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7/30/2019 Chemical Processes2 Jrw
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Chemical Processes
What is Engineering?
July 25, 2007
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Chemical Processes Outline
Motivations
Reactions Separations
Calculations using Conservation of
Mass and Energy Distillation
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Chemists vs Chemical Engineers
Chemists
Design reaction
pathways to producea chemical from rawmaterials
Work in the laboratorysetting to producematerial on the gramto kilogram scale
Chemical Engineers
Design a process toscale the chemistsprocess to massproduce the product
Work in a chemicalplant to producematerial in the ton andbeyond range
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Why do we care about Chemical
Engineering?
Shampoo
Soap
Toothpaste
Dyes
Gasoline
DecaffeinatedCoffee
Sugar
Chemicals Are All Around
Cosmetics
Paint
Foodadditives
Hydrogen
Fertilizer
Polymers
Pharmaceuticals
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If that isnt reason enough
In the United States
170 Major ChemicalCompanies
$400 Billion a year
Employs more than amillion workers
http://money.cnn.com/2006/02/13/pf/college/starting_salaries/index.htm
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Molecules that Chemicals Engineers
work with
Small and Simple
Helium (He)
Ammonia (NH3)
Hydrogen Flouride (HF)Trinitrotoluene (C6H2(NO2)3CH3)
Large and ComplicatedInsulin C257H383N65O77S6
Large and SimplePolyvinyl Chloride (-CH2-CHCl-)n
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How to Produce Chemicals
Two methods to obtain a desiredchemical
Design a reactor to produce a chemicalfrom raw materials
To isolate the compound that exists incombination with other substancesthrough separation processes
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Chemical Reactions
Raw Materials
Energy
Catalysts
ReactorProducts
Raw Materials
Byproducts
EnergyCatalysts
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Possible Problem with Exothermic
Reactions
Reactor
Water Bath
A+B->C
Energy Produced byreaction is proportional to
reactor volume L3
Energy Removed isproportional to surfacearea L2
L
Possible Scale up Problem
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Separations
Molecular Property
Boiling Point
Freezing Point
Particle size
Affinity to astationary phase
Density
Selective affinity tosolid particles
Separation Process
Distillation
Crystallization
Filtration
Chromatography
Centrifuge
Adsorption
Exploits Differences of Material Properties
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Separations: Unit OperationsUse separation processes to:
Purify raw materials Purify products Purify and separate unreacted feed.
Most common types: Distillation
Flash distillation Batch distillation Column distillation
Absorption
Stripping
Extraction
Chromatography
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Mass and Energy Balances
Balance Equation
Input + generation Output =
Accumulation
ControlVolume
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Mass and Energy Balances
For non-reacting systemsGeneration = 0
For systems operated at steadystate Accumulation = 0
Mass and Energy Balances reduce to
Input = Output
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Separations Calculation
Magic
Separating
Machine
100 moles
10% C2H5OH
90% H2O
V moles
40% C2H5OH
80 moles
x % C2H5OH
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Separation Calculation
Magic
SeparatingMachine
100 moles
10% C2H5OH
90% H2O
V moles
40% C2H5OH
80 moles
x % C2H5OH
Conservation of total Moles 100 (V+80) = 0
V =20
Conservation of moles of C2H5OH 100*.1 (.4*V+x*80) = 0
x = 2.5%
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Separations: Distillation
Magic
Separating
Machine
Equilibrium Stages
(Distillation Column)
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Distillation
Separates liquids based on differences in volatility!
Consider a liquid mixture of A and B:
Boiling point of A: 70 C
Boiling point of B: 100 C
What would be the composition of the vapor phase if the entireliquid mixture vaporized?
As mixture begins to boil, the vapor phase becomesricher in A than the liquid phase!
As temperature increases, the concentration of B inthe vapor phase increases.
Condense vapor phase to get a mixture with a higherconcentration of A!
