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ChE 131: Transport Processes

ChE 131: Transport Processes. Outline. Class Policies Introduction Review. Course Assessment. 3 Long Exams 60% Final Exam 20% 3 Machine Problems 15% Classwork 5%. Policies to Remember. Submit 12 sheets of colored pad paper at least the day before an exam.

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ChE 131: Transport Processes

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  1. ChE 131: Transport Processes

  2. Outline • Class Policies • Introduction • Review

  3. Course Assessment 3 Long Exams 60% Final Exam 20% 3 Machine Problems 15% Classwork 5%

  4. Policies to Remember Submit 12 sheets of colored pad paper at least the day before an exam. Get an official excuse slip from the Collegeif you miss an exam and you have a valid excuse. No exemptions will be given for the final exam.

  5. Policies to Remember Quizzes may be given from time to time. All quizzes shall be written in bluebooks. No makeup shall be given to missed quizzes.

  6. Outline • Class Policies • Introduction • Review

  7. Transport Phenomena What exactly are "transport phenomena"? Transport phenomena are really just a fancy way that Chemical Engineers group together three areas of study that have certain ideas in common. These three areas of study are: • Fluid mechanics • Heat transfer • Mass transfer Transport processes

  8. Transport Processes Momentum Transport – transfer of momentum which occurs in moving media (fluid flow, sedimentation, mixing, filtration, etc.) Heat Transport – transfer of energy from one region to another (drying, evaporation, distillation) Mass Transport – transfer of mass of various chemical species from one phase to another distinct phase (distillation, absorption, adsorption, etc.)

  9. Why Study Transport Phenomena?

  10. Why Study Transport Phenomena?

  11. Why Study Transport Phenomena? Chemical Engineering Thermodynamics Transport Phenomena Materials Science PROCESS EQUIPMENT DESIGN Chemical Reaction Kinetics Process Economics

  12. Levels of Analysis MACROSCOPIC MICROSCOPIC MOLECULAR

  13. Levels of Analysis MACROSCOPIC Use of macroscopic balances Overall assessment of a system MICROSCOPIC MOLECULAR

  14. Levels of Analysis MACROSCOPIC Small region/volume element is selected Use of equations of change Velocity, temperature, pressure and concentration profiles are determined MICROSCOPIC MOLECULAR

  15. Levels of Analysis MACROSCOPIC Molecular structure and intermolecular forces become significant Complex molecules, extreme T and P, chemically reacting systems MICROSCOPIC MOLECULAR

  16. Review LET’S REVIEW!!!

  17. Dimensional Analysis Check the dimensional consistency of the following empirical equation for heat transfer between a flowing fluid and the surface of a sphere: h – heat transfer coefficient (W/m2-K) D – diameter of sphere (m) k – thermal conductivity of fluid (W/m-K) G – mass velocity of fluid (kg/m2-s) μ – viscosity (kg/m-s) cp – heat capacity (J/kg-K)

  18. Dimensional Analysis We use the following convention: Energy unit – E Mass unit – M Length unit – L Time unit – t Temperature unit – T

  19. Dimensional Analysis For the heat transfer coefficient: For thermal conductivity: For diameter: For viscosity:

  20. Dimensional Analysis For mass velocity: For heat capacity: Combining:

  21. Dimensional Analysis Simplifying:

  22. Dimensional Analysis Simplifying:

  23. Material Balance An evaporator is fed continuously with 25 metric tons/h of a solution consisting of 10% NaOH, 10% NaCl, and 80% H2O. During evaporation, water is boiled off, and salt precipitates as crystals, which are settled and removed from the remaining liquor. The concentrated liquor leaving the evaporator contains 50% NaOH, 2% NaCl, and 48% H2O. Calculate the MT of water evaporated per hour, the MT of salt precipitated per hour, and MT of liquor produced per hour.

