Chemical processes 500.101

whooshdisguisingBiotechnologie

14 déc. 2012 (il y a 8 années et 10 mois)

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Chemical processes

500.101







Chemical engineering products, processes, and challenges



Commodities

Molecules


Nanostructures


Key



cost


speed to market


function


Basis


unit operations

discovery



properties

Chemical processes

500.101

A commodity: TiO
2
(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*10
3

*4*10
6

$/yr =
$80 million/yr.

Chemical processes

500.101

Molecules

Small and simple:


ammonia (NH
3
)




sulfuric acid (H
2
SO
4
)




ethylene (C
2
H
4
)




sugar (C
12
H
22
O
11
)

Large and complex: insulin

C
257
H
383
N
65
O
77
S
6

Large and simple (polymers): polyethylene[
-
CH
2
-
CH
2
]
n


See
www.psrc.usm.edu/macrog

for a very

good introduction to polymers.

Chemical processes

500.101

Polymers
, e.g. polyethylene

is made up of many monomers:



n
2
2
CH
CH



Chemical processes

500.101

Copolymers


are made up of two kinds of monomers, say A and B

Chemical processes

500.101

SBS rubber (tires, shoe soles)

The polystyrene is tough;


the polybutadiene is rubbery

Chemical processes

500.101

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?

Chemical processes

500.101

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/cm
2

= 1.7*10
11

Computer application:

Chemical processes

500.101

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 processes

500.101

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”

Chemical processes

500.101

Reactors

raw materials

energy

energy

product + contaminants

byproducts

catalyst

catalyst

Reactor

fermenters in a brewery

pharmaceuticals reactor

Chemical processes

500.101

Separations

Based on differences between individual substances:



Boiling point


Freezing point


Density


Volatility


Surface Tension


Viscosity


Molecular Complexity



Size



Geometry



Polarization

Chemical processes

500.101

Separations

Based on differences in the presence of other materials



Solubility


Chemical reactivity

Chemical processes

500.101

Separations: Garbage

Chemical processes

500.101

Garbage separation (cont.)

Chemical processes

500.101

Garbage separation (cont.)

Chemical processes

500.101

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.
Chemical processes

500.101

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 liquid-
liquid 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.
Chemical processes

500.101

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 he
avier solids from liquids with
centrifugation.
D. Differential anything. . .
Unit operations—more details:
A) The transfer of energy and/or material through physical (sometimes physical-
chemical) 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) con
tinuous
Chemical processes

500.101

Single
-
stage

counter
-
current 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


V
1
V
2
L
0
L
1
stage 1



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):





L
0

+ V
2

= L
1

+ V
1

= M

Chemical processes

500.101

Assume three components: A = dye, B = oil, C = water
x
A
= mass fraction of A in stream L
y
A
= mass fraction of A in stream V
(e.g., L
0
x
A0
= mass of component A in stream L
0
)
Component mass balance (mass/time):
L
0
x
A0
+ V
2
y
A2
= L
1
x
A
1
+ V
1
y
A1
= M
x
AM
L
0
x
C0
+ V
2
y
C2
= L
1
x
C1
+ V
1
y
C1
= M
x
CM
(equation for B not necessary because
x
A
+
x
B
+
x
C
= 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).
Chemical processes

500.101

stage 1



V
1

=
oil + less dye

V
2

=
oil + dye

L
0

=
water

L
1
=
water + some dye

Oil flow = V(1
-
y
A
) = V
′ = constant (conservation of oil)

Water flow = L(1
-
x
A
) = L
′ = constant (conservation of water)

Then, for mass balance of the A component (dye)












































1
1
1
1
2
2
0
0
1
1
1
1
A
A
A
A
A
A
A
A
y
y
V
x
x
L
y
y
V
x
x
L
Mass of dye contained in oil

and coming from stage 2.

Mass of dye contained in water

and leaving stage 1.

Chemical processes

500.101

Assume that the dye concentrations in the mixing stage come

into equilibrium according to Henry’s Law that defines the relative

concentration of dye in the oil and the water:


y
A1

= H x
A2
, where H depends on the substances A, B, C

Chemical processes

500.101

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 V’ = 100 ( 1 - .01) = 99 kg/hr
x
A0
= 0 (no dye in incoming water)
y
A2
= .01 (initial contamination in oil)
y
A1
= 4 x
A1
(equilibrium concentration of dye between oil and water)
100
0
1
0
99
01
1
01
100
1
99
1
1
1
1
1
































.
.
x
x
y
y
A
A
A
A
1
100
25
1
25
99
1
008
1
1
1
1
1


















.
.
.
y
y
y
y
y
A
A
A
A
A
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 V’ = 100 ( 1 - .01) = 99 kg/hr
x
A0
= 0 (no dye in incoming water)
y
A2
= .01 (initial contamination in oil)
y
A1
= 4 x
A1
(equilibrium concentration of dye between oil and water)
100
0
1
0
99
01
1
01
100
1
99
1
1
1
1
1
































.
.
x
x
y
y
A
A
A
A
1
100
25
1
25
99
1
008
1
1
1
1
1


















.
.
.
y
y
y
y
y
A
A
A
A
A
Chemical processes

500.101

Single stage countercurrent centrifugal extractor

(Rousselet
-
Robatel)

Chemical processes

500.101

Counter
-
current heat exchangers in nature

Chemical processes

500.101

Counter
-
current heat exchangers

How do they work?

limited

heat exchange

good

heat exchange

appendage

body

T
b
-
out

T
b
-
in

heat loss

exchanger

body

appendage

T
b
-
out

T
b
-
in

exchanger