CHEG411: Chemical Reaction Engineeirng. Fall 2010

Chemical Reaction Engineering CHEG 411 Fall 2010 Ahmed Abdala Chemical Engineering program The Petroleum Institute Abu Dhabi, UAE

CHEG411: Chemical Reaction Engineering

Introduction CHEG411: Chemical Reaction Engineering

Chemical Plants: The Refinery EXAample A

refinery converts crude oil into valuable petroleum products such as gasoline, kerosene, diesel, aviation fuel, ….etc

Catalyst/ H2 Crude Oil

Cracking, reforming,

Distillation LPG


Blending

Petrochemical Feedstock, Coke, Sulfur

Gasoline Kerosine Diesel

$$$

Processes in Chemical Plants 

The aim of any chemical plant is to produce valuable products form available raw materials



A Chemical plant involves chemical and physical processes



A chemical process is a process that involves chemical reactions



A physical process includes separation, miXAing, or blending of different materials Reactants Catalyst Raw Materials

Physical Processes

Chemical Processes Products


products

Physical Processes

products

$$$

Course Overview  This

course focuses in the chemical processes part where chemical reactions take place

 It

focuses mainly on the design of the vessels where these reactions occur, i.e. the chemical reactors

 We

do not, however, focus on the mechanical design of the reactor


Course Objectives 

The aim of the course is to answer the following questions: 1. Will a certain reaction take place under certain conditions (temperature, pressure and catalyst)? 2. If the answer for question 1 is yes, how fast will the reaction proceed? 3. What type of reactor(s) should we use? 4. What is the size of the reactor(s) and, if more than one reactor is to be used, how those reactors should be arranged?


Approach 1.

To answer the first question we need information about thermodynamic functions for the products and reactants at the reaction conditions 

2.

To answer the second question we need to have the necessary information to determine the reaction rate equation 

3.

Chemical reaction equilibria [thermodynamics]

Kinetics

To answer the third and fourth questions we need information about the reaction rate and reactants/products flow rates 

Kinetics, transport (fluid, heat and mass)


Chemical Reaction Equilibria 

Consider the gas phase reaction: aA



+

 cC + d D

bB

⇒ A

From the Law of Mass Action: 

+

b

a

B



c

a

C +

aC a D a K = a A aB a a



(1)

ai is activity of species i

fi = ai = γ i Pi 0 fi 

D

d

b



a

The true (dimensionless) equilibrium constant K is: c



d

fi is the fugacity of species i fi0 is the fugacity of species i at the standard state  For gases, standard state is 1 atm γi is the activity coefficient


(2)

Chemical Reaction Equilibria 

The true equilibrium constant, K aC a a D a = = K a A aB a c

d

b

γ C a γ Da γ A γ Ba c

d

b



c

d

PC a PD a = b PA PB a   

activity pressure equilibrium constant equilibrium const ant





d c b   + − −1  atm  s a a 

The pressure equilibrium constant, KP KP 

d

PC a PD a = b PA PB a

(4 )

Pi is the partial pressure of species i, Pi=Ci RT

The concentration equilibrium constant, KC c

KC



d c b  −  + − −1  atm  s a a 

KP 

Kγ=1 for ideal gas c



Kγ 

d

C C a C Da = b C A C Ba

(5)

For ideal gases: K P = K C ( RT

)

δ


; δ =

c

a

+

d

a

−

b

a

−1

(6 )

(3)

Equilibrium Constant versus Temperature  KP is

a function of temperature only d ln K P ∆ H Rx ,T = dT RT 2

van' t Hoff ' s equation

0  d ln K P ∆ H Rx ,T R + ∆C P (T − T R ) = dT RT 2



(8)

Integrating

K  ln  P ,T  =    K P ,T1 



(7 )

0   ∆ H Rx  1 ,T R − ∆C P (T − T R )  1   − R   T R T 

 + 

∆C P R

T  ln    TR 

If CP is assumed to be constant (∆CP=0), then K P ,T

 ∆ H Rx ,T R K P ,T R exp   R


 1 1  −    T R T  

( 10 )

(9)

Equilibrium Constant versus Temperature  For 

eXAothermic Reactions:

As T increases, the equilibrium shifts to the left. i.e. both K and XAe decreases

K

exothermic reactions

 For 

T

endothermic Reactions:

As T increases, the equilibrium shifts to the right. i.e. both K and XAe increases

endothermic reactions

K

T


Equilibrium Constant and Free Energy  The

equilibrium constant at temperature T can be calculated from the change in the Gibbs free energy as follow: 0 −RT ln K ( T )  = ∆G Rx (11) (T ) 

where

∆GRx0 = cGC0 + dGD0 - aGA0 - bGB0

(12)

 ∆G

is related to the reaction enthalpy and entropy as follow: ∆G =∆H − T ∆S


(13)

Example: Calculation of Equilibrium Constant 

Acetic acid is esterified in the liquid phase with ethanol at 100° C and atmospheric pressure to produce ethyl acetate and water according to the reaction: CH 3 COOH + C 2 H 5 OH  CH 3 COOC 2 H 5 + H 2 O





For this reaction ∆G0298 = -4.65 kJ/mol and ∆H0298 = 3.64 kJ/mol Estimate: 1.

The equilibrium constant for the reaction at 100° C. For this reaction ∆G0298 =-4.65 kJ and ∆H0298 = 3.64 kJ

2.

The mole fraction of the reaction components at equilibrium, if initially there is one mole of each of acetic acid and ethanol.


Solution: Equilibrium Constant  Calculation 



K298

of the equilibrium constant:

 K ( 298 )  ln =

0 ∆G Rx ( 298 )

4650 = = 1.876 − Rx 298 8.314 x 298.5

= K ( 298 ) exp = ( 1.876 ) 6.527

K373 

Assume constant Cp between 298 and 373° K K P (T

)

 ∆ H Rx (T R )  1 1   K P (T 1 ) exp   −  R   T 1 T  

0  K ( 373 )  ∆ H 298 1  3640  1 1   1 ln  0.295 = − = − =      R  298 373  8.314  298 373   K ( 298 ) 

= K ( 373 ) K= ( 298 ) exp (0.295 ) 8.77 CHEG411: Chemical Reaction Engineering

Solution: Equilibrium Composition CH 3 COOH + C 2 H 5 OH  CH 3 COOC 2 H 5 + H 2 O aCH 3COOC 2 H 5 aH 2O K = aCH 3COOH aC 2 H 5 OH

  The

activity coefficient (γ) is assumed to be 1 K= K= C

 Calculation

  ) 

C CH 3COOH C C 2 H 5 OH

of equilibrium composition

C CH 3COOC 2 H 5 C H 2O = KC = C CH 3COOH C C 2 H 5 OH

 X = 8.77   (1 − X

C CH 3COOC 2 H 5 C H 2O

X2 = 2 (1 − X )

 X   (1 − X

2

⇒

= X 0.75

  ) 

CH3COOH

C2H5OH

CH3COOC2H5

Initial, Ni0

1

1

0

0

Change, dNi

-XA

-XA

XA

XA

Equilibrium, Nie

1-XA

1-XA

XA

XA

(1-XA)/2

(1-XA)/2

(XA)/2

(XA)/2

(1-0.75)/2 = 12.5%

(0.75)/2 = 37.5%

(0.75)/2 = 37.5%

Mole fraction


H 2O

2

(1-0.75)/2 = 12.5%

EXample 2: Gas phase 

The water-gas shift reaction:

CO 2 +H 2 CO + H 2 O 

is carried out under different conditions. Calculate the equilibrium conversion of steam following cases: a) b) c) d) e)

The reactants consist of 1 mol of H2O vapors and 1 mol of CO. the temperature is 1100° K and the pressure is 1 bar Same as in a but the total pressure is 10 bar The reactants consist of 2 mol of H2O vapors and 1 mol of CO. the temperature is 1100° K and the pressure is 1 bar The reactants consist of 1 mol of H2O vapors and 2 mol of CO. the temperature is 1100° K and the pressure is 1 bar The reactants consist of 1 mol of H2O vapors, 1 mol of CO, and 1 mol of CO2. the temperature is 1100° K and the pressure is 1 bar

The equilibrium constant (K) for this reaction at 1100° K is 1. Assume ideal gas. CHEG411: Chemical Reaction Engineering

Chapter 2: Conversion and Reactor Sizing

Chemical Reaction Engineering

Batch Reactor Design Equation 

The reaction



Can be rewritten as:

aA + bB → cC + dD c d A+b B → C + D a a a

Conversion, Xi, is used to quantify how far the reaction proceeds in the right direction  Conversion, XA, is the number of moles of A that have reacted per mole of A fed to the system 

XA = 

Moles of A reacted Moles of A fed

The maximum conversion  

For irreversible reaction is 1 For reversible reaction is the equilibrium conversion, Xe


Batch Reactor Design Equation 

Assume number of moles of A initially fed to the reactor is NA0



Number of moles of A reacted after a time t  Mole of A   reacted  =  Mole of A  Mole of A reacted     fed   Moles of A Fed   ( consumed ) =N A0 X



Number of moles of A that remain in the reactor after a time t (NA):  Mole of A   in reactor  =    at time t 

 Mole of A    initially fed   Mole of Athat   −  have been consumed   to reactor at      by chemical reaction  = t 0  

N A0  − N A0 X   N A  = Chemical Reaction Engineering

⇒

NA = N A0 ( 1 − X

)

Batch Reactor Design Equation  For 

ideal batch reactor

No spatial variations in concentration nor reaction rate

 The

mole Balance for species A: dN A = r AV dt

 Since

N A =− N A0 ( 1 X

)

dN A = r AV dt

⇒

N A0

dX = −r A V dt

t = N A0 ∫

X

0


dX −r A V

⇒

dN A dX 0 − N A0 = dt dt

dX − N A0 = r AV dt

Design equation

Design equation

( differential form )

( Integral )

Constant Volume Batch Reactor  For

constant volume batch reactor V=V0 dN A = ⇒ r AV = r AV 0 dt

d ( N A / V0 ) dt

= rA

dC A = − rA dt

 The 

Design Equation becomes:

The differential form N A0



dX = − rA V dt

⇒

C A0

dX = − rA dt

The integral form = t N A0 ∫

X

0


X dX dX = ⇒ t C A0 ∫ 0 −r − rA V A

Conversion for Flow Reactors  For

flow reactors Mole of A reacted Moles of A fed Moles of A fed Mole of A reacted Mole of A reacted •X = • tim e Moles of A fed time

X = FA0

Molar rate at  Molar Flow rate     Molar flow rate   at which A is  − which A is consumed  =  at which A        fed to the system   within the system   leaves the system 

FA0 

−

FA0 X  = FA FA0 ( 1 − X


=

)

FA 

Molar Flow Rate for Ideal Gas System FA = C A0 ν 0  For

ideal gas system C = A0

PA0 = RT0

FA0 C= = A0ν 0

y A0 P0 RT0

PA0 y P = ν 0 ν 0 A0 0 RT0 RT0


ideal gas system

ideal gas system

Design Equation for CSTR 

For the reaction:

c d A+b B → C + D a a a



The volume of batch reactor, V



With



The reactor volume becomes

= FA FA0 ( 1 − X

V =

FA0

−

( FA0 − FA0 X ) −r A

V =


FA0 X ( −r A )exit

V =

)

FA0

− FA −r A

Tubular Flow Reactor PFR 

The differential Equation for PFR: -dF −r A = A dV FA = FA0 ( 1 − X

)

⇒

− FA0 dX dFA =



But



Hence, the differential Equation becomes FA0 −r A =



dX dV

The integral Equation for PFR: Fj

dFj

Fj 0

rj

V =∫

X dx -FA0 dx = V ∫= FA0 ∫ 0 0 −r rA A X


Volume of Plug Flow Reactor  The

volume of PFR can be obtained by plotting The parameter FA0/(-rA) versus conversion X

 The 

volume correspond to the area under curve

Dashed Area

 The

area of equivalent CSTR reactor is the area under curve plus the dotted area 

CSTR volume= the area of



the rectangle PFR volume = area under curve (dashed area)


Packed-bed Reactor (PBR) 

The differential form of the design Equation for PBR: FA = FA0 ( 1 − X

)

-dF −r A' = A dW



But



Hence, the differential Equation becomes

⇒

dFA = − FA0 dX

−r A' = FA0 

dX dW

The integral Equation for PBR: W =

= W


Fj

dFj

Fj 0

r A'

∫

X -F dx X dx A0 = F A0 ∫0 ∫0 rA' − rA'

CSTRs in Series 

Consider two CSTRs connected in series: 

For the first reactor V1 =



Mole Balance around the second reactor In FA1

FA0 ( 1 − X

− −

)

Out FA 2

+ +

Generation V 2 rA 2

− FA0 ( 1 − X 2 ) +V 2 r A 2

V2 =



FA0 X 1 ( − r A )1

FA0 ( X 2 − X 1 )

( −r A ) 2

The volume of 2 CSTRs connected in series is smaller than the volume of one CSTR to achieve the same conversion


= 0 = 0 = 0

CSTRs in Series 

For n CSTRs connected in series we have: = Vn



As the number of CSTRs increases: 



FA0 (X − X ) ( − r A )1 n n − 1

The volume of the CSTRs becomes equivalent to that of PFR to achieve the same conversion

PFR can be modeled with a large number of CSTRs connected in series


PFRs in Series 

For two PFRs in series we have: 

