

X

0 , Xe

KC1 KC

XEB iCPi

T T0 HRx

CPA

ICPI T T0 HRx

T0

XXEB

Xe

T T0 HRx X

CPA ICPI

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Adibatic EquilibriumConversion on Temperature Exothermic ΔH is negativeAdiabatic Equilibrium temperature (Tadia) and conversion (Xeadia)

PA

Rx0 C

XHTT

C

Ce K1

KX

X

Xeadi

a

Tadia T28

X2

FA0 FA1 FA2 FA3

T0X1 X3T0 T0

Q1 Q2

29

X

T

X3

X2

X1

T0

Xe XEB

RX

0Pii

HTTC

X

30

T

X

Adiabatic T and Xe

T0

exothermic

T

X

T0

endothermic

PIIPA

R0 CC

XHTT

Gas Phase Heat EffectsTrends:-Adiabatic:

32

33

Effects of Inerts In The Feed

What happens when we vary

k1 I

34

Endothermic

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:• 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.

Gas Phase Heat Effects

35

X

T

Xe

T0

X

T

X

T

Adiabatic:

Gas Phase Heat Effects

36

Effect of adding Inerts

X

T

V1 V2X

TT0

I

0I

Xe

X

X T T0 CpA ICpI

HRx

37

Exothermic Adiabatic

As θI increase, T decrease and

dXdV

k

0 H I

k

θI

Adiabatic

KC

V

XeX

V

T

V

Xe

V

k

V

Xe

XFrozen

V

OR

Endothermic

T

V

KC

V

Xe

V

XeX

V

k

V

Exothermic

40

Heat Exchange

T

V

KC

V

Xe

V

XeX

V

Endothermic

T

V

KC

V

Xe

V

XeX

V

Exothermic

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End of Web Lecture 21 Class Lecture 19

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lecture 21 (chapter 12)

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