Chemical Reaction Engineering

Chapter 5-6

Isothermal Reactor Design- Heterogeneous Reactions- Pressure Drop

• Describe the algorithm that allows the reader to solve chemical reaction engineering problems through logic rather than memorization.

• Sizing batch reactors, semi-batch reactors, CSTRs, PFRs, and PBRs for isothermal operation given the rate law and feed conditions.

• Studying a liquid-phase batch reactor to determine the specific reaction rate constant needed for the de-sign of a CSTR.

• Design of a tubular reactor for a gas-phase reaction.• Account for the effects of pressure drop on

conversion in packed bed tubular reactors and in packed bed spherical reactors.

• The principles of unsteady operation and semi-batch reactors.

Objectives

Fig. 5-1 Isothermal Reaction Design Algorithm for Conversion

Algorithm for isothermal reactor design1. Mole balance and design equation2. Rate law3. Stoichiometry4. Combine5. Evaluate

4.1 Design structure for isothermal reactors

We can solve the equations in the combine step either

A. Graphically (Chapter 2)

B. Numerical (Appendix A4)

C. Analytical (Appendix A1)

D. Software packages (polymath)

Algorithm for Isothermal Reactors

Pressure Drop in Reactors

In liquid-phase reactions - the concentration of reactants is insignificantly affected by even relatively large change in the total pressure => - Hence, the effect of pressure drop on the rate of reaction can be ignored when sizing liquid-phase chemical reactors• In gas-phase reactions, • the concentration of the reacting species is proportional to the total pres-

sure - the effects of pressure drop on the reaction system are a key factor in the success or failure of the reactor operation - that is, pressure drop may be very important for gas-phase reactions

Pressure Drop and Rate Law

• For an ideal gas,

- determine the ratio P/P0 as a function of V or W

- combine the concentration, rate law, and design equation

- the differential form of the mole balance (design equation) must be used

)/)(/)(1( 000

0

TTPPXv

XvF

v

FC iiAi

i

00 1 P

P

X

XvCC ii

Ai

To

T(4-18)

Pressure Drop and Rate Law

• For example, the second order isomerization reaction in a packed-bed reactor

2A B + C

-the differential form of the mole balance

- rate law

-from stoichiometry for gas-phase reactions

mincatalyst g

gmoles0 AA r

dW

dXF

2AA kCr

T

T

P

P

X

XCC AA

0

00 1

1

Pressure drop and the rate law

• Then, the rate law

- the larger the pressure drop from frictional losses, the smaller the reaction rate

• Combining with the mole balance and assuming isothermal

operation (T=To) in a Packed-bed reactor,

• Dividing by FA0

20

00 1

1

T

T

P

P

X

XCkr AA

2

0

20

0 1

)1(

P

P

X

XCk

dW

dXF A

A

2

0

2

0

0

1

1

P

P

X

X

v

kC

dW

dX A

(5-20)

Pressure Drop and Rate Law

• For isothermal operation (T =T0), a function of only conversion

and pressure

-Another equation is needed to determine the conversion as a

function of catalyst weight, - that is, we need to relate the pres-

sure drop to the catalyst weight

),( PXfdW

dX

)(WfP

(5-21)

Flow Through a Packed Bed

• The majority of gas-phase reactions are catalyzed by pass-ing the reactant through a packed of catalyst particles

• Ergun equation: to calculate pressure drop in a packed por-ous bed

• The gas density is the only parameter that varies with pres-sure on the right-hand side. So, calculate the pressure drop through the bed

G

DDg

G

dz

dP

ppc

75.1)1(1501

3

laminar turbulent

(4-22)

G=ru=superficial mass velocity [g/cm2s]; u=superficial velocity [cm/s]; Dp=diameter of particle in the bed [cm]; f=porosity=volume of void/total bed volume; 1- f =volume of solid/total bed volume

Flow through a Packed Bed

• Equation of continuity

- steady state the mass flow rate at any point is equal to the entering

mass flow rate

• Gas-phase volumetric flow rate

• Then,

vv

mm

00

0

00

00

T

T

F

F

T

T

P

Pvv

T

T

F

F

T

T

P

P

v

v 00

00

00

mo = m

(3-41)

(4-23)

Pressure Drop in a Packed Bed Reactor

• then, Ergun equation

• Simplifying

• The catalyst weight,

00

00

T

T

F

F

T

T

P

P

dz

dP

cc zAW )1(

00

03

0

75.1)1(150)1(

T

T

ppc F

F

T

T

P

PG

DDg

G

dz

dP

GDDg

G

ppc

75.1)1(150)1(

30

0

Volume ofsolid

Density of solid catalyst

dzAdW cc )1(

(4-24)

dW

dPWe need

(4-26)