The volume of the first reactor V1 = FA0 ∫

X1

0



The volume of the second reactor V2 = FA0 ∫



X2 X1

dx − rA

The total volume of the two reactors = V FA0 ∫ 1 + V2

X1

0



dx − rA

X 2 dx X 2 dx dx + FA0 ∫= FA0 ∫ X1 − r 0 − rA − rA A

The volume of n PFRs in series is the same as the volume of one PFR to achieve the same conversion


Combination of CSTRs and PFRs in Series  CSTRs

and PFRs can be connected in series

 The

arrangement of the reactors in this case is of high importance

 For 

the case of two CSTRs and one PFR:

Option 1: CSTR-CSTR-PFR FA0, X=0

FA1, X1 FA2, X2



Option 2: CSTR-PFR-CSTR

FA2, X2

FA0, X=0 FA1, X1



Option 3: PFR-CSTR-C,STR FA0, X=0


FA1, X1

FA3, X3

FA2, X2 FA3, X3

V1

FA0, X=0

V3

V2 X1

FA0/(-rA)

FA0/(-rA)

Combination of CSTRs and PFRs in Series

X1

X

V1 X1

X1

FA1, X1 FA2, X2

FA0, X=0


X1

X FA1, X1

FA3, X3

V3

V2

X1

FA2, X2 FA3, X3

Space Time and Space Velocity 

Space time is the time necessary to process one reactor volume of fluid based on the entrance conditions τ≡





V v0

The space time is equal to the mean residence time, tm Space Velocity (SV) is the reciprocal of the space time v0 1 SV ≡


V

=

τ

Reactor

Mean Residence Time

Batch

1 5 min - 20 h

CSTR

1 0 min- 4 h

Tubular

0. 5 s - 1 h

Space Velocity 

Space velocity commonly reported as either: 

Liquid Hourly Space Velocity, LHSV LHSV ≡



v0

liquid @ 60°F

V

Gas Hourly Space Velocity, GHSV

LHSV ≡


v0

Gas @ STP

V

Chemical Reaction Engineering CHEG 411 Fall 2010 Chapter 3: Rate Laws and Stoichiometry

Ahmed Abdala Chemical Engineering program The Petroleum Institute Abu Dhabi, UAE

Rate Laws and Stoichiometry

Chapter 3


Types of Reactions 

Reaction can be classified according to: Phases

• Homogeneous • Involves one phase • Heterogeneous • Involves more than one phase • Interfacial • Occurs on the interface of two or more phases Chemical Reaction Engineering

Direction • Irreversible • Reversible

Molecularity • Unimolecular • Bimolecular • Termolecular

The Rate of Reaction and the Rate Law  The

rate of reaction is usually reported as the rate of disappearance of the limiting species A, -rA

 -rA 

is function of temperature and concentration k (T )  fn (C A ,C B , ..... )  −r A =

k(T) is the rate “constant” and is a function of temperature

 The

rate equation is an algebraic equation that relates –rA to the species concentration

 For 

.

the reaction:

aA + bB → cC + dD

The rate constant for different species are related as follow: − rA − rb rC rD ri = = = = a b c d νi


⇒

k A k b kC k D = = = = a b c d

ki vi

Reaction Rate Equation: Power Law Models Zero Order

•

mole = −rA k ; = [k ] volume ⋅ time

1 −rA k C A= ; [k ] First = Time Order Volume Second = −rA k C = ; [k ] Mole ⋅ Time Order 2 A

1− n

n-th Order

 Mole  ; [k ]  = −rA k C =  V olume  


n A

 1    Time  

Reaction Rate Constant: k k

is the specific reaction rate (constant) and is given by the Arrhenius Equation: k = A e  −E   RT   

E is the activation energy, kcal/mol



R is the gas constant



T is the Absolute temperature

k

Products

T

Reaction coordinates Chemical Reaction Engineering

Ln(k)

Low E ∆Hr

E



Reactants

A is the frequency factor (same units as k)

Potential Energy



Slope=-E/R

High E 1/T

Reaction Rate Constant: In Class Exercise  The

rule of thumb that the rate of reaction doubles for a 10° C increase in temperature occurs only at a specific temperature for a given activation energy. 

Develop a relationship between the temperature and activation energy for which the rule of thumb holds.



Determine the activation energy and the frequency factor from the following data: k (min-1) T (° C)


0.001

0.050

0.500

2.00

0

100

200

300

Power Law Model and Elementary Laws  The

dependence of reaction rate on concentration is usually determined by experimental observation  It can also be postulated from theory  The rate law is the product of concentration of the individual reacting species each of which is raised to a power : −r = k C αC β A



A

A

B

Where  α is the reaction order with respect to species A  β is the reaction order with respect to species A  n= α+β is the overall reaction order

 The

dimension of the rate constant k will vary depending on the reaction order (Concentration ) {k } ≡


1 −n

Time

Elementary Reactions  Elementary

reaction are reactions that involves a single step

 The

stoichiometric coefficients in elementary reaction are identical to the powers of the Rate Law

 The

equation can be written based on the stoichiometric equation aA + bB → cC + dD



For the reaction:



The rate equation is: −r A = k AC AaC Bb


“Pseudo” Elementary Reactions  The

powers of the rate law for some reactions are identical to the stoichiometric coefficients but it involves more than one step 

These reactions are nonelementary but “follow elementary rate law”



Example is the oxidation of nitric oxide: 2 NO + O 2 → 2 NO

which has the rate equation 2 −r NO = k NO C NO C O2

but the mechanism involves more than one step


Nonelementary Reactions 

Many reactions does not follow simple rate laws: 

Reactions with non-integer reaction order CO + Cl 2 → COCl 2 ;



3

− rCO = kCO CCO CCl2 2

Reaction with rate equation that cannot be separated into temperature dependent term and concentration dependent term: 2 N 2 O → 2 N 2 + O2 ;

k N 2O C N 2O − rN 2O = 1 + k' CO2

The overall reaction order cannot be stated  At the beginning of the reaction: 

 



low concentration of O2, the reaction is apparent first order  1st order w.r.t. N2O and zero order w.r.t. O2

At the end of reaction:  

high concentration of O2 and low concentration of N2O The reaction has an apparent order of 0  1st order w.r.t. N2O and order of -1 w.r.t. O2


Heterogeneous Reaction Rate  For

catalyzed reactions the reaction rate is expressed as the rate of disappearance of species A per unit mass of moles the catalyst: −r ≡ ' A

catalyst mass ⋅ time

 For

gas catalyzed reaction the rate law is usually written in terms of partial pressure Catalyst C 6 H 5 CH 3 + H 2 → C 6 H 6 + CH 4 ;

− rC' 6 H 5 CH 3 =

 Where  

kPH 2 PT 1 + K B PB + K T PT

k is the rate constant, [mole/(kg cat.s.kPa2)] KB, KT are the adsorption constant, [kPa-1]

 -rA’ and

-rA are related through the bulk density ρb −r = ρ ( −r ) A


b

' A

Reversible Reactions  For

the reversible reaction: kf  → cC + dD aA + bB ←  kr



The equilibrium constant, KC c d C Ce C De kA = = KC a b C Ae C Be k −A

 Assuming

elementary reaction



k AC AaC Bb The rate of disappearance of A: −r A ,forward =



The rate of formation of A: r A ,reverse = k − AC Cc C Dd



The overall rate of formation of A,

(

)

= r A r A ,reverse − −r A ,forward = k − AC Cc C Dd − k AC AaC Bb 

The overall rate of disappearance of A: = −r A k AC AaC Bb − k − AC Cc C Dd


Reversible Reactions  a b k −A c d  c d C k C CC D  = −r A k AC AaC Bb − k − AC = C D A  C AC B − k A    a b C Cc C Dd  = k A  C AC B −  KC  

 Similarly

 a b C Cc C Dd  = − r B k B  C AC B −  KC    a b C Cc C Dd = rC k C  C AC B − KC   a b C Cc C Dd r D = k D  C AC B − KC 

 Remember:

−r A = a kA = a

     

−r B rC r D = = b c d k B k C k D k − A k − B k −C k − D = = ; = = = b c d a b c d


In Class Exercise:  Write

the rate law for the following reactions assuming each reaction follows an elementary rate law: C2 H6  → C2 H4 + H2

C2 H4 + 12 O2  →

( CH )

3 3

CH2 − CH2 O

 → C H + 2CH COCH COOC ( CH3 ) 3 ←  2 6 3 3

 → n − C4 H10 ← 

i − C4 H10

 → CH COOC H + C H OH C4 H9 OH + CH3 COOC2 H5 ←  3 4 9 2 5


Stoichiometry  → cC + dD aA + bB ← 



For the reaction



The relative rates of reaction are related through the stoichiometric coefficients as follow −r A −r B rC r D = = = a b c d



We can rewrite the reaction as follow: A+ 



b  c d → B ← C + D  a a a

Everything is put on a basis of “per mole of A”

We can now setup a stoichiometric table for different reaction systems


Stoichiometric Table: Batch System 

For the reaction



Which can be written As A+



NA0

 → cC + dD aA + bB ← 

NB0 NC0

b  →c C + d D B ←  a a a

ND0 NIO

The Stoichiometric is as follow: Species

Initial Moles Ni0

Change ∆Ni

A

NA0

- (NA0X)

B

NB0

- b/a (NA0X)

C

NC0

c/a (NA0X)

D

ND0

d/a (NA0X)

I, inert

NI0

0

Totals

NT0

Remaining moles Ni

NA

= N A N A0 − N A0 X b = N B N B 0 − N A0 X a c = N C N C 0 + N A0 X a d = N D N D 0 + N A0 X a

t=t

c

NC NI

N i = N i0

d

NB ND

b



NA0X(d/a+ c/a- b/a- 1 ) N T= N T 0 +  + − − 1  N A0 X a a a 


t=0

Stoichiometric Table  For 

the reaction

 → cC + dD aA + bB ← 

Moles reacted of A ∆NA = NA0X



Moles reacted of species i ∆Ni= (νi/νA)NA0X



Total number of moles, NT 

Where ,

= N T N T 0 + δ N A0 X

δ =


d c b + − −1 a a a

Concentration Equations for Batch System Ci =

N A N A0 − N A0 X C = = A V V

NB = = C B V

N B0 −

b N X a A0 V

NA V

= C C

NC = V

c N X a A0 V

NC 0 +

= C D

ND = V

N D0 +

d N X a A0 V

N i0 C i0 y i0  Let = θi = = N A0

C A0

y A0

  ν N A0 θ B − i X  νA   Ci = V N A N A0 − N A0 X C = = A V V

b   N A0 θ B − X  NB a   C = = B V V


c   N A0 θC + X  NC a   C = = C V V

d   N A0 θ D + X  ND a   C = = D V V

Concentration Equations for Constant Volume Batch System 

For constant volume system, V=V0  



Liquid phase reaction Gas phase reaction with δ=0

The concentration of species i becomes, Ci=Ni/V0 C = A



N A N A0 ( 1 − X ) = = C A0 ( 1 − X V0 V

)

b   N A0 θ B − X  NB a   C = = = C A0 B V0 V0

b    θB − a X   

c   N A0 θC + X  NC a   C = = = C A0 C V0 V0

c    θC + a X   

d   N A0 θ D + X  ND a   C = = = C A0 D V0 V0

d   θD + a X   

Now the rate equation can be reported in terms of CA0 and X


Stoichiometric Table: Flow Systems 

For the reaction: A+

Species

FA0 FB0 FC0 FD0 FI0

b  →c C + d D B ←  a a a

Feed rate to reactor

Leaving

Entering

A +

Change within reactor

a c d B → C + D b a a

FA FB FC FD FI

Effluent rate from reactor

A

FA0

−FA0 X

B

FB0 = FA0 θ B

C

FC0 = FA0 θ C

D

FD0 = FA0 θ D

b − FA0 X a c + FA0 X a d + FA0 X a

I

FI0 = FA0 θ I

0

FI = FA0 θ I

FT= FA0 (1 + θ B + θ C + θ D + θ I ) 0

d c b  FA0 X  + − − 1  = δ FA0 X a a a 

d c b  FT = FT 0 + FA0 X  + − − 1  = FT 0 + δ FA0 X a a a 

Total


= FA FA0 (1 − X )  b  = FB FA0  θ B − X  a   c FC FA0  θ C + X  = a   d FD FA0  θ D + X  = a  

Stoichiometric Table: Flow Systems 

For the reaction:

A+

( ( ) )

b  c d → B ← C D +  a a a

i Fi FA0 θ i − a X C= = i

ν

ν

C (θ − ( i ) X ) ; ( ( a) X ) = a

FA0 Ci = θi − i

ν0

Species


A

FA0

B

FB0 = FA0 θ B

C

FC0 = FA0 θ C

D

FD0 = FA0 θ D

I

FI0 = FA0 θ I

Total

A0

Change −FA0 X

i

constant ν = ν0 =

Effluent Concentration

Effluent rate from reactor = FA FA0 (1 − X )

CA =

 b  b FB FA0  θ B − X  C= − FA0 X= a  B a   c  c FC FA0  θ C + X  C= FA0 X = a  C a   d  d FD FA0  θ D + X  C= FA0 X = a  D a  0

  FT 0 FA0  1 + ∑ θ i  δ FA0 X = i  


FI = FA0 θ I F= FT 0 + δ FA0 X T

C= I

FA FA0 (1 − x )

ν FB = ν FC = ν FD = ν FI = ν

ν

(

Effluent Conc, v=constant C A C A0 (1 − x ) =

)

FA0 θ B − ( b a ) X CB C A0 θ B − ( b a ) X =

(

ν

)

(

ν

)

(

)

