3. Design of CSTRs I

Apr/04 METU-NCC 1

o Single CSTR 1

- Design equation

- Substitute FA0 = v0CA0

- Space time τ

- 1st order rxn assume

- Rearranging

exitA

A0

)( r

XFV

A

A00

r

XCvV

A

A0

0

τr

XC

v

V

X

X

kr

XC

1

1τ

A

A0

k

kX

τ1

τ

3. Design of CSTRs II

Apr/04 METU-NCC 2

o Single CSTR 2

- CA = CA0(1 - X)

- Damköhler number ⇒ dimensionless number

• quick estimate of the degree on conversion

achieved by continuous reactors

k

CC

τ1

A0A

imereaction t chemical ticCharateris

timefluid sticCharacteri

rate convectionA

raterxn A

A of rate flow Entering

entranceat rxn of RateDa

A0

A0

F

Vr

3. Design of CSTRs III

Apr/04 METU-NCC 3

o Single CSTR 3

- Damköhler number for a 1st order irrev. rxn

- Damköhler number for a 2nd order irrev. rxn

- Rule of thumb

• if Da < 0.1, then X < 0.1

• if Da > 10, then X > 0.9

☞ 1st order rxn, X = Da/(1 + Da)

kCv

VkC

F

VrτDa

A00

A0

A0

A0

A0

A00

2

A0

A0

A0 τDa kCCv

VkC

F

Vr

3. Design of CSTRs IV

Apr/04 METU-NCC 4

o CSTR in series 1

- 1st order irrev. rxn with no change in volumetric flow

rate, effluent of the first reactor

- For 2nd reactor

- Solving for CA2

- For n CSTRs in series

- n tank in series

11

A0A1

τ1 k

CC

A22

A2A10

A2

A2A12

)(

Ck

CCv

r

FFV

)τ1)(τ1(τ1 1122

A0

22

A1A2

kk

C

k

CC

nnn

C

k

CC

)Da1()1(

A0A0A

nnkX

)Da1(

11

)τ1(

11

3. Design of CSTRs V

Apr/04 METU-NCC 5

o CSTR in series 2

nnnk

CkkCr

)τ1(

A0AA

3. Design of CSTRs VI

Apr/04 METU-NCC 6

o CSTR in parallel 1

- One large reactor of volume V

o 2nd order reactor in a CSTR

- Dividing by v0

- For conversion X

22

A0

A00

2

A

A0

A

A0

)1( XkC

XCv

kC

XF

r

XFV

2

A00 )1(τ

XkC

X

v

V

2Da

4Da1-2Da)1( X

Ex 4-2,

p 163

4. Tubular Reactors I

Apr/04 METU-NCC 7

o Design equation

- Differential form

• Q or ∆P

- Integral form

• no Q or ∆P

o 2nd order reactor in a PFR 1

AA0 rdV

dXF

X

r

dXFV

0A

A0

X

kC

dXFV

0 2

A

A0

4. Tubular Reactors II

Apr/04 METU-NCC 8

o 2nd order reactor in a PFR 2

- Liquid phase reaction (v = v0)

• combining MB & rate law

• conc. of A,

• combining & integrating

• solving for X

A0

2

A

F

kC

dV

dX

)1(A0A XCC

V

X

X

kC

v

X

dx

kC

FV

0A0

0

22

A0

A0

1)1(

2

2

A0

A0

Da1

Da

τ1

τ

kC

kCX

4. Tubular Reactors III

Apr/04 METU-NCC 9

o 2nd order reactor in a PFR 3

- Gas phase reaction (T = T0, P = P0)

• conc. of A,

• combining

X

XCC

ε1

1A0A

dXXkC

XFV

X

2

A00

A0)1(

)ε1(

X

dXX

X

kC

vV

0 2

2

2

A0

0

)1(

)ε1(

X

XXX

kC

vV

1

)ε1(ε)1ln()ε1(ε2

22

A0

0

4. Tubular Reactors IV

Apr/04 METU-NCC 10

o 2nd order reactor in a PFR 4

- Conversion as a function of distance down the

reactor

4. Tubular Reactors V

Apr/04 METU-NCC 11

o 2nd order reactor in a PFR 5

- Three types of reactions

• ε = 0 (δ = 0) ⇒ v = v0

• ε < 0 (δ < 0)

⇒ gas molecule spends

longer time

☞ higher conv.

