Key Basis

Chemical engineering products, processes, and challenges

Commodities

cost unit operations

Molecules

speed to market discovery

Nanostructures

function properties

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.

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.

Polymers

, e.g. polyethylene is made up of many monomers:

  CH 2  CH 2   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/cm 2 = 1.7*10 11

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 Reactor energy catalyst product + contaminants byproducts pharmaceuticals reactor 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 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.

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 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) 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 V 1 V 2 stage 1 L 0 L 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

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 A1 + 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).

V 1 = oil + less dye V 2 = oil + dye stage 1 L 0 = water L 1 = water + some dye Oil flow = V(1 - y A ) = V’ = constant Water flow = L(1 - x A ) = L’ = constant Then, for mass balance of the A component:

x A

0

L

'  1 

x A

0 

V y A

2 '  1 

y A

2 

L

' 

x A

1 1 

x A

1 

V

'  1 

y A

1

y A

1  Another assumption: dye concentrations y A1 , x A1 come into equilibrium according to Henry’s Law: y A1 = H x A1 , 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 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    99   .

01 .

   100  1

x A

1 

x A

1  99 

y A

1 1 

y A

1  1  100  .

25

y A

1 .

y A

1  99  1 

y A

1

y A

1 

y A

1  .

008

Counter-current heat exchangers in nature

Counter-current heat exchangers How do they work?

T b-out heat loss T b-in body exchanger appendage

limited heat exchange

T b-out T b-in body exchanger appendage

good heat exchange