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 18

Today’s lectureSolution to in-class problem

User friendly Energy Balance DerivationsAdiabaticHeat Exchange Constant TaHeat Exchange Variable Ta Co-currentHeat Exchange Variable Ta Counter

Current

2

Adiabatic Operation CSTR

The feed consists of both - Inerts I and Species A with the ratio of inerts I to the species A being 2 to 1.

A B

FA0

FI

Elementary liquid phase reaction carried out in a CSTR

3

Adiabatic Operation for CSTRa) Assuming the reaction is irreversible for CSTR, A

B, (KC = 0) what reactor volume is necessary to achieve 80% conversion?

b) If the exiting temperature to the reactor is 360K, what is the corresponding reactor volume?

c) Make a Levenspiel Plot and then determine the PFR reactor volume for 60% conversion and 95% conversion. Compare with the CSTR volumes at these conversions.

d) Now assume the reaction is reversible, make a plot of the equilibrium conversion as a function of temperature between 290K and 400K.

4

CSTR Adiabatic Example

A B

T X

FA0 5 molmin

T0 300K

FI 10 molmin

H Rx 20,000cal mol A (exothermic)

Mole Balance:

exitA

0A

rXFV

5

CSTR Adiabatic Example

T1

T1

RHexpKK

ekk

KCCkr

2

Rx1CC

T1

T1

RE

1

C

BAA

1

Rate Law:

XCC

X1CC

0AB

0AA

Stoichiometry:

6

CSTR Adiabatic ExampleEnergy Balance - Adiabatic, ∆Cp=0:

X

36164000,20300X

182164000,20300T

CCXHT

CXHTT

IAi PIP

Rx0

Pi

Rx0

X 100300T

7

CSTR Adiabatic ExampleIrreversible for Parts (a) through (c)

)K (i.e., X1kCr C0AA

(a) Given X = 0.8, find T and V

Given X Calc T Calc k Calc rACalc V

Calc KC(if reverible)

8

CSTR Adiabatic Example

3

A

0A

0A

0A

exitA

0A

dm 82.28.01281.3

8.05rXFV

81.3380

12981

989.1000,10exp1.0k

K3808.0100300T

X1kCXF

rXFV

Given X, Calculate T and V

9

CSTR Adiabatic Example

(b) VrkTXGiven CalcA

CalcCalcCalc

CK Calc(if reverible)

3

1

0AA

dm 05.24.0283.1

6.05V

min83.1k

6.0100

300TX

K360Tble)(Irreversi X1kCr

Given T, Calculate X and V

10

CSTR Adiabatic Example(c) Levenspiel Plot

X1kCF

rF

0A

0A

A

0A

Choose X Calc T Calc k Calc rACalc

FA 0

rA

X100300T

11

CSTR Adiabatic Example(c) Levenspiel Plot

12

CSTR     X = 0.95     T = 395

0

5

10

15

20

25

30

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

X

-Fa0/Ra

CSTR 60%

CSTR     X = 0.6     T = 360

13

CSTR Adiabatic Example

PFR     X = 0.6

PFR     X = 0.95

14

CSTR Adiabatic Example

CSTR X = 0.6 T = 360 V = 2.05 dm3

PFR X = 0.6 Texit = 360 V = 5.28 dm3

CSTR X = 0.95 T = 395 V = 7.59 dm3

PFR X = 0.95 Texit = 395 V = 6.62 dm3

Summary

15

CSTR Adiabatic Example

Choose T Calc KCCalc Xe , repeat

(d) At Equilibrium

e

e

e0A

e0A

Ae

BeC X1

XX1C

XCCCK

C

Ce K1

KX

T1

T1

RHexpKK

2

R2CC

Calculate Adiabatic Equilibrium Conversion and Temperature:

16

(d) At Equilibrium

T1

2901

987.1000,20exp000,1KC

12.0X355.0X72.0X

937.0X999.0X

136.0K95.0K6.2K9.14K

000,1K

390T370T350T330T290T

e

e

e

e

e

C

C

C

C

C

17

Calculate Adiabatic Equilibrium Conversion and Temperature:

300T 1.0X

300T000,20

200HTTCCX

Rx

0PIP IA

(e) Te = 358 Xe = 0.5918

Calculate Adiabatic Equilibrium Conversion and Temperature:

TTa

V+ΔVV

mc , HC

FA, Fi

In - Out + Heat Added = 0

PFR Heat Effects

D/4a

L4DV

LDA

turbine)(no 0W

2

S

TTAUQ a

V+ΔVVFi

T

Ta

Fi

19

dVdFH

dVdHF

dVHFd

0TTUdV

HFd

0VTTUHFHF

ii

ii

ii

aaii

aaVViiVii

VTTUQ aa

PFR Heat Effects

20

aiii

RPi0ii

rrdVdF

TTCHH

dVdTC

dVdH

Pii

RiiaiiPiiii HHrH

dVdTCF

dVHFd

PFR Heat Effects

21

0TTUrHdVdTFC aaaRiPi

aaaRPii TTUrHdVdTCF

Pii

aaaR

CFTTUrH

dVdT

PFR Heat Effects

22

Pii

rg

CFQQ

dVdT

Heat removed

Heat generated

XCCFCXFCF PiPii0APiii0APii

PFR Heat Effects

XCCFTTUrH

dVdT

PPii0A

aaaR

23

User Friendly Equations Relate T and X or Fi

3. PBR in terms of molar flow rates

dTdW

rAHRx T

Uab

T Ta FiCPi

24

4. For multiple reactions

€

dTdV

=

Uaρ B

Ta − T( ) + rijΔHRxij∑

FiCPi∑

5. Coolant Balance

€

dTA

dV=

Ua T − Ta( )˙ m cCPc

Heat Exchange ExampleElementary liquid phase reaction carried out in a PFR

The feed consists of both inerts I and Species A with the ratio of inerts to the species A being 2 to 1.25

