Lecture 13 - University of Michiganelements/5e/powerpoints/2013lectures/Lec13_PDF… · Lecture 13...

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 13

Lecture 13 – Tuesday 2/26/2013 � Complex Reactions:

A +2B à C A + 3C à D

� Example A: Liquid Phase PFR � Example B: Liquid Phase CSTR � Example C: Gas Phase PFR � Example D: Gas Phase Membrane Reactors

Sweep Gas Concentration Essentially Zero Sweep Gas Concentration Increases with Distance

�  Example E: Semibatch Reactor

2

Gas Phase Multiple Reactions

3

New things for multiple reactions are:

4

1. Number Every Reaction 2. Mole Balance on every species 3. Rate Laws (a) Net Rates of Reaction for every species (b) Rate Laws for every reaction (c) Relative Rates of Reaction for every reaction For a given reaction i: (i) aiA+biB àciC+diD:

∑=

=N

iiAA rr

1

3222

211

CACC

BAAA

CCkrCCkr

−=

−=

i

iD

i

iC

i

iB

i

iA

dr

cr

br

ar

==−

=−

Reactor Mole Balance Summary

5

Reactor Type

Gas Phase Liquid Phase

Vr

dtdN

AA = A

A rdtdC

=Batch

VrdtdN

AA = V

CrdtdC A

AA 0υ−=Semibatch

0BBB FVr

dtdN

+= [ ]VCCr

dtdC BB

BB −

+= 00υ

Reactor Mole Balance Summary

6

Reactor Type

Gas Phase Liquid Phase

A

AA

rFFV

−

−=

0 ( )A

AA

rCCV

−

−= 0

0υCSTR

AA r

dVdC

=0υAA r

dVdF

=PFR

AA r

dWdC

ʹ′=0υAA r

dWdF

ʹ′=PBR

Note: The reaction rates in the above mole balances are net rates.

VNC B

B = υB

BFC =

TT

PP

NNVVT

T 00

00=

TT

PP

VN

NNC T

T

BB

0

00

0=

TT

PP

NNCCT

BTB

0

00=

TT

PP

FF

T

T 00

00υυ =

TT

PP

FFCCT

BTB

0

00=

TT

PPF

FFC T

T

BB

0

00

0

υ=

Batch Flow

7

Note: We could use the gas phase mole balances for liquids and then just express the concentration as: Flow: Batch:

Stoichiometry

8

Concentration of Gas:

CA =CT 0FAFT

!

"#

$

%& p

T0T

!

"#

$

%& FT = FA +FB +FC +FD

0υA

AFC =

0VNC A

A =

Example A: Liquid Phase PFR

9

NOTE: The specific reaction rate k2C is defined with respect to species C.

23)2( DAC →+ 2322 ACCC CCkr =−

2)1( CBA →+ 211 BAAA CCkr =−

NOTE: The specific reaction rate k1A is defined with respect to species A.

The complex liquid phase reactions follow elementary rate laws:


10

Complex Reactions

1) Mole Balance on each and every species

€

(1) A+ 2B→C

(2) A+ 3C→D

DD

CC

BB

AA

rdVdFr

dVdF

rdVdFr

dVdF

==

==

)4( )3(

)2( )1(


11

2) Rate Laws: Net Rates

Rate Laws

Relative Rates Reaction 1

DDCCC

BBBAAA

rrrrrrrrrrr

221

2121

0 )8( )6( )7( )5(

+=+=

+=+=

3222

211

)10(

)9(

CACC

BAAA

CCkrCCkr

−=

−=

11 1

1 1

1 1

1 2 1

(11) 2(12)