FA0 θ C + ( c a ) X = CC C A0 θ C + ( c a ) X

(

FA0 θ D + ( d a ) X = CD C A0 θ D + ( d a ) X

FA0 θ I

ν

ν

(

CI = C A0 θ I

)

)

In Class Exercise 

For the following liquid-phase elementary reaction:

2 A + B  → 2C 

Write the rate law in terms of conversion, XA

Species


Change

A B C I Total Chemical Reaction Engineering

Effluent rate from reactor

Effluent Concentration

Effluent Conc, v=constant

Variable Volume Batch Reactor  For

gas- phase reaction in a batch reactor where the volume of reactor is not constant  The concentration of individual component can be determined by expressing the volume at any time t as follow: PV = ZN RT  At

T

time t=0 this equation becomes P0V0 = Z 0NT 0RT0

 The

reaction volume at time t, V, as function of the initial volume, V0, becomes  P0 V = V0  P


 T  Z  NT     T0  Z 0  NT 0

Variable Volume Batch Reactor  But

N = NT 0 + δ N A0 X T

 Then

NT N 1 + A0 δ X = 1 + y A0δ X = 1+ ε X = NT 0 NT 0 = ε y= δ A0

 Where 

NT 0 X

At X=1, NT=NTf and ε becomes ε= =



N T − NT 0

N Tf − NT 0 NT 0 Chnage in total numbers of Moles at complete conversion total moles fed

And the volume at time t becomes (assuming Z ≈ Z0)  P0  T = V V0   (1 + ε X ) T0 P


Variable Volumetric Flow Rate Flow Reactors  The

total concentration at any point of the reactor, CT is given by FT =

C= T

ν

 The

P ZRT

concentration at the entrance of the reactor, CT0 F P C= T0

T0 =

ν0

0

Z 0 RT0

 The

volumetric flow rate at some point along the reactor, v as function of the volumetric flow rate at the entrance, v0  FT  P0  T     F  T 0  P  T0 

ν = ν0  Chemical Reaction Engineering

Concentrations in Terms Other Than Conversion  The

concentration of species j at any point of the reactor, Cj is given by Fj C= = j

ν

Fj  FT P0 T   F P T 0   T0

ν0 

 The

concentration at the entrance of the reactor, CT0 FT 0  Fj  ν 0  FT

  P   T0      P T   0 

 Fj C j = CT 0  F  T

  P   T0      P T   0 

Cj =


Variable Volumetric Flow Rate Reactors 

To express the concentration as function of conversion for variable v reactors 

We have



Then

 FT ν = ν0   FT 0

= ν

 P0  P

FT 0 + FA0δ X  P0  FT 0 P

T     T0 

F= FT 0 + FA0δ X T

and

T P0 = + ν δ 1 y X ( )  0 A0 T P  0

T     T0 

P0  T  ν == ν 0 (1 + ε X )   P  T0  

The concentration of species j, Cj becomes Fj C= = j

ν

(

)

(

)

C A0 θ j + υ j X  P =  P0   ε 1 + X P0 T ( )  ν 0 (1 + ε X )   P  T0 


FA0 θ j + υ j X

 T0  T

Summary


In-Class Exercise (Fogler P3-13) 

The gas-phase reaction

1 2

N2 + 23 H2  → NH3

is to be carried out isothermally. The molar feed is 50% H2, and 5O% N2, at a pressure of 16.4 atm and 227 °K. a) Construct a complete stoichiometric table. b) What are CA0, δ, and ε ?. Calculate the concentrations of ammonia and hydrogen when the conversion of H2 is 60%. c) Suppose by chance the reaction is elementary with kN2 = 40 dm3/mol/s. Write the rate of reaction solely as a function of conversion for: i. A flow system ii. A constant volume batch system Chemical Reaction Engineering

Chapter 3 Review “Rate Laws and Stoichiometry”


Reaction Rate Equation  -rA

is function of temperature and concentration k (T )  fn (C A ,C B , ..... )  −r A =

 The

rate constant for different species are related as aA + bB → cC + dD follow:

 Rate 



constant

 −E k = A exp   RT

Units

⇒

k A k b kC k D = = = = a b c d

E 1  ln k ln A ; = −  RT  1− n

mole  [ k ] =    volume 


ki vi

Low E Ln(k)

− rA − rb rC rD ri = = = = a b c d νi

Slope=-E/R

 1     time 

High E 1/T

Rate Law  For

the reaction:

 The

power-law rate equation:

aA + bB → cC + dD

−r A = k AC AαC Bβ

 If

the reaction is elementary or pseudo-elementary −r A = k AC AaC Bb

 Non-power-law

forms 2 N 2 O → 2 N 2 + O2 ;


k N 2O C N 2O − rN 2O = 1 + k' CO2

Heterogeneous Reaction Rate moles catalyst mass ⋅ time

−r A' ≡

 Gas

catalyzed reaction the rate law is usually written in terms of partial pressure Catalyst C 6 H 5 CH 3 + H 2 → C 6 H 6 + CH 4 ;

 Where  

− rC' 6 H 5 CH 3 =

kPH 2 PT 1 + K B PB + K T PT

k is the rate constant KB, KT are the adsorption constant, [kPa-1]

 -rA’ and

-rA are related through the bulk density ρb (

−r A = ρb −r A'

k A = ρb k 'A Chemical Reaction Engineering

)

Reversible Reactions  For

the reversible reaction: kf  → cC + dD aA + bB ←  kr



The equilibrium constant, KC c d C Ce C De kA = = KC a b C Ae C Be k −A

 Assuming

elementary reaction



k AC AaC Bb The rate of disappearance of A: −r A ,forward =



The rate of formation of A: r A ,reverse = k − AC Cc C Dd



The overall rate of formation of A,

(

)

= r A r A ,reverse − −r A ,forward = k − AC Cc C Dd − k AC AaC Bb 

The overall rate of disappearance of A: = −r A k AC AaC Bb − k − AC Cc C Dd


Stoichiometry  → cC + dD aA + bB ← 



For the reaction



We can rewrite the reaction as follow: A+ 



b  c d → B ← C + D  a a a

Everything is put on a basis of “per mole of A”

Stoichiometric i   N i N A0  θ i + X  = a  

i   C i C A0  θ + X  = a  

liquid phase Gas Phase

i   Fi FA0  θ i + X  = a  

 F V =V 0  T  FT 0

  P0   P

 T  Z  T  Z   0  0 

i   F  P C i C A0  θ + X   T 0   = a   FT   P0 


 FT  FT 0

ν = υ0 

  P0   P

  T0   Z 0      T  Z 

 T  Z  T  Z   0  0 

Stoichiometric Table  → cC + dD aA + bB ← 



d c b δ = + − −1 a a a



F= FT 0 + FA0δ X T



N = NT 0 + N A0δ X T


F FT 1 + A0 δ X = 1 + y A0δ X = 1+ ε X = FT 0 FT 0 N NT 1 + A0 δ X = 1 + y A0δ X = 1+ ε X = NT 0 NT 0

Chemical Reaction Engineering CHEG 411 Fall 2010 Chapter 4


Isothermal Reactor Design

Chapter 4


End

Start General mole balance equation V

dN FA0 − FA + ∫ rA dV = A dt

Chapter 1

Mole Balance to specific reactor design equations in terms of conversion

is –rA=f(X) given?

Evaluate the design equations numerically or analytically to Estimate the reactor volume the reaction time, or conversion

Chapter 2

Yes Chapter 2

Obtain –rA=f(X)

No

Using stoichiometry, express concentration as function of X Chapter 3

Chapter 3

Elementary reaction

Yes

Determine the rate law in terms of concentration

Chapter 4

Liquid phase or gas phase with P=P0

Combine mole balance, rate law and stoichiometry and transport law, and ∆P terms in an ODE solver

No

Chapter 4

Algorithm for Reaction Design  The 1.

solution of reactor design involve the following steps: Mole balance

 2.

Based on the limiting species

Rate law 

Based on reaction type, mechanism, or experimental data

3.

Stoichiometry

4.

Combine the mole balance, the rate equation, and the Stoichiometry relations

5.

Evaluate to get reactor volume, reaction time, or conversion 

Analytical, numerically, graphically, or using software


Evaluate Combine Stoichiometry Rate Law Mole Balance

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Scale UP of Liquid Phase Batch Reactor Data to the Design of a CSTR 



To design of a full-scale plant: 1.

Microplant, laboratory-bench-scale unit is used to collect data about the process

2.

A pilot-plant maybe designed based on the laboratory data

3.

The full-plant is designed by scaling up the pilot plant data

Step 2 can be surpassed and a full scale plant may be designed based on laboratory data. However, this requires thorough understanding and careful analysis of the laboratory data


Liquid Phase-Batch Operation  For

a batch system, the mole balance equation is: 1  dN A  V  dt

 For

  = rA 

a liquid phase reaction, the volume is constant and the mole balance equation can be written in terms of concentration as follow: 1  dN A  =   V  dt 

1  dN A  =   V0  dt 

d ( N A V0 ) = dt

dC A dt

dC A = rA dt

 This

form will be used to analyze the reaction rate data


Batch Operation  For

batch system, the time calculated to achieve certain conversion is the reaction time

 The

total time required to process one batch, total cycle time, tt Total Cycle time = Filling time + Heat time + Reaction time + Cleaning time tc = tf + te + tR + tc

 The

reaction time is usually optimized with the processing time to produce the maximum number of batches per day


Design of CSTR  The

volume of CSTR reactor to achieve conversion X = V

 And

FA0 X ν 0 C A0 X = −rA ( −rA )exit

the space time, τ = τ

 For

V C A0 X = ν0 −rA

a first order liquid phase reaction: -rA = k CA=k CA0(1-X) X τ= k (1 − X ) CA =

C A0 1 +τk


⇒

τk X= 1 +τk

Design of CSTR 

Damkohler number, Da, is the ratio of the rate of reaction at entrance to the entering flow rate of A = Da

−rA0 V Reatcion rate = FA0 Convection rate

−rA0 V kC A0 V = = kτ FA0 C A0 ν 0

For a first order reaction:  The conversion is related to Da as follow = Da



X =





1 0.9

At high Da, Da ≈ 10, X is greater than 90% At low Da, Da ≈ 0.1, X is less than 10%

For a second order reaction: −rA0 V kC V = Da = = τ kC A0 FA0 C A0 ν 0 2 A0


0.8 0.7 first order reaction

0.6 X



kτ Da = 1 + kτ 1 + Da

0.5 0.4 0.3 0.2 0.1 0

0

10

20 Da

30

CSTR in Series: First Order Reactions 

For two CSTRs in series: C A1



C A0 C A1 = and C A 2 1 + τ 1 k1 1 + τ 2 k2

C A1 = 1 + τ 2 k2

C A0

= + k τ 1 1 ( 1 1 )( + τ 2 k 2 )

CA2 X2

-rA1 V1 -rA2 V2

C A0

(1 + τ k )

2

For n reactors connected in series = C An



CA1 X1

If both reactors are of equal size ( τ1 =τ1 = τ) and operates at same temperature (k1 =k2=k) = CA2



CA0

C A0 C A0 = ( 1 + τ k )n ( 1 + Da )n

If the final conversion based on reactor n is X C An= C A0 (1 − X= )


C A0

(1 + Da )

n

1 1 X= 1− = 1 − n n (1 + Da ) (1 + τ k )

CSTR in Series: First Order Reactions 

For n CSTRs connected in series 1 1 X= 1− = 1 − n n (1 + Da ) (1 + τ k )



The rate of disappearance of species A the nth reactor is −r= = k kC An A

C A0

(1 + τ k )

n

1

Da = 1 Da = 0.5

0.9 0.8 0.7

Da = 0.1

X

0.6 0.5 0.4 0.3 0.2 0.1 0 0

1

2

3

4

5

6

7 n


8

9 10 11 12 13 14

CSTR in Parallel 

For n equal size CSTRs connected in parallel, operates at the same temperature and the feed is distributed equally among them

FA01

-rA1 V1

X= X= ................ = X= X 1 2 n



FA02

The rate in each reactor is −rA1 = −rA 2 = ................ = −rAn = −rA





The volume of each reactor And the total volume V, = V


X1

FA0

V FA0  X i  = V=   i n n  −rAi  FA0n

FA0 X i FA0 X = −rAi −rA

Xn

Design of PFR 

Recall the assumption for PFR: 





No radial variation in velocity, temperature, concentration, nor reaction rate Plug flow assumption is valid when Re > 2300

The design equations of PFR 

Differential:

FA0 

dX = −r A dV

Integral: V = FA0 ∫

X

0


dX −r A

Reactants

Products

Second Order Liquid Phase Reaction in PFR   

→ Products For the second order reaction A  −r = kC The rate law is The differential form of the design equation becomes 2 A

A

kC A2 dX = dV FA0

1.

liquid phase isothermal reaction = C A C A0 ( 1 − X

)



From stoichiometry



Combine the design equation with the stoichiometry

w hy

2 2 kC A0 (1 − X ) dX kC A0 ( 1 − X ) = = dV FA0 C A0 υ0 V X FA0 X υ0 dx dx = = V ∫= dV 2 2 ∫0 0 kC A0 ( 1 − X ) kC A0 ∫0 ( 1 − X 2



Evaluate

τ kC= = Da A0 2

V

υ0

X = Chemical Reaction Engineering

kC= A0

Da 2 1 + Da 2

2

dx

∫ (1 − X ) X

0

2

)

2

Second Order Gas Phase Reaction in PFR   

→ Products For the second order reaction A  −r = kC The rate law is The differential form of the design equation becomes 2 A

A

kC A2 dX = dV FA0

2.