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Distillation
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V1 V2
L0 L1
stage 1
Distillation: Equilibrium Stages
A) Phases are brought into close contactB) Components redistribute between phases toequilibrium concentrationsC) Phases are separated carrying new componentconcentrations
D) Analysis based on mass balance
L is a stream of one phase; V is a stream of another phase. Use subscripts to identify stage of origination (for multiplestage problems) Total mass balance (mass/time): L0 + V2 = L1 + V1 = M
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Distillation
Represent vapor liquid equilibrium data for more volatilecomponent in an x-vs-y graph
Pressure constant, but temperature is changing!
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Distillation: McCabe-Thiele Calculation
Operating Line
Calculation of theoretical number of equilibrium stages
xD
xF
xB
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Distillation: McCabe-Thiele
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Distillation
Benefits
Applicable formany liquidsystems
Technology is welldeveloped
High Throughput
Drawbacks
High heating andcooling costs
Azeotropes
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Azeotrope
Separations limitation
Due to molecular interactions. Composition of vaporequal to composition of liquid mixture.
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Distillation
Batch distillationapparatus only oneequilibrium stage!
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Conclusions
Chemicals are produced by
reactions or separations The driving force for separations are
property differences
Mass and Energy are Conserved
Distillation is the workhorse ofseparations
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Todays Laboratory
Three Parts:
Energy Transfer
Chromatography
Batch Distillation
(One equilibrium stage)
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Todays Laboratory: Energy Transfer
Want efficient transfer and conversionof energy ($$)
In lab, will be examining energytransfer in the form of heat:warming a pot of water with a hotplate what is the efficiency ofenergy transport from electricity tothe water?
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Todays Laboratory: Chromatography
Separation technique that takesadvantage of varying affinities of solutesfor a given solvent traveling up a filterpaper. Solutes: colored dyes Solvents: water, methanol, 2-propanol
Measure the distance traveled by thesolutes and solvents!
**Methanol and 2-propanol are poisons! Wearsafety goggles, do not ingest or inhale andrinse skin immediately if spilled.
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Todays Laboratory: Distillation
Using distillation to separate a liquidmixture of ethanol and water Ethanol is the more volatile material (it will boil
first)
Take samples of distillate with time todetermine the concentration of ethanol inthe mixture!
**Ethanol is a poison! Wear safety goggles, donot ingest or inhale and rinse skin immediatelyif spilled.
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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 xAML0 xC0 + V2 yC2 = L1 xC1 + V1 yC1
= M xCM
(equation for B not necessary becausexA + 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 theoil. When V comes in contact with L, the dye redistributesitself between the V and L. L and V are immiscible (i.e., two
distinct liquid phases).
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V1 = oil + less dye V2 = oil + dye
L0 = water L1 = water + some dye
stage 1
Oil flow = V(1 - yA) = V = constantWater flow = L(1 - xA) = L = constantThen, for mass balance of the A component:Another assumption: dye concentrations yA1, xA1come into equilibrium according to Henrys Law: yA1 =H xA1 , where H depends on the substances A, B, C.
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1000
1 099
01
1 01100
199
1
1
1
1
1
.
.
x
x
y
y
A
A
A
A
1 100
25
1 25 99 1 008
1
1
1
11
.
. .
y
y
y
y y
A
A
A
AA
Specific problem: 100kg/hr of dye-contaminated oil(1% by weight) is mixed with 100 kg/hr of water toreduce the dye concentration in the oil. What is theresulting dye concentration in oil after passing throughthe mixing stage if dye equilibrium is attained and
Henrys constant H = 4?Soln:
L = 100kg/hr V = 100 ( 1 - .01) =99 kg/hr
xA0 = 0 (no dye in incoming water)yA2 = .01 (initial contamination in oil)
yA1 = 4 xA1 (equilibrium concentration of dyebetween oil and water)
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q-line
zF*
*
*
xD
xB
y-int ~ 0.36
Rectifying operating line
Stripping operating line
Nideal = 6 2/3