Lecture 20

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 20

0

0 0 0 0

i iH H

sysi i i i i i S

dEF U PV F U PV Q W

dt

0 0sys

i i i i S

dEF H FH Q W

dt

Last Lecture Energy Balance Fundamentals

2

Substituting for W

000 iiiiS HFHFWQ

dtdE

WQEFEF sysiiii 00

Web Lecture 20Class Lecture 16-Thursday 3/14/2013

3

Reactors with Heat ExchangeUser friendly Energy Balance DerivationsAdiabaticHeat Exchange Constant Ta

Heat Exchange Variable Ta Co-currentHeat Exchange Variable Ta Counter Current

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

4

Adiabatic Operation CSTR

Adiabatic Operation CSTR

5

Assuming the reaction is irreversible for CSTR, A B, (KC = 0) what reactor volume is necessary to achieve 80% conversion?If the exiting temperature to the reactor is 360K, what

is the corresponding reactor volume?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.Now assume the reaction is reversible, make a plot of

the equilibrium conversion as a function of temperature between 290K and 400K.

1) Mole Balances:exitA

0A

rXFV

6

CSTR: Adiabatic Example

A B

FA0

FI

(exothermic)

𝑇=?

T1

T1

RHexpKK

ekk

KCCkr

2

Rx1CC

T1

T1

RE

1

C

BAA

1

2) Rate Laws:

XCC

X1CC

0AB

0AA

3) Stoichiometry:

7


4) Energy Balance Adiabatic, ∆Cp=0

X

36164000,20300X

182164000,20300T

CCXHT

CXHTT

IAi PIP

Rx0

Pi

Rx0

X 100300T

8


Irreversible 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 reversible)

9


3

A

0A

0A

0A

exitA

0A

dm 82.28.01281.3

8.05rXFV

81.33801

2981

989.1000,10exp1.0k

K3808.0100300T

X1kCXF

rXFV

Given X, Calculate T and V

10


(b) VrkTXGiven CalcA

CalcCalcCalc

CK Calc(if reversible)

3

1

0

05.24.0283.1

6.05min83.1

6.0100

300360

ble)(Irreversi 1

dmV

k

TX

KTXkCr AA

Given T, Calculate X and V

11


(c) Levenspiel Plot

X1kCF

rF

0A

0A

A

0A

Choose X Calc T Calc k Calc rACalc

FA 0

rA

X100300T

12


(c) Levenspiel Plot

13


CSTR X = 0.95 T = 395 K

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 K

14


PFR X = 0.6

PFR X = 0.95

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

PFR 60%

15


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

16

CSTR: Adiabatic Example - Summary

dVdF

HdVdH

FdV

HFd

TTUadV

HFd

VTTUaHFHF

ii

ii

ii

aii

aVViiVii

0

0

Energy Balance in terms of Enthalpy

17

RPiii TTCHH 0

dVdTC

dVdH

Pii

xRii

AiiPiiii

HH

rHdVdTCF

dVHFd

PFR Heat Effects

18

Aiii rr

dVdF

0

TTUarH

dVdTFC aARiPi

aARPii TTUarHdVdTCF

Pii

aAR

CFTTUarH

dVdT

19 Need to determine Ta

PFR Heat Effects

Heat Exchange:

iPi

aRxA

CFTTUaHr

dVdT

)B16( CF

TTUaHrdVdT

iPi0A

aRxA

20Need to determine Ta

Energy Balance:

Adiabatic (Ua=0) and ΔCP=0

)A16( C

XHTTiPi

Rx0

Heat Exchange Example: Case 1 - Adiabatic

21

A. Constant Ta e.g., Ta = 300K

)17( 0 , CTTV

CmTTUa

dVdT

aoaP

aa

cool

B. Variable Ta Co-Current

C. Variable Ta Counter Current Guess ? 0

aP

aa TVCm

TTUadVdT

cool

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

User Friendly Equations

22

Coolant Balance:

