[PPT]The Logic of Isothermal Reactor Design - University Of …scs.illinois.edu/~mlkraft/CHBE 424 materials/L6 Pressure... · Web viewReview: Logic of Isothermal Reactor Design In

L6-1

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

XA AA0

A0dXt=N -r V

XA dXAV =FA0 -rA0

Review: 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 AA

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 ZPC1 X P T Z

n

A jr kC

nj0 j A0 A 0 0

AA 0

C C X T ZPr k1 X P T Z

4. Plug rA into design eq and solve for the

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

(We will always look conditions where Z0=Z)

Examples of combining rates & design eqs follow!

L6-2


Review: Batch Reactor Operation

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 CN 1

tX

dXV

d

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

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

Integrate this equation in order to solve for time, tB

e ab

le to

do

thes

e 4

step

s, a

nd

then

inte

grat

e to

sol

ve fo

r tim

e fo

r AN

Y R

EAC

TIO

N

A

A0 A

X1 tkC 1 X

L6-3


Review: CSTR OperationA → B -rA = kCA

Calculate the CSTR volume required to get a conversion of XA

Mole balance

Rate law A Ar kC


A A0 AC C (1 X )

Combine A0

A0 A

F XV

kC 1 X

A0 0 A

A0 A

C XV

kC 1 X

1st order reaction rate

rXF

VA

A0A

Put FA0 in terms of CA0

0 A

A

XV

k 1 X

Volume required to achieve XA for 1st order rxn

Be able to do these steps for any order reaction!

L6-4


A0

A

Xk

V1 X

Review: 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

• 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

0 in the small reactor must be different from 0 in the bigger reactor

Suppose for a 1st order irreversible rxn:

A

0 A

XVk 1 X

Separate variables we will vary from those held constant

So the reactor volume must be proportional to the volumetric flow rate 0

0Vt A

A

Xk 1 X

t

Be able to do this for any order rxn!

Eq is for a 1st order rxn only!

L6-5


Review: Damköhler Number, DaA0

A0

r V rate of reaction reaction rateDaF enterin

at eng flow rate of A convection rat

trancee

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

kVDa

Da kt 1st order irreversible reaction

Second order irreversible reaction:

A02

A

0 A0 0

0

A

r V VD

kCa

F C

A0

0

kC VDa

A0Da kC t

2nd order irreversible reaction

AkX

1 ktt

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

If Da < 0.1, then XA < 0.1

If Da > 10, then XA > 0.9A

DaX1 Da

0Vt Substitute

From slide L6-7:

L6-6


A0 A0

A0

1 2 kC 1 4 kCX

2 kCt t

t

Review: Sizing CSTRs for 2nd Order Rxn

• Mole balance

• Rate laws

• Stoichiometry

• Combine

A

A0 0 A0

Ar rF X C X

V

A2

Ar kC

A A0C C (1 X)

A22

0

0

A0

kC X

C XV

1

or

t

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?t A0Da kC

Eq is for a 2nd order liquid irreversible rxn

Be able to do these steps!

L6-7


Review: n CSTRs in SeriesCA00

CA1 CA2 0

For n identical CSTRs, then:

A0An n

CC

1 kt

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

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

A0

A0 A nC

C 1 X1 kt

A n11 X

1 kt

An

11 X1 kt

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

An

1or 1 X1 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 nC

r kC k1 kt

L6-8


Review: Isothermal CSTRs in Parallel

FA0

FA01

FA02 same T, V,

1 2 n X =X =...=X =XA1 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 CSTRsVn # of CSTRs

Volume of each CSTR

A0A0i

F total molar flow rateFn # 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

L6-9


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

Calculate volume required to get a conversion of XA in a PFR2A → B -rA = kCA

2 2nd order reaction rate

Mole balance

Rate law


AA

A0

d rXdV F

2A Ar kC

A A0 AC C (1 X )

Combine

A0 A

A0

2A

2C 1X XdV F

k

d

X VAA0 A22 0 0A0 A

F dX dVk C 1 X

A0 A2 AA0

F X V1 Xk C

Liquid-phase 2nd order reaction in PFR

Be

able

to d

o th

ese

4 st

eps,

in

tegr

ate

& s

olve

for V

for A

NY

OR

DER

RXN

See Appendix A for integrals frequently used in reactor design

L6-10


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

Calculate catalyst weight required to get a conversion of XA in a PBR2A → B -r’A = kCA

2 2nd order reaction rate

Mole balance

Rate law


AA

A0

rX 'ddW F

A2

Ar ' kC

A A0 AC C (1 X )

Combine

A0 A

A0

2A

2C 1X XdW F

k

d

X WAA0 A22 0 0A0 A

F dXdW

k C 1 X

A0 A

2 AA0

F XW

1 Xk C

Liquid-phase 2nd order reaction in PBRBe

able

to d

o th

ese

4 st

eps,

inte

grat

e &

sol

ve fo

r V fo

r AN

Y O

RD

ER R

XN

L6-11


Isobaric, Isothermal, Ideal Rxns in Tubular Reactors

GAS PHASE: j0 j A0 A 0 0j

A 0

C C X T ZPC1 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

L6-12


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 rXdV F

A2

Ar kC

Combine

22A0 AA

02

AA

C 1 X

1

kdXV FXd

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

L6-13


Effect of on and XATf T0 A

T0

N N Change in total # moles at X 1N total moles fed

: expansion factor, the fraction of change in V per mol A reacted0: volumetric flow rate

