lecture 20 - university of michigan · 2019-08-07 · web lecture 20 class lecture 16-thursday...

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

Fi0∑ Ui0 + P0 !Vi0!" #$

Hi0" #$ %$− Fi∑ Ui + P !Vi#$ %&

Hi" #$ %$+ &Q − &WS =

dEsysdt

Fi0∑ Hi0 − Fi∑ Hi +!Q − !WS =

dEsysdt

Last Lecture Energy Balance Fundamentals

2

Substituting for !W

∑∑ =−+− 000 iiiiS HFHFWQ !!

dtdE

WQEFEF sysiiii =−+−∑∑ 00

!!

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

3

�  Reactors with Heat Exchange �  User friendly Energy Balance Derivations

� Adiabatic � Heat Exchange Constant Ta � Heat Exchange Variable Ta Co-current � Heat 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

Δ𝐻↓𝑅𝑥𝑛 =−20000𝑐𝑎𝑙/𝑚𝑜𝑙 𝐴  (exothermic)

𝑇= ?�𝑋= ?

𝐹↓𝐴0 =5𝑚𝑜𝑙/𝑚𝑖𝑛  𝑇↓0 =300 𝐾 𝐹↓𝐼 =10𝑚𝑜𝑙/𝑚𝑖𝑛 

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−

Δ=

=

⎥⎦

⎤⎢⎣

⎡−=−

⎟⎟⎠

⎞⎜⎜⎝

⎛−

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

( ) ( )

( )( )( )

X36164000,20300X

182164000,20300T

CCXHT

CXHTT

IAi PIP

Rx0

Pi

Rx0

++=⎥

⎦

⎤⎢⎣

⎡

+

−−+=

Θ+

Δ−+=

Θ∑

Δ−+=

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

81.3380

12981

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⎯ → ⎯

FA0

−rA

X100300T +=

12


(c) Levenspiel Plot

13


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 95%

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%

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 95%

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

( )

( )

⎥⎦

⎤⎢⎣

⎡+−=

−

=−+−

=Δ−+−

∑∑∑

∑

∑∑ Δ+

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

Δ=

⎥⎦

⎤⎢⎣

⎡ −+−=−

∑

∑ ∑∑

υ

υ

PFR Heat Effects

18

( )Aiii rr

dVdF

−== υ

− CPiFi∑ dTdV

+ΔHRx −rA( )$

%&'

()+Ua Ta −T( ) = 0

Fi∑ CPidTdV

= ΔHRxrA −Ua T −Ta( )

dTdV

=ΔHRx( ) −rA( )−Ua T −Ta( )

Fi∑ CPi

19 Need to determine Ta

PFR Heat Effects

Heat Exchange:

( )( ) ( )iPi

aRxA

CFTTUaHr

dVdT

∑

−−Δ−−=

dTdV

=−rA( ) −ΔHRx( )−Ua T −Ta( )

FA0 ∑ΘiCPi


Energy Balance:

Adiabatic (Ua=0) and ΔCP=0

( ) )A16( C

XHTTiPi

Rx0 Θ∑

Δ−+=

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 ==

−= a

P

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

=−+−

=−Δ+−Δ+

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 TTVCmTTUa

dVdT

==−

=!


24

( )

( )

( )

0

0

PCC

aa

aC

C

aVCCVVCC

CmTTUa

dVdT

TTUadVdH

m

TTVUaHmHm

!

!

!!

−=

=−+

=−Δ+−Δ+

Heat Exchanger Energy Balance Variable Ta Counter-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

Ta cm!

Heat Exchange Fluid

€

A⇔ BT

Heat Exchanger – Example Case 1 – Constant Ta

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−=

T1

T1

REexpkk )3(

11

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−

Δ=

T1

T1

RHexpKK )4(

2

Rx2CC

FrdVdX )1( 0AA−=1) Mole Balance:

⎥⎦

⎤⎢⎣

⎡−−=

C

BAA K

CCkr )2(2) Rate Laws:

26


( )0CP =Δ

( ) ( )

( )6 XCC

5 X1CC

0AB

0AA

=

−=3) Stoichiometry:

( )9 CCC PIIPAPii θ+=θ∑

( )8 k1

kXC

Ceq +=

4) Heat Effects: ( )( ) ( ) ( )7 0∑

−−−Δ=

PiiA

aAR

CFTTUarH

dVdT

θ


27

Parameters:

A

IPIPAA

AaC

R

rrateCCC

FTUakkTTREH

−=

Δ

, , , ,, , , , ,

, , , , ,

0

021

21

θ


28

Pii

rg

CFQQ

dVdT

∑−

=

Heat removed

Heat generated

FiCPi =∑ FA0 θi +υiX( )CPi = FA0∑ θiCPi +ΔCPX∑#$ %&

dTdV

=ΔHR( ) rA( )−Ua T −Ta( )FA0 θiCPi +ΔCPX∑$% &

'

PFR Heat Effects

29

Energy Balance:

Adiabatic and ΔCP=0 Ua=0

( ) )A16( C

XHTTiPi

Rx0 Θ∑

Δ−+=

Additional Parameters (17A) & (17B)

IAi PIPPi0 CCC ,T Θ+=Θ∑30

Mole Balance:

€

dXdV

=−rAFA 0

Heat Exchanger – Example Case 2 – Adiabatic

31

Adiabatic PFR

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

V

rate

V

T

V

X X

Xeq

Example: Adiabatic

32

( )( ) ( )iPi

aRxA

CFTTUaHr

dVdT

∑

−−Δ−−=

( )( ) ( ) )B16( CF

TTUaHrdVdT

iPi0A

aRxA

Θ∑

−−Δ−−=

Heat Exchange


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

Ua ,C ,TiPia Θ∑

Additional Parameters (18B – (20B):

( ) )17( 0 CTTVCmTTUa

dVdT

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

=−+−

=−Δ+−Δ+

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 Counter-current

35

( )0 0 , aa

PCC

aa TTVCmTTUa

dVdT

==−

=!


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

!!

!!

−==−+

=−Δ+−Δ+

0

0

Heat Exchange Energy Balance Variable Ta Co-current

( ) 0HFHFdWTTUaii0i0ia

W

0 B

=−+−ρ ∑∑∫

( ) 0dWdHFH

dWdF0TTUa i

iii

aB

=−−+−ρ ∑∑

Differentiating with respect to W:

37

Derive the user-friendly Energy Balance for a PBR

⎟⎠⎞⎜

⎝⎛ ʹ′−υ=ʹ′= Aii

i rrdWdF

Mole Balance on species i:

( ) ∫+= οT

TPiRii

R

dTCTHH

Enthalpy for species i:

38


Differentiating with respect to W:

dWdTC0

dWdH

Pii +=

( ) 0dWdTCFHrTTUa

PiiiiAaB

=−υʹ′+−ρ ∑∑

39


( ) 0dWdTCFHrTTUa

PiiiiAaB

=−υʹ′+−ρ ∑∑

( )THH Rii Δ=υ∑( )XFF ii0Ai υ+Θ=

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

Δʹ′+−ρ

=

B: ( )T,Xg

Fr

dWdX

0A

A =ʹ′−

=

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

( )δ=RTKK P

C

Van’t Hoff Equation:

( ) ( ) ( )2

RPRR2

RP

RTTTCTH

RTTH

dTKlnd −Δ+Δ

=Δ

=ο

43


For the special case of ΔCP=0

Integrating the Van’t Hoff Equation gives:

( ) ( ) ( )⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−

Δ=

ο

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 - university of michigan · 2019-08-07 · web lecture 20 class lecture 16-thursday...

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