In - Out + Heat Added = 0

0

0

aC

C

aVVCCVCC

TTUadVdH

m

TTVUaHmHm

Heat Exchanger Energy Balance Variable Ta Co-current

dVdTC

dVdH

TTCHH

aPC

C

raPC0CC

23

0 0 , aa

PCC

aa TTVCm

TTUadVdT


24

0

0

PCC

aa

aC

C

aVCCVVCC

CmTTUa

dVdT

TTUadVdH

m

TTVUaHmHm

Heat Exchanger Energy Balance Variable Ta Co-current

Elementary 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

FA0

FI

Tacm Heat Exchange

Fluid

A BT

Heat Exchanger – ExampleCase 1 – Constant Ta

T1

T1

REexpkk )3(

11

T1

T1

RHexpKK )4(

2

Rx2CC

FrdVdX )1( 0AA

1) Mole Balance:

C

BAA K

CCkr )2(2) Rate Laws:

26


0CP

6 XCC

5 X1CC

0AB

0AA

3) Stoichiometry:

9 CCC PIIPAPii

8 k1

kXC

Ceq

4) Heat Effects: 7 0

PiiA

aAR

CFTTUarH

dVdT


27

Parameters:

A

IPIPAA

AaC

R

rrateCCC

FTUakkTTREH

, , , ,, , , , ,

, , , , ,

0

021

21


28

Pii

rg

CFQQ

dVdT

Heat removed

Heat generated

XCCFCXFCF PiPii0APiii0APii

XCCFTTUarH

dVdT

PPiiA

aAR

0

PFR Heat Effects

29

Energy Balance:

Adiabatic and ΔCP=0Ua=0

)A16( C

XHTTiPi

Rx0

Additional Parameters (17A) & (17B)

IAi PIPPi0 CCC ,T 30

Mole Balance:

dXdV rAFA 0

Heat Exchanger – ExampleCase 2 – Adiabatic

31

Adiabatic PFR

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

V

rate

V

T

V

XX

Xeq

Example: Adiabatic

32

iPi

aRxA

CFTTUaHr

dVdT

)B16( CF

TTUaHrdVdT

iPi0A

aRxA

Heat Exchange

33Need to determine Ta

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

Ua ,C ,TiPia

Additional Parameters (18B – (20B):

)17( 0 CTTVCm

TTUadVdT

aoaP

aa

cool

B. Variable Ta Co-Current

C. Variable Ta Countercurrent

? 0

aP

aa TVCm

TTUadVdT

cool

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

User Friendly Equations

34

Coolant balance:


0

0

aC

C

aVVCCVCC

TTUadVdHm

TTVUaHmHm

dVdTC

dVdH

TTCHH

aPC

C

raPC0CC

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

Heat Exchange Energy Balance Variable Ta Co-current

35

0 0 , aa

PCC

aa TTVCm

TTUadVdT


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

36

PCC

aaa

CC

aVCCVVCC

CmTTUa

dVdTTTUa

dVdHm

TTVUaHmHm

0

0

Heat Exchange Energy Balance Variable Ta Co-current

0HFHFdWTTUaii0i0ia

W

0 B

0dWdHFH

dWdF0TTUa i

iii

aB

Differentiating with respect to W:

37

Derive the user-friendly Energy Balance for a PBR

Aii

i rrdWdF

Mole Balance on species i:

T

TPiRii

R

dTCTHH

Enthalpy for species i:

38


Differentiating with respect to W:

dWdTC0

dWdH

Pii

0dWdTCFHrTTUa

PiiiiAaB

39


0dWdTCFHrTTUa

PiiiiAaB

THH Rii XFF ii0Ai

Final Form of the Differential Equations in Terms of Conversion:

A:

40


Final form of terms of Molar Flow Rate:

Pii

AaB

CF

HrTTUa

dWdT

B: T,Xg

Fr

dWdX

0A

A

41


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:

2

RPRR2

RP

RTTTCTH

RTTH

dTKlnd

43


For the special case of ΔCP=0

Integrating the Van’t Hoff Equation gives:

21

RR1P2P T

1T1

RTHexpTKTK

44


endothermic reaction

exothermic reaction

KP

T

endothermic reaction

exothermic reaction

Xe

T

45


46

End of Lecture 20

Lecture 20

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