Gas phase isothermal reaction with no pressure drop (P=P0)

FA0 ( 1 − X ) FA0 ( 1 − X ) (1 − X ) = = C A0 υ0 ( 1 + ε X ) υ0 ( 1 + ε X ) (1 + ε X )



From stoichiometry



Combine the design equation with the stoichiometry



Evaluate

C A ==

FA0 = V ∫= dV 2 0 kC A0 V

V =

υ0 kC A0


FA

υ

=

( 1 + ε X ) dx ∫0 (1 − X ) 2 2

X

2  1 X ε + ( ) 2  2 ε ( 1 + ε ) ln ( 1 − X ) + ε X +  1−X   

Gas Phase reactions in PFR 



For gas phase reaction, the volume of the PFR reactor depends not only on X but also with ε Therefore, the conversion depends on both the reactor volume and ε V =

υ0 kC A0

2  1 X ε + ( ) 2  2 ε ( 1 + ε ) ln ( 1 − X ) + ε X +  1−X    1.0

ε=-0.5

0.9

ε=0 0.8

ε=1

0.7

ε=2 X

0.6 0.5 0.4 0.3 0.2 0.1 0.0 0


2

4

6 V, m3

8

10

12

Pressure Drop in Reactors 

The effect of pressure on the concentration of reactants in liquid phase reaction is insignificant 



In gas phase reactions, the concentration of reacting species is proportional to the total pressure  



Effect of pressure drop can be ignored

Proper accounting for pressure drop is crucial This is very important in packed bed system, in particular microreactors and reaction in membranes

Recall that the concentration of Species i for variable density, variable pressure system is expressed as: C = C A0


(θ i + ν i X )

P  T0    (1 + ε X ) P0  T 

Pressure Drop in Packed Bed Reactors  

The gas phase second order reaction: is to be carried in a packed bed reactor (PBR) The mole balance for Packed Bed Reactor is:

2 A  → B +C

−r ' dX = A dW FA0



The rate equation is :



From Stoichiometry for gas phase reactions:

−r A' = kC A2 C = C A0

 (1 − X ) P T0  kC A0  − rA' =   (1 + ε X ) P0 T 

2



And the rate equation becomes



Combining the rate equation and the mole balance: dX = kC 2A0 FA0 dW 

2

 (1 − X )   P       (1 + ε X )   P0 

2

(1 − X ) P  T0    (1 + ε X ) P0  T 

2

2

dX kC A0  (1 − X )   P  F1 ( X ,P ) = =   υ 0  (1 + ε X )   P0  dW

To solve this equation, the relation between the pressure drop and the catalyst weight must be known


Flow Through a Packed Bed 



Many gas phase reactions are carried in a tubular reactor packed with catalyst, Packed Bed reactor The pressure drop for flow through a packed porous bed is given by Ergun Equation: dP  G  1 − φ  150 (1 − φ ) µ dZ



= − ρ g c Dp  φ 3   DP

The only parameter that varies with pressure in the right hand side is density, ρ .

.

υ

υ0

P = pressure

φ porosity = =

volume of void = void fraction total bed volume

volume of solid 1− φ = total bed volume

µ = vis cos ity of gas z = length down the packed bed

P T F  ρ = ρ 0  0   T  P0  T   FT 0 


+ 1.75G  

DP = Particle diameter

υ ρυ= ρ 0υ 0 ⇒ ρ= ρ 0 0 υ

β0

L

g c = conversion factor (32.174 lbm .ft / s 2 .lbf or1kg m .m / s 2 .kg f )

. . m0 m ; and m m0 = = ρ = ρ0

 P0  T G (1 − φ ) 150 (1 − φ ) µ dP 1.75G = − +     .dZ ρ 0 g c Dpφ 3  DP P  T0  

z

G = superficial mass velocity ,kg / m2 .s

 FT   FT 0

P T dP = −β0 0  dZ P  T0

 FT   FT 0

Flow Through a Packed Bed 

To relate the pressure drop to the catalyst weight rather than z W=

(1 − φ )

Ac 

CrossSection Area

W = Ac z

z

ρc 

density of catalyst

ρb 

z

bulk density

P T dP = −β0 0  dZ P  T0

 FT   FT 0

β0 P0  T  FT dP = −   dW Ac (1 − φ ) ρ c P  T0  FT 0

2 α P0 dP = − (1 + ε X ) dW 2 P



By solving the two differential equation we can get the relation between W and X


dP = F2 ( X ,P ) dW dX dX == FF11 ( X X ,P ,P) dW dW

L

Analytical Solution for Reaction with Pressure Drop 

for a second order reaction we had: dX dW

ε=-0.5

2

2

kC A0  (1 − X )   P  F1 ( X ,P ) =   υ 0  (1 + ε X )   P0 

ε=0

ε=2

α PP2 dP dP α 0 (1 + ε X ) = = − FF2 (( X ,P )) 1 X X ,P = − + = ε ( ) 2 dW dW 22 PP0



P

For ε=0 we get

2

dX kC A0 2 P  F1 ( X ,P ) = (1 − X )   = υ0 dW P  0 2 dP α P0 = − = F2 ( P ) dW 2 P

W−

αW 2 2

W

0

kC A0 1 = υ0 1 − X


W, kg

= P P0 (1 − α W ) 2 1

dX kC A0 = (1 − X ) 2 (1 − α W ) dW υ0 X 0

ε=1

W

X

kC A0 1 1 W dW dX − α = ( ) ∫0 υ 0 ∫0 (1 − X ) 2

αW 2

W= − 2

kC A0  1  kC A0  X  1 = −   υ 0 1 − X  υ0  1 − X 

In-Class Exercise 

Ethyl acetate is an extensively used solvent and can be formed by the vapor- phase esterification of acetic acid and ethanol. The reaction was studied using a microporous resin as a catalyst in a packed bed reactor [Ind. Eng. Chem. Res. , 26(2), 1 98 (1 987)]. The reaction is first- order in ethanol and pseudo first- order in acetic acid. For an equal molar feed rate of acetic acid and ethanol, the specific reaction rate is 1 . 2 dm3/g cat/min. The total molar feed rate is 1 0 mol/min, the initial pressure is 1 0 atm, the temperature is 1 1 8 ° C, and the pressure drop parameter, α, equals 0. 01 g- 1 .  

Calculate the maximum weight of catalyst that one could use and maintain an exit pressure above 1 atm. (Ans: W = 99 g) Determine the catalyst weight necessary to achieve 90% conversion. What is the ratio of catalyst needed to achieve the last 5% (85 to 90%) conversion to the weight necessary to achieve the first 5% (0 to 5%) conversion in the reactor? Vary α and write a few sentences describing and explaining your findings. What generalizations can you make?


Solution Acetic Acid + Ethanol → Ethyl Acetate + Water



A+B → C+D



Mole Balance PBR

1.

FA0 dX/dW=-rA’



Rate equation

2.

-rA’=k CA=k CA



Stoichiometry

3.



Combine 2 and 3 1.

5.

(why?)

CA=CA0(1-X)/(1+εX) (P/P0)(T0/T)=CA0(1-X)(P/P0)  ε =0 and isothermal conditions CB=CA0 (FB0/FA0-X) (1+εX) (P/P0)(T0/T)=CA



4.

P0=10 atm T=118° C F0=2FA0=10 mol/min

-rA’=kCA0(1-X)(P/P0)

Combine 4 and 1

FA0 dX/dW=kCA0 (1-X) (P/P0) dX/dW=kCA0/FA0 (1-X) (P/P0) =F1(X,P)


Solution, continue 1.

X versus W

2

dX kC A0 2 P  F1 ( X ,P ) = (1 − X )   = υ0 dW  P0 

6.

Pressure drop function α P02 α P02 dP = − − = F2 ( X ,P ) (1 + ε X ) = dW 2 P 2 P



Separating the variables and integrating: P

∫ 2PdP = −

P0

α P02 W 2

= W

2 0

0

α P02

P −P = − 2

∫ dW

2

W

P02 − P 2 =W α P02

100 − 1 = 99 g 0.01x100


Solution, Part B P dX kC = We have (1 − X )   dW F P   And from the pressure drop equation



A0

A0

P02 − P 2 =W α P02



⇒

0

1 P =(1 − α W ) 2 P0

Substitute (1 - αW) for (P/P0)2 dX kC A0 = (1 − X ) (1 − α W ) dW FA0



Separating the variables F

1

A0 dX (1 − α W )0.5 dW = kC A0 (1 − X )



Integrating W

∫ 0

X

FA0 1 1 − α W dW = dX ∫ kC A0 0 (1 − X )

3 FA0 2 (1 − α w ) 2= ( − ln(1 − X ) ) kC A0 −3α


W

3 FA0 2 X − ln(1 − X ) 0 (w − αw )2 = kC A0 −3α 0

 2kC A0 X= 1 − exp   3α FA0

(1 − α W ) 32 − 1     

Analytical Solution for Reaction with Pressure Drop

ε=0

P

P

W, kg

ε=0

W, kg


Production of Ethylene Glycol  Ethylene

Glycol is produced by the first order reaction of ethylene oxide and water in presence of catalyst (H2SO4)  It is required to design a CSTR reactor to produce 200 million pounds of EG per year  The reaction will be carried at 55˚ C to avoid formation of by-products  A laboratory experiment is carried to estimate the reaction rate equation and use that equation to design the CSTR reactor (find the reactor volume)  The laboratory data was obtained using 500 ml batch reactor using 2 mol/liter of ethylene oxide in water which is mixed with mixed with 2 obtained Chemical Reaction Engineering

Chapter 4: Sizing of Isothermal reactor Reaction

Liquid System

Constant Density

Constant density, no pressure drop effect

Variable Density, Constant Pressure

Gas System Evaluate Combine Stoichiometry

Variable Density, Constant Pressure

Rate Law Mole Balance

Variable Density, Variable Pressure


Chapter 4: Sizing of Isothermal reactor 

Reaction Types 

Liquid phase, V=V0 or ν= ν0, and no pressure effect



Gas phase [V=V0 (1+εX) (P0/P) or v=v0 (1+εX) (P0/P)] 1.

∆n = 0, no pressure drop (same as liquid phase)

2.

∆n = 0, pressure drop (V=V0(P0/P) or v= v0 (P0/P)

3.

∆n ≠ 0, no pressure drop (V=V0 (1+εX) or v=v0 (1+εX))

4.

∆n ≠ 0, pressure drop (V=V0 (1+εX) (P0/P) or v=v0 (1+εX) (P0/P)) Evaluate Combine Stoichiometry Rate Law


Mole Balance

Chemical Reaction Engineering CHEG 411 Fall 2008 Chapter 4 Sections 4.6 – 4.10


Overview  Design

of Chemical Plant

 Microreactors  Membrane  Unsteady

Reactors

State Operation of CSTR


Production of Ethylene Glycol Most chemical Plants involve number of reaction and separation steps  For example, production of polyethylene glycol requires the following reaction and separation steps 

H3C

H2C

CH3

CH2

+ O

+

H2C

CH2

+ 1/2

O2

O H2C

CH2

+

Ag

H2SO 4 H2O

H2

H2C

CH2

CH2 CH2 HO HO

Some of the reactors shown will include multiple reactors connected in series or in parallel  For production of 200 tons/year of polyethylene glycol requires the following flow rates and reactor sizes 


Ethylene Glycol Plant  Ethylene

is produced in 100 parallel PFR reactors

 Ethylene

oxide is produced in 1000 parallel tubes packed with catalyst coated with silver

 Ethylene 

Profit

glycol is produced in 197 ft3 CSTR reactor

value of products − cost of reactants – operating cost – Separation costs

 Profit

can be increased by optimizing separation and conversion and using recycles


Microreactor  Microreactors

are characterized by their high surface area-to-volume ratios in their microstructured regions that contain tubes or channels.





A typical channel width might be100 µm with a length of 20000 µm The resulting high surface area-to volume ratio (ca. 10000 m2/m3 reduces or even eliminates heat and mass transfer resistance

 Microreactors

are also used for the production of special chemicals, combinatorial chemical screening, lab-on-a-chip, and chemical sensors


Design of Microreactors  The

gas-phase reaction 2 NOCl  → 2 NO + Cl 2

is carried out at 425°C and 1641 kPa (16.7 atm). Pure NOCl is to be fed and the reaction follows an elementary rate. It is desired to produce 20 tons of NO per year in a microreactor system using a bank of ten microrractors in parallel. Each microreactor has 100 channels with each channel 0.2 mm square and 250 mm length.  Assume 85% conversion, plot the molar flow rates as a function of volume down the length of the reactor.  The reaction rate constant, k, is 0.29 dm3/mol/s at 500° K and the activation energy, E, is 24 kcal/mole Chemical Reaction Engineering

Membrane Reactors  Membrane

reactors are used to:



Increase conversion of thermodynamically limited reactions



Increase selectivity for multiple reactions

 The

membrane can either provide a barrier to certain components, preventing certain components such as particulates from contacting the catalyst or contacting reactive sites and being catalyst in itself


Membrane Reactors 

Like reactive distillation, the membrane reactor is another technique for driving reversible reactions the right toward completion in order to achieve high conversions.



These high conversion can he achieved having one of the reaction products diffuse out of a semipermeable membrane surrounding the reacting mixture. 



As a result, the reverse reaction will not be able to take place and the reaction will continue to proceed to the right toward completion.