  24. Material Balance NaOH bal: 0.10(25) = 0.5M + 0C + 0H NaCl bal: 0.10(25) = 0.02M + 1.0C + 0H H2O bal: 0.80(25) = 0.48M + 0C + 1.0H 25 MT/h 0.1 NaOH 0.1 NaCl 0.8 H2O M (mother liquor) 0.5 NaOH 0.02 NaCl 0.48 H2O EVAPORATOR H (water) 1.0 H2O C (crystals) 1.0 NaCl

  25. Material Balance H = water evaporated per hour = 17.6 MT/h C = salt precipitated per hour = 2.4 MT/h M = liquor produced per hour = 5 MT/h 25 MT/h 0.1 NaOH 0.1 NaCl 0.8 H2O M (mother liquor) 0.5 NaOH 0.02 NaCl 0.48 H2O EVAPORATOR H (water) 1.0 H2O C (crystals) 1.0 NaCl

  26. Material Balance Dry gas containing 75% air and 25% NH3 vapor enters the bottom of a cylindrical packed absorption tower that is 2 ft in diameter. Nozzles in the top of the tower distribute water over the packing. A solution of NH3 in H2O is drawn at the bottom of the column, and scrubbed gas leaves the top. The gas enters at 80°F and 760 mm Hg. It leaves at 60°F and 730 mm Hg. The leaving gas contains, on the solute-free basis, 1.0% NH3. If the entering gas flows through the empty bottom of the column at velocity (upward) of 1.5 ft/s, how many ft3 of entering gas are treated per hour? How many pounds of NH3 are absorbed per hour?

  27. Material Balance G (scrubbed gas) 0.01 NH3 (solute-free) W (water) 1.0 H2O Volume of gas entering = velocity diameter of tower SCRUBBER D (dry gas) 0.75 air 0.25 NH3 S (water + ammonia) x H2O y NH3

  28. Material Balance Convert solute-free basis percentage to mass fraction: We now rewrite our diagram:

  29. Material Balance G (scrubbed gas) 0.0099 NH3 0.9901 air W (water) 1.0 H2O Determine the number of moles of dry gas entering the scrubber. Assuming ideal gas behavior, SCRUBBER D (dry gas) 0.75 air 0.25 NH3 S (water + ammonia) x H2O y NH3

  30. Material Balance Determine the number of moles of dry gas entering the scrubber. Assuming ideal gas behavior and a basis of 1 hour:

  31. Material Balance G (scrubbed gas) 0.0099 NH3 0.9901 air W (water) 1.0 H2O Air balance: 0.75(42.35) = 0.9901G G = amount of dry gas = 32.08 lbmol dry gas SCRUBBER D (dry gas) = 42.35 lbmol 0.75 air 0.25 NH3 S (water + ammonia) x H2O y NH3

  32. Material Balance G (scrubbed gas) 0.0099 NH3 0.9901 air W (water) 1.0 H2O NH3 balance: 0.25(42.35) = 0.0099(32.08) + xS xS = amount of NH3 absorbed = 10.27 lbmol NH3 SCRUBBER D (dry gas) = 42.35 lbmol 0.75 air 0.25 NH3 S (water + ammonia) x H2O y NH3

  33. Material Balance G (scrubbed gas) 0.0099 NH3 0.9901 air W (water) 1.0 H2O Pounds of NH3 absorbed: SCRUBBER D (dry gas) = 42.35 lbmol 0.75 air 0.25 NH3 S (water + ammonia) x H2O y NH3

  34. Energy Balance Air is flowing steadily through a horizontal heated tube. The air enters at 40°F and at a velocity of 50 ft/s. It leaves the tube at 140°F and 75 ft/s. The average specific heat of air is 0.24 Btu/lb-°F. How many Btu’s per pound of air are transferred through the wall of the tube?

  35. Energy Balance Energy Balance:

  36. Energy Balance Energy Balance:

  37. Differential Equation Solve the following differential equation:

  38. Differential Equation This equation follows the form: whose solution is:

  39. Differential Equation

  40. Differential Equation

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