L7-1 Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign. Review: Liquid Phase Reaction in

L7-1

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

Review: Liquid Phase Reaction in PFRLIQUID PHASE: Ci ≠ f(P) → no pressure drop

Calculate volume required to get a conversion of XA in a PFR

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

Mole balance

Rate law

Stoichiometry (put CA in terms of X)

AA

A0

d rX

dV F

2A Ar kC

A A0 AC C (1 X )

Combine

A0 A

A0

2A

2C 1X Xd

V F

k

d

X VAA0 A

220 0A0 A

F dXdV

k C 1 X

A0 A

2AA0

F XV

1 Xk C

Liquid-phase 2nd order reaction in PFR

Be

ab

le t

o d

o t

he

se

4 s

tep

s,

inte

gra

te &

so

lve

fo

r V

fo

r A

NY

O

RD

ER

RX

N

See Appendix A for integrals frequently used in reactor design

L7-2


Review: Liquid Phase Reaction in PBRLIQUID PHASE: Ci ≠ f(P) → no pressure drop

Calculate catalyst weight required to get a conversion of XA in a PBR

2A → B -r’A = kCA2 2nd order reaction rate

Mole balance

Rate law


AA

A0

rX 'd

dW F

A2

Ar ' kC

A A0 AC C (1 X )

Combine

A0 A

A0

2A

2C 1X Xd

W F

k

d

X WAA0 A22

0 0A0 A

F dXdW

k C 1 X

A0 A

2AA0

F XW

1 Xk C

Liquid-phase 2nd order reaction in PBRBe

ab

le t

o d

o t

he

se

4 s

tep

s, in

teg

rate

&

so

lve

fo

r V

fo

r A

NY

OR

DE

R R

XN

L7-3


Review: Isobaric, Isothermal, Ideal Gas-Phase Rxns in Tubular Reactors

GAS PHASE:j0 j A0 A 0 0

jA 0

C C X T ZPC

1 X P T Z

1 1 1

j0 j A0 Aj

A

C C XC

1 X

Gas-phase reactions are usually carried out in tubular reactors (PFRs & PBRs)

• Plug flow: no radial variations in concentration, temperature, & ∴ -rA

• No stirring element, so flow must be turbulent

FA0 FA

Stoichiometry for basis species A:

A0 AA0 A0 AA A

A A

C 1 XC C XC C

1 X 1 X

L7-4


Review: Effect of e on u and XATf T0 A

T0

N N Change in total # moles at X 1

N total moles fed

: expansion factor, the fraction of change in V per mol A reactedu0: volumetric flow rate

00 A

0 0

PZ T1 X

Z T Pu u

u varies if gas phase & moles product ≠ moles reactant, or if a DP, DT, or DZ

occursNo DP, DT, or DZ occurs, but moles product ≠ moles reactant → 0 A1 Xu u

• = 0 (mol product = mol reactants): u u0: constant volumetric flow rate as XA ↑ < 0 (mol product < mol reactants): u < u0 volumetric flow rate ↓ as XA ↑

Q1: For an irreversible gas-phase reaction, how does the residence time and XA change when < 0?

a)They don’tb)The residence time is longer & XA increasesc)The residence time is longer & XA decreasesd)The residence time is shorter, & XA decreasese)The residence time is shorter & XA increases

L7-5


Review: Effect of e on u and XATf T0 A

T0

N N Change in total # moles at X 1

N total moles fed

: expansion factor, the fraction of change in V per mol A reactedu0: volumetric flow rate

00 A

0 0

PZ T1 X

Z T Pu u

u varies if gas phase & moles product ≠ moles reactant, or if a DP, DT, or DZ

occursNo DP, DT, or DZ occurs, but moles product ≠ moles reactant → 0 A1 Xu u

• = 0 (mol product = mol reactants): u u0: constant volumetric flow rate as XA increases

• < 0 (mol product < mol reactants): u < u0 volumetric flow rate decreases as XA increases

• Longer residence time than when u u0

• Higher conversion per volume of reactor (weight of catalyst) than if u u0

• > 0 (mol product > mol reactants): u > u0 with increasing XA

• Shorter residence time than when u u0

• Lower conversion per volume of reactor (weight of catalyst) than if u u0

L7-6


Review: Isobaric, Isothermal, Ideal Rxn in PFRGAS PHASE:

Ci = f( ) → no DP, DT, or DZ

Calculate PFR volume required to get a conversion of XA

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

Mole balance

Rate law


AA

A0

d rX

dV F

2A Ar kC

Combine

A0

2A

A0

2

2A

A

C 1 XdX

d 1 X

k

V F

2XA AA0A22

0A0 A

1 XFV dX

k C 1 X

22 AA0A A2

AA0

1 XFV 2 1 ln 1 X X

1 Xk C

Gas-phase 2nd order rxn in PFR no DP, DT, or DZ

A0 AA

A

C 1 XC

1 X

Integral A-7 in appendix

Be

ab

le t

o d

o t

he

se

5 s

tep

s, &

so

lve

for

V

for

AN

Y O

RD

ER

RX

N

L7-7


Review: Pressure Drop in PBRs

A0 AA

A 0

C 1 X PC

1 X P

AA

A0dX

Fd

r 'W

GAS PHASE: A → B -r’A = kCA2

Calculate dXA/dW for an isothermal ideal gas phase reaction with DP

2nd order reaction rate

Mole balance

Rate law A2

Ar ' kC


Combine

A0 A

A

2 22

2A

A 00

PP

k C 1 X

1 X

dX

dW F

Relate P/P0 to W (Ergun equation)

0A

0 0

PdP T1 X

dW 2 T P P

Ergun Equation can be simplified by using y=P/P0 and T=T0:

Ady

1 XdW 2y

Simultaneously solve dXA/dW and dP/dW (or dy/dW) using Polymath

L7-8


Review: Ergun Equation

A0

dy T1 X

dW 2y T

0

c c 0

2

A 1 P

Differential form of Ergun equation for pressure drop in PBR:

0

Py

P Tf T0

A0T0

N Ny

N

AC: cross-sectional area C: particle density

: constant for each reactor, calculated using a complex equation that depends on properties of bed (gas density, particle size, gas viscosity, void volume in bed, etc)

: constant dependant on the packing in the bed

volume of solid1 : fraction of solid in bed =

total bed volume

0A

0 0

PdP T1 X

dW 2 T P P

Calculates pressure drop in a packed bed.This equation can be simplified to:

L7-9


L7: Unsteady-State Isothermal Reactor Operation: CSTR Start-Up

and Semi-Batch Reactors

V0 Vf

start

CBu0

V0 + u0t

time t end

Semi-batch

• Time required to reach steady-state after CSTR start-up

• Predicting concentration and conversion as a function of time

A+BA

L7-10


Start-Up of a Fixed-Volume CSTRIsothermal (unusual, but simple case), well-mixed CSTR

Unsteady state: concentrations vary with time & accumulation is non-zero

Goal: Determine the time necessary to reach steady-state operation

moles A in CSTR changes with time until steady state is reached

In Out- +Generation = Accumulation

AA0 A A

dNF F r V

dt

CA0u0u0CA

Use concentration rather than conversion in the balance eqs

AA0 0 A 0 A

dNC C r V

dtu u

Divide by V to convert dNA to dCA

A AA0 A0 0

V

C C dN 1d

r

V

V

t

u u

0

V u

A0 AA

AC C

dt

Cr

d

A

A0 A AdC

C C rdt

Multiply

by t

L7-11


CSTR Start-Up: 1st Order Reaction A

AA0 A

dCC C

dtr AAr kC A

AA0 A

dCC C

dtCk

Integrate this eq to find CA (t) while 1st order rxn in CSTR is at unsteady-state:

Combine

Bring variables to one side & factor A

A0 AdC 1

C C 1 kdt

A0A

ACdC 1

1 k Cdt 1 k

C tAA

A00 0A

1 kdCdt

CC

1 k

A0

A

A0

CC

1 k 1 kln 0 t

C0

1 k

1 kt

A

A0

C1 e

C

1 k

t 1 kA0A

C1 e C

1 k

Put like variables with their integrals

L7-12


CSTR Start-Up: 1st Order Reaction

t 1 kA0A

C1 e C

1 k

AA0 A A

dCC C kC

dt We integrated this eq to find CA (t) while

CSTR of 1st order rxn is in unsteady-state:

At steady state, t is large and: 0

A0

ASC

C1 k

AA0 A A

dCC C kC

dt Is this consistent with steady

state balance eq for CSTR? No accumulation at steady state

0

A0

A0 A A ASC

C C kC 0 C1 k

In the unsteady state, when CA = 0.99CAS:

0t 1 kA0 s ACC1 e

1 k0.99

1 k

t 1 k t 1 k t 1 ks s s1 e 0.99 0.01 e ln 0.01 ln e

s1 k

4.6 t

s4.6 t1 k

time to reach 99% of steady-state concentration in terms of tk

Solve for ts to determine time to reach 99% of steady-state concentration

Goal: combine start-up and SS eqs to estimate time to reach SS (ts)

Yes, same!

