This document discusses the recycle structure design level for a chemical process. It addresses questions like how many reactor systems and recycle streams are required. Using examples, it explains how to determine the number of recycle streams based on boiling points of components leaving the reactor. The document also discusses whether excess reactants should be used, if a gas compressor is needed, and how to determine the reactor heat requirements and temperature changes.

1. Chapter 6
Recycle Structure of the Flowsheet

2. Level 3 – Recycle Structure of the Flowsheet
Now we have decided about input-output structure of the
system.
This is the time to consider the recycle structure. Since the
product distribution dominates the design, the detail of reactor
and also gas compressor should be added.
The separation system is only treated as a black box this stage.
Reactor Separation
?
?
?
?
Purge
Benzene
Diphenyl
H2
Toluene

3. Questions that should be answered at this level
1. How many reactor systems are required? Is there any
separation between the reactor systems?
2. How many recycle streams are required?
3. Do we want to use an excess of one reactant at reactor
inlet?
4. Is a gas compressor required? What are the costs?
5. Should the reactor be operated adiabatically, with direct
heating or cooling or a diluent or heat carrier is required?
6. Do we want to shift the equilibrium conversion? How?
7. How do the reactor costs affect the economic potential?

4. Q.1- How many reactor systems are required?
If sets of reactions take place at different temperatures
or pressures, or if they require different catalysts, then
we use different reactor systems.
only one reactor system is needed.
Toluene + H2 Benzene + CH4
2Benzene Diphenyl + H2
1150 -1300 °F , 500 Psia
No Catalysts

5. Two reactor systems is needed.
Acetone Ketene + CH4
Ketene CO + ½ C2H4
Ketene + Acetic acid Acetic Anhydride 80°C , 1 atm
700°C , 1 atm

6. Q.2- How many recycle streams are required?
From the discussion above, we can associate the
recycle streams with the reactor numbers, for example
in Anhydride process, acetone would be recycled to first
reactor (R1) and Acetic Acid would be recycled to
second reactor (R2).

7. List all the components leaving the reactor(s) in order of their
boiling points.
Next, group recycle components having neighboring boiling
points if they have the same destination.
Then the number of groups are the number of recycle streams.
We also distinguish between liquid and gas recycles, because
liquid recycles need pumps which are normally cheap and gas
recycles need compressors which are always expensive.
( gas are components boiling lower than propylene)

8. Example
A. Waste byproduct
B. Waste byproduct
C. Reactant – recycle to R1
D. Fuel byproduct
E. Fuel byproduct
F. Primary product
G. Reactant – recycle to R2
H. Reactant – recycle to R2
I. Reactant – recycle to R1
J. Valuable byproduct
There are four product streams [ A+B, D+E, F, J ]
Three recycle streams [ C, G+H, I ]

9. HDA Process
Three product streams [ Purge , Benzene , Diphenyl ]
Two recycle streams [ H2+CH4 (gas) , Toluene (liquid) ]
Fuel byproduct
253
Diphenyl
Reactant - Recycle
111
Toluene
Primary Product
80
Benzene
Recycle + Purge - gas
-161
CH4
Recycle + Purge - gas
-253
H2
Stream
NBP (°C)
Component

10. Anhydride Process
Two product streams [CH4 + CO + C2H4 , Anhydride ]
Two liquid recycle streams [ Acetone to R1 , Acetic acid to R2]
Unstable reactant – completely converted
-42.1
Ketene
Reactant – recycle to R1 – liquid
133.2
Acetone
Primary product
281.9
Acetic
anhydride
Reactant – recycle to R2 – liquid
244.3
Acetic acid
Fuel byproduct
-154.8
C2H4
Fuel byproduct
-161
CH4
Fuel byproduct
-312.6
CO
Stream
NBP (°F)
Component

12. Q.3- Do we want to use an excess of one reactant at reactor
inlet?
a) Use of an excess reactant can shift the product distribution
Excess of Isobutane leads to improved selectivity to produce
Isooctane.
The larger the excess, the greater the selectivity, but the
higher the cost to recover and recycle Isobutane.
There must be an optimum value for excess Isobutane.
Butene + Isobutane Isooctane
Butene + Isooctane C12