(4-25)

Pressure Drop in a Packed Bed Reactor

• then, Ergun equation

• Simplifying

00

00

)1( T

T

cc F

F

T

T

P

P

AdW

dP

0

0

)1(

2

PA cc

00

0

0 /2 T

T

F

F

PP

P

T

T

dW

dP(4-27)

X

F

FFXFFF

T

ATTTT

0

0000 1 X

F

F

T

T 10

0

00

T

AA F

Fy

)1(/2 0

0

0

XPP

P

T

T

dW

dP

(4-28)

(4-29)

Let y=P/Po 2ydy FT

FTo

Pressure drop in a packed bed reactor

ε < 0, the pressure drop (DP) will be less than that for ε = 0

ε > 0, the pressure drop (DP) will be greater than that for ε = 0

(4-30)

• For isothermal operation

),( PXfdW

dP ),( PXf

dW

dXand

• The two expressions are coupled ordinary differential equations. We can solve them simultaneously using an ODE solver such as Polymath.

• For the special case of isothermal operation and ε = 0, we can obtain an analytical solution.

• Polymath will combine the mole balance, rate law and stoichiometry

(4-31)

)1(/2 0

0

0

XPP

P

T

T

dW

dP

2y(1+X)

dy

Pressure Drop in a Packed Bed Reactor

Analytical Solution

If ε = 0 or ε X ≪ 1, we can obtain an analytical solution to Eq. (4-30) for isothermal operation (i.e., T=T0). For isothermal operation with ε = 0, Eq. (4-30) becomes

)/(2 0

0

PP

P

dW

dP Isothermal (T=To)

with ε = 0

(4-30))1(/2 0

0

0

XPP

P

T

T

dW

dP

2y(1+X)

dy

dy

2y

2ydydW = -

dy2

dW = -

At W=0, y=1 (P/Po=1) y2= 1- w

Pressure Drop in a Packed Bed Reactor

If ε = 0 or ε X ≪ 1, we can obtain an analytical solution to Eq. (4-30) for isothermal operation (i.e., T=T0). For isothermal operation with ε = 0, Eq. (4-30) becomes

0

0

)1(

2

PA cc

Pressure ratio only for ε = 0

WP

P 1

0

0

0

0

21

P

z

P

P

cc zAW )1(

(4-33)

(4-34)

(4-26)

Pressure as a function of reac-tor length, z

y=

y= =f (z)

Pressure Drop in Pipes

D

fG

dL

dG

dL

dP

22

Integrating with P=P0 at L=0, and assuming that f = constant

Pressure drop along the length of the pipe

02 2

2

00

D

fG

PdL

dPG

dL

dP

P

P

P

P

D

Lf

PG

PP 0

0

0222

0 ln22

0

Rearranging, we get

DAP

fGV

P

P

cpp

00

20 4

1

Example 4-4: 1½” schedule 40 x1000-ft L (ap=0.018), DP<10%

Analytical Solution for Reaction with Pressure DropConversion as a function of catalyst weight

AA rdW

dXF 0

A B

2nd-order isothermal reaction

Mole balance:

2AA kCr

Rate law:

00 )1(

P

PXCC AA

Stoichiometry: Gas-phase isothermal with e=0

WP

P 1

0

2/10 )1)(1( WXCC AA

22/12

0

20 1)1( WX

F

kC

dW

dX

A

A

Combining

21

1

0,0

1)1(

0

0

000

220

0

WW

X

X

kC

v

vCFandWXat

dWWX

dX

kC

F

A

AA

A

A

Separating variable and Integrating

211

21

0

0

0

0

W

v

WkC

W

v

WkC

XA

A

(4-38)y=

Reaction with Pressure DropConversion as a function of catalyst weight

211

21

0

0

0

0

W

v

WkC

W

v

WkC

XA

A

(4-38)

2/100 )1/(/)2(11 XXkCv

W A (4-39)

Catalyst weight for 2nd-order isothermal reaction in PFR with DP

Reaction with Pressure DropConversion as a function of catalyst weight

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.

