HIERARCHICAL DESIGN PROCEDURE
Level 1.
Batch versus Continuous
Level 2.
Input-Output Structure of the Flowsheet
Level 3.
Recycle structure and reactor considerations
Level 4.
Separation System Synthesis
4a. Vapor Recovery
4b. Liquid Recovery
Level 5.
Heat Integration
Costly errors are caused by fixing the flowsheet too early in the design.
Input Information Needed to Develop a Flowsheet
A. Reactions
1. Stoichiometry, T, P, phase conditions
2. Product distribution versus conversion
3. catalysts used, deactivation
From chemist. May be incomplete.
B. Products
1. Production capacity, product priority cf with existing plants.
2. Product price (may depend on purity)
3. By product value or disposal cost
C. Raw Materials
1. Composition, T, P.
2. Prices, Purity
3. Effect of Impurities on Reaction.
D. Constraints
1. Explosive Limits
2. Safety limits
3. Temperature Limits (unstable materials)
4. Environmental
E. Plant and Site Data
1. Utility costs (Fuel, steam, cooling water)
2. Waste disposal costs
Level 1. Batch vs. Continuous
1. Depends on Production Rate
A. Consider batch if < 10 million lb/yr
B. Always batch if < 1 million lb/yr
2. Market Conditions
A. Seasonal Production, favours batch
B. Short Product Lifetime, favours batch
3. Scale-up Problems. If expected use batch
A. Slurries
B. Fouling
EXAMPLE
Production of Monochlorodecane:
Evaluation of alternative reaction paths:
Monochloro.
Concept of Economic Potential.
Decane
Process
Reaction Paths
Path 1:
5 C10H22 + 6 Cl2