Types f catalytic membrane reactors 

Inert membrane reactor with catalyst pellets on the feed side (IMRCF)



Catalytic membrane reactor (CMR)


Types of Catalytic Membrane Reactors


Design of Membrane Reactor  The

following reaction

 → C H + 3H C6H12 ←  6 6 2

is to be carried out isothermally in a membrane reactor with no pressure drop. The membrane is permeable to Product C, but it is impermeable to all other species. The total initial concentration is 0.2 mol/dm3 and the molar flow of A is 10 mol/s. The reaction rate constant, kA is 10 dm3/kg-cat/s. the equilibrium constant, KC, is 200 mol3/dm9 and the mass transfer coefficient for H2, 0.5 dm3/kg.cat/s. Plot the concentration of the reacting species as function of catalyst mass. Chemical Reaction Engineering

Unsteady State Operation  Example

of Unsteady state Operations



Batch Reactor



Startup of CSTR



Semibatch reactor


Startup of CSTR: Steady State time  Estimate

the time required to reach steady state after the startup of CSTR reactor for a first-order liquidphase reaction

 Solution: 

Mole balance dN FA 0 − FA + rAV = A dt FA 0 FA dC − + rA = A V V dt



Rate Equation −rA = kC A


( unsteady state ) CA0

τ

−

CA

dC + rA = A τ dt

Semibatch Reactor  The

production of methyl bromide is an irreversible liquid-phase reaction that follows laws an elementary rate Paw. The reaction is carried out isothermally in a semibatch reactor. An aqueous solution of methyl amine (B)at a concentration of 0.025 mol/dm3 is to be fed at a rate of 0.05 dm3/s to an aqueous solution of bromine cyanide (A) contained in a glass-lined reactor  The initial Volume of fluid in the vessel is to be 5 dm3 with a bromine cyanide concentration of 0.05 mol/dm3. The specific reaction rate constant is k = 2.2 dm3/mol/s  Solve for the concentrations of bromine cyanide and methyl bromide and the rate reaction as a function of time. Chemical Reaction Engineering




Collection and Analysis of Rate Data

Chapter 5 Lecture 1: Differential Method


The Rate Expression 

The rate law is an algebraic equation that relates the reaction rate of reaction to the concentration of reacting species = −r A

 −E  A0 exp = C AαC Bβ k C AαC Bβ  RT     Concentation  Dependent

Temperature Dependent Term



term

The temperature dependent term can be evaluated by knowing the equilibrium constant k at different temperatures  −E  k = A0 exp    RT  = ln ( k ) ln ( A0 ) −

E 1 RT

k

1


T

Collecting and Analyzing of Experimental Data  Experimental

data obtained from Batch reactor or flow reactor system

 Data

obtained are:



Concentration-time measurement (Batch Reactor)



Concentration measurements (flow reactor)

 Data

Analysis



Differential Method



Integral Method



Method of half-lives



Method of initial rates



Linear and nonlinear regression


Batch Reactor Data    

Batch reactor is preferred for collecting data for estimating of the rate equation parameters A + B → Product For the irreversible reaction Which is carried at constant volume batch reactor Assume that the rate law takes a power-law form: −r A=





dC A = k AC Aα dt α

A

For Batch reactor and constant density reaction we know that: dC A −r A = − dt



β

k AC C B Rate equation of the form −r A = can also be converted to the previous form 1. Using excess concentration of component B 2. Using equimolar concentration of A and B

Combining with the rate expression −


dC A = kC Aα dt

( how ?)

Analysis of Batch Reactor Data: Integral Method Using the integral method the reaction order and rate constant can be determined:

 1.

Guess the reaction order

2.

Substitute the rate expression into the mole balance equation

3.

Integrate the differential design equation

4.

Linearize the time concentration relation

5.

Plot the appropriate concentration function versus time

6.

Linear relation? 

Yes, find k



No, repeat steps 2-4 Mole balance

Guess − rA = kC A2

−

dC A = kC A2 dt


Integrate CA

Linearize

t 1 1 dC A = − kt −∫ 2 = k ∫ dt C C A A 0 CA C A 0

Plot

Check

Analyzing Batch Reactor Data: 1. Differential Method 

For constant volume batch reactor: −

 

dC A = kC Aα dt

The concentration of a reactant species is measured as function of time Taking the natural logarithm for both sides of the equation  dC  ln  −  = ln ( k ) + α ln (C )  dt  A

A

Plotting ln(-dCA/dt) versus ln(CA) gives straight line  



α = the slope k can be calculated = the intercept

-dCA/dt is obtained from the CA versus t data using one of the following methods:  



Graphical method Numerical method  For equally spaced data point Polynomial fit  CA= a0 + a1t + a2t2 + a3t3 +……+ antn  -dCA/dt = -[ a1+ 2a2t + 3a3t2+……+ nantn-1]


Ln(-dCA/dt)



Ln(CA)

Numerical Estimation of –dCA/dt 

For equally spaced data points 





∆ti = constant i.e. (t1-t0)=t2-t1=t3-t2=……..ti+1-ti t, min

t0

t1

t2

t3

tn

CA, mol/liter

CA0

CA1

CA2

CA3

CAn

-dCA/dt

first

First Point Interior points:

Interior points

−3C A0 + 4C A1 − C A 2  dC A   dt  = 2 ∆t  t 0

C A( i +1 ) − C A( i −1 )  dC A   dt  = 2 ∆t  t i

t, min

0

5

10

15

20

25

C A, mol/liter

1.0

0.92

0.85

0.78

0.74

0.68

-dCA/dt



Last Point:

last

C A( n − 2 ) − 4C A( n −1 ) + 3C An  dC A   dt  = 2 ∆t  t n


Determination of Reaction order and Rate Constant 

The following reaction A → B +C

was carried out in a constant volume batch reactor and the following concentration measurement was estimated



t, min

0

5

10

15

20

25

CA, mole/liter

2

1.6

1.35

1.1

0.87

0.7

Determine the reaction order and the rate constant using the differential Method


Solution: Differential Method A → B +C

t, min

0

5

10

15

20

30

40

60

CA, mole/lit er

2

1.6

1.35

1.1

0.87

0.7

0.53

0.35

1.

Let:

dC A −r A = − = kC Aα dt

2.

Hence:

 dC A ln  −  dt

3.

Estimation of –dCA/dt graphically:

  = ln ( k ) + α ln C A 

( )

t, min

1.

Calculate

0

CA, mole/liter 2.0

 ∆C A  − ∆t   ∆C A  − ∆t 

−∆CA/∆t= C i − C i −1  = − -(Ci-Ci-1)/(ti-ti-1)  t i − t i −1 t i −1 ,t i 1.6 − 2.0  0.12 = − = 0.08  − 5 0 t 1 , t 2

5

10

15

20

30

40

60

1.6

1.35

1.1

0.87

0.7

0.53

0.35

0.08

0.063

0.042

0.033

0.021

0.017

0.009

0.1

Plot

versus time

0.08

-DCA/Dt

2.

 ∆C A   − ∆t   

0.06

0.04

0.02

0


0

10

20

30 40 t, min

50

60

70

Solution, Continued Using equal-area differentiation, t, min C mole/liter the value of –dCA/dt can −∆C /∆t= be estimated -(C -C )/(t -t )

3.

A,

0

5

10

15

22 20

30

40

60

2.0

1.6

1.35

1.1

0.87

0.7

0.53

0.35

0.08

0.063

0.042

0.033

0.021

0.017

0.009

0.07

0.053

0.035

0.0024 0.016

0.006

A

i

i-1

-dCA/dt 0.12

0.1

-DCA/Dt .

0.08

0.06

0.04

0.02

0 0

10

20

30

40 t, m in


50

60

70

i i-1

0.092

Solution, continued 

Plot ln(-dCA/dt) versus ln(CA)



Fit to a straight line



Slope = α =1.58

-1

ln(k)=-3.45 k = 0.032



(what units?)

0.1

y = 1.58x - 3.45 R2 = 1.00

-2 ln(-dCA/dt)



0

-3

-4

y = 0.032x1.580 R2 = 0.996

-dCA/dt

-5

-6 -1

-0.8

-0.6

-0.4

-0.2

0

ln(CA), mole/liter

0.01 0.1

1 CA, mole/liter


10

0.2

0.4

0.6

0.8

Collection and Analysis of Rate Data

Chapter 5 Lecture 2: Integral Method


Integral Method: Zero Order Reactions 

For zero order reaction:



Substitute the rate expression into the design equation for batch reactor dC

−r A= kC A0= k

−

dt

Separate variables and integrate −

CA

∫

C A0



= k

t

dC A = k ∫ dt 0

The concentration time relation

Plot CA versus t  Straight line?

C = C A0 − kt A



 

k = -slope Assume different order and repeat


CA



A

slope=-k time

Integral Method: First Order Reactions 

For first order reaction:



Substitute the rate expression into the design equation for batch reactor

− rA = kC A

−


CA

∫

C A0



t

dC A = k ∫ dt CA 0

The concentration time relation C ln  A0  CA

Plot ln(CA0/CA) versus t  Straight line?

  = kt 



 

k = Slope Assume different order and repeat


Ln(CA0/CA)



dC A = kC A dt

slope=k

time

Integral Method: Second Order Reactions 

For second order reaction:



Substitute the rate expression into the design equation for batch reactor dC A 2

kC A2 − rA =

−


CA

∫

C A0



= kC A

t

dC A = k ∫ dt C A2 0

The concentration time relation



Plot 1/CA versus t



Straight line?

1 1 = + kt C A C A0

 k = Slope  Assume different order and repeat


1/CA



dt

slope=k time

Integral Method: Example  The

irreversible isomerization

A→ B

was carried out in a batch reactor and the following concentration-time data were obtained:



t, min

0

3

4

8

10

CA, mol/dm3

4.0

2.89 2.25 1.45 1.0

12

15

17.5

0.65 0.25 0.07

Determine the reaction order and the reaction rate constant.


Solution

A→ B 1.

Assume zero order reaction: 

CA=CA0-kt



Plot CA versus time



The data does not follow a zero order relation

t, min

0

3

4

CA, mol/dm3

4.0

2.89 2.25 1.45 1.0


8

10

12

15

17.5

0.65 0.25 0.07

Solution, continued

A→ B Assume first order reaction: 

ln(CA0/CA)= k t



Plot ln(CA0/CA) versus time



The relation is not linear



4.5 y = 0.214x - 0.3673 R² = 0.9226

4 3.5 3

The reaction does not follow a first order kinetics

Ln(CA,/CA)

2.

2.5 2 1.5 1 0.5 0

0

5

10 time, min

t, min

0

CA, mol/dm3 4.0 ln(CA0/CA)

3

4

8

10

2.89 2.25 1.45 1.0

12

15

17.5

0.65 0.25 0.07

0.00 0.33 0.58 1.01 1.39 1.82 2.77 4.05


15

20

Solution, continued

A→ B 3. Assume    

second order reaction:

1/CA0 = 1/CA0+ k t Plot 1/CA0 versus time The relation is not linear The reaction does not follow second order kinetics

The reaction does not follow zero, first, or second order reaction  Try differential method or nonlinear regression method 


t, min

0

3

4

8

10

12

15

17.5

CA, mol/dm3

4.0

2.89

2.25

1.45

1.0

0.65

0.25

0.07

1/CA

0.25

0.35

0.44

0.69

1.00

1.54

4.00

14.29

Integral Method: n-order Reaction  For

reaction of n order kC nA − rA =

 Substitute

in the mole balance equation −

 Integrate

−

CA

∫

C A0

dC A = kC nA dt t

dC A = k∫ t n CA 0

1− n 1− n 1  C A0 − C A  t=   k  1 − n 

 Because

n is not know, this expression cannot be

plotted  Using regression method, least squares, the reaction order, n, and the rate constant (k) can be estimated Chemical Reaction Engineering

Nonlinear Regression Analysis kC nA − rA =



Assuming the rate expression is:



The integral methods yield the following relation: C A10− n − C A1 − n = ( 1 − n ) kt



Using regression analysis we can search for values for n and k that minimize the sum of the square of the error (S2): 1   2 1 −n 1 −n   S C C C C 1 n kt = − = − − − ( )   ∑ ∑ A ,c i   A0 A ,m A ,m i 1= i 1 =  N



(

)

N

2

This can be done numerically using one of software packages such as Matlab, Polymath, etc….





