L14: Equilibrium Conversion in Nonisothermal Reactor Design

L13-1

Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.

Review: Nonisothermal Reactor DesignSteady–state total energy balance (TEB):

n nsys

s A0 i0 i i RX A0 Ai 1 i 1

ˆdE0 Q W F H H H T F X

dt

For a SS nonisotherm flow reactor:

Tn

s A0 i p,i RX A0 Ai 1Ti0

0 Q W F C dT H T F X

“Simplified” TEB:

n

s A0 i p,i i0 RX A0 Ai 1

0 Q W F C T T (T)FH X Constant (average) heat capacities :

Can rearrange this equation to solve for T

RX Rn

s A0 i p,i i P R0 A0 Ai 1

H (T )0 Q W F ˆ TC T T XT FC

T = reaction temp Ti0 = initial (feed) temperature TR= reference temp

L13-2


Review: Solve TEB for Conversion

ns A0 i p,i i0 RX R P R A0 A

i 1ˆ0 Q W F C T T H (T ) C T T F X

Solve for XA:

nA0 i p,i i0 s

i 1A

RX R P R A0

F C T T W QX

H (T ) C T T F

Plug in Q for the specific type of reactor

ni p,i i0

i 1A

RX R P R

C T TX

H (T ) C T T

T = reaction temp Ti0 = initial (feed) temperature TR= reference temp

For an adiabatic reaction (Q=0) and shaft work can be neglected (ẆS=0)

L13-3


Review: Application to CSTRa) Solve TEB for T at the exit (Texit = Tinside reactor)b) Calculate k = Ae-E/RT where T was calculated in step ac) Plug the k calculated in step b into the design equation to calculate VCSTR

Case 1: Given FA0, CA0, A, E, Cpi, H°I, and XA, calculate T & V

a) Solve TEB for T as a function of XA (make a table of T vs XA using EB)b) Solve CSTR design equation for XA as a function of T (plug in k = Ae-E/RT )

(use design eq to make a table of XA vs T)c) Plot XA,EB vs T & XA,MB vs T on the same graph. The intersection of these 2

lines is the conditions (T and XA) that satisfies the energy & mass balance

Case 2: Given FA0, CA0, A, E, Cpi, H°I, and V, calculate T & XA

XA,EB = conversion determined from the TEB equationXA,MB = conversion determined using the design equation

XA

T

XA,EB

XA,MB

XA,exit

Texit

Intersection is T and XA that satisfies both equations

L13-4


Review: Application to a SS PFR

PFRFA0 FA

distance

TXA

Negligible shaft work (ẆS=0) and adiabatic (Q=0)

a) Use TEB to construct a table of T as a function of XA

b) Use k = Ae-E/RT to obtain k as a function of XA

c) Use stoichiometry to obtain –rA as a function of XA

d) Calculate:

XA AA0

A AXA0

dXV Fr X ,T

May use numerical methods

L13-5


L13: Equilibrium Conversion in Nonisothermal Reactor Design

• The highest conversion that can be achieved in reversible reactions is the equilibrium conversion– For reversible reactions, the equilibrium conversion is

usually calculated first• The equilibrium conversion increases with increasing

temperature for endothermic reactions• The equilibrium conversion decreases with increasing

temperature for exothermic reactions

L13-6


Review of Equilibrium Kinetics

KC: equilibrium constant (capital K):

c dC D

C a bA B

C CK

C C

kAk A

aA b B c C d D

Gas-phase reaction:

products

reactants

C raised to stoichiometric coefficientsC raised to stoichiometric coefficients

If KC is given at a single temperature T2, & CP can

be neglected then:

RX RC C 2

2

H T 1 1K T K T expR T T

KP: equilibrium constant in terms of partial pressures Pi:

c dC D

P a bA B

P PK

P P i iP CRT

For ideal gases, KP = KC(RT)Δn where n c d b a

Temp dependence of KP is given by van’t Hoff’s equation:

Rx Rx R P RP2 2

H T H T C T TdlnKdT RT RT

RX RP P 2

2


If CP can be

neglected then:

L13-7


Equilibrium Conversion XAe

T

XA,e

0

1

A B heatexothermic

endothermicA heat B

Example) AB CA0=1 CB0=0

A0 AeBeC

Ae A0 Ae

C 0 XCK

C C 1 X

AeC

Ae

XK

1 X

C Ae AeK 1 X X

Rearrange to solve in terms of XAe

C Ae C AeK X K X C Ae CK X 1 K

C

AeC

KX

1 K

This equation enables us to express Xae as a function of T

L13-8


XAe and Temperature

CAe

C

KX

1 K

RX RC C 2

2


Substitute for KC:

RX RC 2

2Ae

RX RC 2

2

H T 1 1K T expR T T

XH T 1 11 K T exp

R T T

Divide numerator & denominator by KC

Ae

RX R

C 2 2

1XH T1 1 1exp 1

K T R T T

XX1 e

e

Changed sign

R

RX RAe

2X

HExot

T 1 1 exp and XR T

hermic: HT

0, when T

R

RX RAe

2X

H TEndo 1thermic: H 0, when T 1 exp and X

R T T

L13-9


XAe and Temperature

Ae

RX R

C 2 2

1XH T1 1 1exp 1

K T R T T

RX RAe

2P RX R

H T 1 1 when T eExothermic & C =0: H T 0 xp &, XR T T

p R

RX R

2R AeX

H T 1 1, when T Endothermic & C 0: H T 0 exp & XR T T

A B heatMakes sense from Le Chatelier’s principle

Exothermic rxn produces heat→ increasing temp adds heat (product) & pushes rxn to left (lower conversion)

Makes sense from Le Chatelier’s principle

Heat is a reactant in an endothermic rxn→ increasing temp adds reactant (heat) & pushes rxn to right (higher conversion)

A heat B

L13-10


For the elementary solid-catalyzed liquid-phase reaction

A B1. Make a plot of equilibrium conversion as a function of temperature.2. Determine the adiabatic equilibrium temperature and conversion when

pure A is fed to the reactor at a temperature of 298 K.

A

B

PA

PB

e

H (298K) 40000 cal / mol

H (298K) 60000 cal / molC 50 cal / mol K

C 50 cal / mol K

K 100,000 at 298K

Adiabatic Equilibrium T Example

BA A

C

Cr k CK

L13-11


Rate law:CBr k CA A Ke

equilibrium

-rA = 0Be

Aee

CC

K

A0 e

eA0 e

C XK (T)

C 1 X

ee

e

K TX

1 K T

e e 1R

1

oX 1 1K K (T )exp -

R T TH

R

TpTX RR

oRXH T C dTH T

B Ap p pC C - C 0

o o oRX B AH H - H -20000

The Van’t Hoff equation: e

2d lnK

dT TH

R

1T =298K

Xe = f (T) Xe only depends on thermodynamics!Nothing to do with the energy balance!

A B A B

P P eA B

H (298K) 40000 cal / mol H (298K) 60000 cal / molC 50 cal / mol K C 50 cal / mol K K 100000 at 298K

L13-12


nA0 i p,i i0 s

i 1A

ARX R P R 0

F C T T W Q

ˆH (T ) C T FTX

Reaction is carried out adiabatically with an inlet temp of 298 K, CPA = 50 cal/mol∙K, & the heat of reaction = 20,000 cal/mol. The energy balance is:

ni pi 0

i 1EB

RX

C T TX

H T

A

ni pi p

i 1C 1 C

AP 0EB

RX

C T TX

H T

EB

50 T 298X

20000

From thermodynamicsX

T

From energy balance

TRX RX R PTR

H T H (T ) C dT

0 0

How to increase the conversion?

C

AeC

KX

1 K

AP 0EB

RX

C T TX

H T

A B

ns A i pi i A RX R p Ri

ˆQ W F C ( T T ) F X H ( T ) C ( T T ) Rearrang e

0 0 01

0

L13-13


Does increasing the entering temperature increase XA?

XAe

XA

T (K)XA,EB for adiabatic operation

(slants up for exothermic rxn)

Equilibrium curve

for exothermic rxn

p,A A0A,EB

RX R

C T TX

H (T )

Ae

RX R

C 2 2

1XH T1 1 1exp 1

K T R T T

Higher T moves XA,EB curve to the right

Adiabatic op, XA,EB at T i0,2

XA,e (Tadiabatic) decreases for an exothermic reaction

XA,e at Ti0,1

XA,e at Ti0,2

L13-14


Optimum Feed TemperatureFor a reversible and exothermic reaction, the feed temperature

should be optimized to maximize the conversion

From thermodynamicsXEB

T600500350

0.15

0.33

0.75

XA

W

T0 = 600T0 = 500

T0 = 350

There is an optimum inlet temperature!

High T0: reaction reaches equilibrium fast, but XA is low

Low T0 would give high XA,e but the specific reaction rate k is so small that most of the reactant passes through the reactor without reacting (never reach XA,e)

L13-15


How does one increase XA for adiabatic operation of an exothermic reaction?

with interstage cooling!

