1 / 35

Chemical Processes

Chemical Processes. What is Engineering? July 25, 2007. Chemical Processes Outline. Motivations Reactions Separations Calculations using Conservation of Mass and Energy Distillation. Chemists Design reaction pathways to produce a chemical from raw materials

Download Presentation

Chemical Processes

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.


Presentation Transcript

  1. Chemical Processes What is Engineering? July 25, 2007

  2. Chemical Processes Outline • Motivations • Reactions • Separations • Calculations using Conservation of Mass and Energy • Distillation

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

  4. Why do we care about Chemical Engineering? Chemicals Are All Around Dyes Hydrogen Toothpaste Shampoo Gasoline Fertilizer Food additives Soap Decaffeinated Coffee Cosmetics Polymers Sugar Paint Pharmaceuticals

  5. 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

  6. Molecules that Chemicals Engineers work with • Small and Simple Helium (He) Ammonia (NH3) Hydrogen Flouride (HF) Trinitrotoluene (C6H2(NO2)3CH3) • Large and Complicated Insulin C257H383N65O77S6 • Large and Simple Polyvinyl Chloride (-CH2-CHCl-)n

  7. 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

  8. Chemical Reactions Reactor Products Raw Materials Byproducts Energy Catalysts Raw Materials Energy Catalysts

  9. Possible Problem with Exothermic Reactions L Energy Produced by reaction is proportional to reactor volume L3 Reactor Energy Removed is proportional to surface area L2 A+B->C Possible Scale up Problem Water Bath

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

  11. 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

  12. Mass and Energy Balances Balance Equation Input + generation – Output = Accumulation Control Volume

  13. 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

  14. Separations Calculation V moles 40% C2H5OH 100 moles 10% C2H5OH 90% H2O Magic Separating Machine 80 moles x % C2H5OH

  15. Separation Calculation V moles 40% C2H5OH Magic Separating Machine 100 moles 10% C2H5OH 90% H2O 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%

  16. Magic Separating Machine Separations: Distillation (Distillation Column) Equilibrium Stages

  17. 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?

  18. Distillation

  19. 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 • 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

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

  21. Distillation: McCabe-Thiele Calculation Calculation of theoretical number of equilibrium stages xD Operating Line xF xB

  22. Distillation: McCabe-Thiele

  23. Benefits Applicable for many liquid systems Technology is well developed High Throughput Distillation • Drawbacks • High heating and cooling costs • Azeotropes

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

  25. Distillation Batch distillation apparatus – only one equilibrium stage!

  26. 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

  27. Today’s Laboratory • Three Parts: • Energy Transfer • Chromatography • Batch Distillation • (One equilibrium stage)

  28. 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?

  29. 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.

  30. 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.

  31. 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).

  32. 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.

  33. 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)

  34. * xD Rectifying operating line * zF q-line y-int ~ 0.36 Stripping operating line * xB Nideal = 6 2/3

More Related