lecture 21 - umich.edu

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 21

Web Lecture 21 Class Lecture 17 – Tuesday 3/19/2013 �  Gas Phase Reactions �  Trends and Optimums

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3

User Friendly Equations relate T, X, or Fi Review Last Lecture

1. Adiabatic CSTR, PFR, Batch, PBR achieve this:

0ˆ =Δ= PS CW!

XEB =Θi∑ CPi

T −T0( )−ΔHRx

X =Θi∑ CPi

T −T0( )−ΔHRx

( )∑Θ

Δ−+=

iPi

Rx

CXHTT 0

4

User Friendly Equations relate T, X, or Fi 2. CSTR with heat exchanger, UA(Ta-T) and a large coolant flow rate:

XEB =

UAFA0

T −T a( )"

#$

%

&'+ Θi

!CPi∑ T −T 0( )

−ΔHRx

T Ta

Cm!

Fi0

X

5

User Friendly Equations relate T, X, or Fi 3. PFR/PBR with heat exchange:

FA0 T0

Coolant Ta

3A. In terms of conversion, X

( ) ( )

( )∑ Δ+Θ

Δʹ′+−=

XCCF

THrTTUa

dWdT

pPiA

RxAaB

i

~0

ρ

6

User Friendly Equations relate T, X, or Fi 3B. In terms of molar flow rates, Fi

( ) ( )

∑

Δʹ′+−=

i

ij

Pi

RxAaB

CF

THrTTUa

dWdT ρ

4. For multiple reactions ( )

∑

∑ Δ+−=

i

ij

Pi

RxijaB

CF

HrTTUa

dVdT ρ

5. Co-Current Balance ( )

cPc

aA

CmTTUa

dVdT

!−

=

7

Reversible Reactions

endothermic reaction

exothermic reaction

KP

T

endothermic reaction

exothermic reaction

Xe

T

Heat Exchange

8

Example: Elementary liquid phase reaction carried out in a PFR

FA0 FI

Ta cm! Heat Exchange

Fluid

BA⇔

The feed consists of both inerts I and Species A with the ratio of inerts to the species A being 2 to 1.

Heat Exchange

9

a)  Adiabatic. Plot X, Xe, T and the rate of disappearance as a function of V up to V = 40 dm3.

b)   Constant Ta. Plot X, Xe, T, Ta and rate of disappearance of A when there is a heat loss to the coolant and the coolant temperature is constant at 300 K for V = 40 dm3. How do these curves differ from the adiabatic case.

Heat Exchange

10

c)  Variable Ta Co-Current. Plot X, Xe, T, Ta and rate of disappearance of A when there is a heat loss to the coolant and the coolant temperature varies along the length of the reactor for V = 40 dm3. The coolant enters at 300 K. How do these curves differ from those in the adiabatic case and part (a) and (b)?

d)   Variable Ta Countercurrent. Plot X, Xe, T, Ta and rate of disappearance of A when there is a heat loss to the coolant and the coolant temperature varies along the length of the reactor for V = 20 dm3. The coolant enters at 300 K. How do these curves differ from those in the adiabatic case and part (a) and (b)?

Heat Exchange

11

Example: PBR A ↔ B

5) Parameters

•  For adiabatic:

• Constant Ta:

• Co-current: Equations as is

• Counter-current:

0dWdTa =

T)-T toT-T flip(or )1(dWdT

aa−⋅

0=Ua

Reversible Reactions

12

1) Mole Balances

)1( FrdWdX

0AAʹ′−=

0A

A

0A

BA

b

Fr

Fr

dVdX

VW

−=ρʹ′

−=

ρ=

Reversible Reactions

13

2) Rate Laws

)2( ⎥⎦

⎤⎢⎣

⎡−−=

C

BAA KCCkr

)3( 11exp1

1 ⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−=TTR

Ekk

)4( 11exp2

2 ⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−

Δ=

TTRHKK Rx

CC

Reversible Reactions

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3) Stoichiometry

5( ) CA =CA0 1− X( ) p T0 T( )

6( ) CB =CA0Xp T0 T( )

FT = FT 0dpdW

=αpFTFT0

TT0

!

"#

$

%&= −

α2p

TT0

!

"#

$

%&

W = ρV

dpdV

= −αρb2p

TT0

!

