L3b reactor sizing example problems

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

rXF

VA

A0A

Vr dt

dXN AA

0A 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

'rdWdXF 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

rFV

rXFV

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

Aexit,AX

0 A

0APFR

exit,AX

0 A

A0APFR dX

rF

V 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

Aexit,AX

0 A

0APBR

exit,AX

0 A

A0APBR dX

rFW

rdXFW

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

3hdxxf

hXX 2

XXh 0102

Trapezoidal rule (2-point):

101X

0XfXf

2hdxxf

01 XXh

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

32103X

0XfXf3Xf3Xfh

83dxxf

3XXh 03

h2XX hXX 0201

Simpson’s five-point quadrature :

432104X

0XfXf4Xf2Xf4Xf

3hdxxf

4XXh 04

L3b-5


Review: Reactors in Series2 CSTRs 2 PFRs

CSTR→PFR

VCSTR1 VPFR2

VPFR2VCSTR1

VCSTR2

VPFR1

VPFR1

VCSTR2

VCSTR1 + VPFR2

≠ VPFR1 + CCSTR2

PFR→CSTR

A

A0r-

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, X0X1=0.3

X2=0.8

Config 1

X1=0.3FA0, X0 X2=0.8

Config 2

Aexit,AX

in,AX A

0AnPFR dX

rFV

←Use numerical methods to solve

in,Aout,AnA

0AnCSTR XX

rFV

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

Convert to seconds→minmol52F 0A

00152 860

67

Amol minm

mol. Fsin s

-rA is in terms of mol/dm3∙s

L3b-8


A(00)

AFr

3

3mol0.0053

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, X0X1=0.3

X2=0.8

Config 1

X1=0.3FA0, X0 X2=0.8

Config 2

Aexit,AX

in,AX A

0AnPFR dX

rFV


in,Aout,AnA

0AnCSTR XX

rFV


164


minmol52F 0A

00152 860

67

Amol minm

mol. Fsin s

Convert to seconds→

L3b-9


A(00)

AFr

3

3mol0.0053

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, X0X1=0.3

X2=0.8

Config 1

X1=0.3FA0, X0 X2=0.8

Config 2

Aexit,AX

in,AX A

0AnPFR dX

rFV


in,Aout,AnA

0AnCSTR XX

rFV


164


minmol52F 0A

00152 860

67

Amol minm

mol. Fsin 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=0.3FA0, X0

A( 0.85)3A0

3

mol0.867F smolr 0.001

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, X0X1=0.3

X2=0.8

Config 1

X2=0.8

Config 2

Aexit,AX

in,AX A

0AnPFR dX

rFV


in,Aout,AnA

0AnCSTR XX

rFV

Convert to seconds→minmol52F 0A



00152 860

67

Amol minm

mol. Fsin 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 164 167 173 193 217 263 347 482 694 867

FA0, X0X1=0.3

X2=0.8

Config 1

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

AA AA

A A0A

0.3 A0PFR1 A

0A0

X0A X

A0A X 0.30.20A .X 1A 0

F 3 0.3 0V dX 3 F F3 rrF

rr8 3F

r

4-pt rule:

10.3 A0

PFR A03

A16

F 3V dX 0.1 3 3 1r 8

934 173 5167 1.6 dm

A,out2CSTR

A0A,o A i

X, nut

A

FXV X

r 2

3CSTR 694 0.8 3470.3 dmV

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


PFR1 CSTR20

XA,exit A

PFRn AXA,in A

FV dXr

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

AA .

A,outCSTR AF XV Xr

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

X1=0.3FA0, X0 X2=0.8

Config 2

CSTR30. 583 0193 dmV

A0PFR2 A

A

0.8

0.3

FV dX

r

PFRV ... . 263 263 34217 34 3 3

8 33 2482193 694

0 08 57

0 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


ACSTRA

AV XC

r

0

0

CSTRA

AA

V Cr X

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 h V time

0

5

=5 h 0=2 ft3/min

ftmin h hV min

3 60 52 3V ft 600

VSV

0 1Space velocity:

-1hSV . 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

AFr XV

0 A

ACSTR

A

Cr

V X

0

0

00

V V

L3b-14


XA,exitPFR

AA

X AA,in

CV dXr

0

0

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, X0X1=0.3

X2=0.7

Config 2

X1=0.3FA0, X0 X2=0.7

Config 1

ACSTR A,out A,in

AV X Xr

C 0

0

Shown on graph

XA,exitPFRn A

AA,in

AX

V dXFr

0

CSTRA

AA

V XrF

0

• Since 0 is the same in both reactors, we can use this graph to compare the 2 configurations

• PFR- volume is 0 multiplied by the area under the curve between XA,in & XA,out

• CSTR- volume is 0 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, X0X1=0.3

X2=0.7

Config 2

X1=0.3FA0, X0 X2=0.7

Config 1

• PFR- V is 0 multiplied by the area under the curve between XA,in & XA,out

• CSTR- V is 0 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

L3b reactor sizing example problems

Engineering

Transcript of L3b reactor sizing example problems