L5-1 Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign. Relate all V( ) to XA Put together

L5-1

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

Relate all

V(u) to XA

Put together

Review: Derive –rA = f(XA)Relate all rj to Cj

• nj ≡ stoichiometric coefficient• for products, ⊖ for reactants

Relate all

Cj(X

A) to V(u)

•

jA

j

rr

j j0 j A0 Aj

N N N XC

V V

j j0 j A0 Aj

F F F XC

Batch: Flow:

00 A

0 0

P T ZV V 1 X

P T Z

Batch: Flow:

00 A

0 0

PZ T1 X

Z T P

j0 j A0 A 0 0j

A 0

C C X T ZPC

1 X P T Z

Batch &

Flow:

Now that Cj is in terms of XA, we can write the rate law in terms of XA

L5-2


Review: Stoichiometric Tables

SpeciesFeed rate (mol/time)

Change in reactor (mol/time)

Effluent rate from reactor (mol/time)

A FA0 -FA0XA FA = FA0 (1–XA)

B FB0 = QBFA0 nBFA0XA FB = FA0 (QB+ nBXA)

C FC0 = QCFA0 nCFA0XA FC = FA0 (QC+ nCXA)

D FD0 = QDFA0 nDFA0XA FD = FA0 (QD+ nDXA)

I FI0 = QIFA0 --- FI =FI0

Total FT0 dFA0XA FT = FT0 + dFA0XA

FA0

FB0

FC0

FD0

FI0

FA

FB

FC

FD

FI

D ad

C ac

B ab A

j0 j0 0 j0 j0J

A0 A0 0 A0 A0

F C C y

F C C y

d c b= 1

a a a

nj ≡ stoichiometric coefficient for products, ⊖ for reactants

In Out

L5-3


L5: Reactor Design Recipe and Reactor Scale-Up (Sizing)

Goal: Develop an algorithm that combines reactor design equations with reaction rates for the design of different reactors

L5-4


XA A

A0A0

dXt=N

-r V XA dXAV =FA0 -rA0

The Logic of Isothermal Reactor Design

V jj0 j j

dNF F r dV

dt

In Out- +Generation =Accumulation1. Set up mole balance for specific reactor

2. Derive design eq. in terms of XA for each reactor

Batch

A0 A

A

F XV =

-r

CSTR PFR

3. Put Cj is in terms of XA and plug into rA

j0 j A0 A 0 0j

A 0

C C X T ZPC

1 X P T Z

n

A jr kC

nj0 j A0 A 0 0

AA 0

C C X T ZPr k

1 X P T Z

4. Plug rA into design eq and solve for the

time (batch) or volume (flow) required for a specific XA

Today and next week!

(We will always look conditions where Z0=Z)

L5-5


Batch Reactor Operation (1)

Batch Volume is constant, V=V0

AAA0

dXN V

dtr Mole balance

Rate law A2

Ar kC

Stoichiometry (put CA in terms of X)

A A0 AC C (1 X )

Combine AA02

AA 2

0C 1d

Vdt

kN XX

22A0 A

AA0 0k C

dN

dt1 VX

X

A → B -rA = kCA2 2nd order reaction rate

Calculate the time required for a conversion of XA in a constant V batch reactor

L5-6




Evaluate 2A

AA0 0

20 A

dXkC 1 XN

tV

d

22A

0

A0

AA0

dX1 X

t

VC

Nk

d 22A

A0A

A0

dXkC 1 X

1Cdt

2AA0 A

dXkC 1 X

dt Rearrange to get like variables together

A

A02A

dXkC dt

1 X

A

2A0 A

dX1dt

kC 1 X

k is constant for an isothermal reaction

X tAA

2A0 0 0A

dX1dt

kC 1 X

A

A0 A

X1t

kC 1 X

Time required to achieve XA for 2nd order rxn

Integrate

A → B -rA = kCA2 2nd order reaction rate

L5-7



Batch Volume is constant, V=V0

AAA0

dXN V

dtr


Mole balance

Rate law A ACr k


A A0 AC C (1 X )

Combine AA0 0 AA k

dXC 1N V

dtX

AA0 00 AA kC 1

dXN V

dtX

A → B -rA = kCA 1st order reaction rate

Mole balance as a function of conversion

L5-8




Evaluate to solve for time AA A0 0A0

dXkC 1N X

tV

d

AA0

0A

A0

dXkC 1 X

V

Ndt A

A0A0

AdX

kC Xt

1C

1d

AA

dXk 1 X

dt Rearrange to get like variables together

A

A

dXkdt

1 X

A

A

dX1dt

k 1 X

k is constant for an isothermal reaction

X tA

A

A0 0

dX1dt

k 1 X A

1 1ln t

k 1 X

Time required to achieve XA for 1st order rxn

Integrate

Mole balance as in terms of XA:


1ln ln 1 x

1 x

Remember:Confused about the integration? See the next slide

L5-9




X tA

A

A0 0

dX1dt

k 1 X

Integrate


X tAA A00

1 1 1- ln 1 X t ln 1 X ln 1 0 t 0

k k k

0=ln(1)

AA

1 1 1ln 1 X t ln t

k k 1 X

1ln ln a

a

Remember:

L5-10


Typical Cycle Time for a Batch Polymerization

Activity Time (h)

1. Charge feed to the reactor and agitate (tf) 1.5 - 3.0

2. Heat to reaction temperature (te) 0.2 – 2.0

3. Carry out reaction (tR) (varies)

4. Empty and clean reactor (tc) 0.5 – 1.0

Total time excluding reaction 3.0 – 6.0

Total Cycle Time tt = tf + te + tR + tc

Total Cycle Time tt for a batch process is much longer than the reaction time because it takes time to set up, heat, and clean the reactor each time it is used

L5-11


CSTR Operation (1)

Calculate the CSTR volume required to get a conversion of XA

Mole balance

Rate law A ACr k


A A0 AC C (1 X )

Combine A0 A

A0

Ck

F XV

1 X

A0 0 A

A0 A

C XV

kC 1 X

A → B -rA = kCA Liquid-phase 1st order reaction rate

A

A

A0F XV

r

Put FA0 in terms of CA0

0 A

A

XV

k 1 X

Volume required to achieve XA for 1st order rxn (u0=u)

L5-12


0 A

A

XV

k 1 X

A0

A

X

kV

1 X

Scaling CSTRs

0 A 0 A

small biggerA A

X Xknown: V want: V

k 1 X k 1 X

Space time t (residence time) required to achieve XA for 1st order irreversible rxn

• Chemical engineers are involved in scaling up a laboratory scale reaction to the pilot plant scale or full-scale reactor

• If one knows the volume of the pilot-scale reactor required to achieve XA, how is this information used to achieve XA in a larger reactor?

k in the small reactor is the same as k in the bigger reactor

Want XA in the small reactor to be the same as XA in the bigger reactor

u0 in the small reactor must be different from u0 in the bigger reactor

Suppose for a 1st order irreversible liquid-phase reaction:

A

0 A

XVk 1 X

Separate variables we will vary from those held constant

So the reactor volume V must be proportional to the volumetric flow rate u0

How?

0V A

A

X

k 1 X

L5-13


Scaling CSTRs with Spacetime t

A → B -rA = kCA

So if you know the spacetime t required to get a conversion of XA in a CSTR, you can use that to achieve the same XA in a different size CSTR

1st order reaction rate

A

A

Xk

1 X

What t is required to achieve a specific XA?

A

A

X

k 1 X

A Ak kX X A Ak X kX Ak X 1 k

Ak

X1 k

CSTR relationship between t and XA for 1st order liquid-phase rxn (isothermal and V = V0)

Space time t (residence time) required to achieve XA for 1st order irreversible rxn

A

A

X

k 1 X

Rearrange to get XA in terms of :t

L5-14


Damköhler Number, Da

A0

A0

r V reaction rateDa

F enterinrate of reacti at entrance

g flow rate of A convectioon

n rate

Estimates the degree of conversion that can be obtained in a flow reactor

First order irreversible reaction:

A0

A0 A0 0

A0kr V VCDa

F C

0

kDa

V Da k

1st order irreversible reaction

Second order irreversible reaction:

A02

A

0 A0 0

0

A

r V VD

kCa

F C

A

0

0kCa

VD

A0Da kC

2nd order irreversible reaction

Ak

X1 k

How is XA related to Da in a first order irreversible reaction in a flow reactor?

ADa

X1 Da

0V Substitute

From previous

slide:

L5-15


If Da<0.1 for this 1st order irreversible rxn in a flow reactor, then

Damköhler Number, DaA0

A0

r V rate of reaction reaction rateDa

F enterinat en

g flow rate of A convection rattrance

e

Estimates the degree of conversion that can be obtained in a flow reactor

A

kX

1 k

Relate XA to Da for a 1st order irreversible rxn in a flow reactor:

ADa

X1 Da

A ADa 0.1

X X 0.0911 Da 1 0.1

If Da>10 for this 1st order irreversible rxn in a flow reactor, then

A ADa 10

X X 0.911 Da 1 10

Da k1st order

irreversible rxn

L5-16


A0 A0

A0

1 2 kC 1 4 kCX

2 kC

Sizing CSTRs for 2nd Order Rxns

• Mole balance

• Rate laws

• Stoichiometry

• Combine

A

A0 0 A0

Ar r

F X C XV

A2

Ar kC

A A0C C (1 X)

A22

0

0

A0

kC X

C XV

1

or

0 A

20kC 1

V

X

X

1 2Da 1 4DaX

2Da

Calculate the CSTR volume required to get a conversion of XA

A → B -rA = kCA2 Liquid-phase 2nd order reaction rate

In terms of conversion?