L7-13


CSTR Start-Up: 1st Order Reaction t 1 kA0

AC

1 e C1 k

99% of the steady-state concentration is achieved at: A AS4.6 C 0.99C

1 k

When k is very small (slow rxn), 1>>k: st 4.6

When k is very big (fast rxn), 1<<k s

4.6t

k

63% of the steady-state concentration is achieved at: 1 k

CA = 0.63CAS

k

t 4.61

In the unsteady state, the time to reach CA = 0.99CAS is:

L7-14


Better Selectivity in a Semi-Batch Reactor

To enhance selectivity of desired product over side product

kPA B P 2

P p A Br k C C Desired product P

kSA B S 2

S S A Br k C C Undesired side product S

Instantaneous selectivity, SP/S, is the ratio of the relative rates*: 2

P P A BP/S 2

S S A B

r k C CS

r k C C

Higher concentrations of A favor formation of the desired product P

Higher concentrations of B favor formation of the undesired side product S

Slowly feed B into the reactor containing A

Commonly used in bioreactors, when the enzyme is inhibited by excess substrate

P A

S B

k C

k C

To maximize the formation of the desired product:

*We’ll look at this concept of instantaneous selectivity in more detail in L9

L7-15


Semi-Batch Reactor Design EquationCBu0

V0 + u0t

Do a mole balance on A since it does not enter or leave the reactor (assume the reactor is well-mixed)

In Out- + Generation = Accumulation

AA0 A A

dNF F r V

dt

AAd

0 0 r V N

dt

Use whatever units are most convenient (NA, CA, XA, etc)

AAA A

NN

VC C V A

Ad

VC

rV

dt

AA A

dC

dr

tV V C

dVdt

2 parts: how CA changes with t and how V changes with t

Convert NA to CA using:

L7-16



V0 + u0t

Do a mole balance on A since it does not enter or leave the reactor (assume the reactor is well-mixed)


AA

dN0 0 r V t

dt

AA A

dCr V V C

dtdVdt

2 parts: how CA changes with t and how V changes with t

Reactor volume at any time can be found with a mole balance


0 0

d V 0 0

dt

u

u = u0

0 0

dVdt

u0 0V t Vu

Substitute: AA 0A

dCr V V C

dtu Rearrange to get in terms of dCA/dt

AA A 0

dCr V C V

dtu A 0 A

AC dC

rV dt

u Balance on A

L7-17



V0 + u0t

Mole Balance on B

BB B0

dNr V F

dt


BB0 B

dNF 0 r V

dt

0dVdt

u

BB B0

dNr V F

dt

B B0B

BdVC d

C V rt

Vd

Fdt

BB 0 B B0 0

dCC V r V C

dtuu

B B0 0B

BdC

C Vd

r V Ct t

dVd

u

0 B0 BBB

C CdCr

dt V

u

Substitute

Rearrange to get in terms of dCB/dt

Balance on B

L7-18


Semi-Batch Reactor Design Equation: in Terms of NACBu0

V0 + u0t

AA

dNr V

dt

AA 0 0

dNr V t

dtu

AA

dN0 0 r V

dt


0 0Substitute V V tu

Reactor design eq. provided that rA is a function of NA

AAdN

Vdt

r

A B

0

A

0

NdN

dk

Vt

N

tu

NB comes from BMB: BA B0

dNr V F

dt B A B

B00 0

dN N Nk F

dt V tu

The design eq in terms of XA can be messy. Sometimes it gives a single

equation when using Nj or Cj gives multiple reactor design equations.

A B A BA A 2

0 0

N N N Nr k r k

V V V tu-rA = kACACB

L7-19


+ H2O

V0 Vf

FD

V0 - u0t

Semi-batch

To improve product yield in a reversible reaction: A l B l C l D g • Start with A(l) and B(l) in the reactor• D(g) bubbles out of the liquid phase, pushing the equilibrium to the right

and forcing the reaction to go to completion

Common industrial reaction:

n

+ n

Boil off water to produce high MW polymer

nylon

A+BC+DA+B

Improving Yields of Reversible Rxns with Semi-Batch Reactors

L7-20


How do we account for the loss of product D in the material balance?

V0 Vf

FD

V0 - u0t

Semi-batch

To improve product yield in a reversible reaction: A l B l C l D g • Start with A(l) and B(l) in the reactor• D(g) bubbles out of the liquid phase, pushing the equilibrium to the right

and forcing the reaction to go to completion

A+BC+DA+B

Improving Yields of Reversible Rxns with Semi-Batch Reactors

L7-21


Loss of Mass in Semi-Batch Reactor

In Out- + Generation = AccumulationOverall Mass balance:

0dm

0 dt

mu

↑want in terms of dV/dt

V0 + u0t

D(g)u = u0 0 A l B l C l D g elementary rxn

ggas leaving reactor

timem

mV

mV

Divide mass balance by

00

d 1 ddt dt

Vmm m uu

Relate ṁ to a rate: From stoichiometry, rD = -rA

Amoles

rvolume time

Next, convert units to:

masstime

Amoles

r Vtime

DD D D D

D

massmoles moles MW mass

MW A

massti

V MWme

r m

Substitute for ṁ

0A Dr VdV

dt

MWu

One of the diff. eq. that are simultaneously solved (by Polymath)

Conversion 1:

Conversion 2:

Documents

L7-1 Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign. Review: Liquid Phase Reaction in