Chemical Reaction Engineering - PowerPoint Presentation

Slide1

Chemical Reaction Engineering (CRE) is the field that studies the rates and mechanisms of chemical reactions and the design of the reactors in which they take place.

Lecture

18Slide2

Today’s lecture

Solution to

in-class

problemUser friendly Energy Balance DerivationsAdiabaticHeat Exchange Constant TaHeat Exchange Variable Ta Co-currentHeat Exchange Variable Ta Counter Current

2Slide3

Adiabatic Operation CSTR

The feed consists of both -

Inerts

I and Species A with the ratio of inerts I to the species A being 2 to 1.

Elementary liquid phase reaction carried out in a CSTR

3Slide4

Adiabatic Operation for CSTR

Assuming the reaction is irreversible for CSTR, A



B, (KC = 0) what reactor volume is necessary to achieve 80% conversion?If the exiting temperature to the reactor is 360K, what is the corresponding reactor volume?Make a

Levenspiel

Plot and then determine the PFR reactor volume for 60% conversion and 95% conversion. Compare with the CSTR volumes at these conversions.

Now assume the reaction is reversible, make a plot of the equilibrium conversion as a function of temperature between 290K and 400K.

4Slide5

CSTR Adiabatic Example

Mole Balance:

5Slide6

CSTR Adiabatic Example

Rate Law:

Stoichiometry

:

6Slide7

CSTR Adiabatic Example

Energy Balance - Adiabatic, ∆C

p

=0:

7Slide8

CSTR Adiabatic Example

Irreversible for Parts (a)

through

(c)

(a) Given X = 0.8,

find

T and V

(

if

reverible

)

8Slide9

CSTR Adiabatic Example

Given X, Calculate T and V

9Slide10

CSTR Adiabatic Example

(b)

(

if

reverible

)

Given T, Calculate X and V

10Slide11

CSTR Adiabatic Example

(c)

Levenspiel

Plot

11Slide12

CSTR Adiabatic Example

(c)

Levenspiel

Plot12Slide13

CSTR     X = 0.95     T = 395

CSTR     X = 0.6     T = 360

13

CSTR Adiabatic ExampleSlide14

PFR     X = 0.6

PFR     X = 0.95

14

CSTR Adiabatic ExampleSlide15

CSTR

X = 0.6

T = 360

V = 2.05 dm

3

PFR

X = 0.6

T

exit

= 360

V = 5.28 dm

3

CSTR

X = 0.95

T = 395

V = 7.59 dm

3

PFR

X = 0.95

T

exit

= 395

V = 6.62 dm

3

Summary

15

CSTR Adiabatic ExampleSlide16

(d) At

Equilibrium

Calculate Adiabatic Equilibrium Conversion and Temperature:

16Slide17

(d) At

Equilibrium

17

Calculate Adiabatic Equilibrium Conversion and Temperature:Slide18

(e) Te = 358

Xe

= 0.59

18

Calculate Adiabatic Equilibrium Conversion and Temperature:Slide19

T

T

a

V+

Δ

V

V

m

c

, H

C

F

A,

F

i

In - Out + Heat Added = 0

PFR

Heat

Effects

V+

Δ

V

V

F

i

T

T

a

F

i

19Slide20

PFR

Heat

Effects

20Slide21

PFR

Heat

Effects

21Slide22

PFR

Heat

Effects

22Slide23

Heat removed

Heat generated

PFR

Heat

Effects

23Slide24

User Friendly Equations Relate T and X or

F

i

3. PBR in terms of molar flow rates

24

4. For multiple reactions

5. Coolant BalanceSlide25

Heat Exchange Example

Elementary

liquid phase reaction carried out in a PFRThe feed consists of both

inerts I and Species A with the

ratio

of inerts to the species A

being

2 to 1.

25

F

A0

F

I

T

a

Heat Exchange Fluid

TSlide26

1)

Mole

Balance

:

2) Rate

Law

:

Heat Exchange Example

26Slide27

3)

Stoichiometry

:

4) Heat Effects:

Heat Exchange

Example

:

Case 1-

Constant

T

a

27Slide28

Parameters:

28

Heat Exchange

Example

:

Case 1-

Constant

T

aSlide29

Heat removed

Heat generated

PFR Heat

Effects

29Slide30

Energy

Balance

:

Adiabtic and ΔCP=0Ua=0

Additional

Parameters (17A) & (17B)

30

Heat Exchange

Example

:

Case 2

Adiabatic

Mole

Balance

:Slide31

Adibatic

PFR

31Slide32

Find conversion,

X

eq

and T as a function of reactor volume

V

rate

V

T

V

X

X

X

eq

Example

:

Adiabatic

32Slide33

Heat Exchange:

33

Need

to

determine

T

aSlide34

A.

Constant

Ta (17B) Ta = 300K

Additional Parameters (18B – (20B):

B. Variable T

a

Co-Current

C. Variable T

a

Counter

Current

Guess

T

a

at V = 0 to match T

a0

= T

a0

at

exit

, i.e., V =

V

f

34

User

Friendly

EquationsSlide35

Coolant balance:

In - Out + Heat Added = 0

Variable

T

a

Co-current

All equations can be used from before except T

a

parameter, use differential T

a

instead, adding

m

C

and C

PC

35

Heat

Exchanger

Energy

BalanceSlide36

In - Out + Heat Added = 0

All equations can be used from before except

dT

a/dV which must be changed to a negative. To arrive at the correct integration we must guess the T

a

value at V=0, integrate and see if T

a0

matches; if not,

re-guess

the value for T

a

at V=0

Variable

T

a

Counter-current

36

Heat

Exchanger

Energy

BalanceSlide37

Derive the User Friendly Energy

Balance

for

a PBRDifferentiating

with

respect

to W:

37Slide38

Mole

Balance

on species i:Enthalpy

for species i:

38

Derive the

User

Friendly

Energy

Balance

for

a PBRSlide39

Differentiating

with

respect

to W:39

Derive the

User

Friendly

Energy

Balance for a PBRSlide40

Final Form of the Differential

Equations

in Terms of

Conversion:

A:

40

Derive the User Friendly Energy Balance for a PBRSlide41

Final Form of terms of Molar Flow Rate:

B:

41

Derive the User Friendly Energy Balance for a PBRSlide42

Reversible Reactions

The rate

law

for this reaction will follow an elementary rate law.

Where

K

e

is the

concentration

equilibrium

constant

.

We

know

from Le

Chaltlier’s

law

that

if

the

reaction

is

exothermic

,

K

e

will

decrease

as the

temperature

is

increased

and the

reaction

will

be

shifted

back to the

left

.

If the reaction is endothermic and the temperature is increased, K

e

will

increase

and the

reaction

will

shift

to the right.

42Slide43

Reversible Reactions

Van’t

Hoff

Equation:

43Slide44

Reversible Reactions

For the special

case

of ΔCP=0Integrating the Van’t Hoff Equation gives:

44Slide45

Reversible Reactions

endothermic

reaction

exothermic

reaction

K

P

T

endothermic

reaction

exothermic

reaction

X

e

T

45Slide46

End of Lecture 1846