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

Last Lecture

0 0sys

i i i i S

dEF H FH Q W

dt

2

Substituting for W

000 iiiiS HFHFWQ

dtdE

WQEFEF sysiiii 00

Lecture 20 – 3/10/2011User friendly Energy Balance Derivations

AdiabaticHeat Exchange Constant TaHeat Exchange Variable Ta Co-currentHeat Exchange Variable Ta Counter

Current

3

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

4

Adiabatic Operation 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.

5

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

6


T1

T1

RHexpKK

ekk

KCCkr

2

Rx1CC

T1

T1

RE

1

C

BAA

1

Rate Law:

XCC

X1CC

0AB

0AA

Stoichiometry:

7

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

X

36164000,20300X

182164000,20300T

CCXHT

CXHTT

IAi PIP

Rx0

Pi

Rx0

X 100300T

8

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)

9


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

10


(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

11

CSTR Adiabatic Example(c) Levenspiel Plot

X1kCF

rF

0A

0A

A

0A

Choose X Calc T Calc k Calc rACalc

FA 0

rA

X100300T

12

CSTR Adiabatic Example(c) Levenspiel Plot

13

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

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

Summary

16


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

PFR Heat Effects

19 Need to determine Ta

Heat Exchange:

iPi

aRxA

CFTTUaHr

dVdT

)B16( CF

TTUaHrdVdT

iPi0A

aRxA

20Need to determine Ta

Energy Balance:

Adiabtic (Ua=0) and ΔCP=0

)A16( C

XHTTiPi

Rx0

21

Heat Exchange Example: Case 1 Adiabatic

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 ?T 0V

CmTTUa

dVdT

aP

aa

cool

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

22

User Friendly Equations

Coolant balance:

In - Out + Heat Added = 0

0

0

aC

C

aVVCCVCC

TTUadVdH

m

TTVUaHmHm

Heat Exchanger Energy BalanceVariable Ta Co-current

dVdTC

dVdH

TTCHH

aPC

C

raPC0CC

23

0 0 , aa

PCC

aa TTVCmTTUa

dVdT



24

0

0

PCC

aa

aC

C

aVCCVVCC

CmTTUa

dVdT

TTUadVdH

m

TTVUaHmHm

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

HR ra Ua T Ta

FA 0 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

HR ra Ua T Ta FA 0 iCPi CPX

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

rAFA 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

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



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

aaaaa

CC

aaVCCVVCC

CmTTU

dVdT 0TTU

dVdHm

0TTVUHmHm

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


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


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


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


endothermic reaction

exothermic reaction

KP

T

endothermic reaction

exothermic reaction

Xe

T

45

End of Lecture 20

46

Lecture 20

Documents

heat exchange example2626

cstr adiabatic example14pfrx

cstr adiabatic example15cstrx

pfr heat effects19need

adiabatic3232heat exchange

ta19heat exchange

adiabatic21constant

constant ta28 heat removedheat