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# Review: Equilibrium Conversion XAe

## 1. Review: Equilibrium Conversion XAe

L14-1Review: Equilibrium Conversion XAe

1

endothermic

A heat

XA,e

B

exothermic

B heat

A

0

Example) A⇌B

KC

T

CA0=1 CB0=0

CBe CA0 0 X Ae

X Ae

KC

CAe CA0 1 X Ae

1 X Ae

K C 1 X Ae X Ae KC XAe KC XAe

Rearrange to solve

in terms of XAe

K C X Ae 1 K C

KC

X Ae This equation enables us to

express Xae as a function of T

1 K C

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

## 2. Review: XAe and Temperature

L14-2Review: XAe and Temperature

X Ae

Clicker question

material

1

H RX TR 1 1

1

exp

1

K C T2

R

T T2

H RX TR 1 1

Exothermic & CP =0: H RX TR 0, when T exp

T T & X Ae

R

2

Makes sense from Le Chatelier’s principle A

B heat

Exothermic rxn produces heat→

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

H RX TR 1 1

Endothermic & Cp 0: H RX TR 0, when T exp

& X Ae

R

T T2

Makes sense from Le Chatelier’s principle A heat

B

Heat is a reactant in an endothermic rxn→

increasing temp adds reactant (heat) & pushes rxn to right (higher conversion)

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

## 3. Review: Optimum Feed Temperature

L14-3Review: Optimum Feed Temperature

For reversible, exothermic rxns, optimize feed temperature to maximize XA

High T0: moves XA,EB line to the right. Rxn reaches equilibrium fast, but low XA

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)

XEB

From thermodynamics

XA

0.75

T0 = 500

T0 = 600

T0 = 350

0.33

0.15

350

500

600

T

W

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

## 4. Review: Interstage Cooling

L14-4Review: Interstage Cooling

Adiabatic operation of each reactor simplifies the energy balance

Higher feed temp- reaction reaches equilibrium quickly but XA,e is low

Lower feed temp- higher XA,e but reaction rate is too slow to be practical

Cooling between reactors shifts XA,EB line to the left, increasing XA

XA,EB4

XEB

XA,EB3

final conversion

Each reactor

operates

adiabatically

XA,EB2

XA,EB1

cooling process

Cooling, C1

T0

Reactor 1

Reactor 2

C2

T

Reactor 3

C3

Reactor 4

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

## 5. Review: Endothermic Reactions

L14-5Review: Endothermic Reactions

The equilibrium conversion increases with increasing temperature, so use

interstage heating to increase the conversion

XEB

final conversion

heating process

Red lines are from

the energy balance,

slant backwards

because H°RX >0 for

endothermic reaction

T

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

## 6. L14: Nonadiabatic PFR/PBR Operation and Reactor Stability

L14-6L14: Nonadiabatic PFR/PBR

Operation and Reactor Stability

1. T changes with distance down reactor- differential form of EB must be

used

2. Multiple steady states: more than one set of conditions satisfies both the

energy balance & mole balance

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

## 7. Review: Application to a SS PFR

L14-7Review: Application to a SS PFR

FA0

FA

XA

T

distance

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

a)

b)

c)

d)

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

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

Use stoichiometry to obtain –rA as a function of XA

XA

Calculate:

dX A

may use numerical

V FA0

rA X A ,T methods

X

A0

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

## 8. Steady-State PFR/PBR w/ Heat Exchanger

L14-8Heat is added or removed through the cylindrical walls of the reactor

A

Q U A Ta T Ua Ta T V

a

V

T

a

FA0

T0

FAe

FH

i i

T

FH

i i

Te

Heat exchange

area per volume

of reactor

V V+ V

Energy balance on small volume of SS PFR:

