ERT 316

ISOTHERMAL REACTOR DESIGN

PART 1

Copyright Cheng 05

START

1. The general mole equation

2. Design Equations: • Batch • CSTR • PFR

3. Is –rA=f(X) given?

4. Determine the rate law in terms of the concentration

of the reacting species

5. Use Stoichiometry to express concentration as a function of conversion • Liquid phase or Gas phase • Constant Volume Batch • Constant P and T

Evaluate the algebraic (CSTR) or integral (PFR)

equations

6. Combine steps 4 and 5 to obtain –rA=f(X)

END

YES

NO

Algorithm for Isothermal Reactor

4.1 Design Algorithm for Isothermal Reactors

• To design an isothermal reactors, the following sequence is

highly recommended.

4.1 Design Algorithm for Isothermal Reactors

To carry out the evaluation, the following method can

be used:

Graphically (Chapter 2 plot)

Numerical (Quadrature Formulas Chapter 2 and

Appendix A4)

Analytical (Integral Tables)

Software (Polymath)

Copyright Cheng 05

Algorithm for isothermal

Reactor (PFR) reactor

volume for 1st order gas-

phase reaction

Scale up of Batch Reactor to

the design of CSTR

- scale up of a pilot plant - costly

- Instead - build microplant (laboratory

bench scale)

- example: analyze data from a

laboratory batch reactor-determine

specific reaction rate, k - use it to

design full-scale flow reactor

Batch Operation

A B

Step 1: Write the mole balance

Step 2: Write the rate law

Step 3: Derive concentration term from stoichiometry

Vrdt

dXN AA 0

- Calculation of time taken to achieve a given conversion X

2

AA kCr Example for second order

reaction

Batch Operation

A B

Step 4: Combine equation from step 1,2,3

Step 5: Evaluate

2

0 1 XkCdt

dXA

- Calculation of time taken to achieve a given conversion X

t X

A X

dX

kCdt

0 0

2

0 1

1

This is the reaction time or tR

Algorithm to estimate reaction time

Mole Balance

Rate Law First order Second order

Stoichiometry

Combine

Evaluate

(integrate)

Vrdt

dXN A

R

A 0

2

AA kCr AA kCr

2

0 1 XkCdt

dXA

R

Xkdt

dX

R

1

Batch Operation

A B

To reach 90% conversion in a constant-volume batch reactor scales:

if k = 10-4 s-1

- Calculation of time taken to achieve a given conversion X

For first order

4.2 Design of CSTR

Step 1: Write the mole balance of CSTR

Design equation for CSTR is

If volumetric flow rate does not change with the

reaction, (i.e. v = v0), then

exitA

AO

exitA

AO

r

XCv

r

XFV

)()(

0

A

A

A

A

r

XC

vr

XCv

v

V

0

0

00

0

where is the space time

Step 2: Write the rate law

For 1st order irreversible reaction,

Step 3: Derive concentration in terms of conversion

(from stoichiometry)

Step 4: Combine eq from step 1,2, 3

Rearranging;

4.2 Design of CSTR

k

kX

1

AA kCr

X

X

k 1

1

is often referred to

as Damköhler number

(for 1st order)

k

Step 2: Write the rate law

For 1st order irreversible reaction,

Step 3: Derive concentration in terms of conversion

(from stoichiometry)

Step 4: Combine eq from step 1,2, 3

Rearranging;

4.2 Design of CSTR

k

CC A

A

1

0

k

kX

1

AA kCr

X

X

k 1

1

Damköhler number

• Is the ratio of the rate of reaction of A to the rate of convective transport of A at the entrance to the reactor.

rate of reaction at entrance

entering flow rate of A

• For first order irreversible reaction;

• For second order irreversible reaction;

0

0

A

A

F

VrDa

kCv

VkC

F

VrDa

A

A

A

A

00

0

0

0

0

00

2

0

0

0A

A

A

A

A kCCv

VkC

F

VrDa

How to estimate degree of conversion for a CSTR?

By using Damkohler number,

0

0

A

A

F

VrDa

If Da 0.1, X < 0.1 If Da 10, X > 0.9

If first degree order, Da = k If second degree order, Da =kCA0

4.2 Design of CSTR (for first order)

For CSTRs in series, conversion as a function of

the number of tanks in series:

For CSTRs in parallel, conversion is:

nkX

)1(

11

k

kX

1Just like a single

CSTR

Example: Producing 200 Million

Pounds per Year in a CSTR It is desired to produce 200 million pounds per

year of ethylene glycol (EG). The reactor is to

be operated isothermally. A 1lb mol/ft3 solution

of ethylene oxide (EO) in water is fed to the

reactor shown in figure together with an equal

volumetric solution of water containing 0.9 wt%

of the catalyst H2SO4. The specific reaction rate

constant is 0.311 min-1 .

(a)If 80% conversion is to be achieved,

determine the necessary CSTR volume.

(b)If 800-gal reactors were arranged in parallel,

what is the corresponding conversion?