13. b) Use of an excess reactant can force another component to close to
complete conversion.
The product must be free of Cl2 . An excess of CO will force the
Cl2 to almost complete conversion.
c) Use of an excess reactant can shift the equilibrium conversion.
We want to obtain equilibrium conversion close to unity because the
separation of benzene from cyclohexane in a distillation column is
very difficult (close B.P.). We can shift the equilibrium conversion
to the right by using an excess of H2.
CO + Cl2 COCl2 Phosgene (an intermediate in production of di-isocyanate)
Benzene + 3H2 Cyclohexane

14. Therefore the molar ratio of reactants at the rector inlet is
often a design variable.
Unfortunately there is no rule of thumb available to make a
reasonable guess of the optimum amount of excess.
We need to carry out our economic analysis in terms of this
additional design variable.

15. Q.4- Is a gas compressor is required? what are the costs ?
Whenever a gas recycle stream is present, we will need a gas
recycle compressor.
The design equation for theoretical horsepower (hp) for a
centrifugal gas compressor is:
Pin = inlet pressure (lbf/ft2)
Qin = volumetric flow rate (ft3/min)






−
×
=
−
1
)
(
)
10
03
.
3
(
5
γ
γ in
out
in
in
P
P
Q
P
hp
V
P
V
P
C
C
C
C 1
−
=
γ , Cp and Cv are heat capacities (Btu/mol.°F)

17. The exit temperature from a compressor stage is:
Values for γ (for first estimate)
Efficiency
For first design we assume a compressor efficiency of 90% and
also we assume a drive efficiency of 90%.
R/Cp
Other gases
0.23
More complex gases (CH4,CO2)
0.29
Diatomic
0.4
Monotonic
γ
)
(
in
out
in
out
P
P
T
T
= ( Temperatures and pressures are absolute )

18. Spare
Compressors are so expensive that spares are seldom provided for
centrifugal compressors. (although reciprocating compressors may
have spares because of a lower service factor). In some cases two
compressors may be installed with each providing 60% of the load,
so that partial plant can be maintained in case of one compressor failure.
Multistage Compressors
It is common practice to use multistage compressors. The gas is cooled
to cooling water temperature (100°F) between stages. Also knockout
drums are installed between stages to remove any condensate.
For a three-stage compressor with inter cooling, the work required is:

19. The intermediate pressures that minimize the work are determined from
Which leads to the result
we obtain another design heuristic:
The compression ratio for each stage in a multi-stage
gas compressor should be equal.
3
2 P
Work
P
Work
∂
∂
=
∂
∂
3
4
2
3
1
2
P
P
P
P
P
P
=
=
]
1
)
(
)
(
)
[(
3
4
2
3
1
2
−
+
+
= γ
γ
γ
P
P
P
P
P
P
T
R
M
Work in
M = Molecular weight
R = Gas constant

20. Annualized Installed Cost
Guthrie’s correlation or some equivalent correlation can be used
to find the installed cost for various types of compressors:
Where FC is correction factor and given in table E.2-4 (Appendix E).
MS = Marshall and Swift inflation index (published in Chemical
Engineering)
To put the installed cost on an annualized basis, use CCF.
)
11
.
2
(
)
)(
5
.
517
)(
280
( 82
.
0
C
F
bhp
S
M
Cost
Installed +
=
9
.
0
hp
bhp =
Compressor efficiency

21. Operating Cost
gives us the utility requirement.
Then from utility cost and using 8150 hr/yr we can calculate the
annual operating cost.
TAC = Annualized capital cost + annual operating cost
9
.
0
bhp
drive efficiency

22. Q.5- Adiabatic reactor with direct heating or cooling ?
or heat carrier required?
First we have to have a material balance of recycle flows
in order to find flow to the reactor. The goal here is to
obtain a quick estimate of recycle flows, rather than
rigorous calculations.
We have no details of separation system yet.
We only have the heuristic: “more than 99% recovery of
valuable material is desired”.
At this stage assuming 100% recovery of reactants will
help to make a quick balance. This will usually introduce
only a little error to the stream flows.

