Recycle Streams

It is common for the conversion in a chemical reactor to be less than 100%.

Multi-systems: Large scale processes

• Recycle

• Purge

• Bypass

• System boundaries for multi-systems

chemical reactor to be less than 100%. Equilibrium constants for each reaction dictate the maximum conversion possible.

ReactorReactants

Products

+Unconverted Reactants

Conversion < 100%

Consider the following process:

The obvious solution is to recycle the unconverted reactants back to the reactor until all is converted. I.e. the stream leaving the reactor is passed through a separatorwhere the products are separated and the unreacted feed is recycled backinto the reactor.

In this processes there is a significant wastage of unconverted reactants. If conversion was 60% then 40% of the feed would be wasted. This makes for poor process economics.

Note: If we do an overall mass balance, all feed is converted to product.

Products

SeparatorReactor

Recycle stream(unreacted feed)

Feed

Simple example:

Consider a reaction in which A → B and a reactorconversion of 60% .

CASE 1:

Reactor100 mol A X mol A

Y mol B

Component balances:

A: 0 = 100 – X + 0 – 100 x 0.60 X = 40mol

B: 0 = 0 – Y + 100 x 0.60 x 1/1 – 0 Y = 60mol

CASE 2:

100 mol B100 mol A

Recycle stream

SeparatorReactor

100 mol B100 mol A

W mol A

SeparatorReactor

X mol A

Y mol B

Balance on the separator:

B: 0 = Y – 100 Y = 100 mol

Balance on the reactor: Balance on the reactor:

B: 0 = 0 – 100 + (100 +W) x 0.60 x 1/1

W = 66.67 molBalance on the separator:

A: 0 = X – 66.67 X = 66.67 mol

100 mol B100 mol A

66.67 mol A

SeparatorReactor

66.67 mol A

100 mol B

Note:

• Same amount of feed: 100 mol A

• Improved reagent to product conversion with recycle

• Larger reactor volume with recycle

• Requires a separator with recycle

CASE 1:

Reactor100 mol A 40 mol A

60 mol BReactor 60 mol B

( ) %60%10010040100 A of conversion pass Single =⋅−=

CASE 2:

100 mol B100 mol A

66.67 mol A

SeparatorReactor

( ) %100%100100100100 A of conversion Overall =⋅−=

REACTOR

Conversion around reactor is 60% This is called the single-pass conversion

Conversion around overall system is 100% -This is called overall conversion.

Purge streams

In previous examples of Recycleof unreacted chemicals, overall conversion is 100%.

For example: water shift reaction

CO, H2O

CO,

222 CO H OH CO ++

Products: H2, CO2

Reactor SeparatorH2O

CO

FeedCO,H2O

H2

CO

If equimolar amounts of CO and H2O are placed in the feed (i.e. in the right stoichiometric ratio), then 100% overall conversion is possible.

However, what if the amounts of CO and H2O are not equimolar (i.e. not in their stoichiometric ratio)?

Consider feeding 100% excess of H2O:

CO2, H2

ProductFeed

50 mol CO100 mol H2O

Element balance:

The excess H2O will continue to be recycled as well as being continuously fed and not all of it will be converted.

∴ H2O must be removedfrom the system as it will accumulate.

C: moles (CO 2) out = 50 mol

H: moles (H 2) out = 100 mol

O: moles (CO2) out = 75 mol

Not possible

What if inert species (eg. N2) enter with the feed?

CO2

H2

50 mol CO

50 mol H2O10 mol N2

Not possible. N2 (or any other inert compounds) must be removed or purged from system.

2

must be removed or purged from system.

If N2 is not purged, the recycle stream builds up N2 continuously (accumulates) and steady state is never achieved.

To remove excess reactants or inert species from a system (if these species cannot be removed selectively: separator) usually a proportion of the recycle stream must be split off.

Example:

Methanol is synthesized from carbon monoxide and hydrogen in a catalytic reactor. The fresh feed to the process contains 32.0% CO, 64% H2 and 4.0% N2.

This stream is mixed with a recycle stream in a ratio of 5 mol recycle/mol fresh feed to produce the feed to the reactor, which contains 13.0 mol% N2.

The reactor effluent goes to a condenser from which two streams emerge: A liquid product stream containing all the methanol formed in the reactor and a gas stream containing all the H2, CO and N2 leaving the reactor. The gas stream is split into two fractions: One is removed from the process as a purge stream, and the other is the recycle stream that combines with the fresh feed to the reactor.

Calculate: (a) production rate of methanol (mol/h); (b) molar flow rate and composition of the purge gas; (c) overall and single pass conversions.

