Chemical Reaction Engineering

Lecture 7

Home problem: nitroaniline synthesis• the disappearance rate of orthonitrochlorobenzene

[ ] [ ][ ]3ad ONCB

r k ONCB NHdt

= − = −

• Stoichiometric table:

CA0(ΘD+2X)FA0(ΘD+2X)2FA0XFD0DNH4Cl

CA0(ΘC+X)FA0(ΘC+X)FA0XFC0CNACA0(ΘB-X)FA0(ΘB-X)-2FA0XFB0BNH3

CA0(1-X)FA0(1-X)-FA0XFA0AONCB

Concentration**Remaining*Change

*) ONCB is a limiting reagent (from stoichiometry)**) reaction in liquid, thus the volume is constant

Home problem: nitroaniline synthesis• the disappearance rate of orthonitrochlorobenzene in terms of

conversion

[ ][ ] ( )( )23 0 1a A Br k ONCB NH kC X X= − = − − Θ −

• rate of the reaction at 188ºC, 25ºC, 288ºC when X=0.9:

/( ) E RTk T Ae−= 2 11 1

2 1( ) ( )ER T Tk T k T e⎡ ⎤

− −⎢ ⎥⎣ ⎦=

11273 / 4.187 / 1 10 0 68.31 25 273 25 188(25 ) (188 ) 2 10

cal mol J cal

k C k C e⋅ ⎡ ⎤− −⎢ ⎥ −+ +⎣ ⎦= = ⋅

0 2(288 ) 1.5 10k C −= ⋅ 32.85 10ar−= ⋅

43.2 10ar−= ⋅0 3(188 ) 1.7 10k C −= ⋅

73.7 10ar−= ⋅

Home problem: nitroaniline synthesis• find the reactor volume for CSTR at X=0.9: 0 0A A

a a

F X vC XVr r

= =− −

• find the reactor volume for PFR at X=0.9:1.14m32.9*10-3288ºC10.2 m33.2*10-4188ºC8.6*103 m33.7*10-725ºCVolume (X=0.9)Reaction rate

0 0.2 0.4 0.6 0.80

0.5

1

1.5Levenspilplot, T=288C

Conversion, X

F/-r

a

1.136

0

dv x( )

0.900 x

at 288ºC: V=0.15 m3 at 25ºC: V=1.14*103 m3

0 0.2 0.4 0.6 0.80

5000

1 .104Levenspilplot, T=25C

Conversion, X

F/-r

a

8.627 103×

0

dv x( )

0.900 x

00

X

Aa

dXV vCr

=−∫

Hippo’s stomach• Our starting point:

X0=0; X1=0.34 X2=0.45;

• Fitting the data

• flow rate 40 kg/day, with density of grass 365kg/m3, volumetric rate 0.13 m3/day

V1=0.45m3

Hippo’s digesting problem• The hippo has picked up a river fungus and now the effective volume of

the CSTR stomach compartment is only 0.2 m3. The hippo needs 30% conversion to survive. Will the hippo survive?

0 1A

A

F XVr

=−CSTR:

02 011

01

0.2 0.45 0.260.34

V XXV

⋅= = =

PFR:

0.26 0.28 0.3 0.32 0.340.05

0.1

0.15

0.2

0.250.201

0.08

V_pfr X1( )

0.350.26 X1

0.26

0.075 (1 16.5(1 )( )1

X xV X dxx

⋅ + −=

−∫

if Vpfr=0.15 m3 as before, X1~0.31, so the hippo survives

Hippo’s digesting problem• The hippo had to have surgery to remove a blockage. Unfortunately, the surgeon, Dr.

No, accidentally reversed the CSTR and PFR during the operation. Oops!! What will be the conversion with the new digestive arrangement? Can the hippo survive?

( )0 2 1AA

F X XV

r−

=−

CSTR:

PFR:1

0.0

0.075 (1 16.5(1 )( )1

X xV X dxx

⋅ + −=

−∫

if Vpfr=0.15 m3 as before, X1~0.11

0.06 0.08 0.1 0.12 0.140.05

0.1

0.15

0.20.198

0.066

V_pfr X1( )

0.150.05 X1

if Vpfr=0.45 m3 as before, X2~0.420.2 0.25 0.3 0.35 0.4

0.1

0.2

0.3

0.4

0.5

0.519

0.149

v3 x( )

0.450.2 x

Design structure for isothermal reactors

Example: batch operation• Calculate time necessary to achieve given conversion X for

irreversible 2nd order reaction.