23. Limiting Reactant
First we make a balance on the limiting reactant. For HDA process:
If FT is the toluene flow in the reactor inlet
In some cases, some of the limiting reactant might leave the process in
gas recycle and purge stream (Ammonia process), or it
may leave with the product (Ethanol process)
X
F
F
X
F
F FT
T
FT
T =
⇒
−
= )
1
(

24. Ethanol Production
Suppose that we want to produce 783 mol/hr of an EtOH-H2O
azeotrope that contains 85.4% mol EtOH, from an ethylene feed
stream containing 4% CH4, and pure water (with complete recycle
of by-product, diethyl ether)
CH2CH2 + H2O CH3CH2OH
2CH3CH2OH (CH3CH2)2O + H2O

25. Pazeo = 783 mol/hr
yazeo Pazeo = PEtOH
or PEtOH = 0.854 × 783 = 669 mol/hr EtOH
Amount of water in the product stream:
PH2O = Pazeo – PEtOH = 783 – 669 = 114 mol/hr H20
From stoichiometry of the reaction the amount of feed rate of
water which is the limiting reactant, is:
FH2O = yazeo Pazeo + (1 – yazeo) Pazeo = 669 + 114 = 783 mol/hr
reacted left with product

26. Let the water leaving with the product be FW,P = 114 mol/hr
and fresh feed water required for the reaction be FW,R and the
amount of water entering the reactor be FW
Thus, the material balance of the limiting reactant (water) at the
mixing point is: (FW,P + FW,R) + [FW (1 – X) – FW,P ] = FW
FW = FW,P / X
This result is identical to what we found for HDA process.

27. Other Reactants
After we have estimated the flow of the limiting reactant, we use
a specification of the molar ratio at the reactor inlet to calculate the
recycle flows of the other components.
HDA Process
RG = flow of the recycle gas
MR = molar ratio of H2/Toluene of reactor inlet
Once we specify design variables X, MR and yPH the
recycle flow can be found from the above equation.
)
(
X
F
MR
R
y
F
y FT
G
PH
G
FH =
+
[ ]






−
−
+
+
−
=
)
(
2
)
1
(
)
1
(
PH
FH
PH
FH
PH
B
G
y
y
S
y
S
y
X
MR
y
S
P
R
or

28. Heuristics
There is no rule of thumb available to select X for the case
of complex reactions and there is no rule of thumb for
selecting the purge composition yPH or the molar ratio MR.
For the cases of single reactions, choose X = 0.96 or X =
0.98Xeq as a first guess.
Reversible byproduct
If we recycle a byproduct produced by a reversible reaction and let
the component build up to its equilibrium level, such as Diphenyl
in HDA process or diethyl-ether in ethanol process, then we will
find the recycle flow using the equilibrium relationship at the
reactor exit.
2
2
)
(
)
)(
(
Benzene
H
Diphenyl
Keq =

29. The H2 and benzene flows have been determined by using the
first reaction and the purge calculations, so we can use the
equilibrium equation to calculate the Diphenyl flow at the reactor
exit.
Reactor Heat Effects
We need to make a decision as to whether the reactor can be
operated adiabatically with direct heating or cooling, or whether a
diluent or heat carrier is needed. If we need to introduce a diluent
or heat carrier, then our recycle balance and perhaps our overall
material balance will change. This decision should be made before
any specification of separation system because these diluents or heat
carriers will affect the design of separation system.

30. To make the decision on reactor heat effects first we estimate
The reactor heat load and the adiabatic temperature change.
Reactor Heat Load
For single reactions we know that all the fresh feed of the
limiting reactant usually gets converted (conversion per-pass
might be small so that there is a large recycle flow but all the
fresh feed is converted except the loses in product or purge).
Thus , for single reactions:
Reactor heat load = Heat of reaction × Fresh feed rate
Note: The heat of reaction is to be calculated at reactor conditions.

31. For complex reactions the extent of each reaction will
depend on the design variables (conversion, molar ratio,
temperature, pressure, ..).
Once we select the design variables we can calculate the
extent of each reaction and calculate the heat load
corresponding to the side reactions.