In terms of space time?

In terms of XA as a function of Da? A0Da kC

2nd order liquid irreversible reaction

L5-17


CSTRs in SeriesCA0u0

CA1uCA2u

u0 = u Effluent of reactor 1 is input for reactor 2, no change in u

A first order reaction is carried out isothermally using 2 CSTRs that are the same size, and u and k are the same in both reactors (t1 = t2 = t & k1 = k2 = k)

Relate CA2 to CA1, k, & t

1. Mole balance CSTR2A1 A2

A2

F FV

r

2. Rate law CSTR2 A2 A2r kC

3. Stoichiometry CSTR2

4. Combine for CSTR2

0 A1 A2 A1 A2

A2 A2

C C C CV or =

kC kC

A1 A2A2 A1 A2 A2 A2 A1

A2

C C = kC C C kC C C

kC

A1A2 A1 A2

C C k 1 C C

k 1

Determine V1 for 1st CSTR using our standard procedure. For 2nd CSTR:

Skip this step for now.

L5-18


CSTRs in Series, CA1CA0u0CA1u

CA2u

A first order reaction is carried out isothermally using 2 CSTRs that are the same size, and u and k are the same in both reactors (t1 = t2 = t & k1 = k2 = k) What is CA1 in terms of t and k?

A

kX

1 k

We know for a single CSTR:

A1A1 A0 A1 A1

A0

A1A1

A0

CX 1

CC C (1 1 X

C CX )

Put XA for 1st CSTR in

terms of CA1:

Substitute:

A1A1

A0

k k1 k 1

CX 1

Ck

Solve for CA1:

A1

A0

Ck 1 1 k

C

A1 A1

A0 A0

C Ck 1 k k

C C

A1 A1

A0 A0

C Ck 1

C C

A1

A0

C1 k 1

C

A1

A0

C 1C 1 k

A0A1

CC

1 k

u0 = u Effluent of reactor 1 is input for reactor 2, no change in u

L5-19


CSTRs in Series, CA2CA0u0

CA1uCA2u u0 = u Effluent of reactor 1 is

input for reactor 2

A first order reaction is carried out isothermally using 2 CSTRs that are the same size, and u and k are the same in both reactors (t1 = t2 = t & k1 = k2 = k)

A0A1

AA2 1C &k

1

C

k

CC

1 Relate CA2 to k & t

Substitute

AA

20 1

k 11C

C

k

A0

A2 2

CC

1 k

1st order irreversible rxn with V1 = V2, t1 = t2 and k1 = k2

L5-20


n CSTRs in SeriesCA0u0

CA1uCA2u u0 = u

For n identical CSTRs, then:

A0An n

CC

1 k

How is conversion related to the # of CSTRs in series?

Put CAn in terms of XAn (XA at the last CSTR):

A0A0 An n

CC 1 X

1 k

An n

11 X

1 k

Ann

11 X

1 k1st order irreversible liquid phase rxn run in n CSTRs with identical V, t and k

Ann

1or 1 X

1 Da

1st order irreversible liquid-phase rxn run in n CSTRs with identical V, t and k

Rate of disappearance of A in the nth reactor:

A0

An An n

Cr kC k

1 k

L5-21


Isothermal CSTRs in Parallel

FA0

FA01

FA02same T, V, u

1 2 n X =X =...=X =X

A1 A2 An Ar r ... r r

Subscript i denotes reactor i

Aii A0i

Ai

XV F

r

FA01 = FA02 = … FA0n

iV total volume of all CSTRs

Vn # of CSTRs

Volume of each CSTR

A0A0i

F total molar flow rateF

n # of CSTRs

Molar flow rate of each CSTR

A0 Ai

Ai

F XV

n n r

Mole Balance

AA0

A

XV F

r

Conversion achieved by any one of the reactors in parallel is the same as if all the reactant were fed into one big reactor of volume

V

AA0

A

XV F

r

Documents

L5-1 Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign. Relate all V( ) to XA Put together