XEB

T

final conversion

XA,EB1

cooling process

XA,EB2

XA,EB3

XA,EB4

T0 Reactor 4Reactor 1

Cooling, C1 C2 C3

Reactor 3Reactor 2

Each reactor operates adiabatically

Endothermic reactions are similar, but with heating instead of cooling

L13-16


The equilibrium conversion increases with increasing temperature, so use interstage heating to increase the conversion

Endothermic Reactions

XEB

T

heating process

final conversion

Red lines are from the energy balance, slant backwards because H°RX >0 for endothermic reaction

L13-17


Suppose pure A enters a reactor at 298K . What is the maximum XA achievable in an adiabatic reactor? Assume ẆS=0, and CP = 0, CP.A=60

J/molK, H°RX(TR)= 20,000 J/mol, KC=10 exp[2405T-7.2]

Solve TEB for XA:

ns A0 i p,i i0 RX R P R A0 A

i 10 Q W F C T T H (T ) C T T F X

n

A0 i p,i i0 RX R P R A0 Ai 1

F C T T H (T ) C T T F X

0 0

AA 1 and A is only sp solve ecies for, X :

p,A A0A

RX R P R

C T TX

H (T ) C T T

p,A A0P A

RX R

C T TC 0 so X

H (T )

Plot XEB vs T and XA,e vs T to compute the maximum XA graphically

Ae

RX R

C 2 2

1XH T1 1 1exp 1

K T R T T

L13-18


Suppose pure A enters a reactor at 298K . What is the maximum XA achievable in an adiabatic reactor? Assume ẆS=0, and CP = 0, CP.A=60

J/molK, H°RX(TR)= 20,000 J/mol, KC=10 exp[2405T-7.2]

p,A A0A,EB

RX R

C T TX

H (T )

Ae

RX R

C 2 2

1XH T1 1 1exp 1

K T R T T

T XA,EB

200 0.294250 0.144298 0325 -0.081350 -0.156375 -0.231400 -0.306425 -0.381450 -0.456500 -0.606550 -0.756600 -0.906

T XAe

200 0.000149250 0.001634298 0.007612325 0.014746350 0.024722375 0.038481400 0.056318425 0.078262450 0.10407500 0.165203550 0.234315600 0.305584

L13-19


200 250 300 350 400 450 500 550 600

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

Energy balance conversion

Equilibrium conversion

T (K)

XASuppose pure A enters a reactor at 298K . What is the maximum XA

achievable in an adiabatic reactor? Assume ẆS=0, and CP = 0, CP.A=60 J/molK, H°RX(TR)= 20,000 J/mol (endothermic), KC=10 exp[2504T-7.2]

p,A A0A,EB

RX R

C T TX

H (T )

Ae

RX R

C 2 2

1XH T1 1 1exp 1

K T R T T

XAe

EB for adiabatic operation,

(slants down for endothermic rxn)

Tadiabatic

Nearly 0 conversion, not good!

Tadiabatic: Outlet T if reactor had an infinite volumeXA,e at Tadiabatic is max achievable XA in adiabatic reactor

L13-20


200 250 300 350 400 450 500 550 600

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

Energy balance conversionEquilibrium conversion

T (K)

XADoes increasing the inlet temperature to 600K improve the conversion of this

reaction? ẆS=0, and CP = 0, CP.A=60 J/molK, H°RX(TR)= 20,000 J/mol (endothermic), & KC=10 exp[2504T-7.2]

p,A A0A,EB

RX R

C T TX

H (T )

Ae

RX R

C 2 2

1XH T1 1 1exp 1

K T R T T

XAe

Tadiabatic: Outlet T if reactor had an infinite volume

EB for adiabatic operation (endothermic) T0=298K

Tadiabatic

XA,e at Tadiabatic is max achievable XA in adiabatic reactor

Nearly 0 conversion

L13-21


200 250 300 350 400 450 500 550 600

-1

-0.75

-0.5

-0.25

0

0.25

0.5

0.75

1

Equilibrium conversion

T (K)

XADoes increasing the inlet temperature to 600K improve the conversion of this

reaction? ẆS=0, and CP = 0, CP.A=60 J/molK, H°RX(TR)= 20,000 J/mol (endothermic), & KC=10 exp[2504T-7.2]

p,A A0A,EB

RX R

C T TX

H (T )

Ae

RX R

C 2 2

1XH T1 1 1exp 1

K T R T T

XAe ≈ 0 at T0 = 298K

EB for adiabatic operation (endothermic) T0=298K

Tadiabatic

EB for adiabatic operation T0=600KXAe ≈ 0.2 when T0 = 600K

Tadiabatic

(T0=600K)

Yes, higher conversion is achieved

Documents

L14: Equilibrium Conversion in Nonisothermal Reactor Design