00 A

0 0

PZ T1 XZ T P

varies if gas phase & moles product ≠

moles reactant, or if a DP, DT, or DZ occurs

No DP, DT, or DZ occurs, but moles product ≠ moles reactant → 0 A1 X

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

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

• Longer residence time than when 0

• Higher conversion per volume of reactor (weight of catalyst) than if 0

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

• Shorter residence time than when 0

• Lower conversion per volume of reactor (weight of catalyst) than if 0

L6-14


Pressure Drop in PFRs & PBRsGAS PHASE: j0 j A0 A 0

jA 0

C C X TPC1 X P T

Considering ideal gas phase behavior (Z0=Z)

Concentration is a function of P so pressure drop is important in gas phase rxns

Why? Take a 1st order reaction A → B in a PBR with –r’A = kCA

Substitute concentration of A into the rate law:

A0 A0 A 0

A 0A

C C X TPr1 X T

kP

'

If P drops during the reaction, P/P0 is less than one, so CA ↓ & the rxn rate ↓

A

A0 A 3dX molesF r dV dm min

AA0 A

dX molesF r ' dW g catalyst min

For tubular reactors:

PFR PBRUse the differential forms of the design equations to address pressure drop

Pressure drops are especially common in reactions run in PBRs → we will focus on PBR applications

L6-15


Pressure Drop in PBRs

A0 AA

A 0

C 1 X PC1 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

0A

2 22

2A

A0

C 1 X PP

dXdW

k

F 1 X

22AA0A

20 0A

1 XkCdX PdW P1 X

We need to relate P/P0 to W

This eq. is solved simultaneously with an eq. that describes how the

pressure drops as the reactant moves down the reactor

Function of XA and pressure

L6-16


Ergun Equation relates P to W

A0

dy T1 XdW 2y T

0

c c 0

2A 1 P

Differential form of Ergun equation for pressure drop in PBR:

0

PyP

Tf T0A0

T0

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 T 1 XdW 2 T P P

This equation can be simplified to:

L6-17


Gas Phase Reaction in PBR with ΔP

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

Combine with rate law and stoichiometry

22

2A

0

AA0

0A

1 XC PkP1 X

dXdW

Relate P/P0 to W 0

A0 0

PdP T 1 XdW 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

L6-18


Analytical Solutions to P/P0

Sometimes P/P0 can be calculated analytically. When T is constant and = 0:

A0

dy T 1 XdW 2y T

10

1

dydW 2y

2ydy dW

Evaluate

P yP W0

P 01P0

2ydy dWP P021

y W

2

0

P 1 WP

0

P 1 WP

Only for isothermal rxn where =0

From no pressure change

To pressure change

L6-19


Pressure Drop Example GAS PHASE: A → B 2nd order reaction rate

This gas phase reaction is carried out isothermally in a PBR. Relate the catalyst weight to XA Tf T0

T0

N N 1 1 0N 1

A0 A0 A 0A

A 0

C C X TPC1 X P T

10

A A0 A

0

PP

C C 1 X

= 0 and isothermal, so:0

P 1 WP

Plug into CA

A A0 A0 AC C X 1C W

Plug into PBR design eq:

AA

A0dX

Fd

r 'W

22A A

A0 AA A0 A0dX dX

C C 1 Xk 1F FdW

kdW

W

22AA0 A0 A

dXF kC 1 X 1 W

dW

X WAA0 A2 2

0 0A0 A

F dX1 W dW

kC 1 X

-r’A = kCA2

Simplify, integrate, and solve for XA in terms of W or W in terms of XA:

L6-20


Pressure Drop Example A → B -r’A = kCA

2 2nd order gas phase rxn non-elementary rateThis gas phase reaction is carried out isothermally in a PBR. Relate the catalyst weight to XA

X WA A2

0 0A

A02

A0

dX1 W d

FW

k XC 1

A0

A0

A

A

X WWk X 2C

11

A0A

A 0

kCX WW 11 X 2

Solve for XA

A0 A0

A A0 0

kC kCW WX W 1 W 1 X2 2

A0 A0

A A0 0

kC kCW WX W 1 X W 12 2

A0

0A

A0

0

kCW12

XkCW1 1

2

0 A

A0 A

2 X1 1kC 1 X

W

Rearrange eq. for W:

L6-21


Next Time•Startup of a CSTR under isothermal conditions•Semi-batch reactor

Documents

[PPT]The Logic of Isothermal Reactor Design - University Of …scs.illinois.edu/~mlkraft/CHBE 424 materials/L6 Pressure... · Web viewReview: Logic of Isothermal Reactor Design In