Monolithic Reactors for Environmental Catalysis

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Monolithic Reactors for

Environmental Catalysis

朱信

Hsin Chu

Professor

Dept. of Environmental Eng.

National Cheng Kung University

1

1. Introduction

Minimize the pressure drop associated with high flow rates

Allow the process effluent gases to pass uniformly through the channels of the honeycomb

2

2. Chemical Kinetic Control

To be controlled by chemical kinetics rather than by diffusion to or within the catalyst pore structure while the geses are cold

When the surface becomes sufficiently hot, the rate will be determined by mass transfer.

In the laboratory, when screening a large number of catalyst candidates, it is important to measure activity at low conversion levels to ensure that the catalyst is evaluated in the intrinsic or chemical rate-controlling regime.

Good laboratory practice is to maintain all conversions below

20% for kinetic measurements. (adiabatic)

For highly exothermic reactions (i.e., △ H > 50kcal/mol), measurements should be made at conversions no greater than

10%.

3

A material balance across any reactor gives the following equation assuming one-dimensional, plug flow, steady-state operation:

( )

  r dz where v = velocity (cm/s)

C = molar concentration [(g‧mol)/cm 3 ] z = length (cm) r = rate of reaction [(g‧mol)/(cm 3 ‧s)]

When the conversion or the reactant concentration is low, the reactor is considerd isothermal; hence dc v

  r dz 4

Assume the oxidation of ethane to CO

2 and H

2

O in a large excess of O

2 in a fixed bed of catalyst:

C H

2 6

7

O

2

2 CO

2

3 H O

2

2

We can assume that the rate is independent of O

2

.

It obeys first-order kinetics (pseudo-zero-order in O

2

), so the rate is expressed as: v dC

( C H

6

)  

'

( C H

6

) dz where k’ = the apparent rate constant

Integrating from the reactor inlet (i) to outlet (o) gives: ln

C o

C i

 

  where t = actual residence time (s)

5

volume of catalyst t = volumetric flow rate

Volumetric hourly space velocity (VHSV)

VHSV = volume flow rate at STP

1 volume of catalyst plus void time at STP

The rate expression then becomes: ln

C o

C i

  k "

VHSV

By varying the space velocity, the change in conversion can be determined.

The slop of the plot yields the k” of the reaction at STP.

Next slide

(Fig. 4.1)

Ethane conversion versus temperature at different space

velocities.

6

3. Bulk Mass Transfer

When experiments are conducted with extremely active catalyst or at high temperatures, diffusional effects are introduced, and the intrinsic kinetics of the catalytic material is not determined accurately.

The activation energy will decrease as pore diffusion and bulk mass transfer become more significant.

Stationary environmental abatement processes are designed to operate in the bulk mass transfer regime where maximum conversion of the pollutant to the nontoxic product is desired.

Diffusion processes have small temperature dependency (low activation energies).

Chemical-controlled reactions have a high degree of dependence on temperature (high activation energies).

8

Important benefit of diffusion processes: the physical size and other geometric parameters of the honeycomb for a required conversion can be obtained using fundamental parameters of mass transfer.

v dC dz

  k aC g

Where k g

= mass transfer coefficient (cm/s) a = geometric surface area per unit volume (cm 2 /cm 3 )

C = reactant gas phase concentration [(g‧mol)/cm 3 ]

Integrating, ln

C

  k at g

C i

Fractional conversion = 1- exp[-(k g at)]

9

Some dimensionless numbers

N sh

N sc

 k d ch

D

( sherwood number )

D

( schmidt number )

N

Re

W

( ) ch (

A d

  channel Reynolds number ) where D = diffusivity of pollutant in air (cm 2 /s)

W = total mass flowrate to honeycomb catalyst (g/s)

A = frontal area of honeycomb (cm 2 ) d ch

= hydraulic diameter of honeycomb channel (cm)

ρ = gas density at operating conditions (g/cm 3 )

μ= gas viscosity at operating conditions (g/s‧cm)

ε=void fraction of honeycomb, dimensionless

Equation on last slide becomes:

Fractional conversion = 1- exp

N sh

L

N N sc Re

 where L = honeycomb length (cm)

10

Next slide

(Fig. 4.2)

Correlations for N

sh

, N

sc

, and N

Re

11

Example 1 Calculation for Mass Transfer Conversion

The removal of propane (C

8

) in a stream air at 300℃ and atmospheric pressure with:

3

H

Flow rate, W= 1000 lb/h (126g/s)

Diameter of monolith, D=6 in. (15.24 cm)

Length of monolith, L=6 in.(15.24 cm)

Area of monolith, A=182.4 cm 2

Monolith geometry, 100 cpsi (15.5 cells/cm 2 )

C

3

H

8 feed fraction, X=1000 vppm (volume parts per million)

Sol: From the literature (Lachman and McNally, 1985) d ch

= 0.083 in. (0.21cm)

ε=0.69

a = 33 in.

2 /in.

3 (13 cm 2 /cm 3 )

13

Using Hodgman’s (1960) Handbook of Chemistry and

Physics

The density (ρ) and viscosity (μ) of air :

ρ at 300℃ = 6.16 × 10 -4 g/cm 3

μ at 300℃ = 297 × 10 -6 g/s‧cm

Therefore,

N

Re

( /

) ch

 g s cm

 cm

2

(126 / )(0.21

) /(0.69 182.4

)

 

6

/

)

707.9

To utilize Fig. 4.2, the following term must be determined:

N d

Re ch

L

(707.9)(0.21

cm )

15.24

cm

9.75

14

From Bird et al., 1960, the diffusivity for a binary system:

D

0.0018583

T

3

(1/

P

M

2

A

AB

1/

,

M

B

) where M = molecular weight of species, A=air;

B=C

3

H

8

[g/(g‧mol]

P = total pressure (atm)

σ

AB

= collision diameter for binary system (Å)

T = absolute operating temperature (K)

= collision integral for binary system, dimensionless

Using Table B-1 from Bird et al., 1960:

For air: M

For C

3

H

8

A

: M where σ and

= 28.97, σ

B

/ k

A

=44.09, σ

=3.617Å,

B

=5.061Å,

 k

B k

A

97.0

254

K

K are Lennard-Jones parameters for the single components.

15

The binary system:

AB

 k

AB

1

2

(

A

 

B

)

1

2

(3.617 5.061)

4.339

A o

 

A

 

B

 k

 k

(97)(254)

156.96

K

 kT

AB

573 K

156.96

K

3.65

Using this value and Table B-2 from Bird (1960),

 

0.903

Therefore,

D

0.0018583

0.359

cm

2

/ s and N

Sc

D

 

6

/

 

4 g cm

3

)(0.359

cm

2 s

1.34

16

Using Figure 4.2,

N

Re d ch

/L=9.75→N sh

/N sc

0.56

=3.8

Therefore, N sh

=4.4

Fractional conversion =

1

 exp

N sh

L

N N

Re

=

1

 exp

(4.4)(13 cm

2

/ cm

3

/ 0.69)(15.24

cm )

(1.34)(707.9)

= 0.736 = 73.6% (done)

17

4. Reactor Bed Pressure Drop

Pressure drop ( △ P) a. flow contracts within the restrictive channel diameter b. washcoat on the surface of the honeycomb channel creates friction

The basic equation for △ P derived from the energy balance:

1

 dP

2 f dL

 g d ch

2 where P = total pressure (atm) f = friction factor, dimensionless g c

= gravitational constant (980.665 cm/s 2 )

υ= velocity in channel at operating conditions (cm/s)

ρ= gas density at operating conditions (g/cm 3 )

Next slide

(Fig. 4.3)

Friction factor correlation to N

Re

18

The velocity in the channel (υ)

  w whereε = void fraction (percent open frontal area of the honeycomb)

A = cross-sectional area of honeycomb

Simplify the basic equation for △ P

2 fL

 ch

2 g d c ch

Next slide

(Fig. 4.4)

△ P versus flow rate

To select the optimum honeycomb geometry (volume, crosssectional area, length, cpsi, etc.) for a given application

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