"#

$

%&

Note: Nomenclature change for 5th edition p ≡ y

Reversible Reactions

15

Parameters

bARx

CA

TCTHKTREkF

ρα , , , , ,)15()7( , , , , , ,

002

2110

Δ

−

3) Stoichiometry: Gas Phase

v = v0 1+εX( ) P0

PTT0

5( ) CA =FA0 1− X( )v0 1+εX( )

PP0

T0

T=CA0 1− X( )

1+εX( )pT0

T

6( ) CB =CA0X1+εX( )

pT0

T

7( ) dpdW

=−α2p

FTFT 0

TT0

=−α2p

1+εX( ) TT0

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Example: PBR A ↔ B

Reversible Reactions Gas Phase Heat Effects

KC =CBe

CAe

=CA0XepT0 T

CA0 1− Xe( ) pT0 T

8( ) Xe =KC

1+KC

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Reversible Reactions Gas Phase Heat Effects Example: PBR A ↔ B

18

Exothermic Case: Xe

T

KC

T

KC

T T

Xe ~1

Endothermic Case:

Example: PBR A ↔ B

Reversible Reactions Gas Phase Heat Effects

19

€

dTdV

=−rA( ) −ΔHRx( )−Ua T −Ta( )

∑FiCPi

∑FiCPi= FA0 ∑ΘiCPi

+ΔCPX$% &'

Case 1: Adiabatic and ΔCP=0

€

T = T0 +−ΔHRx( )X∑ΘiCPi

(16A)

Additional Parameters (17A) & (17B)

€

T0, ∑ΘiCPi=CPA

+ΘICPI

Reversible Reactions Gas Phase Heat Effects

Heat effects: ( )( ) ( )

( )9 0∑

−−Δ−−=

PiiA

ab

RxA

CF

TTUaHr

dWdT

θρ

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Case 2: Heat Exchange – Constant Ta

Reversible Reactions Gas Phase Heat Effects

( ) )C17( TT 0V , Cm

TTUadVdT

aoaP

aa

cool

==−

=!

Case 3. Variable Ta Co-Current

Case 4. Variable Ta Countercurrent

( ) ? 0 ==−

= aP

aa TVCm

TTUadVdT

cool!

Guess Ta at V = 0 to match Ta0 = Ta0 at exit, i.e., V = Vf

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Reversible Reactions Gas Phase Heat Effects

22

23

24

25

26

Endothermic

PFR

€

A→←B

€

dXdV

=

k 1− 1+1

KC

⎛

⎝ ⎜

⎞

⎠ ⎟ X

⎛

⎝ ⎜

⎞

⎠ ⎟

υ0 , Xe =

KC1+ KC

€

XEB =∑Θ iCPi

T−T0( )−ΔHRx

=CPA

+Θ ICPI T− T0( )−ΔHRx

T0

XXEB

Xe

€

T = T0 +−ΔHRx( )X

CPA+ΘICPI

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Conversion on temperature Exothermic ΔH is negative Adiabatic Equilibrium temperature (Tadia) and conversion (Xeadia)

( )PA

Rx0 C

XHTT Δ−+=

C

Ce K1

KX+

=

X

Xeadia

Tadia T 28

Adiabatic Equilibrium

X2

FA0 FA1 FA2 FA3

T0 X1 X3 T0 T0

Q1 Q2

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X

T

X3

X2

X1

T0

Xe

( )Rx

PiiEB H

TTCX

Δ−

−=∑ 0θ

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31

T

X

Adiabatic T and Xe

T0

exothermic

T

X

T0

endothermic

PIIPA

Rx

CCXHTT

Θ+

Δ−+= 0

Trends: Adiabatic

Gas Flow Heat Effects

32

Effects of Inerts in the Feed

33

€

k1+ΘI

⎛

⎝ ⎜

⎞

⎠ ⎟

Endothermic

34

As inert flow increases the conversion will increase. However as inerts increase, reactant concentration decreases, slowing down the reaction. Therefore there is an optimal inert flow rate to maximize X.

First Order Irreversible

Adiabatic:

35

As T0 decreases the conversion X will increase, however the reaction will progress slower to equilibrium conversion and may not make it in the volume of reactor that you have.

Therefore, for exothermic reactions there is an optimum inlet temperature, where X reaches Xeq right at the end of V. However, for endothermic reactions there is no temperature maximum and the X will continue to increase as T increases.

X

T

Xe

T0

X

T

X

T

Gas Phase Heat Effects

Adiabatic:

36

Effect of adding inerts

X

T

V1 V2 X

T T0

∞=ΘI

0I =Θ

Xe

X

€

X =T −T0( )CpA +θ ICpI[ ]

−ΔHRx

Gas Phase Heat Effects

Exothermic Adiabatic

37

As θI increase, T decrease and

€

dXdV

=k

υ0 Hθ I( )

k

θI

38

39

KC

V

XeX

V

T

V

Xe

V

k

V

Xe

XFrozen

V

OR

Endothermic

T

V

KC

V

Xe

V

XeX

V

k

V  

Exothermic

40

Adiabatic

Heat Exchange

T

V

KC

V

Xe

V

XeX

V

Endothermic

T

V

KC

V

Xe

V

XeX

V

Exothermic

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End of Web Lecture 21 End of Class Lecture 17

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