(c) If 800-gal reactors were arranged in series,

what is the corresponding conversion?

CBA catalyst

Example: Producing 200 Million

Pounds per Year in a CSTR

Extract the given information:

FC = 2 x 108 lbm/yr x 1 yr/365 days x 1day/24 h x 1hr/60 min x 1lbmol/62lbm

= 6.137 lbmol/min

From reaction stoichiometry,

FC = FA0X

min67.7

8.0

137.60

lbmol

X

FF C

A

Example: Producing 200 Million

Pounds per Year in a CSTR

STEP 1: Design equation of CSTR

STEP 2: Rate Law

STEP 3: Stoichiometry (Liquid phase, v = v0 )

STEP 4: Combining;

exitA

AO

r

XFV

)(

AA kCr

)1()(

00

Xk

Xv

r

XFV

exitA

A

Example: Producing 200 Million

Pounds per Year in a CSTR

STEP 5: Evaluate

The entering volumetric flowrate of stream A, with CA01 = 1 lb mol/ft3 before mixing is;

From the problem statement,

Thus, the total entering volumetric flow rate of liquid is

Substituting all the values to calculate volume of reactor;

min67.7

/1

min/67.7 3

3

01

00

ft

ftlbmol

lbmol

C

Fv

A

AA

00 AB vv

min34.1567.767.7

3

000

ftvvv BA

galft

ft

Xk

Xv

r

XFV

exitA

A 14763.1978.01min311.0

8.0

min34.15

)1()(

3

1

3

00

v0VV

Example: Producing 200 Million

Pounds per Year in a CSTR

b) CSTR in parallel.

Rearranging the equation of volume in part a)

)1(

0

Xk

XvV

)1(0 Xk

X

v

V

)1( Xk

X

)1( X

Xk

)1( k

kX

)1( Da

Da

Example: Producing 200 Million

Pounds per Year in a CSTR

b) CSTR in parallel.

)1( k

kX

)1( Da

Da

min94.13min/67.7

1

48.7

1800

3

3

0

ftgal

ftgal

v

V

34.4min311.0min94.13 1 kDa

81.0)34.41(

34.4

X

Example: Producing 200 Million

Pounds per Year in a CSTR

c) CSTR in series

min97.6min/34.15

1

48.7

1800

3

3

0

ftgal

ftgal

v

V

167.2min311.0min97.6 1 kDa

nkX

)1(

11

90.0

167.21

11

2

X

4.3 PFR Assume no dispersion and no radial gradients in either

temperature, velocity, or concentration and in the absence of

pressure drop or heat exchange.

STEP 1: Write the mole balance of PFR:

STEP 2: Write the rate law

Eg: For second order,

X

A

Ar

dXFV

0

0

2

AA kCr

4.3 PFR STEP 3: Write concentration in terms of conversion

(from stoichiometry)

XCC AA 10

For liquid phase

For gas phase

x

XCC AA

1

10

4.3 PFR STEP 4: Combine all the equations

Rearranging,

X

AA

A

X

X

kC

v

X

dX

kC

FV

0 0

0

22

0

0

11For liquid phase

For gas phase

X

A

A dXX

X

kC

FV

0

2

2

2

0

0

1

1

X

XXX

kC

v

A 1

)1()1ln()1(2

22

0

0

2

2

0

0

11 Da

Da

kC

kCX

A

A

Design a PFR: summary

X

X

kC

Cv

X

dX

kC

FV

A

AX

A

A

112

0

00

0 22

0

0

• mole balance

• rate laws

• Stoichiometry

• combination

AA rdV

dXF 0

2

AA kCr

)1(0 XCC AA

or

2

2

0

0

11 Da

Da

kC

kCX

A

A

No pressure drop

No heat exchange

X

A

Ar

dXFV

00

Damköhler number for 2nd-order reaction

In case of 2nd order rxn, liquid phase, isothermal

Design a PFR: summary

X

A

A dXXkC

XFV

0 22

0

2

01

1

• mole balance

• rate laws

• Stoichiometry

• combination

AA rdV

dXF 0

2

AA kCr

)1(

)1(

)1(

)1(

)1(0

0

0

0 X

XC

Xv

XF

Xv

F

v

FC A

AAAA

No pressure drop

No heat exchange

X

A

Ar

dXFV

00

X

XXX

kC

vV

A 1

1)1ln()1(2

2

2

0

0

In case of 2nd order rxn, gas phase, isothermal

Pressure Drop In Reactors

Isothermal Reactor Design – Part 2

Design a PBR

2

0

2

0 1

1

P

P

X

X

v

kC

dW

dX AO

• mole balance

• rate laws

• stoichiometry

• combination

AA rdW

dXF 0

2

AA kCr

T

T

P

P

X

XCC AA

0

0

0)1(

)1(

PXFdW

dX,1

In case of 2nd order rxn, gas phase, isothermal

Need to relate pressure

drop to catalyst weight

(in order to determine

conversion)

= pressure (kPa) = inlet pressure (kPa) = temperature (K) = inlet temperature (K)

