Chemical Processes

What is Engineering?

July 25, 2007

Chemical Processes Outline

     Motivations Reactions Separations Calculations using Conservation of Mass and Energy Distillation

Chemists vs Chemical Engineers

Chemists Chemical Engineers  Design reaction pathways to produce a chemical from raw materials  Design a process to scale the chemist’s process to mass produce the product  Work in the laboratory setting to produce material on the gram to kilogram scale  Work in a chemical plant to produce material in the ton and beyond range

Why do we care about Chemical Engineering?

Chemicals Are All Around Dyes Toothpaste Shampoo Fertilizer Gasoline Decaffeinated Coffee Paint Soap Cosmetics Sugar Pharmaceuticals Hydrogen Food additives Polymers

If that isn’t reason enough

 In the United States  170 Major Chemical Companies  $400 Billion a year  Employs more than a million workers http://money.cnn.com/2006/02/13/pf/college/starting_salaries/index.htm

Molecules that Chemicals Engineers work with

 Small and Simple Helium (He) Ammonia (NH 3 ) Hydrogen Flouride (HF) Trinitrotoluene ( C 6 H 2 (NO 2 ) 3 CH 3 )  Large and Complicated Insulin C 257 H 383 N 65 O 77 S 6  Large and Simple Polyvinyl Chloride (-CH 2 -CHCl-) n

How to Produce Chemicals

 Two methods to obtain a desired chemical  Design a reactor to produce a chemical from raw materials  To isolate the compound that exists in combination with other substances through separation processes

Chemical Reactions

Reactor Raw Materials Energy Catalysts Products Raw Materials Byproducts Energy Catalysts

Possible Problem with Exothermic Reactions

L Reactor A+B->C Water Bath Energy Produced by reaction is proportional to reactor volume L 3 Energy Removed is proportional to surface area L 2 Possible Scale up Problem

Separations

Exploits Differences of Material Properties  Molecular Property  Separation Process       Boiling Point Freezing Point Particle size Affinity to a stationary phase Density Selective affinity to solid particles     Distillation Crystallization Filtration Chromatography   Centrifuge Adsorption

Separations: Unit Operations

Use 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

Mass and Energy Balances

Balance Equation

Input + generation – Output = Accumulation

Control Volume

Mass and Energy Balances

 For non-reacting systems Generation = 0  For systems operated at steady state Accumulation = 0 Mass and Energy Balances reduce to Input = Output

Separations Calculation

V moles 40% C 2 H 5 OH 100 moles 10% C 2 H 5 OH 90% H 2 O Magic Separating Machine 80 moles x % C 2 H 5 OH

Separation Calculation

V moles 40% C 2 H 5 OH 100 moles 10% C 2 H 5 OH 90% H 2 O Magic Separating Machine 80 moles x % C 2 H 5 OH Conservation of total Moles 100 – (V+80) = 0 V =20 Conservation of moles of C 2 H 5 OH 100*.1 – (.4*V+x*80) = 0 x = 2.5%

Separations: Distillation

(Distillation Column) Magic Separating Machine Equilibrium Stages

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 As mixture begins to boil, the vapor phase becomes richer in A than the liquid phase!

Condense vapor phase to get a mixture with a higher concentration of A!

As temperature increases, the concentration of B in the vapor phase increases.

What would be the composition of the vapor phase if the entire liquid mixture vaporized?

Distillation

Distillation: Equilibrium Stages

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

Distillation

Represent vapor liquid equilibrium data for more volatile component in an x-vs-y graph Pressure constant, but temperature is changing!

x B

Distillation: McCabe-Thiele Calculation

Calculation of theoretical number of equilibrium stages x D Operating Line x F

Distillation: McCabe-Thiele

Distillation

 Benefits    Applicable for many liquid systems Technology is well developed High Throughput  Drawbacks  High heating and cooling costs  Azeotropes

Azeotrope

Separations limitation Due to molecular interactions. Composition of vapor equal to composition of liquid mixture.

Distillation

Batch distillation apparatus – only one equilibrium stage!

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 of separations

Today’s Laboratory

 Three Parts:  Energy Transfer   Chromatography Batch Distillation  (One equilibrium stage)

Today’s Laboratory: Energy Transfer

Want efficient transfer and conversion of energy ($$) In lab, will be examining energy transfer in the form of heat: warming a pot of water with a hot plate – what is the efficiency of energy transport from electricity to the water?

Today’s Laboratory: Chromatography

  Separation technique that takes advantage of varying affinities of solutes for a given solvent traveling up a filter paper.

  Solutes: colored dyes Solvents: water, methanol, 2-propanol Measure the distance traveled by the solutes and solvents!

**Methanol and 2-propanol are poisons! Wear safety goggles, do not ingest or inhale and rinse skin immediately if spilled.

Today’s Laboratory: Distillation

  Using distillation to separate a liquid mixture of ethanol and water  Ethanol is the more volatile material (it will boil first) Take samples of distillate with time to determine the concentration of ethanol in the mixture!

**Ethanol is a poison! Wear safety goggles, do not ingest or inhale and rinse skin immediately if spilled.

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).

V 1 = oil + less dye V 2 = oil + dye stage 1 L 0 = water L 1 = water + some dye Oil flow = V(1 - yA) = V’ = constant Water flow = L(1 - xA) = L’ = constant Then, for mass balance of the A component: 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 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 dye between oil and water) 100   0    99   .

01 .

   100 

x

1 

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1

x A

1  99  1 

y A

1

y A

1  1  100  .

25

y A

1 .

y A

1  99  1 

y A

1

y A

1 

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008

Rectifying operating line

*

q-line

z F

y-int ~ 0.36

* x B

Stripping operating line N ideal = 6 2/3

* x D