Chemical engineering products, processes, and challenges
Commodities
Key
Basis
cost
unit operations
Molecules
Nanostructures
speed to market
function
discovery
properties
A commodity: TiO2 (titanium oxide)
Extremely white, opaque, edible, dirt resistant. Used in paper, food,
cosmetics, paint, textiles, plastics. World consumption: 4 million tons/yr.
Cost: $2,000/ton. Total world value = $8 billion/yr.
A 1% increase in production efficiency = 0.01*2*103 *4*106 $/yr =
$80 million/yr.
Molecules
Small and simple:
ammonia (NH3)
sulfuric acid (H2SO4)
ethylene (C2H4)
sugar (C12H22O11)
Large and complex:
insulin C257H383N65O77S6
Large and simple (polymers):
polyethylene[-CH2-CH2]n
See www.psrc.usm.edu/macrog for a very
good introduction to polymers.
Polymers, e.g. polyethylene
is made up of many monomers:
 CH2  CH2 n
Copolymers
are made up of two kinds of monomers, say A and B
SBS rubber (tires, shoe soles)
The polystyrene is tough;
the polybutadiene is rubbery
Nano applications of polymers
Organized block copolymer of PMMA (polymethylmethacrylate)
and PS (polystyrene).
Spin casting in electric field
produces cylinders of PS embedded
in the PMMA which are oriented
in the direction of the electric field
PMMA cylinders are 14nm diameter,
24nm apart.
PS can be dissolved with
acetic acid to leave holes.
Use as a microscopic filter?
Computer application:
Cylindrical holes are electrochemically
filled with magnetic cobalt. Each cylindrical
hole can then store 1 “bit” of information.
bit/cm = 1 / (2.4*10-7)
bit/cm2 = 1.7*1011
Genetic engineering: production of synthetic insulin
1) Extract a plasmid (a circular molecule of DNA) from the
bacterium E-coli
2) Break the circle
3) Insert a section of human DNA
containing the insulin-producing
gene
4) Insert this engineered gene
back into the E-coli bacterium
5) The E-coli and its offspring
now produce insulin
Chemical Engineering
Two strategies for obtaining chemical compounds and materials:
1) Create the desired compound from raw materials
via one or more chemical reactions in a “reactor”
2) Isolate the compound where it exists in combination
with other substances through a “separation process”
Reactors
raw materials
energy
catalyst
pharmaceuticals reactor
Reactor
energy
catalyst
product + contaminants
byproducts
fermenters in a brewery
Separations
Based on differences between individual substances:
Boiling point
Freezing point
Density
Volatility
Surface Tension
Viscosity
Molecular Complexity
Size
Geometry
Polarization
Separations
Based on differences in the presence of other materials
Solubility
Chemical reactivity
Separations:
Garbage
Garbage separation (cont.)
Garbage separation (cont.)
Separation processes-- “Unit operations”:
A. Evaporation—the removal of a valueless component from a mixture through
vaporization. Mixture is usually a nonvolatile solid or liquid and a volatile
liquid. E.g., evaporation of sea-water to obtain salt
B. Distillation—extraction by vaporization and condensation. Depends on
different boiling points of components. E.g., distillation of wine to produce
brandy.
C. Gas absorption
1. gas absorption—the transfer of a soluble component of a gas mixture to
a liquid, e.g. bubbler in a fish tank to oxygenate the water.
2. desorption or stripping—the transfer of a volatile component from a
liquid to a gas.
D. Solvent extraction
1. liquid-liquid extraction—requires two immiscible phases—an “extract”
layer and a “raffinate” layer. Solute partitions between two phases.
2. washing—the removal of soluble substance and impurities
mechanically holding on to insoluble solids.
3. precipitative extraction—a liquid solution can be split into a liquidliquid or liquid-solid by adding a third substance.
4. leaching—the extraction of a component in solid phase by a liquid
solvent—e.g., making coffee.
E. Filtration—the process of removing a solid from a liquid/solid or gas/solid
mixture.
F. Chromatography—the process of separating fluid components by capillary
transport.
Bases for separation:
A. Differential boiling points, e.g., reducing alcohol content in wine-based sauce
by cooking.
B. Differential freezing points, e.g., separating fat from broth by refrigeration
C. Differential densities, e.g., separating heavier solids from liquids with
centrifugation.
D. Differential anything. . .
Unit operations—more details:
A) The transfer of energy and/or material through physical (sometimes physicalchemical) means.
B) Involves multiple phases: gas-liquid, liquid-liquid, solid-gas, etc.
C) Phases consist of mixtures of components
D) Under the right conditions, one phase is enriched with a component as another
is depleted of that component.
E) Component transfer
1) single stage
2) multiple stage
3) continuous
Counter-current processes
Single-stage process
A) Phases are brought into close contact
B) Components redistribute between phases to equilibrium concentrations
C) Phases are separated carrying new component concentrations
D) Analysis based on mass balance
V2
V1
stage 1
L0
L1
L is a stream of one phase; V is a stream of another phase.
Use subscripts to identify stage of origination (for multiple stage problems)
Total mass balance (mass/time):
L0 + V2 = L1 + V1 = M
Assume three components: A = dye, B = oil, C = water
xA = mass fraction of A in stream L
yA = mass fraction of A in stream V
(e.g., L0 xA0 = mass of component A in stream L0 )
Component mass balance (mass/time):
L0 xA0 + V2 yA2 = L1 xA1 + V1 yA1 = M xAM
L0 xC0 + V2 yC2 = L1 xC1 + V1 yC1 = M xCM
(equation for B not necessary because xA + xB + xC = 1)
Suppose the following: V is oil (B) contaminated with dye (A). L is
water (C) which is used to extract the dye from the oil. When V comes in
contact with L, the dye redistributes itself between the V and L. L and V
are immiscible (i.e., two distinct liquid phases).
V2 = oil + dye
V1 = oil + less dye
stage 1
L0 = water
Oil flow
L1 = water + some dye
= V(1 - yA) = V’ = constant
Water flow = L(1 - xA) = L’ = constant
Then, for mass balance of the A component:
 x

 y A2 
 x

 y A1 
  L'  A1   V ' 

L'  A0   V ' 
 1  x A0 
 1  y A2 
 1  x A1 
 1  y A1 
Another assumption: dye concentrations yA1, xA1 come into equilibrium
according to Henry’s Law: yA1 = H xA1 , where H depends on the
substances A, B, C.
Specific problem: 100kg/hr of dye-contaminated oil (1% by weight) is
mixed with 100 kg/hr of water to reduce the dye concentration in the oil.
What is the resulting dye concentration in oil after passing through the
mixing stage if dye equilibrium is attained and Henry’s constant H = 4?
Sol’n:
L’ = 100kg/hr
xA0 = 0
V’ = 100 ( 1 - .01) = 99 kg/hr
(no dye in incoming water)
yA2 = .01 (initial contamination in oil)
yA1 = 4 xA1 (equilibrium concentration of dye between oil and water)
 x A1 
 y A1 
 0 
 .01 
  99

100
  99
  100
 1  0
 1.01
 1  x A1 
 1  y A1 
 .25 y A1 
 y A1 
  99

1  100
 1.25 y A1 
 1  y A1 

y A1  .008
Counter-current heat exchangers in nature
Counter-current heat exchangers
How do they work?
Tb-out
Tb-out
heat loss
Tb-in
Tb-in
body
exchanger
limited
heat exchange
appendage
body
exchanger
appendage
good
heat exchange