Solution: use element balances

Basis: 100 mol fresh feed

100 mol/h

N2

COH2

CH3OH

CondenserReactor

COH2

N2

7

3

4

1 2

6

Purge

CH3OH

CH3OH32% CO64% H2

4% N2

5

let n(i)j = number of moles of species ‘i’ in stream ‘j’

)n(N )n(N )n(N

:N

mol/h 600

100 500

n(T) n(T) n(T)

:Total

:point mixing around balance

mol/h 500

5n(T) n(T) ,now

2

172

17

+=

=+=

+=

∴==

0.148 be alsomust 6 and 5

streamsin N offraction mol The

0.148 500

74 is 7

streamin N offraction mol

mol/hr 74 )n(N

100 0.04 )n(N 600 0.13

)n(N )n(N )n(N

2

2

72

72

127222

=

∴=⇒

×+=×⇒

+=

Mass balanced around the entire system:

As nitrogen is an inertspecies, we can do an overall mass balance on N2

mol/hr 27.0 n(T)

n(T) 0.148 100 0.04

out N in N

6

6

2 2

=⇒

=×⇒

=

We must do element balances to determine We must do element balances to determine the flow of H2, CO, CH3OH as these are reactants/products.

Element Balances:

Around entire system:

Carbon:

OH)4n(CH )2n(H )2n(H

out H in H

balance H

1 - OH)n(CH n(CO) 100 0.32

out C in C

4362 12

436

+==

+=×⇒

=

n(CO) - 23 )n(H

23 4 - 27 n(CO) )n(H

mol/hr 4.00 27 0.148 )n(Nbut

3- 27 )n(N n(CO) )n(H

2- OH)2n(CH )n(H 100 0.64

OH)4n(CH )2n(H )2n(H

662

662

62

62662

4362

4362 12

=⇒

==+∴=×=

=+++=×⇒

+=

Note there is no ‘O’ balance. This would be identical to the ‘C’ balance. C and O appear in the same ratio (1:1) in the same components: CO & CH3OH

15.4 )n(H

24.3 OH)n(CH

OH)3n(CH 73

1equation to thisAdd

n(CO) -OH)2n(CH 14

2equation into substitute

n(CO) - 23 )n(H

43

43

643

662

=⇒

=⇒

=

=

=⇒

76% 100 32

24.3 conversion Overall

OHCHH2CO

7.6 n(CO)

15.4 )n(H

32

6

62

=×=∴

↔+

=⇒

=⇒

For single pass conversion, we need to know amount of CO entering the reactor.

141 32

n(CO) n(CO) n(CO)

mol/h 141

500 0.282 n(CO)

0.282

stream) purgein as (same 27

7.6 is 7 Stream

in CO offraction mol ,Now

712

7

+=+=∴

=×=∴

=

14% 100 173

24.3

is conversion pass Single

mol/h 24.3

OH)n(CH

OH)n(CH reactor leaving

OHCH ofamount reactor

in Converted CO ofAmount

mol/h 173

141 32

43

33

3

=×

∴===

=

=+=

Bypass

UnitOperation

For reasons of product quality, or other issues, some times a fraction of a stream will be diverted around a unit operation only to be diverted around a unit operation only to rejoin on the other side.

Example:

In the preparation of feedstock to a plant manufacturing hydrocarbon product, iso-pentane is removed from a petroleum stream. Given the data on the below diagram, what fraction of the petroleum is passed through the iso-pentane tower?

3

Iso-PentaneTower

1 2

4

Feed80% n - P20% i - P

100% i - P

90% n – P10% i - P

100% n - P

n – P = n – pentane

i – P = iso-pentane

5 6

Basis: Feed Flow of 100 kg

Over Entire System:

Overall Balance:

64

641

m(T) m(T) 100

m(T) m(T) )T(m

+=⇒

+=

kg 89 100 0.9

0.8 m(T)

m(T) 0.9 m(T) 0 0.8(100)

Balance P -n

6

64

=×=⇒

+×=

kg 11 89- 100 m(T)

kg 89 100 0.9

m(T)

4

6

==⇒

=×=⇒

5

542

m(T) 11

m(T) m(T) m(T)

Balance Overall

:Tower Around Balance

+=+=

52 m(T) m(T) 0.8

Balance P -n

=

kg 55 m(T)

11 m(T) 0.2

m(T) 0.8 11 m(T)

2

2

22

=⇒

=⇒

+=⇒

∴ 45 kg of feed by-passes iso-pentane tower.

∴ 45% of the feed by-passes tower.

55% of the feed passes through the 55% of the feed passes through the tower