32. Example: HDA Process
If we want to estimate the reactor heat load for a case where, X=0.75 (then s = 0.9694),
PB= 265 mol/hr FFT = 273 mol/hr
We might neglect the second reaction and write:
QR = ∆HR FFT = (-21530) (273) = -5.878 × 106 Btu/hr
Where ∆HR is the heat of reaction @ 1200°F and 500 Psia
Heat is generated by
exothermic reaction
Toluene + H2 Benzene + CH4
2Benzene Diphenyl + H2

33. Example: Acetone Process (IPA)
If we design to produce 51.3 mol/hr of acetone, then 51.3 mol/hr of IPA is required, the
heat of reaction at 570°F and 1 atm is 25800 Btu/mol
QR = 25800 (51.3) = 1.324 × 106 Btu/hr
Adiabatic Temperature Change
Once we have determind the reactor heat load and the flow rate through the reactor as a
function of design variables, we can estimate the adiabatic temperature change:
QR = F CP ( TRin
– TRout
)
(CH3)2CHOH (CH3)2CO + H2
Heat is generated by
exothermic reaction

34. Example: HDA Process
for cases of X = 0.75 and yPH = 0.4
Toluene + H2 Benzene + CH4
2Benzene Diphenyl + H2
48.7
91
Toluene recycle
48.7
273
Toluene feed
0.4 (7) + 0.6 (10.1) = 8.86
3371
Recycle gas
0.95 (7) + 0.05 (10.1) = 7.16
492.2
Make up gas
Cp, (Btu/ mol °F)
Flow, (mol/hr)
Stream

35. If TRin
= 1150°F
QR = -5.878 × 106 = [(273 + 91) 48.7 + 496 (7.16) + 3371 (8.86)] ( TRin
– TRout
)
TRout
= 1150 + 115 = 1265°F
This value is below the constraint on the reactor exit temperature of 1300°F. Also, the
calculation is not very sensitive to CP values or to flows.
Thus, an adiabatic reactor can be used. (see figure 6.3-1)

36. Example: Acetone Process (IPA)
If the feed stream to acetone process is an IPA-H2O azeotrope (70 mol% IPA) and we
recycle an azeotropic mixture, then it is easy to show that 22 mol/hr of water enters
with feed. Also, for a conversion of X = 0.96, the recycle flow will be 2.1 mol/hr of
IPA and 0.9 mol/hr of water.
(CH3)2CHOH (CH3)2CO + H2

37. If the reactor inlet temperature is 572°F:
(51.3 / 0.7) × 0.3 = 22 mol/hr H2O
51.3 × 0.04 = 2.1 mol/hr IPA (recycle)
(2.1 / 0.7) × 0.3 = 0.9 mol/hr H2O (recycle)
QR = 1.324 × 106 = [(51.3 + 22) + (2.1 + 0.9)] (22) ( 572 – TRout
)
or TRout
= 572 - 788 = -216°F
Clearly, this is not a reasonable result. Therefore, we can not use adiabatic reactor and
we try to achieve an isothermal reaction.

38. Heat Carriers
The adiabatic temperature change depends primarily on the flow through the reactor.
Hence, we can always moderate the temperature change by increasing the flow
through the reactor. To do this, we prefer to recycle more of a reactant or byproduct
or product. However, where this is not possible, we may add an extraneous component.
Of course, this will complicate the separation task. Thus, we normally try to avoid this
situation.
In HDA process, the methane in the gas recycle stream (60%) acts as a heat carrier.
Thus, if we purify the hydrogen in the recycle stream we have decreased the recycle
gas flow but this will increase the exit temperature of the reactor. If this exceeds the
1300°F limit we could no longer use an adiabatic reactor. Instead, we would have to
cool the reactor, increase the recycle flow or introduce an extraneous component as a heat
carrier.

39. A similar behavior happens in many oxidation reactions. If pure oxygen is use as a
reactant, the adiabatic temperature rise would be very large. However, if air is used as
the reactant, the presence of nitrogen moderate the temperature change.
Question No.6, Do we want to shift the equilibrium conversion? How?
We can use our pervious procedure for calculating the process flows as a fraction
of the design variables. Then we can substitute these flows into the equilibrium
relationship to see whether the conversion we selected is above or below the
equilibrium value. Of course, if it exceeds the equilibrium value the result has no
meaning.
In most of cases it is necessary to determine the exact value of equilibrium conversion
as a function of design variables. This normally requires a trial and error solution.

