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ChE 131: Transport Processes. Outline 1.Class Policies 2.Introduction 3.Review.

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Presentation on theme: "ChE 131: Transport Processes. Outline 1.Class Policies 2.Introduction 3.Review."— Presentation transcript:

1 ChE 131: Transport Processes

2 Outline 1.Class Policies 2.Introduction 3.Review

3 Course Assessment 3 Long Exams60% Final Exam20% 3 Machine Problems15% Classwork5%

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 College if 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 1.Class Policies 2.Introduction 3.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?

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

12 Levels of Analysis MACROSCOPIC MICROSCOPIC MOLECULAR

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

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

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

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/m 2 -K) D – diameter of sphere (m) k – thermal conductivity of fluid (W/m-K) G – mass velocity of fluid (kg/m 2 -s) μ – viscosity (kg/m-s) c p – 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% H 2 O. 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% H 2 O. 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 H 2 O bal: 0.80(25) = 0.48M + 0C + 1.0H EVAPORATOR 25 MT/h 0.1 NaOH 0.1 NaCl 0.8 H 2 O M (mother liquor) 0.5 NaOH 0.02 NaCl 0.48 H 2 O H (water) 1.0 H 2 O 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 EVAPORATOR 25 MT/h 0.1 NaOH 0.1 NaCl 0.8 H 2 O M (mother liquor) 0.5 NaOH 0.02 NaCl 0.48 H 2 O H (water) 1.0 H 2 O C (crystals) 1.0 NaCl

26 Material Balance Dry gas containing 75% air and 25% NH 3 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 NH 3 in H 2 O 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% NH 3. If the entering gas flows through the empty bottom of the column at velocity (upward) of 1.5 ft/s, how many ft 3 of entering gas are treated per hour? How many pounds of NH 3 are absorbed per hour?

27 Material Balance Volume of gas entering = velocity  diameter of tower SCRUBBER D (dry gas) 0.75 air 0.25 NH 3 S (water + ammonia) x H 2 O y NH 3 G (scrubbed gas) 0.01 NH 3 (solute-free) W (water) 1.0 H 2 O

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

29 Material Balance Determine the number of moles of dry gas entering the scrubber. Assuming ideal gas behavior, SCRUBBER D (dry gas) 0.75 air 0.25 NH 3 S (water + ammonia) x H 2 O y NH 3 G (scrubbed gas) NH air W (water) 1.0 H 2 O

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 Air balance: 0.75(42.35) = G G = amount of dry gas = lbmol dry gas SCRUBBER D (dry gas) = lbmol 0.75 air 0.25 NH 3 S (water + ammonia) x H 2 O y NH 3 G (scrubbed gas) NH air W (water) 1.0 H 2 O

32 Material Balance NH 3 balance: 0.25(42.35) = (32.08) + xS xS = amount of NH 3 absorbed = lbmol NH 3 SCRUBBER D (dry gas) = lbmol 0.75 air 0.25 NH 3 S (water + ammonia) x H 2 O y NH 3 G (scrubbed gas) NH air W (water) 1.0 H 2 O

33 Material Balance Pounds of NH 3 absorbed: SCRUBBER D (dry gas) = lbmol 0.75 air 0.25 NH 3 S (water + ammonia) x H 2 O y NH 3 G (scrubbed gas) NH air W (water) 1.0 H 2 O

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

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