L3b-1 Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign. Ideal CSTR Design Eq with X A :

L3b-1

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

Ideal CSTR Design Eq

with XA:

Review: Design Eq & ConversionD

ad

C ac

B ab A

fed A moles reacted A moles

XA

BATCHSYSTEM: A0Aj0jj XNNN

jA0A

jj0TjT XNNNN

FLOW SYSTEM: A0Aj0jj XFFF

jA0A

jj0TjT XFFFF

r

XFV

A

A0A

Vr dt

dXN A

A0A Ideal Batch Reactor

Design Eq with XA:

AX

0 A

A0A Vr

dXNt

AA

0A rdV

dXF Ideal SS PFR

Design Eq with XA:

AX

0 A

A0A r

dXFV

'rdW

dXF A

A0A Ideal SS PBR

Design Eq with XA:

AX

0 A

A0A 'r

dXFW

j≡ stoichiometric coefficient; positive for products, negative

for reactants

L3b-2


Review: Sizing CSTRsWe can determine the volume of the CSTR required to achieve a specific conversion if we know how the reaction rate rj depends on the conversion Xj

AA

0ACSTR

A

A0ACSTR X

rF

V rXF

V

Ideal SS CSTR

design eq.

Volume is product of FA0/-rA and XA

• Plot FA0/-rA vs XA (Levenspiel plot)

• VCSTR is the rectangle with a base of XA,exit and a height of FA0/-rA at XA,exit

FA 0 rA

X

Area = Volume of CSTR

X1

V FA 0 rA

X1

X1

L3b-3


FA 0 rA

Area = Volume of PFR

V 0

X1FA 0 rA

dX

X1

Area = VPFR or Wcatalyst, PBR

dX'r

FW

1X

0 A

0A

Review: Sizing PFRs & PBRsWe can determine the volume (catalyst weight) of a PFR (PBR) required to achieve a specific Xj if we know how the reaction rate rj depends on Xj

A

exit,AX

0 A

0APFR

exit,AX

0 A

A0APFR dX

r

FV

r

dXFV

Ideal PFR design eq.

• Plot FA0/-rA vs XA (Experimentally determined numerical values) • VPFR (WPBR) is the area under the curve FA0/-rA vs XA,exit

A

exit,AX

0 A

0APBR

exit,AX

0 A

A0APBR dX

r

FW

r

dXFW

Ideal PBR

design eq.

dXr

FV

1X

0 A

0A

L3b-4


Numerical Evaluation of Integrals (A.4)Simpson’s one-third rule (3-point):

2102X

0XfXf4Xf

3h

dxxf

hXX 2

XXh 01

02

Trapezoidal rule (2-point):

101X

0XfXf

2h

dxxf

01 XXh

Simpson’s three-eights rule (4-point):

32103X

0XfXf3Xf3Xfh

83

dxxf 3

XXh 03

h2XX hXX 0201

Simpson’s five-point quadrature :

432104X

0XfXf4Xf2Xf4Xf

3h

dxxf 4

XXh 04

L3b-5


Review: Reactors in Series

2 CSTRs 2 PFRs

CSTR→PFR

VCSTR1 VPFR2

VPFR2VCSTR1

VCSTR2

VPFR1

VPFR1

VCSTR2

VCSTR1 + VPFR2

≠

VPFR1 + CCSTR2

PFR→CSTR

A

A0

r-

F

i j

CSTRPFRPFR VVV

If is monotonically

increasing then:

CSTRi j

CSTRPFR VVV

L3b-6


Chapter 2 Examples

L3b-7


XA 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.85

-rA 0.0053 0.0052 0.0050 0.0045 0.0040 0.0033 0.0025 0.0018 0.00125 0.001

1. Calculate FA0/-rA for each conversion value in the tableFA0/-rA

Calculate the reactor volumes for each configuration shown below for the reaction data in the table when the molar flow rate is 52 mol/min.

