Presentacin de PowerPoint

1.4.2 Tubular Reactor

- The reaction rate will also vary axially.

- To develop the PFR design equation, we shall divide (conceptually) the reactor into a number of sub-volumes so that within each sub-volume DV, the reaction rate may be considered spatially uniform.

Let Fj(y) represent the molar flow rate of species j into volume DV at y

Fj(y+D y) represent the molar flow rate of species j out of volume DV at (y+D y)

In a spatially uniform sub-volume DV,

1.4.2 Tubular Reactor

0

rj is a function of reactant concentration, which is function of the position y down the reactor.

Steady state

(1-8)

Substitute DV into Eq. 1-8 and divide by Dy to yield

Taking the limit as Dy approaches zero, we obtain

Tubular Reactor

It is usually most convenient to have the reactor volume V rather than the reactor length y as the independent variable. Using dV=Ady,

The tubular reactor design equation can be applied to the PFR with variable and constant cross-sectional area.

Tubular Reactor

It is usually most convenient to have the reactor volume V rather than the reactor length y as the independent variable. Using dV=Ady,

The tubular reactor design equation can be applied to the PFR with variable and constant cross-sectional area.

Tubular Reactor

V

FA

FA0

A B

V

FB

FB1

V1

V1

FA1

Packed-Bed Reactor (PBR)

For a fluid-solid heterogeneous system, the rate of reaction of a substance A is defined as

The mass of solid is used because the amount of the catalyst is what is important to the rA

In - out + generation = accumulation

No pressure drop

No catalyst decay

The first-order reaction (liquid phase rxn)

A B

is carried out in a tubular reactor in which the volumetric flow rate, v0, is constant.

(1) Derive an equation relating the reactor volume (V) to the entering concentration of A (CA0), the rate constant k, and the volumetric flow rate v0.

(2) Determine the reactor volume necessary to reduce the exiting concentration (CA) to 10% of the entering concentration (CA0) when the volumetric flow rate (v0) is 10 /min and the specific reaction rate, k, is 0.23 min-1.

Example 1-1 How large is it? (PFR)

CA0 v0

CA

V

rA=-kCA

Example 1-1 How large is it? (PFR)

(GMBE for tubular reactor)

(first-order reaction)

Tubular, 1st order rxn

The first-order reaction (liquid phase rxn)

A B

is carried out in a CSTR in which the volumetric flow rate, v0, is constant.

(1) Derive an equation relating the reactor volume (V) to the entering concentration of A (CA0), the rate constant k, and the volumetric flow rate v0.

(2) Determine the reactor volume necessary to reduce the exiting concentration (CA) to 10% of the entering concentration (CA0) when the volumetric flow rate (v0) is 10 /min and the specific reaction rate, k, is 0.23 min-1.

P1-6B How large is it? (CSTR)

V

rA=-kCA

Fj0

Fj

For CSTR,

the mole balance on species A was shown to be

P1-6B How large is it? (CSTR)

V

rA=-kCA

Fj0

Fj

The CSTR is almost 4 times larger than the PFR for getting 90% conversion

Reactor Differential Algebraic Integral

Mole Balance on Different Reactor

Batch

CSTR

PFR

PBR

Homework

P1-7A

P1-9A

P1-11B

P1-15B

Due Date: Next Week

0

j

F

exit

F

j

,

)

(

y

F

j

)

(

y

y

F

j

D

+

V

D

y

D

y

y

y

D

+

V

r

dV

r

V

j

j

D

=

D

dt

dN

dV

r

F

F

j

V

j

j

j

=

+

-

0

0

)

(

)

(

=

D

+

D

+

-

V

r

y

y

F

y

F

j

j

j

y

A

V

D

=

D

j

j

j

Ar

y

y

F

y

y

F

-

=

D

-

D

+

-

)

(

)

(

j

j

Ar

dy

dF

=

j

j

r

dV

dF

=

-

=

0

1

1

A

A

F

F

A

A

r

dF

V

A

A

r

dV

dF

=

catalyst

g

reacted

A

gmol

r

A

=

-

sec

'

0

)

(

)

(

'

=

D

+

D

+

-

W

r

W

W

F

W

F

A

A

A

=

A

A

F

F

A

A

r

dF

W

0

'

0

A

F

A

F

)

(

W

F

A

)

(

W

W

F

A

D

+

W

r

A

D

'

W

D

W

W

W

D

+

dV

C

dC

k

v

kC

r

dV

dC

v

r

dV

dC

v

dV

v

C

d

dV

dF

kC

r

r

dV

dF

A

A

A

A

A

A

A

A

A

A

A

A

A

=

-

=

-

=

-

=

=

=

=

-

=

0

0

0

0

)

(

l

l

C

C

l

V

C

C

k

v

V

dV

C

dC

k

v

A

A

A

A

V

C

C

A

A

A

A

100

10

ln

23

.

0

10

1

.

0

ln

min

23

.

0

min

/

10

ln

0

0

1

0

0

0

0

0

=

=

=

=

=

-

-

l

l

l

3

.

391

)

min

23

.

0

/(

min)

/

10

)(

9

(

1

.

0

9

.

0

1

.

0

min

23

.

0

min,

/

10

,

1

.

0

1

0

0

0

0

0

1

0

0

0

0

0

0

=

=

=

-

=

=

=

=

-

=

-

-

=

-

-

k

v

kC

v

C

v

C

V

k

and

v

C

C

kC

v

C

v

C

r

F

F

V

A

A

A

A

A

A

A

A

A

A

A

V

r

dt

dN

A

A

=

=

A

A

F

F

A

A

r

dF

V

0

A

A

r

dV

dF

=

A

A

r

dW

dF

=

=

A

A

F

F

A

A

r

dF

W

0

=

A

A

N

N

A

A

V

r

dN

t

0

A

A

A

r

F

F

V

-

-

=

0