= porosity = volume of void = void fraction = volume of solid

total bed volume total bed volume

= cross sectional area (m2)

=diameter of particle in the bed, ft (m)

=viscosity of gas passing through the bed, (kg/m.s)

=length down the packed bed of pipe, ft (m)

=superficial velocity = volumetric flow ÷ cross-sectional area of pipe (m/s)

=gas density (kg/m3) = solid density (kg/m3) = inlet gas density

= = superficial mass velocity, (kg/m2.s)

Design a PBR

P

1

pD

G

cA

z

u

Ergun equation

c0

0P 0TT

XPP

P

T

T

dW

dP

1

2 0

0

0

G

DDg

G

ppc

75.111501

3

0

0

0

0

1

2

PA cc

cg = 32.174lbmft/s2lbf

(conversion factor)

For isothermal operation, we have two sets of equation with

two unknowns, X & P

(1) (2)

Special case: if ε=0, an analytical solution to second equation is

obtained as follows

Design a PBR

2

00 1

1

P

P

X

X

v

kC

dW

dX AO

X

PP

P

T

T

dW

dP

1

2 0

0

0

21

0

1 WP

P Used only when ε=0

Design a PBR

2

0

2

0 1

1

P

P

X

X

v

kC

dW

dX AO

• mole balance

• rate laws

• stoichiometry

• Combination

• Solve

AA rdW

dXF 0

2

AA kCr

T

T

P

P

X

XCC AA

0

0

0)1(

)1(

In case of 2nd order rxn, gas phase, isothermal

WXv

kC

dW

dX AO 112

0

When ε=0

21

1

0 WW

X

X

kC

v

AO

By integration;

Solving for conversion gives:

Solving for catalyst weight,

Design a PBR

21

1

0 WW

X

X

kC

v

AO

211

21

0

0

0

0

W

v

WkC

W

v

WkC

XA

A

21

00 1//211 XXkCvW A

For gas phase reactions, as the pressure drop increases, the

concentration decreases, resulting in a decreased rate of

reaction, hence a lower conversion when compared to a

reactor without a pressure drop.

↑ W, ↓P, ↑ΔP

↑ W, ↓P, ↓CA

↑ ΔP, ↓P, ↓CA, ↓-rA

↑W, ↑X

Effect of pressure drop on the conversion profile Consider a packed bed column with a second order reaction is taking place in 20 meters of a 1 ½ schedule 40 pipe packed with catalyst.

2A B + C

The following data are given:

Inlet pressure, P0 = 10 atm=1013 kPa

Entering flowrate, v0 = 7.15 m3/h

Catalyst pellet size, Dp = 0.006 m

Solid catalyst density: ρc = 1923 kg/m3

Cross sectional area of 1 ½ -in schedule 40 pipe: AC =0.0013 m2

Pressure drop parameter, β0 = 25.8 kPa/m

Reactor length, L = 20 m

Void fraction = 45%

(a) Calculate the conversion in the absence of pressure drop.

(b) Calculate the conversion accounting for pressure drop.

(c) What is conversion in part (b) if the catalyst particle diameter were doubled.

The entering concentration of A is 0.1 kmol/m3 and the specific reaction rate is

.

hcatkgkmol

mk

612

(a) Conversion for ΔP = 0

α = 0 thus,

[Volume of catalyst] x [catalyst density]

211

21

0

0

0

0

W

v

WkC

W

v

WkC

XA

A

Effect of pressure drop on the conversion profile

0

0

0

0

1v

WkC

v

WkC

XA

A

ccLAW 1

32 /1923200013.045.01 mkgmm

kg5.27

6.4/15.7

15.27/1.0

123

36

0

0

hmkgmkmol

hcatkgkmol

m

v

WkCA

82.06.41

6.4

X

(b) Conversion with pressure drop

211

21

0

0

0

0

W

v

WkC

W

v

WkC

XA

A

Effect of pressure drop on the conversion profile

0

0

1

2

PA ccc

693.049.06.41

49.06.4

X

kPamkgm

mkPa

101345.01/19230013.0

/8.25232

1037.0 kg

59.0

2

5.27037.01

21

1

kgkgW

(c) Conversion when catalyst diameter were doubled.

(increase by a factor of 2, )

From , thus

Effect of pressure drop on the conversion profile

dominant

3

0

2

0

175.1

pcDg

G

pD

10

0

0

1

2

PA ccc

0

pD

10

122 pp DD

G

DDg

G

ppc

75.111501

3

0

0

(c) Conversion when catalyst diameter were doubled.

Thus,

Conversion increases from 0.693 to 0.774 by increasing catalyst diameter by a factor of 2.

Effect of pressure drop on the conversion profile

122 pp DD

11

12 0185.02

1037.0

2

1

kgkg

D

D

p

p

774.0

2

5.270185.016.41

2

5.270185.016.4

kg

kg

X

Increasing particle size

decrease the pressure

drop parameter, increase

conversion & reaction

rate