A B⎯⎯→• Mole balance:

0 0A AdXN r Vdt

= −

• Rate law: 2A Ar kC− =

• Stoichiometry: ( )0 1A AC C X= −

• Combining: ( )20 1AdX kC Xdt

= −

( )01

1A

XtkC X

=−

Typical reaction times

• 2nd order reaction: ( )01

1A

XtkC X

=−

• 1st order reaction: ( )1 1ln

1t

k X=

−

CSTR• single CSTR mole balance ( )

0A

A exit

F XVr

=− ( )

0

0

A

A exit

C XVv r

τ = =−

• rate law (1st order) A Ar kC− =

• Stoichiometry: ( )0 1A AC C X= −

11

Xk X

τ ⎛ ⎞= ⎜ ⎟−⎝ ⎠• Combining or 0

1A

ACC

kτ=

+

Dahmköler number

• Dahmköler number is a ration of the rate of the reaction to the rate of the convective transport at the entrance 0

A

A

r VDaF−

=

• For the 1st order: 00 0 0

AA

A A

kC Vr VDa kF v C

τ−= = =

• For the 2nd order:2

00

0 0 0

AAA

A A

kC Vr VDa kCF v C

τ−= = =

CSTR in Series

• concentration flowing to the 2nd reactor 011 11

AA

CCkτ

=+

• design equation the 2nd reactor ( )0 1 21 22 2 2

A AA A

A A

v C CF FVr k C

−−= =

−

• so, ( )( )0

21 1 1 11 1

AA

CCk kτ τ

=+ +

• if the reactors have the same size and temperature: ( )

02 21

AA

CCkτ

=+

CSTR in Series

• in terms of Damköler number

• for n reactors

• conversion:

• when the Damköler number is above 1, a high conversion is achieved in few reactors

( )0

2 21 DaA

ACC =+

( )0

21 Da

AA n

CC =+

( )( ) ( )

00

11 ; 11 Da 1 Da

AA n n

CC X X− = = −+ +

CSTRs in parallel• Iet’s consider identical reactors with the feed

equally distributed, than conversion factors and the reaction rates are the same

0i A iA

XV Fr

⎛ ⎞= ⎜ ⎟−⎝ ⎠

0A

A

FV Xn n r

⎛ ⎞= ⎜ ⎟−⎝ ⎠

or

• So, the situation is identical to a single reactor with the size equal to the total volume of all reactors.

Tubular reactors: liquid phase• Design equation 0A A

dXF rdV

= −

00

X

AA

dXV Fr

=−∫

• in the absence of pressure drop

• Let’s consider a 2nd order reaction2

A Ar kC− =

A products→

• rate law:

• stoichiometry: ( )0 1A AC C X= −

( ) ( )0 0

220 00 11

XA

A A

F vdX XVkC kC XX

= =−−∫

0

0

Da1 1 Da

A

A

kCXkC

ττ

= =+ +

Tubular reactors: gas phase• For T, P constant, the concentration is a function of conversion

2A Ar kC− =• rate law:

• stoichiometry: ( )

( )( )

0

0

11 1

AA AA

C XF FCv v X Xε ε

−= = =

+ +

( )( )

20

220 0

11

XA

A

X dXFVkC X

ε+=

−∫• combining:

if e0: i.e. number of moles ↑ : flow rate ↑, the residence time ↓, so X ↓

Pressure drop in Reactors

• in the liquid phase pressure drop doesn’t lead to any significant volume and therefore concentration changes can be neglected.

• in the gas phase can be an important factor

( ) 000

1 P Tv v XP T

ε= +volumetric flow:

( )( )

( )( )

0 0 0

0 00

0

11

A j j A j jjj

F X C XF TPC P Tv X P Tv XP T

η ηεε

Θ + Θ += = =

++

Pressure drop in Reactors• to account for a pressure drop we have to use differential

form of the equation

0A AdXF rdW

= −

2A Ar kC− =• rate law:

• stoichiometry: ( )( )

0 0

0

11A

A

C X TPCX P Tε−

=+

( )( )

2

0 0

0

11A

A

C X TPr kX P Tε

⎡ ⎤−= ⎢ ⎥+⎣ ⎦

• for isothermal operation:

( )( )

( )( )

2 2

0 0 0

0 0 0 0

1 111 1A A

A

C X XT kCdX P PkdW F X P T v X Pε ε

⎡ ⎤ ⎡ ⎤− −= =⎢ ⎥ ⎢ ⎥+ +⎣ ⎦ ⎣ ⎦

Ergun equation• pressure drop in a packed porous bed is described by Ergun

equation:

( )3

150 11 1.75c p p

dP G Gdz g D D

φ µφρ φ

⎡ ⎤−⎛ ⎞−= − +⎢ ⎥⎜ ⎟

⎝ ⎠ ⎢ ⎥⎣ ⎦

dominant for laminar flow dominant for turbulent flow

φ porosity=volume of void/total volumeparticle diameterpD

superficial mass velocityG uρ=conversion factor, =1 for metric systemcg

• in terms of catalyst weight: (1 ) cW Azφ ρ= −

Pressure drop in a pipe• for the flow in pipes:

22dP du fGGdL dL Dρ

= − −

• where: u – average velocity; f – Fanning friction factor; G-mass flow rate. 2

20

0

2 220 0 0

0

2 0

2 ln2

P dP dP fGGP dL PdL D

P P P PLG fD P

ρ

ρ

− + =

− ⎛ ⎞= +⎜ ⎟⎝ ⎠

1/ 22

0 0 0

41c

P fG VP DP Aρ

⎡ ⎤= −⎢ ⎥⎣ ⎦