Multiple Reaction

Chapter 6 Lecture 1: Parallel Reactions


Multiple Reactions 

Parallel Reactions A →B A →C



Series Reaction A →B →C





Complex Reactions A + B

→C +D

A + C

→E

Independent Reactions A →B +C D →E +F


Multiple Reactions 

Parallel Reactions CH2=CH2 + 3 O2

→ 2 CO2 + 2 H2O

CH2=CH2 + 0.5 O2 → Ethylene Oxide 

Series Reaction -H2

-H2

Paraffins → Naphthenes → Aromatics

Complex Reactions Ethylene Oxide + NH3 → HOCH2CH2NH2 Ethylene Oxide + HOCH2CH2NH2 → (HOCH2CH2)2NH 

Independent Reactions C15H32 → C12H26 + C3H6 C8H18

→ C6H14 + C2H4


Selectivity 

For the following Parallel Reaction: A → D (Desired ) A →U

(Undesired)



Minimization of undesired compound U relative to desired product D is always desirable



Selectivity is used to quantify formation of desired product D with respect to undesired



Instantaneous Selectivity, SD/U: ~



S= D U

SD : Overall selectivity,= U S ~

D U


rate of formation ofD rD = rU Rate of formation of U ND = NU 

BatchReactor

FD FU 

Flow Reactor

Reaction Yield 

Instantaneous Yield, YD: = YD

~



Overall Yield, Y D :

~

U= Y D

rate of formationof D rD = −rA Rate of reactionof key reactant ND FD = NA 0 − NA FA 0 − FA      Batchreactor



Flow Reactor

For CSTR reactor: 

The instantaneous yield is equivalent to the overall yield



The instantaneous selectivity is equivalent to the overall yield


Parallel Reactions: 



For the following Parallel Reactions kD A → D

rD = kDC αA1

kU A → U

rU = kU C αA1

Rate of disappearance of reactant A, -rA − rA = rD + rU = kDC αA1 + kU C αA1



The instantaneous selectivity rD kDC αA1 kD α1 −α 2 S= CA = = D α2 rU kU C A kU U


Parallel Reactions: Maximizing Selectivity 

For the parallel reactions:

α

rD = kDC A1

A  → D α

rU = kU C A 2

A  → U

kD α1 −α 2 SD = C A U kU





We can maximize the selectivity by choosing the right reactor system (reactor type, arrangement, feed) Case 1: α1 > α2 

(α1 - α2 = + ve)

To make SD/U large , CA must be maintained as high as possible  Inert A

or diluents should be kept minimum

Batch or plug-flow reactor should be used


Parallel Reactions: Maximizing Selectivity 

Case 2: α1 < α2 

(α1 - α2 = -ve)

To make SD/U large , CA must be maintained as low as possible  Use

Inert or diluents and/or use recycle to dilute reactant with products

 Low

pressure for gas phase

 Use

CSTR

 Case 

3: α1 = α2

Case 3-1: ED >EU  Operate



S= D U

kD AD = e kU AU

 −( ED − EU )      RT  

at highest possible T

Case 3-2: ED < EU  Operate

at low T (high enough to have significant rate)


Parallel Reactions: Example Trambouze Reaction (P6-6)  

Consider the following system of gas-phase reactions: The reactions are to be carried at 27° C and 4 atm. Pure A enters the system at volumetric flow rate of 10 dm3/min. a. b. c. d. e. f.

g.

1

k1 → X (Undesired) A 

rX = 0.004 C A 2

k2 → B (Desired) A 

rB = 0.3 C A

k2 → R (Undesired) A 

rR = 0.35 C A2

mol dm3 .min mol dm3 .min mol dm3 .min

Sketch the instantaneous selectivity SB/X, SB/R, SB/XR as function of CA Consider a series of reactors. What should be the volume of the first reactor? What is the conversion of A in the first reactor? What are the effluent concentration of A, B, X, and R from the first reactor. If 99% conversion of A is desired, what reaction scheme and reactor sizes should we use to maximize SB/XR. Suppose that E1=20 kcal/mol, E2=10 kcal/mol, E3=30 kcal/mol. What temperature would you recommend for a single CSTR with a space time of 10 min and entering concentration of A of 0.1 mol/dm3? If you could vary the pressure between 1 and 100 atm, what pressure would you choose?


Solution: Part a B

2

X

1

k1

0.3 1−0.5 75C A0.5 CA = 0.004

k2 α 2 −α3 0.3 1− 2 0.86C A−1 S C CA = = = B  A R 0.35 k3 r rX + rR

B  = = S B

k2 → B (Desired) A 

rB = 0.3 C A

k2 → R (Undesired) A 

rR = 0.35 C A2

35

180 160 120 100

CA*=0.032 mol/dm3


25 20

SB/R

80

15

60

10

40 20 0 -5.55E-170.02

5

SB/XR 0.04

0.06

0.08 CA

0.1

dSB XR CA = 0.3 0.004C A0.5 + 0.35C A2 + 0.3C A 0.5 * 0.004C A−0.5 + 2 * .35= 0 dC A

)

30

SB/X

140

We can find CA* by setting dSB/XS/dCA=0

(

mol dm3 .min mol dm3 .min mol dm3 .min

200

note that SB/XR curve have a maxima at CA* 

rX = 0.004 C A 2

SB/R



XR

0.3C A 0.004C A0.5 + 0.35C A2

1

k1 → X (Undesired) A 

SB/X or SB/XR

k2 α −α = S CA =

(

)

0.12

0.14

0 0.16

Solution: Part b, c 

To maximize the selectivity SB/XR, we need to operate at the maximum selectivity 

Use the first reactor as CSTR operates at final CA=CA*

PA0 1* 4 3 C 0.16 mol / dm = = = A0  RT 0.082* ( 27 + 273 ) V CSTR =

*

Reaction Direction

*

FA0 X FA0 X = −r A* k 1C A0.5 + k 2C A + k 3C A2

C A0 − C *A 0.16 − 0.032  X = = = 0.8 C A0 0.16

10

*

 VCSTR =

0.004* 0.032 = 119 dm 3

0.5

1.6 * 0.8 + 0.3* 0.032 + 0.35* 0.032 2

6

SB/XR

= FA0 υ= 10*= 0.16 1.6mol / min 0 C A0

8

CSTR

4

PFR

2

0 0.16


CA0

0.14

0.12

0.1

0.08 CA, mol/liter

0.06

0.04

0.02

0

Solution: Part d 

CA=0.032 mol/dm3



CB, CR, CX? FR − FR0 FX − FX 0 FA0 − FA FB − FB0 = VCSTR = = = − rA rR rX rB

VCSTR =

CB =

υ0C B

υ0C R υ0C X = = 0.3C A 0.35C A2 0.004c 0.5 A 0.3C AV 0.3* 0.032* 119 = = 0.114 mol / dm 3 υ0 10

0.004C 0.5 V 0.004* 0.032 0.5 * 119 A CX = = = 0.0085 mol / dm 3 10 υ0 0.35C A2V 0.35* 0.032 2 * 119 CR = = = 0.004 mol / dm 3 υ0 10

Check : C A + C B + C X + X R must equal toC A0


Part e  For

99% conversion:



The first reactor is CSTR (X=0 to X=0.8, V=119 dm3)



The second reactor is PFR: FA 2

dFA = = V PFR ∫ rA FA 1

( 1 −0.8 )C A0

∫

( 1 −0.99 )C A0

k 1C A0.5

FA 2

( 1−0.99 )C A

1

0

v 0 dC A + k 2C A + k 3C A2

0 dFA dC A 3 VPFR ∫= 10 ∫ 9.1dm = = 0.5 2 rA FA ( 1−0.8 )C A 0.004C A + 0.3C A + 0.35C A


Solution Part f  For

 For

single CSTR: k2C Af rB = = SB XR rX + rR k1C Af0.5 + k3C Af2

max selectivity dS = k ( k C + k C ) − k C ( 0.5 * k C dC B

XR

2

1

0.5 A

3

2 A

2

A

1

−0.5 A

A

k1C A0.5 + k3C A2 − 0.5 * k1C A0.5 − 2 * k3C 2A = 0 0.5k1C A0.5 − k3C A2 = 0 2 3

 0.5k1  = C Af =  k  3 

 A01 exp ( −E1 / RT )     A exp ( −E / RT )  3  03 


2 3

)

+ 2 * k3C A= 0

Solution Part f  For

max selectivity 2 3

 0.5k1  = C Af =  k  3 

2 3

 0.5 A01 exp ( −E1 / RT )  =    A exp ( −E / RT )  3  03 

 0.5 A01  E3 − E1   exp     A RT    03 2/3

2 3

 k  0.1 −  1  C A0 − C A C A0 − C A  2 k3  = = = τ = 10 1/ 3 2/3 4/3 −rA k1C A0.5 + k2C A + k3C A2  k   k   k  k1  1  + k2  1  + k3  1   2 k3   2 k3   2 k3  2/3 2/3 ( 0.2k ) − k substitute expressions for k1 , k2 , and k3 as function of T and solve for T 10 = 4 /3 1/3 3 2 /3 1 4 /3 k1 k3 + k2 k1 + k1 k3

(

dm3 / mol

)

0.5

 20 x10  12 k= E RT exp / . exp 0 004 = ( )   1.49x10 1 1 min  1.987 x 300   10 x103  6 −1 A02 k= exp ( E2 / RT ) 0.3 exp=   5.79x10 min 2  1.987 x 300   30 x103  21 3 −1 −1 A03 k= exp ( E3 / RT ) 0.35 exp=   2.52x10 dm mol min 3  1.987 x 300  Solving for T using Polymath givesT = 315 K 3

A01


Solution part f and g  The

optimum pressure should be selected such that the initial concentration of A, CA0=0.1 mol/dm3 

This will ensure that the exit condition of the CSTR reactor will correspond to maximum selectivity = P0 C= RT 0.1x 0.08205= x 315 2.6 atm A0


Parallel Reactions: Reactor Selection 



For the following parallel reactions k1 A + B  → D

rD = k1C αA1 C Bβ1

k2 A + B  → U

rU = k2C αA 2 C Bβ2

The instantaneous Selectivity SD = /U



rD k1 α1 −α 2 β1 − β 2 = C CB rU k 2 A

Depending on the values of α1, α2, β1, and β2 we have the following cases 

Case 1: α1 > α2 and β1 > β2



Case 2: α1 > α2 and β1 < β2



Case 3: α1 < α2 and β1 < β2



Case 4: α1 < α2 and β1 > β2


Case 1: α1 > α2, β1 > β2 k1 A + B  → D rD = k1C αA1C Bβ1


k2 A + B  → U

SD = /U



High CA and High CB is preferred



Use PFR or Batch Reactor



Use high Pressure (gas reaction)



Do not use inert or diluents

rD k1 α1 −α 2 β1 − β 2 = C A CB rU k 2

A B


AB

Case 2: α1 > α2, β1 < β2 k1 A + B  → D rD = k1C αA1C Bβ1 k2 A + B  → U




SD = /U

rD k1 α1 −α 2 β1 − β 2 = C A CB rU k 2

B

High CA and Low CB is preferred  Semibatch reactor 



A present initially and B is continuously fed to the reactor

A

PFR with side streams or membrane A B



Series of Small CSTR with A fed to the first reactor and B is divided between the reactors B A

.


Case 3: α1 < α2, β1 < β2 k1 A + B  → D rD = k1C αA1C Bβ1 k2 A + B  → U

B

Low CA and Low CB is preferred 

CSTR



PFR with large recycle



Diluted feed or inert



Low pressure

Recycle




rD k1 α1 −α 2 β1 − β 2 = C A CB rU k 2

SD = /U

A

B

B A Recycle Chemical Reaction Engineering

Case 4: α1 < α2, β1 > β2 k1 A + B  → D rD = k1C αA1C Bβ1


k2 A + B  → U



SD = /U

rD k1 α1 −α 2 β1 − β 2 = C A CB rU k 2

Low CA and High CB is preferred  Semibatch reactor





B present initially



A is continuously fed to the reactor

PFR with side streams or membrane

A

B

B A 

Series of Small CSTR with B fed to the first reactor and A is divided between the reactors A B


Series Reactions: Batch System 

For the series reactions: k1 k2 → B (Desired )  → C (undesired ) A   rC k2CB = − rA k1C A= = rB k1C A − k2C B



The exact reaction time to maximize B is very important C C



For Batch System: dC − A = k1C A ⇒ C A = C A 0 exp( −k1t ) −rA = dt dCB rB =k1C A − k2C B =k1C A 0 exp( −k1t ) − k2C B = dt CB = k1C A 0

e − k1t − e − k2 t k2 − k1

CA k1/k2=2

CB

dCC e − k1t − e − k2 t = rC= k2C B= k2 k1C A 0 k2 − k1 dt

CC =

C A0  k2 (1 − e − k1t ) − k1 (1 − e − k2 t )   k2 − k1 


remember C C = C A 0 − C A − C B

Series Reactions: Maximum Yield of B 

The maximum CB is at dCB/dt=0 CB = k1C A 0

e − k1t − e − k2 t k2 − k1

kC dCB = 0 =1 A 0 ( −k1e − k1t + k2 e − k2 t ) dt k1 − k2 topt =

k  1 ln  1  k1 − k2  k2 

CC

C

CA k1/k2=2

CB

topt


Series Reaction: Example 

For the following liquid-phase series reaction rC = 0.01C B − rA = 0.4 C A → B  → C (undesired ) A 

is carried in a 500-dm3 batch reactor. The initial concentration of A is 1.6 mol/dm3. The desired product is B, and separation of the undesired product C is very difficult and costly. The given rate constants are in h-1 and are at 100°C. 

Plot the CA, CB, CC as function of time for batch system.



What is the optimum reaction time?