FA0, X0

FA0, X0

X1=0.3

X1=0.3 X2=0.8

X2=0.8

Config 1

X1=0.3

X1=0.3FA0, X0

FA0, X0

X2=0.8

X2=0.8

Config 2

A

exit,AX

in,AX A

0AnPFR dX

rF

V

←Use numerical methods to solve

in,Aout,AnA

0AnCSTR XX

rF

V

XA,out and XA,in respectively, are the conversion at the outlet and inlet of reactor n

Convert to seconds→minmol

52F 0A

001

52 860

67

Amol minm

mol. F

sin s

-rA is in terms of mol/dm3∙s

L3b-8


A(

0

0)

AF

r

3

3

mol0.0053

d

mol0.867

s

s

m

m

d

164

1. Calculate FA0/-rA for each conversion value in the table

XA 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.85

-rA 0.0053 0.0052 0.0050 0.0045 0.0040 0.0033 0.0025 0.0018 0.00125 0.001

FA0/-rA 164


FA0, X0

FA0, X0

X1=0.3

X1=0.3 X2=0.8

X2=0.8

Config 1

X1=0.3

X1=0.3FA0, X0

FA0, X0

X2=0.8

X2=0.8

Config 2

A

exit,AX

in,AX A

0AnPFR dX

rF

V


in,Aout,AnA

0AnCSTR XX

rF

V


164


minmol

52F 0A

001

52 860

67

Amol minm

mol. F

sin s

Convert to seconds→

L3b-9


A(

0

0)

AF

r

3

3

mol0.0053

d

mol0.867

s

s

m

m

d

164


XA 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.85

-rA 0.0053 0.0052 0.0050 0.0045 0.0040 0.0033 0.0025 0.0018 0.00125 0.001

FA0/-rA


FA0, X0

FA0, X0

X1=0.3

X1=0.3 X2=0.8

X2=0.8

Config 1

X1=0.3

X1=0.3FA0, X0

FA0, X0

X2=0.8

X2=0.8

Config 2

A

exit,AX

in,AX A

0AnPFR dX

rF

V


in,Aout,AnA

0AnCSTR XX

rF

V


164


minmol

52F 0A

001

52 860

67

Amol minm

mol. F

sin s

Convert to seconds→ For each –rA that corresponds to a XA value, use FA0 to calculate

FA0/-rA & fill in the table

L3b-10


X1=0.3

X1=0.3FA0, X0

FA0, X0

A( 0.85)

3A0

3

mol0.867F s

molr0.001

dm s

867 dm


XA 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.85

-rA 0.0053 0.0052 0.0050 0.0045 0.0040 0.0033 0.0025 0.0018 0.00125 0.001

FA0/-rA 164 167 173 193 217 263 347 482 694 867


FA0, X0

FA0, X0

X1=0.3

X1=0.3 X2=0.8

X2=0.8

Config 1

X2=0.8

X2=0.8

Config 2

A

exit,AX

in,AX A

0AnPFR dX

rF

V


in,Aout,AnA

0AnCSTR XX

rF

V

Convert to seconds→minmol

52F 0A



001

52 860

67

Amol minm

mol. F

sin s

L3b-11


XA 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.85

-rA 0.0053 0.0052 0.0050 0.0045 0.0040 0.0033 0.0025 0.0018 0.00125 0.001

FA0/-rA 164 167 173 193 217 263 347 482 694 867FA0, X0

FA0, X0

X1=0.3

X1=0.3 X2=0.8

X2=0.8

Config 1

Reactor 1, PFR from XA0=0 to XA=0.3:

A

A AA

A A0

A

0.3A0

PFR1 A0

A0

X

0

A X

A0

AX 0.30.20A .X 1A 0

F 3 0.3 0V dX 3

F F3

rr

F

rr8 3

F

r

4-pt rule:

1

0.3 A0PFR A0

3

A16

F 3V dX 0.1 3 3 1

r 8934 173 5167 1.6 dm

A,out2CSTR

A0A,o A i

X, nut

A

FXV X

r

23

CSTR 694 0.8 3470.3 dmV

Total volume for configuration 1: 51.6 dm3 + 347 dm3 = 398.6 dm3 = 399 dm3


PFR1 CSTR2

0

XA,exit APFRn AXA,in A

FV dX

r

L3b-12


XA 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.85

-rA 0.0053 0.0052 0.0050 0.0045 0.0040 0.0033 0.0025 0.0018 0.00125 0.001

FA0/-rA 164 167 173 193 217 263 347 482 694 867

Reactor 1, CSTR from XA0=0 to XA=0.3:

Need to evaluate at 6 pts, but since there is no 6-pt rule, break it up

0

01 0

3

A

A .A,outCSTR A

FXV X

r

Total volume for configuration 2: 58 dm3 + 173 dm3 = 231 dm3

X1=0.3

X1=0.3FA0, X0

FA0, X0

X2=0.8

X2=0.8

Config 2

CSTR30. 583 0193 dmV

A0PFR2 A

A

0.8

0.3

FV dX

r

PFRV... .

263 263 342173

4 3 38 33 2

482193 6940 08 5

70 30 5

3 point rule 4 point rule

3173 dm

PFR2CSTR1

0.A0 A0

PF0.3

R2 A AA

05

.

.

5

8

A0

F FV dX dX

r r

Must evaluate as many pts as possible when the curve isn’t flat

L3b-13


ACSTR

AA

VX

C

r

0

0

CSTRA

AA

VC

rX

00

For a given CA0, the space time needed to achieve 80% conversion in a CSTR is 5 h. Determine (if possible) the CSTR volume required to process 2 ft3/min and achieve 80% conversion for the same reaction using the same CA0. What is the space velocity (SV) for this system?

space time holding time mean residence hV

time

0

5

=5 h 0=2 ft3/min

ftmin h

hVmin

3 605

23V ft 600

VSV

0 1Space

velocity:-1

hSV . h

0 2

51 1

Notice that we did not need to solve the CSTR design equation to solve this problem.Also, this answer does not depend on the type of flow reactor used.

XA=0.8

ACSTR A

AF

rXV

0 AA

CSTR

A

C

r

VX

0

0

00

VV

L3b-14


XA,exit

PFRA

AX AA,in

CV dX

r

00

A product is produced by a nonisothermal, nonelementary, multiple-reaction mechanism. Assume the volumetric flow rate is constant & the same in both reactors. Data for this reaction is shown in the graph below. Use this graph to determine which of the 2 configurations that follow give the smaller total reactor volume.

FA0, X0

FA0, X0

X1=0.3

X1=0.3 X2=0.7

X2=0.7

Config 2

X1=0.3

X1=0.3FA0, X0

FA0, X0

X2=0.7

X2=0.7

Config 1

ACSTR A,out A,in

AV X X

r

C 0

0

Shown on graph

XA,exit

PFRn AAA,in

A

XV dX

F

r

0

CSTRA

AA

V Xr

F

0

• Since u0 is the same in both reactors, we can use this graph to compare the 2 configurations

• PFR- volume is u0 multiplied by the area under the curve between XA,in & XA,out

• CSTR- volume is u0 multiplied by the product of CA0/-rA,outlet times (XA,out - XA,in)

L3b-15


A product is produced by a nonisothermal, nonelementary, multiple-reaction mechanism. Assume the volumetric flow rate is constant & the same in both reactors. Data for this reaction is shown in the graph below. Use this graph to determine which of the 2 configurations that follow give the smaller total reactor volume.

FA0, X0

FA0, X0

X1=0.3

X1=0.3 X2=0.7

X2=0.7

Config 2

X1=0.3

X1=0.3FA0, X0

FA0, X0

X2=0.7

X2=0.7

Config 1

• PFR- V is u0 multiplied by the area under the curve between XA,in & XA,out

• CSTR- V is u0 multiplied by the product of CA0/-rA,outlet times (XA,out - XA,in)

Config 1 Config 2

Less shaded areaConfig 2 (PFRXA,out=0.3 first, and CSTRXA,out=0.7 second) has the smaller VTotal

XA =

0.3

XA =

0.7

XA =

0.3

XA =

0.7

Documents

L3b-1 Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign. Ideal CSTR Design Eq with X A :