FA0FI

Tacm Heat

Exchange Fluid

A BT

T1

T1

REexpkk )3(

11

T1

T1

RHexpKK )4(

2

Rx2CC

FrdVdX )1( 0AA1) Mole

Balance:

C

BAA K

CCkr )2(2) Rate Law:

Heat Exchange Example

26

0CP

6 XCC

5 X1CC

0AB

0AA

3) Stoichiometry:

9 CCC PIIPAPii

8 k1

kXC

Ceq

4) Heat Effects:

€

dTdV

=ΔHR( ) −ra( ) −Ua T − Ta( )

FA 0 θ iCPi∑ 7( )

Heat Exchange Example: Case 1- Constant Ta

27

Parameters:

a

IPIPA0A

0Aaa2C1

21R

rrate

, ,C ,C ,C

,F ,T ,U ,k ,k

,T ,T ,R ,E ,H

28

Heat Exchange Example: Case 1- Constant Ta

Pii

rg

CFQQ

dVdT

Heat removed

Heat generated

XCCFCXFCF PiPii0APiii0APii

PFR Heat Effects

€

dTdV

=ΔHR( ) ra( ) −Ua T − Ta( )

FA 0 θ iCPi + ΔCP X∑[ ]29

Energy Balance:Adiabtic and ΔCP=0Ua=0 )A16(

CXHTT

iPi

Rx0

Additional Parameters (17A) & (17B)

IAi PIPPi0 CCC ,T 30

Heat Exchange Example: Case 2 AdiabaticMole Balance:

€

dXdV

=−rA

FA 0

Adibatic PFR

31

Find conversion, Xeq and T as a function of reactor volume

V

rate

V

T

V

XX

Xeq

Example: Adiabatic

32

Heat Exchange:

iPi

aRxA

CFTTUaHr

dVdT

)B16( CF

TTUaHrdVdT

iPi0A

aRxA

33Need to determine Ta

A. Constant Ta (17B) Ta = 300K

Ua ,C ,TiPia

Additional Parameters (18B – (20B):

)C17( TT 0V , Cm

TTUadVdT

aoaP

aa

cool

B. Variable Ta Co-Current

C. 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

34

User Friendly Equations

Coolant balance:In - Out + Heat Added = 0

0TTUdV

dHm

0TTVUHmHm

aaC

C

aaVVCCVCC

Variable Ta Co-current

dVdTC

dVdH

TTCHH

aPC

C

raPC0CC

All equations can be used from before except Ta parameter, use differential Ta instead, adding mC and CPC

35

0aa

PCC

aaa TT 0V ,Cm

TTUdVdT

Heat Exchanger Energy Balance

In - Out + Heat Added = 0

All equations can be used from before except dTa/dV which must be changed to a negative. To arrive at the correct integration we must guess the Ta value at V=0, integrate and see if Ta0 matches; if not, re-guess the value for Ta at V=0

Variable Ta Counter-current

36

PCC

aaaaa

CC

aaVCCVVCC

CmTTU

dVdT 0TTU

dVdHm

0TTVUHmHm

Heat Exchanger Energy Balance

Derive the User Friendly Energy Balance for a PBR

0HFHFdWTTUaii0i0ia

W

0 B

0dWdHFH

dWdF0TTUa i

iii

aB

Differentiating with respect to W:

37

Aii

i rrdWdF

Mole Balance on species i:

T

TPiRii

R

dTCTHH

Enthalpy for species i:

38

Derive the User Friendly Energy Balance for a PBR

Differentiating with respect to W:

dWdTC0

dWdH

Pii

0dWdTCFHrTTUa

PiiiiAaB

39

Derive the User Friendly Energy Balance for a PBR

0dWdTCFHrTTUa

PiiiiAaB

THH Rii XFF ii0Ai

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

40

Derive the User Friendly Energy Balance for a PBR

Final Form of terms of Molar Flow Rate:

Pii

AaB

CF

HrTTUa

dWdT

B: T,Xg

Fr

dWdX

0A

A

41

Derive the User Friendly Energy Balance for a PBR

Reversible Reactions

The rate law for this reaction will follow an elementary rate law.

DCBA

C

DCBAA K

CCCCkr

Where Ke is the concentration equilibrium constant. We know from Le Chaltlier’s law that if the reaction is exothermic, Ke will decrease as the temperature is increased and the reaction will be shifted back to the left. If the reaction is endothermic and the temperature is increased, Ke will increase and the reaction will shift to the right.

42

Reversible Reactions

RTKK P

C

Van’t Hoff Equation:

2RPRR

2RP

RTTTCTH

RTTH

dTKlnd

43

Reversible ReactionsFor the special case of ΔCP=0

Integrating the Van’t Hoff Equation gives:

21

RR1P2P T

1T1

RTHexpTKTK

44

Reversible Reactions

endothermic reaction

exothermic reaction

KP

T

endothermic reaction

exothermic reaction

Xe

T

45

End of Lecture 18

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