CHEG411: Chemical Reaction Engineeirng. Fall 2010 PDF

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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 valu...

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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

CHEG411: Chemical Reaction Engineering

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

CHEG411: Chemical Reaction Engineering

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

CHEG411: Chemical Reaction Engineering

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?

CHEG411: Chemical Reaction Engineering

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)

CHEG411: Chemical Reaction Engineering

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

CHEG411: Chemical Reaction Engineering

(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

CHEG411: Chemical Reaction Engineering

; 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

CHEG411: Chemical Reaction Engineering

 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

CHEG411: Chemical Reaction Engineering

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

CHEG411: Chemical Reaction Engineering

(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.

CHEG411: Chemical Reaction Engineering

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

CHEG411: Chemical Reaction Engineering

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

Chemical Reaction Engineering

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

Chemical Reaction Engineering

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

Chemical Reaction Engineering

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

Chemical Reaction Engineering

=

)

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

Chemical Reaction Engineering

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 =

Chemical Reaction Engineering

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

Chemical Reaction Engineering

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)

Chemical Reaction Engineering

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

Chemical Reaction Engineering

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

Chemical Reaction Engineering

= 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

Chemical Reaction Engineering

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

Chemical Reaction Engineering

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

Chemical Reaction Engineering

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

Chemical Reaction Engineering

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 ≡

Chemical Reaction Engineering

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 ≡

Chemical Reaction Engineering

v0

Gas @ STP

V

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

Ahmed Abdala Chemical Engineerin...