Q Ws FH

i i V FH

i i V V 0

0

Plug in Q: Ua Ta T V Ws FH

i i V FH

i i V V 0

d FiHi

0

Take limit as V→∞: Ua Ta T

dV

dHi

dFi

Ua

T

T

H

F

0

a

Expand:

i i

dV

dV

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

## 9. TEB for PFR/PBR w/ Heat Exchanger

L14-9TEB for PFR/PBR w/ Heat Exchanger

Ta

FA0

T0

FH

i i

T

FH

i i

FAe

Te

V V+ V

dHi

dFi

Ua Ta T

H Fi

0

dV i

dV

Substitute the differentials:

dFi

dHi

dT

r i rA and

Cpi

dV i

dV

dV

dT

Ua Ta T Hi i rA FiCPi

0

dV

iHi HRX

dT

Ua Ta T HRX rA FC

0

i Pi

dV

Solve for dT/dV:

dT HRX rA Ua Ta T

dT

FC

HRX rA Ua Ta T

i Pi

dV

dV

FC

i Pi

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

## 10. Energy Balance for Tubular Reactors

L14-10Energy Balance for Tubular Reactors

Heat

generated

Heat

removed

dT HRX rA Ua Ta T

dV

FiCPi

Substitute and multiply

out the denominator

dT H RX (TR ) CP T TR rA Ua Ta T C

i Pi

dV

FA0 iCPi CPi i X A

dT H RX (TR ) CP T TR rA Ua Ta T

dV

FA0 iCPi CP X A

dX A rA

dV

FA0

Heat

removed

Multiply Ua

and (Ta-T) by

dT HRX rA Ua T Ta

-1 (-1 x -1 = 1)

dV

FiCPi

HRX H RX (TR ) CP T TR

Fi FA0 i i X A

Heat

generated

Switched

sign & order

in bracket

for A B

b

CPB CPA CP

a

Energy balance for

SS PFR, Ẇs=0

PFR energy balance is coupled to the PFR design eq, and

PFR design eq is coupled to Arrhenius eq for k or Kequil

(these are the 3 equations that must be simultaneously solved)

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

## 11. Liquid Phase Reaction in PFR

L14-11Liquid Phase Reaction in PFR

A

B liquid phase rxn carried out in PFR; WS 0 & pure A enters the PFR

dX A rA

Mole balance

dV

FA0

Rate law

E 1 1

k k1 exp

R T1 T

Stoichiometry

Combine

C

rA k C A B

with

KC

H RX TR 1 1

K C T K C T2 exp

R

T2 T

C A C A0 (1 X A )

CB C A0 X A

dX A k 1 X A X A K C

dV

0

H RX (TR ) CP T TR rA Ua Ta T

dT

Energy balance

dV

FA0 iCPi CP X A

Solve these equations simultaneously with an ODE solver (Polymath)

If this were a gas phase rxn w/ pressure drop, change

stoichiometry accordingly & include an equation for d P/dW

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

## 12. Review: Nonisothermal CSTR

L14-12Review: Nonisothermal CSTR

Isothermal CSTR: feed temp = temperature inside the CSTR

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

a) Solve TEB for T at the exit (Texit = Tinside reactor)

b) Calculate k = Ae-E/RT where T was calculated in step a

c) Plug the k calculated in step b into the design equation to calculate VCSTR

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

a) Solve TEB for T as a function of XA

b) Solve CSTR design equation for XA as a function of T (plug in k = Ae-E/RT )

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

XA,EB = conversion determined from the TEB equation

XA,MB = conversion determined using the design equation

XA,exit

XA,MB

XA

XA,EB

T

Intersection is T and XA that

satisfies both equations

Texit

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

## 13. Multiple Steady States in CSTR

L14-13Multiple Steady States in CSTR

1

XA,EB

XA,MB

0,8

XA

0,6

0,4

0,2

0

0

100

200

300

400

500

600

T (K)

• Plot of XA,EB vs T and XA,MB vs T

• Intersections are the T and XA that satisfy both the mass balance and

energy balance

• Multiple sets of conditions are possible for the same rxn in the same

reactor with the same inlet conditions!

Reactor must operate near one of these steady statesthis requires knowledge of their stability!