40. Example: Cyclohexane Production
Production rate : 100 mol/hr C6H12 with 99.9% purity
Feed : Pure benzene, Hydrogen make up (97.5% H2, 2% CH4, 0.5% N2)
C6H6 + 3H2 C6H12

41. Overall balance
Assume 100% recovery, then
FE = excess H2 feed to process
Total H2 feed = 3PC + FE = 0.975 FG
Purge composition of H2 ,
Make up gas rate,
Purge rate, PG = FE + 0.025 FG
G
E
E
PH
F
F
F
y
025
.
0
+
=
)
975
.
0
1
(
3
PH
PH
C
G
y
y
P
F
−
−
=
Production of C6H12 : PC = 100
Benzene fresh feed : FFB = PC = 100

42. Recycle balance
Benzene fed to reactor,
Recycle gas flow,
Reactor Exit Flows
Cyclohexane = PC
Benzene =
Hydrogen = MR FB – 3PC =
Inert = 0.025FG + (1- yPH) RG =
Total flow =
C
P
X
MR
)
3
( −
X
P
X
F
F C
FB
B =
=
)
975
.
0
(
1
)
(
975
.
0 G
C
PH
G
C
G
PH
G F
X
P
MR
y
R
X
P
MR
R
y
F −
=
⇒
=
+
X
X
P
P
X
P C
C
C )
1
( −
=
−
MR = molar ratio of H2 to benzene
C
PH
PH
P
X
MR
y
y
)
3
(
1
−
−
]
1
)
3
(
1
[
PH
C
y
X
MR
X
P −
+

43. Equilibrium Relationship
From the Washington University Design Case Study No.4, p. 4-3, part П,
vH = 1 vC / vB = 1.13
Then
Having Ptot, Keq, MR and yPH from the above equation we can find the equilibrium
conversion.
3
3
)
3
(
3
1
)
1
(
13
.
1








−
−
+
−
=
PH
eq
eq
eq
eq
eq
tot
y
X
MR
X
MR
X
X
K
P
3
3
3
3
H
B
H
B
tot
C
c
H
B
C
eq
y
y
v
v
P
y
v
f
f
f
K =
=
Keq :
Equilibriu
m constant
fi =
fugacity
vi =
fugacity
coefficient

44. Discussion
Since benzene and cyclohexane are very close boilers, we would like to avoid benzene
cyclohexane distillation separation. This can be accomplished by operating the reactor
at a sufficiently high conversion that we can leave any unconverted benzene as an
impurity in the product. However, to obtain high benzene conversions, we must force
the equilibrium conversion to be very close to unity. (shifting equilibrium conversion).
The equilibrium conversion equation shows that it can be done by number of ways.
Shifting Equilibrium Conversion and Economic Trade-offs
From the equation we can see that the equilibrium conversion can be increased by
increasing the reactor pressure, Ptot or by increasing molar ratio of hydrogen to
benzene at the reactor inlet, MR ,or by decreasing the reactor temerature (the reaction is
exothermic).

45. However, increasing reactor pressure corresponds to a large feed compressor and more
expensive equipments because of increased wall thickness. Large molar ratio of
hydrogen to benzene corresponds to larger gas-recycle compressor. Lower reactor
temperature corresponds to larger reactors because of the decreased rate of reaction.
Thus, an optimization analysis is required to determined the value of Ptot, X and Treact,
molar ratio and yPH. For this optimization an approximate model can be useful:
3
3
)
3
(
2
13
.
1
1
1 





−
−
−
≈
PH
eq
tot
eq
y
MR
MR
K
P
X

46. Separator Reactors
If one of the products can be removed while the reaction is taking place, then an
apparently equilibrium-limited reaction can be forced to go to complete conversion.
Example 1 - Acetone production
Acetone can be produced by the dehydrogenation of isopropanol In the liquid phase as
well as gas phase. At 300°F the equilibrium conversion for the liquid phase process is
about Xeq = 0.32 . By suspending the catalyst in high boiling solvent and operating the
reactor at a temperature above the boiling point of the acetone, both H2 and acetone
can be removed as a vapor from the reactor. Thus, the equilibrium conversion is shifted
to the right. A series of three CSTRs with a pump-around loop containing a heating
system that supplies the endothermic heat can be used.
Isopropanol Acetone + H2

