Lecture 19

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 19

Today’s lectureGas Phase ReactionsTrends and Optimums

2

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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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.

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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….

Gas 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

=αyFTFT0

TT0

⎛ ⎝ ⎜

⎞ ⎠ ⎟= −

α2y

TT0

⎛ ⎝ ⎜

⎞ ⎠ ⎟

W = ρV

dydV

= −αρ b2y

TT0

⎛ ⎝ ⎜

⎞ ⎠ ⎟

14


b00A2Rx

2C110A

, ,T ,C ,T ,H

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

Parameters:

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3) Stoich (gas):

€

v = v0 1+εX( )P0

PTT0

5( ) CA =FA 0 1 − X( )v0 1+εX( )

PP0

T0

T=CA 0 1− X( )

1+εX( )yT0

T

6( ) CB =CA 0X1+εX( )

yT0

T

7( ) dydW

=−α2y

FTFT 0

TT0

=−α2y

1+εX( )TT0


16


€

KC =CBeCAe

=CA 0X eyT0 T

CA 0 1− X e( )yT0 T

8( ) Xe =KC

1+KC


17

Gas Phase Heat EffectsReversible Reactions



18

Exothermic Case: Xe

T

KC

T

KC

T T

Xe~1

Endothermic Case:


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 T19

X2

FA0 FA1 FA2 FA3

T0X1 X3T0 T0

Q1 Q2

20

X

T

X3

X2

X1

T0

Xe XEB

RX

0Pii

HTTC

X

21


22

€

dTdV

=−rA( ) −ΔHRx( ) −Ua T −Ta( )

∑FiCPi

€

∑FiCPi = FA 0 ∑ΘiCPi + ΔCPX[ ]

Case 1: Adiabtic and ΔCP=0

€

T = T0 +−ΔHRx( )X∑ΘiCPi

(16A)

Additional Parameters (17A) & (17B)

€

T0, ∑ΘiCPi =CPA + ΘICPI

Example A ↔ B

Heat effects:

€

dTdW

=−ra( ) −ΔHR( ) −

Ua

ρ bT −Ta( )

FA 0 θ iCPi∑ 9( )

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Case 2: Heat Exchange – Constant Ta

Gas Phase Heat EffectsExample A ↔ B

)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

24

Example A ↔ B

25

26

27

28

29

Endothermic

PFR

A B

dXdV

k 1 1 1KC

X

0 , Xe

KC1KC

XEB iCPi

T T0 HRx

CPA

ICPI T T0 HRx

T0

XXEB

Xe

TT0 HRx X

CPAICPI

30

31

Effects of Inerts In The Feed

What happens when we vary

k1I

32

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

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T

X

Adiabatic T and Xe

T0

exothermic

T

X

T0

endothermic

PIIPA

R0 CC

XHTT

Gas Phase Heat EffectsTrends:-Adiabatic:

34

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.


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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 +θ ICpI[ ]

−ΔHRx

37

Exothermic Adiabatic

As θI increase, T decrease and

€

dXdV

=k

υ 0 Hθ I( )

k

θI

38

39

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

41

Derive the Steady State Energy Balance (w/o Work)

0HFHFdWTTUaii0i0ia

W

0 B

0dWdHFH

dWdF0TTUa i

iii

aB

Differentiating with respect to W:

42

Aii

i rrdWdF

Mole Balance on species i:

T

TPiRii

R

dTCTHH

Enthalpy for species i:


43

Differentiating with respect to W:

dWdTC0

dWdH

Pii

0dWdTCFHrTTUa

PiiiiAaB


44

0dWdTCFHrTTUa

PiiiiAaB

THH Rii

XFF ii0Ai

Final Form of the Differential Equations in Terms of Conversion:A:

45

Final Form of terms of Molar Flow Rate:

Pii

AB

CF

HrUa

dWdT

B: T,Xg

Fr

dWdX

0A

A

46

(d) Gas Phase Counter Current Heat Exchange Vf = 20 dm3

Matches

47

End of Lecture 19

48

Example: PBR A ↔ B VW b

2) Rates:

4 T1

T1

RHexpkk

3 T1

T1

REexpkk

2 kCCkr

2

R2CC

11

CBA'A

1 F

rdWdX

0A

'A1) Mole balance:


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Lecture 19