Download - Slide 1mlkraft/CHBE 424 materials/L3... · PPT file · Web view2015-08-23 · L3-Slides courtesy of Prof M L Kraft, Chemical & Biomolecular EngrDept, University of Illinois at Urbana-Champaign.

L3-1

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

Rate of generation of reactant A in reactor due to rxn

Rate of accumulation ofreactant A in reactor =

Review: Batch Reactor Basic Molar Balance

• No material enters or leaves the reactor• In ideal reactor, composition and temperature are spatially

uniform (i.e. perfect mixing)• No flow in or out of reactor. Fj0 and Fj = 0.

dt

dNdVr jVj Batch Reactor

Design Equation

dt

dNVr jj

Ideal Batch Reactor Design Equation

Ideal (perfectly mixed) reactor: spatially uniform

temp, conc, & reaction rate

L3-2


Review: CSTR Basic Molar Balance

Accumulation = In - Out + Generation by rxn

0 = Fj0 - Fj +

Vrj

V

jdVr

No spatial variation:

0 0 0

j Cj j A Aj

j A

F F C CV F V

r r

• Continuously add reactants and remove products• In an ideal reactor, composition and temperature

are spatially uniform (i.e. perfect mixing) • At steady state- no accumulation

Fj0 Fj

Ideal Steady State CSTR Design Equation:

in terms of concentration

in terms of flow

(upsilon)

L3-3


ΔV

FA0 FA

Review: Molar Balance – PFR

jVjVVj r

VFF

0Vlim

VrjFj0 Fj dtdNj+- =

0VrFF jVVjVj

jj r

dVdF

Ideal SS PFR Design Eq.

• Flow reactor operated at steady state (no accumulation per Δ)• Composition of fluid varies down length of reactor (material

balance for differential element of volume V

L3-4


• Heterogeneous rxn: reaction occurs at catalyst particle surface • Concentration gradient of reactant and product change down

length of the reactor• Rxn rate based on the mass of catalyst W, not reactor volume V

Review: Molar Balance- Packed Bed Reactor (PBR)

jj r

dVdF

Similar to PFR, but expressed in terms of catalyst weight instead of reactor volume

Units for the rate of a homogeneous rxn (rj) :

Units for the rate of a catalytic rxn (rj’) : catalyst kgs

mol3ms

mol

So in terms of catalyst weight instead of reactor volume:

catalyst the of weightthe is W where'rdWdF

jj

L3-5


L3: Conversion and Reactors in Series

FA0 X0

FA1, X1

V1 V2

FA2 X2

FA0 X0

FA1, X1

V1

V2

FA2, X2

VCSTR1 VPFR2

XA

FA0/

-rA

(m3)

L3-6


Conversion, XAConversion is convenient for relating: rj, V, υ, Nj, Fj, and Cj

fed A moles reacted A moles Aon based conversionXA

D d Cc B b A a

Choose limiting reactant A as basis of calculation and normalize:

D ad C

ac B

ab A

BATCHSYSTEM: “Moles A fed” is the amount of A at the start of the reactor (t=0)

FLOWSYSTEM: “Moles A fed” is the amount of A entering the reactor

Usually pick the basis to be the limiting reagent

L3-7


Conversion ExampleA + 2B → 2C

Start with 1 mole of A & 1 mole of B

If A is the basis and at the end we have:

1 mole A, 1 mole B ↔ XA = 0/1 = 0 (no reaction)½ mole A, 0 mole B ↔ XA = 0.5/1 = 1/20 mole A, -1 mole B ↔ XA = 1/1 = 1 (complete reaction)

Not possible!

The correct approach is to take B as the basis because B is the limiting reagentAt the end we have:

1 mole A, 1 mole B ↔ XB = 0/1 = 0 (no reaction)½ mole A, 0 mole B ↔ XB = 1/1 = 1 (complete reaction)

L3-8


Expressing other Components in Terms of Conversion of A (XA)

D ad C

ac B

ab A

BATCHSYSTEM:

Longer reactant is in reactor, more reactant is converted to product (until reactant is consumed or the reaction reaches equilibrium)∴ Conversion (Xj) is a function of time (t) in the batch reactor

A0A0AA XN N N Moles A in

reactor at time t = Moles A fed - Moles A

consumed

A0AA X 1NN

A0A0BB X NabNN A0A0CC X N

acNN

A0A0DD X NadNN component inert NN 0II

reactant product

fed A moles reacted A moles XA

L3-9


Expressing other Components in Terms of Conversion of A (XA)

jA0A0TjT XN1

ab

ac

adNNN

ad

ac 1

ab

dcAB

j≡ stoichiometric coefficient; positive for products, negative for reactants

A0Aj0jj XNNN

jA0A

jj0TjT XNNNN

Total moles in reactor at time t = Total

moles fed + total moles products formed minus reactants consumed

D ad C

ac B

ab A


L3-10


Batch Reactor Design Equation with Xj

A0A0AA XN N N In terms of A:

Vr dt

dNA

A Ideal Batch Reactor Design Eq:

Want to determine how long to leave reactants in reactor to achieve a desired value for the conversion

A0A0AA XN N dtd N

dtd

dtdXN 0

dtdN A

0AA

dt

dXN dt

dN A0A

A ←Substitute into batch reactor design eq

Vr dt

dXN AA

0A Ideal Batch Reactor Design Eq with Xj:

AX

0 A

A0A Vr

dXNt

D ad C

ac B

ab A


→ take derivative of “NA” equation w/ respect to time

L3-11


Flow and Conversion

A0A0AA XF F F

Molar flow rate that A leaves the reactor =

Molar flow rate A is fed to reactor

- Molar rate A is consumed in reactor

D ad C

ac B

ab A


FLOW SYSTEM:

For a given flow rate, the larger the reactor, the more time it takes the reactant to pass through the reactor, the more time to react∴ Conversion (Xj) is a function of reactor volume (V)

A0AA X 1FF

A0Aj0jj XFFF :general in

jA0A

jj0TjT XFFFF

L3-12


CSTR Design Equation & Xj

r

FFVj

A0A

Ideal SS CSTR:

A0A0AA XFFF Substitute for FA

r

XFFFVA

A0A0A0A

rXFVA

A0A

Ideal CSTR design eq in terms of XA

V ≡ CSTR volume required to achieve a specified conversionNote: XA and –rA are evaluated at the exit of the CSTR

L3-13


PFR Design Equation & Xj

Ideal SS PRF:

A0A0AA XFFF

AA r

dVdF

Want to determine the reactor volume required to achieve a desired amount of conversion

A0A0AA XF F dVd F

dVd

dVdXF 0

dVdF A

0AA

dV

dXF dVdF A

0AA ←Substitute into PFR design eq

AA

0A rdV

dXF Ideal SS PFR Design Eq with Xj:

AX

0 A

A0A r

dXFV

Applies for no pressure drop down PFR!

→ take derivative of “FA“ expression with respect to volume

L3-14


PBR Design Equation & Xj

Ideal SS PBF:

A0A0AA XFFF

'rdWdF

AA

Want to determine the weight of catalyst that is required to achieve a desired amount of conversion

A0A0AA XF F dWd F

dWd

dWdXF 0

dWdF A

0AA

dWdXF

dWdF A

0AA ←Substitute into PBR design eq

'rdWdXF A

A0A Ideal SS PBR

Design Eq with Xj:

AX

0 A

A0A 'r

dXFW

Applies for no pressure drop down PBR!

→ take derivative of FA expression with respect to W

L3-15


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

FA 0 rA

X

Area = Volume of CSTR

X1

V FA 0 rA

X1

X1

L3-16


Sizing a CSTR with a Levenspiel PlotXA 0 0.1 0.2 0.4 0.6 0.7 0.8

FA0/-rA 0.89 1.08 1.33 2.05 3.56 5.06 8

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10123456789

XA

FA0/

-rA

(m3)

VCSTR for XA = 0.4?

AA

0ACSTR X

rFV

3

CSTR

m 82.0

4.005.2V

VCSTR for XA = 0.8?

3

CSTR

m 4.6

8.08V

Value of FA0/-rA for XA=0.4

L3-17


Sizing PFRsWe can determine the required volume of a PFR to achieve a specific conversion if we know how the reaction rate rj depends on the conversion Xj

Aexit,AX

0 A

0APFR

exit,AX

0 A

A0APFR dX

rFV

rdXFV

Ideal PFR

design eq.

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

FA 0 rA

Area = Volume of PFR

V 0

X1FA 0 rA

dX

X1

L3-18


Sizing a PFR with a Levenspiel PlotXA 0 0.1 0.2 0.4 0.6 0.7 0.8

FA0/-rA 0.89 1.08 1.33 2.05 3.56 5.06 8

VPFR for XA = 0.4?

Aexit,AX

0 A

0APFR dX

rFV

We do not have an expression for –rA(XA)

L3-19



FA0/-rA 0.89 1.08 1.33 2.05 3.56 5.06 8

VPFR for XA = 0.4?