CA B

B A

C A

rr r

r rr r

= =− −

=

= −


12

Relative Rates Reaction 2

3 )14(

32 )13(

132

22

22

222

CD

CA

DCA

rr

rr

rrr

−=

=

=−

=−

322

3221

21

322

21

3

232

CAC

D

CACBAAC

BAAB

CACBAAA

CCkr

CCkCCkrCCkr

CCkCCkr

=

−=

−=

−−=


13

3) Stoichiometry Liquid

( ) 0 else then 00001.0 ~ )19(

)18( )17( )16( )15(

0

0

0

0

⎟⎟⎠

⎞⎜⎜⎝

⎛>=

=

=

=

=

D

CDC

DD

CC

BB

AA

FFVifS

FCFCFCFC

υ

υ

υ

υ


14

Others 4) Parameters

NeededNot – Liquid )20(NeededNot – Liquid )19(

NeededNot – Liquid

0 =

=

=

T

T

C

Fα

100 )26(200 )28(200 )26(

2500 )25(Liquid )24(

Liquid )23(20 )22(

10 )21(

0

0

0

0

2

1

=

=

=

=

=

=

=

=

υ

α

B

A

f

T

C

A

FFVC

kk

Example B: Liquid Phase CSTR

15

Same reactions, rate laws, and rate constants as Example A

2)1( CBA →+ 211 BAAA CCkr =−



23)2( DAC →+ 2322 ACCC CCkr =−


16

The complex liquid phase reactions take place in a 2,500 dm3 CSTR. The feed is equal molar in A and B with FA0=200 mol/min, the volumetric flow rate is 100 dm3/min and the reaction volume is 50 dm3.

Find the concentrations of A, B, C and D existing in the reactor along with the existing selectivity.

Plot FA, FB, FC, FD and SC/D as a function of V


17

(1) A + 2B →C (2) 2A + 3C → D

3222

211

CACC

BAAA

CCkrCCkr

−=

−=

00)4(00)3(

0)2(0)1(

0

0

000

000

=+−

=+−

=+−

=+−

VrCDVrCC

VrCCBVrCCA

DD

CC

BBB

AAA

υ

υ

υυ

υυ

1) Mole Balance


18

2) Rate Laws: (5)-(14) same as PFR

3) Stoichiometry: (15)-(18) same as Liquid Phase PFR

4) Parameters:

0001.00001.0 )19(

0

0/ +

=+

=D

C

D

CDC C

CF

FSυ

υ

00021 , , , , , υVCCkk BACA


19

(1) A + 2B →C (2) 2A + 3C → D

3222

211

CACC

BAAA

CCkrCCkr

−=

−=

00001.0)19(

)18()17()16()15(

0

0

0

0

+=

=

=

=

=

D

CDC

DD

CC

BB

AA

FFS

FCFCFCFC

υ

υ

υ

υ

1) Mole Balance (1–4) 2) Rates (5–14) 3) Stoichiometry: (15–19)

( ) VrFFFf AAAA +−= 0)1( (=0)

( ) VrFFFf BBBB +−= 0)2( (=0)

( ) VrFFf CCC +−= 0)3( (=0)

( ) VrFFf DDD +−= 0)4( (=0)

In terms of molar flow rates

Same as Example A


20

(1) A + 2B →C (2) 2A + 3C → D

3222

211

CACC

BAAA

CCkrCCkr

−=

−=

1) Mole Balance (1–4) 2) Rates (5–14) 3) Stoichiometry: (15–19)

In terms of concentration

( ) VrCCCf AAAA +−= 000)1( υυ (=0)

( ) VrCCCf BBBB +−= 000)2( υυ (=0)

( ) VrCCf CCC +−= 00)3( υ (=0)

( ) VrCCf DDD +−= 00)4( υ (=0)

00001.0 )15(

+=

D

CDC F

FSSame as Example A

Example C: Gas Phase PFR, No ΔP

21

Same reactions, rate laws, and rate constants as Example A:

2)1( CBA →+ 211 BAAA CCkr =−



23)2( DAC →+ 2322 ACCC CCkr =−


22

1) Mole Balance

2) Rate Laws: (5)-(14) same as CSTR

)4( (2)

)3( )1(

DD

BB

CC

AA

rdVdFr

dVdF

rdVdFr

dVdF

==

==


23

3) Stoichiometry: Gas: Isothermal T = T0

Packed Bed with Pressure Drop

(15) CA =CT 0FAFT

p (16) CB =CT 0FBFT

p

(17) CC =CT 0FCFT

p (18) CD =CT 0FDFT

p

(19) FT = FA +FB +FC +FD

dpdW

= −α2p

FTFT 0

"