47. Example 2 - Production of ethyl acrylate
Both acrylic acid and ethyl acrylate are monomers, which tend to polymerize in the
reboilers of distillation column. We can eliminate a column a column required to purify
and recycle acrylic acid from the process if we can force the equilibrium limited reaction
to completion, say, by removing the water. Hence, we use excess of ethanol to shift the
equilibrium to the right, and we carry out the reaction in reboiler of rectifying
column. With this approach, the ethanol, water and ethyl acrylate are taken overhead,
and the acrylic acid conversion approaches unity.
Acrylic Acid + Ethanol Ethyl Acrylate + H2O

48. Reversible Exothermic Reaction
There are sevral important indastrial reactions that are reversible and exothermic.
For example,
In ammonia synthesis,
High temperature (high reaction rate) corresponds to small reactor volumes, but for
these reactions the equilibrium conversion decreases as the reactor temperature
increases. Hence, these reactions are often carried out in a series of adiabatic beds with
either intermediate heat exchangers to cool the gases or a bypass of cold feed to
decrease the temperatures between the beds. With this procedures we obtain a
compromise between high temperatures (small reactor volumes) and high equilibrium
conversion.
Sulfuric acid process: SO2 + ½ O2 SO3
Amonia synthesis: N2 + 3H2 2NH3
Water gas shift: CO + H2O CO2 + H2

49. Diluents
From the discussion above we have found that the temperature, pressure and molar ratio
can all be used to shift the equilibrium conversion. However, in some cases an extraneous
component (a diluent) is added which also shift the equilibrium conversion. For example,
in styrene process:
The addition of steam (or methane) at the reactor inlet lowers the partial pressure of
styrene an H2 and so decreases the reverse reaction rate. The steam serves in part as a heat
carrier to supply endothermic heat of reaction.
Ethylbenzene Styrene + H2
Ethylbenzene Toluene + Methane
Ethylbenzene Benzene + Ethylene

50. Steam is often used as a diluent because water - hydrocarbon mixture are usually
immiscible after condensation. Hence, the separation of water can be accomplished by
a decanter (and usually stripper to recover the hydrocarbons dissolved in the water, if
the water is not recycled).
Question No.7, How do the reactor costs affect the economic potential ?
Reactor Design
At the very early stages in a new design, a kinetic model normally is not available.
Thus, we base our material balance calculations as correlation of the product distribution.
Also we assume that we will use the same type of the reactor in the plant that the chemist
used in the laboratory and we often base a first estimate of the reactor size on the reaction
half-life measured by chemist.

51. For adiabatic reactors we might base the design on an isothermal temperature which is the
average of the inlet and out let temperature or an average of the inlet and outlet rate
constant.
We estimate the costs of plug flow reactors in the same way as we do for pressure vessels
(Appendix E.2) with CCF = 1/3.
Reactor Configuration
Since the product distribution can depend on the reactor configuration, we need to
determine the best configuration. A set of design guidelines has been published by
Levenspiel (Table 6.6-1). as this table indicates in some cases we obtain complex reactor
configurations; see Fig. 6.6-1

52. Recycle Economic Evaluation
For HDA process, higher composition of hydrogen in the recycle stream means higher
raw material loss in purge stream but, lower cost of recycle compressor, recycling less
methane. Lower composition of hydrogen in the recycle stream means lower raw
material loss in purge stream but, higher cost of recycle compressor, recycling more
methane.
The value for optimum shown in Fig. 6.7-1 are not the true optimum values because we
have not considerate any separation cost yet.

53. yPH
-2,500,000
-2,000,000
-1,500,000
-1,000,000
-500,000
0
500,000
1,000,000
1,500,000
2,000,000
0.1 0.2 0.3 0.4 0.5 0.6 0.7
Conversion x
E
P
3
($/yr)
0.2
0.4
0.6
0.8
Economic potential – level 3