Aexit,AX

0 A

0APFR dX

rFV

We do not have an expression for –rA(XA)

Numerically evaluate (Appendix A.4) to estimate the area under the curve

Volume of PFR

L3-20


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

L3-21


XA 0 0.1 0.2 0.4 0.6 0.7 0.8FA0/-rA 0.89 1.08 1.33 2.05 3.56 5.06 8


FA0/-rA 0.89 1.08 1.33 2.05 3.56 5.06 8

VPFR for XA = 0.4?A

exit,AX

0 A

0APFR dX

rFV

Use Simpson’s one-third rule (3-point):

31 33 20 890 2 4 0 553

05 PFR .V ... m.

2102X

0XfXf4Xf

3hdxxf hXX

2XXh 01

02

0 00 4

3 0 2 00 4

AA

AF

AA

P RA

Fr X

hV Fr. .

Fr X X

2.02.00X 0.22

04.0h 1

= area under the curve

XA increments must be equal

0.89 1.33 2.05

L3-22


Reactors in SeriesIn practice, reactors are usually connected so the exit stream of one reactor is the feed stream for the next reactor

Conversion up to point i (no side streams): reactor 1st into fed A Moles

i point to up reacted A of moles totalXi

FA0

FA1

i=1X1

V1 V3

FA3 i=3 X3

V2

FA2

i=2X2

i0A0AAi XFFF

L3-23


2 CSTRs in Series

FA0 X0

FA1, X1

V1 V2

FA2 X2

Materials balance reactor 1:In Out- + Gen. = Accum.

A1A0 A1 1F r VF 0

A1 A0 A0 1F F F X

A0 AA 10 A1 10FF r V- F X 0

0VrXF 11A10A

VXr

F1CSTR1

1A

0A

Need to express FA1 in terms of X1

L3-24


2 CSTRs in Series

FA0 X0

FA1, X1

V1 V2

FA2 X2

122A

0A2CSTR XX

rFV

Materials balance reactor 2:In Out- + Gen. = Accum.

0Vr F F 22A2A1A

20A0A2A XFFF

2A

2A1A2CSTR r

FFV

2A

20A0A10A0A2CSTR r

XFFXFFV

Value of FA0/-rA at X2

VXr

F1CSTR1

1A

0A

10A0A1A XFFF Materials balance reactor 1:

Need to express FA2 in terms of X2

L3-25


2 CSTRs in Series

Usually for the same overall conversion, VTOTAL, 2 CSTRs IN SERIES < VSINGLE CSTR

FA0 X0

FA1, X1=0.4

V1 V2

FA2 X2=0.8

XA 0 0.4 0.8FA0/-rA 0.89 2.05 8

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.80

2

4

6

8

XAFA

0/-r

A (m

3)

VCSTR1 for XA1 = 0.4?

31CSTR m 82.04.005.2V

VCSTR2 for XA2 = 0.4 to 0.8?

122A

0A2CSTR XX

rFV

32CSTR m 2.34.08.08V

33321CSTR m 02.4m 2.3m 82.0V

VCSTR of single CSTR with XA = 0.8?

3CSTR m 4.68.08V <

L3-26


2 PFRs in Series

A2X

1X A

0AA

1X

0 A

0AA

2X

0 A

0APFR dX

rFdX

rFdX

rFV

FA0, X0

FA1 X1 FA2, X2

XA 0 0.2 0.4 0.6 0.8FA0/-rA 0.89 1.33 2.05 3.56 8

VPFR2 for XA2 = 0.4 to 0.8?

32PFR m 61.3856.3405.2

32.0V

33321PFR m 17.2m 61.1m 55.0V

8.0Xr

F6.0Xr

F44.0Xr

F3hV

A

0A

A

0A

A

0A

Same volume as 1 PFR with XA=0.8

2 PFRs in series, X1=0.4 and X2=0.8

When XA1= 0.4, VPFR1 =0.55 m3 (slide L3-20)

L3-27


Combinations of CSTRs & PFRs in Series

FA0 X0

FA1, X1

V1

V2

FA2, X2

VCSTR1 VPFR2 VPFR1 VCSTR2

FA0, X0

FA1 X1

V1

V2

FA2, X2

( )VCSTR1 + VPFR2 ≠ VPFR1 + CCSTR2

L3-28


Reactors in Series

for any combination of PFRs & CSTRs in series

then ,increasinglly monotonica is r-

F IfA

A0

CSTR onei j

)j(CSTR)i(PFR PFR one VVVV

In general, 1 PFR = any number of PFRs in series 1 PFR = ∞ number of CSTRs in series

Definitions:Space time (t): time necessary to process one reactor volume, also called mean residence time or holding time

Space velocity (SV): inverse of space time, but vo may be measured under different conditions than the space time

0

V

t

t 1V

SV 0

0 liquid @ 60 F or 75 FLHSVV

0 STPGHSVV

Liquid-hourly space

velocity

Gas-hourly space

velocity0| is the volumetric flow rate measured at specified condition