Chemical processes 500.101 Chemical engineering products, processes, and challenges...

Chemical processes500.101

Chemical engineering products, processes, and challenges

Commodities Molecules Nanostructures

Key cost speed to market function

Basis unit operations discovery properties

Chemical processes500.101 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 verygood introduction to polymers.


Polymers, e.g. polyethylene

is made up of many monomers: n22 CHCH


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

Chemical processes500.101 Nano applications of polymers

Organized block copolymer of PMMA (polymethylmethacrylate)and PS (polystyrene).

Spin casting in electric fieldproduces cylinders of PS embeddedin the PMMA which are orientedin the direction of the electric fieldPMMA cylinders are 14nm diameter,24nm apart.

PS can be dissolved withacetic acid to leave holes.

Use as a microscopic filter?


Cylindrical holes are electrochemicallyfilled with magnetic cobalt. Each cylindricalhole can then store 1 “bit” of information.

bit/cm = 1 / (2.4*10-7)

bit/cm2 = 1.7*1011

Computer application:


Genetic engineering: production of synthetic insulin

1) Extract a plasmid (a circular molecule of DNA) from thebacterium E-coli

2) Break the circle

3) Insert a section of human DNAcontaining the insulin-producinggene

4) Insert this engineered geneback into the E-coli bacterium

5) The E-coli and its offspringnow produce insulin


Chemical Engineering

Two strategies for obtaining chemical compounds and materials:

1) Create the desired compound from raw materialsvia one or more chemical reactions in a “reactor”

2) Isolate the compound where it exists in combinationwith other substances through a “separation process”


Reactors

raw materials energy

energy

product + contaminants

byproducts

catalyst

catalyst Reactor

fermenters in a brewerypharmaceuticals reactor


Separations

Based on differences between individual substances:

Boiling pointFreezing pointDensityVolatilitySurface TensionViscosityMolecular Complexity

SizeGeometryPolarization


Separations

Based on differences in the presence of other materials

SolubilityChemical reactivity


Separations: Garbage


Garbage separation (cont.)


Garbage separation (cont.)


Separation processes-- “Unit operations”:

A. Evaporation—the removal of a valueless component from a mixture throughvaporization. Mixture is usually a nonvolatile solid or liquid and a volatileliquid. E.g., evaporation of sea-water to obtain salt

B. Distillation—extraction by vaporization and condensation. Depends ondifferent boiling points of components. E.g., distillation of wine to producebrandy.

C. Gas absorption1. 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 extraction1. 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/solidmixture.

F. Chromatography—the process of separating fluid components by capillarytransport.


Bases for separation:

A. Differential boiling points, e.g., reducing alcohol content in wine-based sauceby cooking.

B. Differential freezing points, e.g., separating fat from broth by refrigerationC. 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 componentsD) Under the right conditions, one phase is enriched with a component as another

is depleted of that component.E) Component transfer

1) single stage2) multiple stage3) continuous


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

V1 V2

L0 L1

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): L0 + V2 = L1 + V1 = M


Assume three components: A = dye, B = oil, C = water

xA = mass fraction of A in stream LyA = 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 iswater (C) which is used to extract the dye from the oil. When V comes incontact with L, the dye redistributes itself between the V and L. L and Vare immiscible (i.e., two distinct liquid phases).


stage 1 V1 = oil + less dye V2 = oil + dye

L0 = water L1 = water + some dye

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

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

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

1

1

1

1

2

2

0

0

1111 A

A

A

A

A

A

A

A

y

yV

x

xL

y

yV

x

xL

Mass of dye contained in oiland coming from stage 2.

Mass of dye contained in waterand leaving stage 1.


Assume that the dye concentrations in the mixing stage comeinto equilibrium according to Henry’s Law that defines the relativeconcentration of dye in the oil and the water:

yA1 = H xA2 , where H depends on the substances A, B, C


S p e c i f i c p r o b l e m : 1 0 0 k g / h r o f d y e - c o n t a m i n a t e d o i l ( 1 % b y w e i g h t ) i sm i x e d w i t h 1 0 0 k g / h r o f w a t e r t o r e d u c e t h e d y e c o n c e n t r a t i o n i n t h e o i l .W h a t i s t h e r e s u l t i n g d y e c o n c e n t r a t i o n i n o i l a f t e r p a s s i n g t h r o u g h t h em i x i n g s t a g e i f d y e e q u i l i b r i u m i s a t t a i n e d a n d H e n r y ’ s c o n s t a n t H = 4 ?

S o l ’ n :

L ’ = 1 0 0 k g / h r V ’ = 1 0 0 ( 1 - . 0 1 ) = 9 9 k g / h r

x A 0 = 0 ( n o d y e i n i n c o m i n g w a t e r )

y A 2 = . 0 1 ( i n i t i a l c o n t a m i n a t i o n i n o i l )

y A 1 = 4 x A 1 ( e q u i l i b r i u m c o n c e n t r a t i o n o f d y e b e t w e e n o i l a n d w a t e r )

1 0 00

1 09 9

0 1

1 0 11 0 0

19 9

11

1

1

1

.

.

x

x

y

yA

A

A

A

1 1 0 02 5

1 2 59 9

10 0 81

1

1

11

.

..

y

y

y

yyA

A

A

AA


Single stage countercurrent centrifugal extractor(Rousselet-Robatel)


Counter-current heat exchangers in nature


Counter-current heat exchangersHow do they work?

limitedheat exchange

goodheat exchange

appendagebody

Tb-out

Tb-in

heat loss

exchanger body appendage

Tb-out

Tb-in

exchanger

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