Assuming the reaction is carried in CSTR of a space time 0.5 h. What temperature would you recommend to maximize B? (E1=10 kcal/mol, E2=20 kcal/mol)


Solution: Batch System −rA = 0.4C A rC = 0.01C B

rC = 0.01C B 0.4 C A − rA = A  → B  → C (undesired )

 Mole

Balance for Batch Reactor −

dC A =−rA =. 0 4C A dt

dC B = rB= 0.01C B − 0.4C A dt

dC C = rC= 0.01C B dt

2

Ci

C= C A 0 exp( −k1= t ) 1.6 exp( −0.4t ) A

( )

e − k1t − e − k2t CB k1C A 0= 2= k2 − k1

( 3)

(

1.64 e −0.01t − e −0.4 t

(

)

) (

)

C A0  k2 1 − e − k1t − k1 1 − e − k2t   k2 − k1  1.6 0.01 1 − e −0.4 t − 0.4 1 − e −0.01t  =  0.01 − 0.4 

CB

1.5

(1)

= rC 0.01C B − 0.4C A

= CC

(

)

(

1

0.5 0

CC

CA

0

20

40

time, min


= topt

60

k   0.4  1 1 ln  1  ln  = = 9.46 min k1 − k2  k2  0.4 − 0.01  0.01 

)

Solution Batch System Using PolyMath


Solution: CSTR  Mole

Balance for CSTR

FA 0 X FA 0 − FA ν 0 ( C A 0 − C A ) = V = = −rA −rA −rA

τ=

τ

V

ν0

=

1.6 − C A 1.6 ⇒ CA = 1 + τ k1 k1C A

CB0 − CB −C B = = 5 1.6 −rB k2C B − k1 1 + τ k1

5k2C B −

8k1 8k1 = −C B ⇒ C B = 1 + 5k1 (1 + 5k1 )(1 + 5k2 )

 E1  1 1    E2  1 1   = k1 k1,T1 exp   −   ; = k2 k2 ,T1 exp   −   R T T  R T T      1   2

Toptimum = 388° K = 115° C Chemical Reaction Engineering

T

k1

k2

CB

200

3.4E-06

7.3E-13

2.7E-05

225

5.6E-05

2.0E-10

4.5E-04

250

5.2E-04

1.7E-08

4.2E-03

275

3.3E-03

6.7E-07

2.6E-02

300

1.5E-02

1.4E-05

1.1E-01

325

5.5E-02

1.9E-04

3.4E-01

350

1.6E-01

1.7E-03

7.2E-01

375

4.3E-01

1.2E-02

1.0E+00

388

6.7E-01

2.8E-02

1.1E+00

400

9.9E-01

6.2E-02

1.0E+00

425

2.1E+00

2.7E-01

6.2E-01

450

4.0E+00

1.0E+00

2.5E-01

475

7.3E+00

3.3E+00

8.9E-02

500

1.2E+01

9.5E+00

3.3E-02

525

2.0E+01

2.5E+01

1.3E-02

550

3.1E+01

5.9E+01

5.4E-03

1.200 1.000 0.800

CB 0.600 0.400 0.200 0.000 200

300

T, K

400

500

In Class Exercise A large fully automated municipal incinerator is being designed. A survey estimates the garbage load to be 1440 tons/day. This will be harvested by a fleet of compaction trucks which will disgorge their loads into an underground storage bin. A conveyor will then feed the garbage to the Incinerator. The proposed daily collection route is such that at the beginning of the working day (6 A.M. sharp!) relatively large quantities of garbage (average of 6 tons/min) are returned from nearby commercial areas. Subsequently, the supply will diminish as more remote suburban areas are serviced. It is assumed that the collection rate is proportional to the amount of garbage still to be collected, the initial rate being one truck load/min. The conveyor, on the other hand, will transport garbage at a uniform 1 ton/min to the incinerator. At the beginning of the working day, the trucks will work faster than the conveyor; later in the day, slower. Thus, each day the bin will accumulate material, then lose material. To evaluate this operation, we need information. Please help us with this. a) b) c) d)

At what time of day will the trucks have collected 95% of the day's garbage? How much garbage should the storage bin be designed for? At what time of day will the bin be fullest? At what time of day will the bin be empty?





Non-Isothermal Reaction

Chapter 8


Non-Isothermal Reactions 

Consider the following liquid-phase exothermic reaction: A  →B



The reaction takes place in a PFR under adiabatic, non-isothermal conditions



We need to calculate the volume of flow reactor to achieve conversion X



The mole balance: kCA0 (1 − X ) dX = = dV FA0


− rA dX = dV FA0

 E  1 1  1 k1exp   −   CA0 (1 − X FA0  R  T1 T  

)

Energy Balance for Flow Reactors 

Since the reaction is non-isothermal, the temperature will vary along the reaction length



The temperature is related to conversion through energy ^ balance equation: d= E δ Q −δ W



For a flow reactor:  rate of   Rate of flow   Rate of work  Accumulation  =  of heat to  −  done by       of energy   the system   the system

  Rate of energy   Rate of energy   +  added to the system  − leaving the system         by mass flowin   by mass flowout 

Q dEˆ sys dt





= Q − W + Fin Ein − Fout Eout

.


Fi0

Fi

Hi0

Hi

Energy Balance for Flow Reactors dEˆ sys dt





n

n

= Fin Ein ∑ = Fi Ei ; Fout Eout Where:

∑FE

i 1 =i 1 = in 

n

i 1 =i 1 = in

But

i out

dEˆ sys dt



Fi

Hi0

Hi

out





Fi0

+ Ws

  n n   = Q − −∑ Fi PVi + ∑ Fi PVi  i1 = dt = i 1 in 

dEˆ sys

Therefore:

Remember

i



n

W= −∑ Fi PVi + ∑ Fi PVi

Rearranging

Q

= Q − W + Fin Ein − Fout Eout

n

(

=Q − Ws + ∑ Fi Ei + PVi

 n n  + Ws + ∑ Fi Ei − ∑ Fout Eout  i 1 out=i 1 = in 

) − ∑ F ( E + PV ) n

in =i 1 =i 1 negligible

i

i

i

out

 u i2 Ei = U i + + gzi + others = U i 2 H i =U i + PVi =Ei + PVi dEˆ sys dt





n

=Q − Ws + ∑ Fi H i in − ∑ Fi Fi =i 1 =i 1




n

out



n

n

=Q − Ws + ∑ Fi 0 H i 0 − ∑ Fi H i =i 1 =i 1

out

The Energy Balance equation 

For the following reactions: n

∑F

H =F H

+F H

+F H

+F H

A+ =F

b c d B→ C+ D a a a

n

∑θ H

i0 i0 A0 A0 B0 B0 C0 C0 D0 D0 A0 i 1 =i 1

 

and

n

∑FH

= F H +F H +F H +F H = F

then ∑ F H − ∑ F= H n

i0 i0 =i 1 =i 1 n

∑F

n

F

i0

n

∑ H (θ i

i

+ν i X )

c b d  − H i ) θi − FA0  H D + H C − H B − H A  X a a a  n

H −= ∑ Fi H i FA0 ∑ ( H i 0 − H i )θi − ( ∆H Rx )T FAo X

i0 i0 =i 1 =i 1



∑(H

i i A0 =i 1

i0

n

i i A A B B C C D D A0 i 1 =i 1

n

i

=i 1

For steady state conditions, the energy balance dEˆ sys dt





n

= Q − W + FA0 ∑ ( H i 0 − H i ) θi − ( ∆H Rx )T FAo X = 0


i =1

1. Adiabatic Reactions with constant CP 



n

Q − W + FA0 ∑ ( H i 0 − H i ) θi − ( ∆H Rx )T FAo X =0 i =1



For Adiabatic reactor with negligible shaft work:   = Q 0= and W S 0



The energy balance equation becomes n

FA0 ∑ ( H i 0 − H i ) θi − ( ∆H Rx )T FAo X =0 i =1



But

T

Hi0 −Hi = − ∫ C Pi dT = C Pi (T −T 0 )

No phase Change and constant C Pi

Ti 0

n



Then



And the temperature is now related to conversion

∑ θ C (T − T ) − ( ∆H ) i =1

i

Pi

0

T= T0 +

Rx T

( −∆H Rx )T X n

∑θ C i =1


X =0

i

Pi

1. Adiabatic Reactions with constant CP 



From energy balance find T at different X X −∆H rx ) X ( T= T0 + 1 0 θ C ∑ i Pi 0.1 Find the rate equation at each X 0.2

rA kCA0 (1 − X −= = )

 E  1 1  k1exp   −   CA0 (1 − X  R  T1 T  



Estimate FA0/-rA



Plot FA0/-rA versus X



Find the reactor volume from the corresponding area


)

1

T T0

-rA

-rA/FA0

Example: F P-8-6 

The elementary irreversible organic liquid phase reaction A + B →C is carried out adiabatically in a flow reactor. An equimolar feed in A and B enters at 27° C, and the volumetric flow rate is 2 dm3/s and CA0=0.1 kmol/m3. a)

Calculate the PFR and CSTR volume necessary to achieve 85% conversion.

b)

What is the maximum inlet temperature one could have so that the boiling point of the liquid (550° K) would not be exceeded even for complete conversion?

c)

Plot the conversion and temperature as function of PFR volume.

d)

Calculate the concentration achieved in one 500 dm3 and in two 250-dm3 CSTRs in series.

HA°(273)= -20 kcal/mol, HB°(273)= -15 kcal/mol, HC°(273)= -41 kcal/mol CP,A=CPB=15 cal/mol.K, CP,C=30 cal/mol.K k = 0.01 dm3/mol.s at 300° K, E =10000 cal/mol Chemical Reaction Engineering

Solution Part a: A + B →C 

∆CP=CPC-CPA-CPB=0 ∆HRX = constant T1

H A° + H B° − H C° ) ( −∆H rx ) X ( T + 300 + = = 0

∑ θi C P

i

T1 = 300 +

k

C PA + C PB

6000X = 300 + 200X 30

E k 1 exp  R

 1 1  10000  1 1  = − − 0.01exp    1.987  300 T       T1 T  

= -rA kC A20 (1 − X ) 2


X

T

k

CA

-rA

FA0/-rA

0

300

0.01

0.1

1.00E-04

2000

0.1

320

0.03

0.09

2.31E-04

865

0.2

340

0.07

0.08

4.61E-04

434

0.3

360

0.16

0.07

8.03E-04

249

0.4

380

0.34

0.06

1.23E-03

163

0.5

400

0.66

0.05

1.66E-03

121

0.6

420

1.21

0.04

1.93E-03

104

0.7

440

2.08

0.03

1.87E-03

107

0.8

460

3.42

0.02

1.37E-03

146

0.9

480

5.40

0.01

5.40E-04

371

0.91

482

5.64

0.009

4.57E-04

438

0.92

484

5.88

0.008

3.77E-04

531

0.93

486

6.14

0.007

3.01E-04

665

0.94

488

6.41

0.006

2.31E-04

867

0.95

490

6.68

0.005

1.67E-04

1197

0.96

492

6.97

0.004

1.12E-04

1794

0.97

494

7.26

0.003

6.54E-05

3059

0.98

496

7.57

0.002

3.03E-05

6607

0.99

498

7.88

0.001

7.88E-06

25372

Solution: Part a  Volume 

VCSTR = 0.85 x 206 = 175 dm3

 Volume 

of CSTR equals the area of rectangle

of PFR equals the entire Area under the curve

VPFR ~ 305 dm3

3000

2500

FA0/-rA

2000

1500

1000 VPFR

VCST

500

0 0


0.1

0.2

0.3

0.4

0.5 X

0.6

0.7

0.8

0.9

1

Solution: Part a Using Excel 

Volume of CSTR at X FA0 X FA0 V = = X CSTR −r A −r A



Volume of PFR at X dV PFR =

FA0 dX −r A

FA0 ∆ X FA0 ∆V PFR = ∆X = −r A −r A



∆X has to be small X

V PFR = ∑ ∆V PFR 0


X

T

k

CA

-rA

FA0/-rA

VCSTR

∆VPFR

VPFR

0

300

0.01

0.1

1.00E-04

2000

0

0

0

0.05

310

0.02

0.095

1.55E-04

1290

64

82.25

82

0.1

320

0.03

0.09

2.31E-04

865

87

53.88

136

0.15

330

0.05

0.085

3.32E-04

602

90

36.69

173

0.2

340

0.07

0.08

4.61E-04

434

87

25.92

199

0.25

350

0.11

0.075

6.18E-04

324

81

18.95

218

0.3

360

0.16

0.07

8.03E-04

249

75

14.32

232

0.35

370

0.24

0.065

1.01E-03

198

69

11.18

243

0.4

380

0.34

0.06

1.23E-03

163

65

9.02

252

0.45

390

0.48

0.055

1.45E-03

138

62

7.51

260

0.5

400

0.66

0.05

1.66E-03

121

60

6.46

266

0.55

410

0.90

0.045

1.82E-03

110

60

5.76

272

0.6

420

1.21

0.04

1.93E-03

104

62

5.33

277

0.65

430

1.59

0.035

1.95E-03

102

67

5.15

282

0.7

440

2.08

0.03

1.87E-03

107

75

5.23

288

0.75

450

2.68

0.025

1.68E-03

119

89

5.65

293

0.8

460

3.42

0.02

1.37E-03

146

117

6.64

300

0.85

470

4.32

0.015

9.71E-04

206

175

8.80

309

0.9

480

5.40

0.01

5.40E-04

371

334

14.41

323

0.95

490

6.68

0.005

1.67E-04

1197

1137

39.19

362

Reactor Volume versus Conversion 500

3000

450 2500 400 350 2000

VPFR 250

1500

200 1000 150 100

VCSTR 50

500

FA0/-rA

0

0 0

0.1

0.2

0.3

0.4

0.5 X


0.6

0.7

0.8

0.9

1

Volume

FA0/-rA

300

Solution : Part b What is the maximum inlet temperature one could have so that the boiling point of the liquid (550 K) would not be exceeded even for complete conversion?

( −∆H rx ) X = T= T0 + T 0 + 200X θ C ∑ i Pi T0,max = 550 − 200* X max = 550 − 200*1 = 350 °K


Part c: Plot the conversion and temperature as function of PFR volume.