#$

%

&'TT0

"

#$

%

&'= −

α2p

FTFT 0


24

4) Selectivity

p =1 21( )

( ) ( ) ( )20 0 else then 00001.0 if ⎟⎟⎠

⎞⎜⎜⎝

⎛>==

D

C

D

C

FFV

FFS

Example D: Membrane Reactor with ΔP

25


2)1( CBA →+ 211 BAAA CCkr =−



23)2( DAC →+ 2322 ACCC CCkr =−


26

Because the smallest molecule, and the one with the lowest molecular weight, is the one diffusing out, we will neglect the changes in the mass flow rate down the reactor and will take as first approximation: 1) Mole Balances

( ) ( )

( ) ( )4 2

3 1

DD

BB

CCC

AA

rdVdFDr

dVdFB

RrdVdFCr

dVdFA

==

−==

CCsg RdVdF

=

We also need to account for the molar rate of desired product C leaving in the sweep gas FCsg

mm !! =0

We need to reconsider our pressure drop equation. When mass diffuses out of a membrane reactor there will be a decrease in the superficial mass flow rate, G. To account for this decrease when calculating our pressure drop parameter, we will take the ratio of the superficial mass velocity at any point in the reactor to the superficial mass velocity at the entrance to the reactor.

⎥⎥⎦

⎤

⎢⎢⎣

⎡

⋅

⋅==

∑∑

ii

ii

MWFMWF

GG

00

00 ααα

27


The superficial mass flow rates can be obtained by multiplying the species molar flow rates, Fi, by their respective molecular weights, Mwi, and then summing over all species:

( )( )

( )( )∑

∑∑∑ =

⋅

⋅==

ii

ii

Cii

Cii

C

C

MWFMWF

AMWFAMWF

AmAm

GG

0000 1

1

1

1


28


29

2) Rate Laws: (5)-(14) same as Examples A, B, and C. 3) Stoichiometry: (15)-(20) same as Examples A and B

(T=T0)

4) Sweep Gas Balance:

( )CSweepCCC CCkR −=

dpdW

= −α2p

FTFT 0

dpdV

= −ρα2p

FTFT 0

21( )

CCsg

CVVCsgVCsg

RdVdF

VRFF

=

=Δ+−Δ+

0

Example E: Liquid Phase Semibatch

30


2)1( CBA →+ 211 BAAA CCkr =−



23)2( DAC →+ 2322 ACCC CCkr =−

The complex liquid phase reactions take place in a semibatch reactor where A is fed to B with FA0= 3 mol/min. The volumetric flow rate is 10 dm3/min and the initial reactor volume is 1,000 dm3.

The maximum volume is 2,000 dm3 and CA0=0.3 mol/dm3 and CB0=0.2 mol/dm3. Plot CA, CB, CC, CD and SS/D as a function of time.

31


1) Mole Balances:

(1) A + 2B →C (2) 2A + 3C → D

0AAA FVr

dtdN

+=

VrdtdN

BB =

VrdtdN

CC =

VrdtdN

DD =

00 =AN

000.2000 == VCN BB

00 =CN

00 =DN32

B

FA0


2) Rate Laws: (5)-(14)

( )

( ) ( )

( ) ( )19 18

17 16

15 00

VNC

VNC

VNC

VNC

tvVV

DD

CC

BB

AA

==

==

+=

Net Rate, Rate Laws and relative rate – are the same as Liquid and Gas Phase PFR and Liquid Phase CSTR

( )20 )0( else then )0001.0( if/ ⎟⎟⎠

⎞⎜⎜⎝

⎛>=

D

CDC N

NtS

mindm10 30 =υ 3

0 dm100V = minmol3F 0A =33


3) Selectivity and Parameters:

End of Lecture 13

34

Lecture 13 - University of Michiganelements/5e/powerpoints/2013lectures/Lec13_PDF… · Lecture 13...

Documents

Transcript of Lecture 13 - University of Michiganelements/5e/powerpoints/2013lectures/Lec13_PDF… · Lecture 13...