Spherical Packed-Bed Reactors

Spherical Ultraformer Reactor (Amoco) for dehydrogenation reaction such as

Paraffin Aromatic + 3 H2

Spherical reactor

- minimize pressure drop

- inexpensive

- the most economical shape for high pressure

Coordinate system and variables used with a spherical reactor

Synthesizing a Chemical Plant

Always challenge the assumptions, constraints, and boundaries of the problem

The profit from a chemical plant will be the difference between income from sales and the cost to produce the chemical

Profit = (value of products) – (cost of re-actants)

– (operating costs) – (separation costs)

The operating cost: energy, labor, overhead, and de-preciation of equipment

Production of Ethylene Glycol

C2H4O(aq)

7 8 H2O, 0.9wt% H2SO4

200 millionlb EG/yr

9

V=197 ft3, X=0.8

CH2OHC2H4O + H2O CH2OH

Cat.

5

O2, C2H4, N2, C2H4O

6

separator

2

H2, C2H4

C2H6

separator

H2O C2H4O

absorber

C 2H 6 C 2H 4 + H 21

V=81 ft3, X=0.8

402 million lbC2H6 /yr

3C2H4

4 Air

W=45,440 lb, X=0.6

C2H4+ ½ O2 C2H4OAg

Synthesizing a Chemical Plant

Ethylene glycol = $0.38/lb (2x108 lb/yr)

Ethane = $0.04/lb (4x106 lb/yr)

Sulfuric acid = $0.043/lb (2.26x108 lb/yr)

Operating cost = $8x106/yr

Profit = ($0.38/lb x 2x108 lb/yr) – ($0.04/lb x 4x108 lb/yr)

-($0.043/lb x 2.26x106 lb/yr) – ($8x106/yr)

= $52 million

How the profit will be affected by conversion, separation, recycle stream, and operating costs?

Using CA (liquid) and FA (gas) in the mole balance and rate laws

More convenient to work in terms of the number of moles or molar flow rate rather than conversion.

Must write a mole balance on each species when molar flow rates (Fi) and concentrations (Ci) are used as variables

Membrane reactor, multiple reaction, and unsteady state

BAAA

DCBA

CCkr

d

r

c

r

b

r

a

r

Da

dC

a

cB

a

bA

Liquid Phase

Mole balance for liquid-phase reactions

AA

AA

A

AA

AA

rdW

dCv

rdV

dCv

r

CCvV

rdt

dC

0

0

00 )(

AB

AB

A

BB

AB

ra

b

dW

dCv

ra

b

dV

dCv

rab

CCvV

ra

b

dt

dC

0

0

00

)/(

)(

Batch

CSTR

PFR

PBR

DCBA dcba

DCBAa

d

a

c

a

b

For liquid-phase reaction with no volume change Concentration is the preferred variable

Gas Phase For gas phase reactions

need to be expressed in terms of the molar flow rates

T

T

P

P

F

FCC

T

jTj

0

00

Total molar flow rate

n

jjT FF

1

jj r

dV

dF

y

(4-28, T=To)dydW

=2y-

FTo

FT

(1-11)

(3-42)

= FA + FB + FC + FD + FI

Algorithm for Gas Phase Reaction

dDcCbBaA

Mole balances:

Vrdt

dN

Vrdt

dN

Vrdt

dN

Vrdt

dN

DD

CC

BB

AA

D

DD

C

CC

B

BB

A

AA

r

FFV

r

FFV

r

FFV

r

FFV

0

0

0

0

DD

CC

BB

AA

rdV

dF

rdV

dF

rdV

dF

rdV

dF

Batch CSTR PFR

Algorithm for Gas Phase Reaction

Rate law: BAAA CCkr

Stoichiometry:

DCBAT

T

DTD

T

CTC

T

BTB

T

ATA

ADACAB

DCBA

FFFFF

T

T

P

P

F

FCC

T

T

P

P

F

FCC

T

T

P

P

F

FCC

T

T

P

P

F

FCC

ra

drr

a

crr

a

br

d

r

c

r

b

r

a

r

0

00

0

00

0

00

0

00

Relative rate of reaction:

Concentration:

Total molar flow rate:

y

Algorithm for Gas Phase Reaction

Combine:

T

B

T

ATA

B

T

B

T

ATA

C

T

B

T

ATA

B

T

B

T

ATA

A

F

F

F

FCk

a

d

dV

dF

F

F

F

FCk

a

c

dV

dF

F

F

F

FCk

a

b

dV

dF

F

F

F

FCk

dV

dF

00

00

- Specify the parameter values: kA, CT0, , , a b T0, a, b, c, d

- Specify the entering number: FA0, FB0, FC0, FD0, and final value: Vfinal

Use an ODE solver