4 C10H21Cl + 6 HCl + C10H20 + Cl2

Path 2:
C10H20 + HCl
Path 3:
C10H21OH + HCl
C10H21Cl

C10H21Cl + H2O
Prices
Decane
Decene
Decanol
Chlorine
HCl
MCD
$4.80/lbmole
$12.00/lbmole
$14.00/lbmole
$ 1.77/lbmole
$ 1.00/lbmole
$16.00/lbmole
Which reaction path is most attractive?
PATH 1
PROFIT 
4  $16  6  $1.00 - 5  $4.80 - 6  $1.77
4 lbmoles of MCD
 $8.845 / lbmole MCD
PATH 2
PATH 3
P = 1 $16.00  1 $12.00  $1.00 1  $3.00 / lbmole MCD
P = 1 $16.00  1 $1.00  $1.00  $14.00  $1.00 / lbmole MCD
RANKING: Path 1 > Path 2> Path 3
Pitfalls to Avoid
1. Make comparisons on same basis
2. Reaction should be balanced.
3. Include by products in analysis
Conversion, Yield and Selectivity
Usually we have multiple reactions
R
P
R
Reactants  Products
+WByproducts
Reactant
1mole
Conversion = X
X
moles of Reactant consumed
mole of Reactant fed
Yield or Selectivit y  S 
Mole of Product formed
Moles of Reactant consumed
Product
Waste
Product 5x
Unconverted
Reactant
R (1-X) moles
Recycle Structure
Level 2: Calculations
Do a preliminary material balance
(Assume a conversion for complex reactions)
Assume complete recovery of valuable raw materials and products
With gas recycle and purge, some valuable raw material is last
Purge rate is based on the following:
inert (impurity in feed) = inert in purge
= purge comp. (mole %) x purge rate
Recycle rate:
depends on conversion
Recycle/Feed ratio = (1-x)/x where x is conversion in reactor.
Note as x 0 Recycle  very large
Level 3: Decisions
1.
2.
3.
4.
5.
6.
How many reactors are required?
How many recycle streams are there?
Specify reactor structure.
Do recycle material balances.
Estimate reactor heat effects.
Reactor costs.
1. # of Reactors:
Use multiple reactors for
- multiple step reactors
- for interstage cooling
2. # and Destination of Recycle Streams:
- do not separate streams returning to the same point.
- return reactants to the first reactor where it is used.
3. Recycle Material Balance
Single Reactions:
Conversion = 96% conversion for irreversible reactions
= 98% of equilibrium for reversible vxns
= 99% if no recycle stream
Trade off
high x  large reactor cost
low recycle cost
possibly more byproduct or waste product esp. for
series type reactions
Multiple Reactions:
high x  more undesirable products less recycle
important design parameter
4. Reactor Heat Effects
- compute adiabatic temp rise.
Is it reasonable?
If not use cooling/heating.
May need a diluent to carry excess heat.
Direct fired furnaces to very high temp.
Packed tubes for catalysts with moderate heat effects.
Level 1. Decision
Batch vs. Continuous
Production Rate  10 lbmoles/hr
 lbs/hr 8000 hr
yr
6
 8 x 10 lbs/year
Can use a batch or continuous process.
Closer to limit for continuous processes.
Decision:
Use continuous process.
Cyclohexanol Process Example
A: Cyclohexanol
B:. Cyclohexanone
C: High boiler
Reaction Stoichiometry
A  B + H2
A+ BC
2A  C + H2
Given:
Conversion in Reactor
Yield of desired product
Costs: Raw Material:
Product:
H2:
C:
Basis: 10 lbmoles/hr of raw material
95% A, 5% B
Production rate = 1000 lb./hr
= 8 million lb/year
Favor continuous process in this case
= 80%
= 99%
$58.64/lbmole
$125.00 lb/mole
$0.00 (use as fuel gas)
Cost of handling = fuel gas credit
Waste material
Hazardous $145/ton
Level 2. Decision
Input/Output Structure
Using given yield:
(Assume all A is converted)
Moles of A Converted
Moles of desired product
= .95 x 10
= 9.5 lbmoles/hr
= 9.5 x 0.99
= 9.405 lbmoles/hr
Moles of B in feed = .5 lbmole/hr
Total B = 9.905 lbmoles/hr
Moles of A going to produce C
= 9.5-9.405 = .095 lbmoles/hr
Moles of C produced
= .095/2
= .0475 lbmoles/hr
Moles of H2 produced
= 1 x 9.405 + 2 x .0475
= 9.5 lbmoles/hr
Economic Potential
Economic Potential = Value of Products and Byproducts – Cost of Raw Materials
– Cost of Waste Disposal
Cost of transformation is neglected (i.e., manufacturing costs)
Choose process with large economic potential to investigate further.
Eliminate all processes with low or regular E.P.
Use U.P. as a guide in making decisions later.
Any cost << 1% of E.P. is negligible
Costs  1% of E.P. is significant.
Economic Potential of Cyclohexanone Process
E.P. = Value of B + value of H2Cost of A- Cost of disposing of C
= 9.905 lbmoles x $125.00/lbmole
hr
- 10 lbmole x $58.64/lbmole
hr
-.0475 lbmole x 198 lbs x 1 tm
x $145
hr
lbmole 2000 lbs
tm
= $1238.125 - $586.4 - $.68
= 4651.045/hr.
= $5.208 million/year
We have to keep cost of manufacturing less than $5.2 million to make a profit.
Any costs more than 50,000 will be considered as significant in our preliminary
analysis.
Recycle Structure
Basis: 100 mole of A entering Reactor
Conversion = 80%
Moles of A conversion
= 80 moles
Moles of B produced
= 80 x .99
= 79.2 moles
Moles of A going to C
= 80-79.2
= .8 moles
Moles of C produced
= .8/2 = 4 moles
Moles of the produced
= 79.2 + 4 x 2
= 80 moles
Moles of A in recycle
= 20 moles
Moles of fresh feed (A) required = 100-20 = 80 moles
Order of actual fresh feed
= 80 = 84.21 moles
.95
Order of B in feed to reactor
= 4.21 moles
Total moles of B produced
= 79.2 + 4.32
= 83.41 moles
Scaling up to Production Capacity (10 lbmole/hr of fresh feed)
Actual fresh feed
Scale down factor
Moles of B produced
Moles of H2 produced
Moles of C produced
Recycle
= 10 lbmole/hr
= (100 ) = 0.11875
84.21
= 83.41 x .11875
9.905 lbmoles/hr
= 80 x .11875
= 9.5 lbmoles/hr
= .4 x .11875
= .0475 lbmoles/hr
= 20 x .11875 = 2.375 lbmoles/hr
Feed to Separation Unit
9.905 lbmoles/hr B
9.5 lbmoles/hr H2
.0475 lbmoles/hr C
2.375 lbmoles/hr A
21.8275 lbmoles/hr
. Separation Synthesis