1

600

0.9 500 0.8 0.7 400

T

0.5

300

0.4 200 0.3 0.2

X

100

0.1 0

0 0

50

100

150

200 VPFR


250

300

350

400

T

X

0.6

Part d Calculate the concentration achieved in one 500 dm3 and in two 250-dm3 CSTRs in series. V = CSTR

=



For 1 CSTR V = CSTR



FA0 X FA0 = = X 2 −r A kC A0 ( 1 − x )

= VCSTR ,2

E k 1 exp  R

 1 1  2 −  C A0 (1 − x )   T1 T  

FA0 E  1 1 − k 1 exp    R  T 1 .300 + 200 X

 2  C A0 ( 1 − x ) 

0.2

 10000 0.01 exp   1.987

For 2 CSTRs V CSTR= ,1

FA0

FA0 X 1 = −r A

X

X

= X 500 , 1  1  2  300 − 300 + 200 X   0.1 ( 1 − x )  

= ⇒ X 0.92

0.2  10000 0.01 exp   1.987

FA0 ( X 2 − X 1 ) = − rA

 1 1 −   300 300 + 200 X 1

0.2

 2   0.1 ( 1 − X 1 ) 

X= 250 1

= X 2 250  10000  1  2 1 0.01exp  −    0.1 ( 1 − X 2 )  1.987  300 300 + 200 X 2  


0.881 ⇒ X= 1

⇒= X 2 0.97

Steady State Reactors with Heat Exchange 

We have studied the operation of reactor under adiabatic condition



In many cases we need to introduce heat or remove heat from the reactor .

UA (T − T a )



Q Let Q be the heat added to the reactor=



The energy Balance around differential volume ∆V: Coolant

.

∆ Q + ∑ Fi H i

V

− ∑ Fi H i

V +∆V

= 0

FA0 T0



= ∆ Q U ∆ A ( Ta −= T ) U a ∆ V ( Ta − T )

where a is heat exchange area per unit volume of the reactor,a =

U a ∆ V (Ta − T ) + ∑ Fi H i

V

− ∑ Fi H i


V +∆V

= 0

Ta

∆V

A V

  d  ∑ Fi H i  = U a (Ta − T ) −  0 dV

FA T


Expanding the energy balance equation: d (∑ F H ) = 0 U a (T − T ) − i

a

FA0 T0

∆V T

i

dV



We get:



The PFR mole Balance equation: dFi = r= ν i ( −r A ) i

U a (Ta − T ) − ∑

Ta

Coolant

dFi dH 0 H i − ∑ i Fi = dV dV

dV



Differentiating the enthalpy equation with respect to V:

dH i d ∫ C pi dT dT = = CPi dV dV dV  Substituting into the

expanded energy balance equation:

H Rx ∆  U a (Ta − T ) − ( ∑ν i H i ) ( −rA ) −


(

)

dT F C 0 ∑ i Pi d= V

⇒

dT rA ∆H Rx − U a (T − Ta ) = dV ∑ FiCPi

(

)

FA T


The design equation for PFR with heat exchange:     rA ∆H Rx − U a (T − T a )



dT = dV

Note: = FC ∑ ∑ F (θ i



Pi

A0

i

T

Heat Removal

Heat Generation

Coolant

(∑ F C ) i

(

)

+ν= FA 0 ∑ θi C Pi + ν= FA 0 i X )C Pi i C Pi X

( ∑θ C i

+ ∆C Pi X

And the “friendly” form of the design equation becomes: r ∆H Rx −U a (T −T a ) dT = A dV FA 0 ∑ θi C Pi + ∆C Pi X

(



Pi

)

For PBR, W=V ρb and the design equation becomes rA' ∆H Rx −

dT = dW FA 0 Chemical Reaction Engineering

U

ρb

( ∑θ C i

Pi

a (T −T a ) + ∆C Pi X

)

FA T

∆V

FA0 T0

Pi

Ta

)

Solving CRE Problem for Reactors with Heat Exchange  The

Differential Equation describes the change in T along the reactor rA ∆H Rx −U a (T −T a )

dT = dV FA 0

 Must  If

( ∑θ C i

Pi

+ ∆C Pi X

)

or

rA' ∆H Rx −

dT = dW FA 0

U

ρb

( ∑θ C i

Pi

a (T −T a )

+ ∆C Pi X

)

be coupled with mole balance equation: −r dX = A dV FA 0

or

rA' dX = dW FA 0

the coolant temperature varies down the reactor we must add coolant balance: dT a U a (T − T a ) = . dV m c C Pc

 This

U

dT a ρ = B dw

a (T − T a ) .

m c C Pc

set of equation must be solved simultaneously


Non-Isothermal Reactor Design: Example 

The elementary irreversible organic liquid phase reaction A + B →C is carried out in a flow reactor. An equimolar feed of A and B enters at 27° C. The volumetric flow rate is 2 dm3/s and CA0 = 0.1 kmol/m3. a)

Plot the conversion and temperature as function of PFR volume when a heat exchange is added, Ua = 20 cal/m3.s.K, and the coolant temperature is constant at Ta= 450 K

b)

Repeat for a coolant flow rate of 50 g/s and CPc = 1 cal/g.K and inlet cooling temperature of Ta0 = 450 K

c)

Vary the cooling rate between 1-1000 g/s

HA°(273)=-20 kcal/mol, HB°(273) = -15 kcal/mol, HC°(273) =-41 kcal/mol CP,A= CP,B =15 cal/mol.K, CP,C= 30 cal/mol.K k = 0.01 dm3/mol.s at 300° K, E =10000 cal/mol Chemical Reaction Engineering

Solution: Part A 

Reaction Kinetic Data:       



Ta Coolant

mol/dm3 mol/dm3



(∆Hrx)273= ∆CP =

Heat Exchange Data  

Ua = Ta =


FA T

∆V

FA0 T0

; T1=

Thermal Data 



-rA = CA0 = CB0 = v0 = T0 = k1 = E=

A + B →C

XA

? VPFR

T

Solution Cont.: 

Mole Balance:



Energy Balance:



Stoichiometry:



FA0 =



−rA =



kT



CA =

;

CB =

 E  1 1  k1 exp   −    R  T1 T  

( ∆H rx )T =

∑θ C i



dX = dV dT = dV

Pi

=

Solve the two differential equations using Polymath


Polymath Solution:


Polymath Solution:


Solution: Part A 

Reaction Kinetic Data:       



-rA = k CA CB CA0 = 0.1 mol/dm3 CB0 = 0.1 mol/dm3 v0 = 2 dm3/s T0 = 300° K k1 = 0.01 dm3/mol. S ; T1= 27+273 = 300° K E = 1x104 cal/mol. ° K



FA0 T0

(∆Hrx)273= -41 + 15 + 20 = - 6 kcal/mol. °K XA ∆CP = 30 - 15 -15 = 0

Heat Exchange Data  

Ta

Coolant

Thermal Data 



A + B →C

Ua = 20 cal/m3.s.° K Ta = 400° K


FA T

∆V

? VPFR

T

Solution Cont.: 

Mole Balance:



Energy Balance:



Stoichiometry:

dX −rA = dV FA0 dT rA ( ∆H Rx )T − U a (T − Ta ) = dV FA0 ∑ θi CPi + ∆CPi X

C A C A0 (1 − X ) =

;

CB= C A0 (1 − X )= C A

 = FA0 C= 0.1 = x 2 0.2 mol / s A 0 v0



 E  1 1  2 2 −rA= f ( X , T )= kC A= k1 exp   −   C A 0 (1 − X )  R  T1 T   −6000 cal / mol.° K ( ∆H rx )T = ( ∆H rx )273 =



∑θ C





2

i

Pi

∆CP ,i =

= θ A CPA + θ BCPB + θ C= CPC 30 cal / mol.° K

∑ν C i

P,i

= C P ,C − C P , A − C P , B = 0


Research

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CSM faculty

PI faculty

Integrated Carbonate Reservoirs Characterization and Simulation of Abu Dhabi Fractured Carbonate Reservoirs

H. Kazemi (CSM),

S. Ghedan (PI)

Fluid Sensitivity of Seismic Properties in Carbonate Reservoirs

M. Batzle (CSM) M. Prasad (CSM)

S. Vega (PI)

Selection and Optimization of Miscible and Immiscible Displacement to Improve Production from Fractured, Carbonate Reservoirs of Abu Dhabi

H. Kazemi (CSM)

S. Ghedan (PI)

Integrated Carbonate Reservoir Characterization

J. Sarg (CSM),

S. Lokier (PI) T. Steuber (PI)

Imaging Two-Phase Flow and Hydrofracturing in Carbonate Rocks by Looking at the Joint Inversion of Passive Seismic and Electrical Signals

Revil (CSM)

S. Vega (PI)


CSM faculty

Upstream Materials and Equipment Engineering Minimizing Corrosion of Petroleum Production Equipment by B. Mishra Hydrogen Sulfide and Carbon Dioxide Damage Diagnosis and Structural Integrity Evaluation of R. Zhang Pipelines and Infrastructural Systems Corrosion Behavior of Expanded Tubes in Harsh B. Mishra Environments Chemical Processing C. Koh Desalination by Hydrate Nucleation and Growth A. Sum B. E. D. Sloan Pyrolysis Reactions of Butene Isomers at Low Temperatures A Direct Internal Reforming Protonic-Ceramic Fuel Cell

A. Dean R. O'Hayre N. Sullivan

PI faculty A. Al-Shoabi A. Seibi A. Seibi

C. Peters I. Economou H. Carstensen A. Al-Shoabi A. Almansoori


CSM faculty

PI faculty

Power Distribution Development of a PV Grid Connected Inverter for Smart Grids

M. Simoes

A. Al Durra

Engineering Education Establishing and Supporting a Center of Teaching Excellence at the PI in Abu Dhabi

R. Miller

Preparing Global Engineers: Developing Engineering Design Education Across Cultures

S. Scott R. Knecht

B. Bielenberg

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UMN Faculty

PI Faculty

Hydrocarbon processing Catalytic Removal of Sulfur from Process Gas Streams without Addition of Hydrogen: Effect of Zeolite Topology, Composition and Mesoporosity Selective Catalytic Synthesis of Branched C4-C7 Hydrocarbons from CH3OH or CH3OCH3 Coatings for Catalytic Processes

Aditya Bhan Michael Tsapatsis

Saleh Al Hashimi Radu Vladea

Aditya Bhan Matteo Cococcioni


Michael Tsapatsis Lorraine Francis


Simulation and optimization Simulation, Optimization and Control of Solid Oxide Fuel Cell Systems Molecular Modelling of Fluid Phase Equilibria

P. Daoutidis Jeffrey J. Derby

A. Almansoori

I. Siepmann

I. Economou C. Peters

RESEARCH UNIVERSITY OF MINNESOTA Projects

UMN Faculty

PI Faculty

Polymer processing Polymeric Membranes for Advanced Process Engineering

Graphene/Polymer Composites Synthesis and Processing of Functionalized Polyolefins

F. Bates E. Cussler M. Hillmyer T. Lodge C. Macosko F. Bates M. Hillmyer C. Macosko

A. Abdala I. Economou C. Peters A. Abdala A. Abdala

Materials Science and Engineering (SEED) Processing Improved Microstructures for the Energy Industry Conversion of CO2 into polymers and chemicals

William W. Gerberich K. Andre Mkhoyan Jeffrey J. Derby Friedrich Srienc

Ahmed Abdala

RESEARCH UNIVERSITY OF MARYLAND Projects

UM Faculty

Energy Recovery and Conversion Sulfur Recovery from Gas Stream using Flameless and Flame A.K. Gupta Combustion Reactor

PI Faculty A. Al Shoaibi

Separate Sensible and Latent Cooling with Solar Energy

R. Radermacher Y. Hwang

I. Kubo

Waste Heat Utilization in the Petroleum Industry

R. Radermacher Y. Hwang

S. Al Hashimi P. Rogers

Energy-Efficient Transport Processes Multidisciplinary Design and Characterization of Polymer Composite Seawater Heat Exchanger Module

A. Bar-Cohen S.K. Gupta D. Bigio

P. Rogers

Study on Microchannel-Based Absorber/Stripper for CO2 Separation from Flue Gas

M. Ohadi S. Dessiatoun

M. Al Shehhi

Enhancement of Polymerization Processes Using Microchannel Reactor

M. Ohadi S. Dessiatoun

E. Al Hajri

RESEARCH UNIVERSITY OF MARYLAND Projects

UM Faculty

PI Faculty

S. Azarm P.K. Kannan

A. Almansoori S. Al Hashimi

Dynamics and Control of Drill Strings

B. Balachandran

Y. Abdelmagid H. Karki

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B. Balachandran N. Chopra

H. Karki

M. Modarres

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Energy System Management Integration of Engineering and Business Decisions for Robust Optimization of Petrochemical Systems

Development of a Probabilistic Model for Degradation Effects of Corrosion-Fatigue Cracking in Oil and Gas Pipelines

William Marsh RICE University

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Modeling Asphaltene behaviour and deposition in Crude Oil Systems Mobility Control for CO2 EOR in heterogeneous high temp, high salinity carbonate reservoirs

Rice Faculty

PI Faculty

Walter Chapman

Francisco Vargas

George Hirasaki, Lisa Biswal, Keith Johnston, Quoc Nguyen

RESEARCH STANFORD UNIVERSITY Stanford Faculty

PI Faculty

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Dynamics and Control of Drill Strings

B. Balachandran

Y. Abdelmagid H. Karki

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B. Balachandran N. Chopra

H. Karki

M. Modarres

A. Seibi M. Chookah

Projects Energy System Management Integration of Engineering and Business Decisions for Robust Optimization of Petrochemical Systems

Development of a Probabilistic Model for Degradation Effects of Corrosion-Fatigue Cracking in Oil and Gas Pipelines

RESEARCH UNIVERSITY OF TEXAS AT AUSTIN Projects

UT Faculty

Energy System Management Kamy Sepehrnoori, Director Enhanced Oil Recovery by CO2 Injection Russell T. Johns Larry W. Lake Kishore K. Mohanty Quoc P. Nguyen

PI Faculty


CHEG411: Chemical Reaction Engineeirng. Fall 2010

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