• ε > 0 (δ > 0)

⇒ gas molecule spends

shorter time

☞ lower conv., p 171 Ex 4-3

4. Tubular Reactors VI

Apr/04 METU-NCC 12

o 2nd order reactor in a PFR 6

- Production of ethylene using PFR

mincatalystgram

molesgramAA0 r

dW

dXF

5. Pressure Drop in Reactors I

Apr/04 METU-NCC 13

o Pressure Drop and the Rate Law

- In PBR in terms of catalyst weight

• rate equation,

• stoichiometry

• isothermal

2

AA kCr

T

T

P

P

X

XCC 0

0

A0A

ε1

)1(

2

0

2

A0A0

ε1

)1(

P

P

X

XCk

dW

dXF

2

0

2

0

A0

ε1

)1(

P

P

X

X

v

kC

dW

dX),(1 PXF

dW

dX

5. Pressure Drop in Reactors II

Apr/04 METU-NCC 14

o Flow through a Packed Bed 1

- Ergun equation

where P = pressureφ = porosity = void fractiongc = conversion factor relating gravityDp = diameter of particle in the bedμ = viscosity of gas passing through the bedz = length down the packed bed of pipeu = superficial velocityρ = gas densityG = ρu = superficial mass velocity

G

DDg

G

dz

dP

pc

75.1μ)1(1501

ρ p

3

Dominant for laminar flow

Dominant for turbulent flow

5. Pressure Drop in Reactors III

Apr/04 METU-NCC 15

o Flow through a Packed Bed 2

- Pressure drop in packed bed 1

where β0 is a constant depending on the properties

of the packed bed and the entrance conditions

00

00β

T

T

F

F

T

T

P

P

dz

dP

XT

T

P

P

dz

dPε1β

0

00

G

DDg

G

PPc

75.1)1(150

ρ

)1(β

3

0

0

5. Pressure Drop in Reactors IV

Apr/04 METU-NCC 16

o Flow through a Packed Bed 3

- Pressure drop in packed bed 2

• interested in more in catalyst weight rather than the

distance z

• catalyst weight, W = zAcρb = zAc(1-φ)ρc

catalyst solid

ofDensity

solids

of Volume

catalyst

ofWeight

ρ φ)-(1 W cc

zA

00

0

c

0

φ)ρ1(

β

T

T

c F

F

T

T

P

P

AdW

dP

5. Pressure Drop in Reactors V

Apr/04 METU-NCC 17

o Flow through a Packed Bed 4

- Pressure drop in packed bed 3

• let

• then

where y = P/P0

00

0

c

0

φ)ρ1(

β

T

T

c F

F

T

T

P

P

AdW

dP

0c

0

φ)1(ρ

2β

PAc

00

0

0

2 T

T

F

F

T

T

PP

P

dW

dP

00

0

0 1

2 T

T

F

F

T

T

PPdW

PPd

00

1

2 T

T

F

F

T

T

ydW

dy

5. Pressure Drop in Reactors VI

Apr/04 METU-NCC 18

o Flow through a Packed Bed 5

- Pressure drop in packed bed 4

• for single reactions

• isothermal operation

• notice that

)ε1(1

2 0

XT

T

ydW

dy

)ε1(1

2X

ydW

dy

),(or ),( and ),( PXfdW

dyPXf

dW

dPPXf

dW

dX

5. Pressure Drop in Reactors VII

Apr/04 METU-NCC 19

o Flow through a Packed Bed 6

- Pressure drop in packed bed 5 – effect of P drop

P 187

Ex 4-5

5. Pressure Drop in Reactors VIII

Apr/04 METU-NCC 20

o Flow through a Packed Bed 7

- Finding optimum particle diameter

Apr/04 METU-NCC 21

6. Synthesizing for Design of a Chemical Plant I

Apr/04 METU-NCC 22

o Manufacturing of ethylene glycol

- Economy

Profit EG cost

year

lb102

lb

$0.38 m8

m

Ethane cost

year

lb10

lb

$0.04 m8

m

4

SA cost

year

lb102.26

lb

$0.43 m6

m

Operating

cost

$8,000,000

= -

- -

= $76,000,000 - $16,000,000 - $54,000 - $8,000,000

= $52 million

6. Synthesizing for Design of a Chemical Plant II