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

## 14.

Consider a jacketed CSTR with constant heat capacity, negligible shaft L14-14work, CP=0, first order kinetics, all feeds at the same temperature (Ti0=T0),

constant Ta in jacket, and an overall heat transfer coefficient

Q UA Ta T

n

TEB : 0 Q FA0 iCp,i T Ti0 HRX (T)FA0 X A

i 1

Substituting for Q UA Ta T and HRX (T) HRX (TR ) since CP 0

n

0 UA Ta T FA0 iCp,i T Ti0 HRX (TR )FA0 XA

i 1

Bring terms that remove heat to other side of equation:

n

FA0 iCp,i T Ti0 UA Ta T HRX (TR )FA0 XA

i 1

n

iCp,i T Ti0

i 1

UA Ta T

HRX (TR )X A

FA0

Heat removed term ≡ R(T)

Heat generated term ≡ G(T)

A steady-state occurs when R(T) = G(T)

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

## 15. Even More Terms…

L14-15Even More Terms…

Heat removed term ≡ R(T)

n

iCp,i T Ti0

i 1

CP0 iCPi

Heat generated term ≡ G(T)

UA Ta T

HRX (TR )XA

FA 0

FA0 X A

rA V

V

XA

rA

FA 0

UA Ta T

rA V

HRX (TR )

Substitute Cp0 T Ti0

FA0

F

A0

More substitutions:

T0FA0Cp0 UATa Ta T0

UA

T

Cp0FA0 c

UA Cp0FA0

1

HRX HRX TR

Cp0 1 T TC HRX

Heat

removed: R T Cp0 1 T TC

rA V

F

A0

Heat

rA V

generated: G T HRX F

A0

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

## 16. Heat Removal Term and T0

L14-16Heat Removal Term and T0

Heat removed: R(T)

Heat generated G(T)

Cp0 1 T TC HRX

rA V

F

A0

R(T) line has slope of CP0(1+ )

UA Cp0FA0 Tc Ta T0

1

R(T)

=∞

=0

R(T)

Increase

For Ta < T0

Increase T0

T

When T0 increases, slope stays

same & line shifts to right

Ta

T0

T

When increases from lowering

FA0 or increasing heat exchange,

slope and x-intercept moves

Ta<T0: x-intercept shifts left as ↑

Ta>T0: x-intercept shifts right as ↑

=0, then TC=T0 =∞, then TC=Ta

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

## 17. CSTR Stability

L14-17CSTR Stability

G(T) & R(T)

G(T)

R(T) > G(T) so

T gradually falls

G(T) > R(T) so T to T=SS3

gradually rises to

T=SS3

R(T)

G(T) > R(T) so

T gradually

rises to T=SS1

1

R(T) > G(T)

so T gradually

falls to T=SS1

3

2

Temperature

T

3 steady states satisfy the TEB and BMB

Suppose a disturbance causes the reactor T to drift to a T between SS1 & SS2

Suppose a disturbance causes the reactor T to drift to a T between SS2 & SS3

Suppose a disturbance causes the reactor T to drop below SS1

Suppose a disturbance causes the reactor T to rise above SS3

SS1 and SS3 are locally stable (return to them after temp pulse)

SS2 is an unstable- do not return to SS2 if there is a temp pulse

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

## 18. Multiple Steady States and T0

L14-18R(T), G(T)

Multiple Steady States and T0

T0,6

T0,5

T0,4

T0,3

Increasing T above these

TS cause a temperature

jump to the higher TS

T0,2 T0,1

T

TT

• Changing the inlet T will shift the steady state temperature (TS)

• Notice that the number of steady state temperatures depends on T0

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

## 19. Temperature Ignition-Extinction Curve

L14-19Plot TS vs T0

Ts, steady-state temp

• Slight increase in T above TS,green

causes reactor T to jump to TS,cyan

• Ignition temp: T where jump from

TS,lower to TS,upper occurs

• Slight decrease in T below TS,magenta

causes reactor T to drop to TS,yellow

• Extinction temp: T where drop from

TS,upper to TS,lower occurs

R(T), G(T)

Temperature Ignition-Extinction Curve

TS along dashed line

are unstable

extinction temperature

TS,upper

TS,lower

T

Upper steady state

Lower steady state

ignition temperature

T0, entering temperature

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

## 20. Runaway Reaction

L14-20R(T), G(T)

Runaway Reaction

T

Ignition temperature is very important:

once T0 exceeds Tignition, transition to the

upper steady state will occur

• undesirable

• dangerous

Runaway reaction

R(T) only intersects with

upper steady state

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