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PFR design. Accounting forpressure drop

Chemical Reaction Engineering IAug 2011 - Dec 2011

Dept. Chem. Engg., IIT-Madras

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Overview

Notation

PFR design equation (mass balance) Pressure drop equation

Accounts for change in number of moles (due to reaction)

Does not consider phase change Methodology, with examples

Liquid phase reaction: Calculations are simple

Gas phase reaction: Calculations are more involved

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Notation

Usual notation applies.

FA = molar flow rate of A, FT= total molar flow rate

V = volume of reactor, Q = volumetric flow rate

D = diameter and A = cross-sectional area of PFR, L = length of PFR, z =

distance (between 0 and L)

T = temperature, R = universal gas constant, P = pressure, PA = partial pressure

of A

x = conversion, e = fractional increase in number of moles for 100% conversion,for a given feed conditions.

k = rate constant, CA = concentration of A

r = density, m = viscosity

ff = friction factor, Re = Reynolds number

= average velocity of the fluid in the PFR. Used sparingly.

Subscript in indicates inlet. E.g. FA-in is molar flow rate of A at the inlet.

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Definitions and formulas

The following definitions are of use here:

Definition of x

Definition of e

Definition of CA

For gas phase only:

From ideal gas law

At constant temperature

(1 )A A inF F x (1 )T T inF F xe

AA

FC

Q

(1 )T inT

F x RT F RTQ

P P

e

(1 ) (1 )T in in inF x RT x P QQP P

e e

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Formulas

For liquid as well as gas phase

4

ReavD V Q

D

r r

m m

2

2 5

32 ff QdP

dz D

r

2

16if Re < 2100

Re

1 if Re > 10,0006.9

12.96 log10Re

ff

You dont need to memorize these formulas

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PFR design equation

Steady state conditions

For a first order reaction

At constant temperature

AA in A

dF dxF r

dV dV

(1 )An A inA i

xFdxk k

dV Q

FF

Q

A in A

dxF kC

dV

(1 ) (1 )(1 )in in

dx x k x P kVd Q P Q xe

(1 )

(1 )in in

dx k x P

P Q x

A

zd e

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PFR design equation

(1 ) (1 ) ,(1 ) (1 )

where

in in

in in

dx A k x P x P dz P Q x x

A k

P Q

e e

For any other order of reaction also, write the equation such

that dx/dz = f(x,P). i.e. The only unknowns on the RHS mustbe x and P i.e. RHS must not have Q or CA. These are not known yet.

But Qin and CA-in are known and hence RHS may have these terms.

For an nth order reaction

1

1

(1 ) (1 )

1 1

n n n n n

A in

n nn n

in in

n

A in

n n

in in

F x Pdx x P kA

dz P Q x x

Fwhere kA

P Q

e e

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Under steady state conditions, Reynolds number is a constant

Even though local velocity changes This is because density also changes with location

We assume that the viscosity of the medium remains the same,even when the reaction occurs

Under steady state conditions, (rQ) = constant

Using Re, Calculate the friction factor ff.

Write pressure drop equation

Pressure Drop Equation

4

Re

Q

Dm

r

2

2 5 2 5 2 5

32 32 32in inf f f

Q Qf Q f Q f QdP

dz D D D

r

rr

2 5 2 5

32 32 (1 )f in in f in in in inf Q Q f Q P QdP x

dz D D P

r r e

2 5

(1 )

32where f in in in in

dP x

dz P

f Q P Q

D

e

r

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Solution

Solve both equationssimultaneously

Initial conditions: At z =0, P = Pin

At z = 0, x = 0

Special case: When e =0, solve the first

equation and find P. Thensubstitute for P in the secondequation and solve for x

1

1

(1 ) (1 )

1 1

n n n n n

A in

n nn n

in in

nA in

n n

in in

F x Pdx x P kA

dz P Q x x

Fwhere kA

P Q

e e

2 5

(1 )

32where

f in in in in

dP x

dz Pf Q P Q

D

e

r

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Example

Consider a gas phase reaction under isothermal conditions.(Isomerization)

A B

Pin = 10 atm, Q = 0.005 m3/s, T = 300 K, Pure A is fed, k =0.1 lit/s, Molecular weight M = 60 g/gmol, Viscosity of thegas = 10-5 Pa-s

What should be the length of the PFR if it is constructed(a) using a 2 cm pipe and the conversion desired is 10%? (b)using a 1.5 cm pipe and the conversion desired is 10% and (c)using a 1.5 cm pipe and the conversion desired is 20%

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Solution

e= 0. This simplifies the equations

2 5

(1 )

32where

f in in in in

dP xdz P

f Q P Q

D

e

r

1dP

dz P 2 2 2inP P z

324.0558 kg/mininP M

RTr

R=0.08206 atm-lit/(gmol-K)

15.9155 /Vavin m s

0.003f

5Re 7.65 10

2.0047 mol/sAinF

9 21.847 10 /Pa m

96.2832 10 1/ ( )Pa m

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Dia = 2 cm

0 10 20 30 40 50 60 70 80 90 1007

8

9

10x 10

5

PFR length / m

Pres

sure/Pa

e = 0

d = 2 cm

0 10 20 30 40 50 60 70 80 90 1000

0.2

0.4

0.6

0.8

Conversion

Conversion and Pressure vs distance

e = 0d = 2 cm

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Dia = 2 cm, expanded

0 2 4 6 8 10 12 14 16 18 209.6

9.7

9.8

9.9

10x 10

5

PFR length / m

Pressu

re/Pa

e = 0d = 2 cm

0 2 4 6 8 10 12 14 16 18 200

0.05

0.1

0.15

0.2

Conversion

Conversion and Pressure vs distance

e = 0d = 2 cm

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Dia = 1.5 cm

10% conversion is possible, but 20% is not

0 10 20 30 40 50 60 700

5

10x 10

5

PFR length / m

Pressur

e/Pa

e = 0d = 1.5 cm

0 10 20 30 40 50 60 700

0.05

0.1

0.15

0.2

C

onversion

Conversion and Pressure vs distance

e = 0d = 1.5 cm

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2nd order reaction, k = 0.01

Coupled equations give correct answer

0 20 40 60 80 100 120 140 160 180 2004

6

8

10x 10

5

PFR length / m

Pressure/Pa

e = 0d = 2 cm

0 20 40 60 80 100 120 140 160 180 2000

0.5

1

Conversion

Conversion and Pressure vs distance

e = 0d = 2 cm

0 1 2 3 4 5 6 7 8 9 100

5

10x 10

5

PFR length / m

Pressure/Pa

e = 0d = 1 cm

0 1 2 3 4 5 6 7 8 9 100

0.1

0.2

0.3

0.4

Conversion

Conversion and Pressure vs distance

e = 0d = 1 cm

0 2 4 6 8 10 120

5

10x 10

5

PFR length / m

P

ressure/Pa

e = -0.5d = 1 cm

0 2 4 6 8 10 120

0.1

0.2

0.3

0.4

Conversion

Conversion and Pressure vs distance

e = -0.5d = 1 cm