1. ChE 131: Transport Processes

2. Outline
Class Policies
Introduction
Review

3. 3 Long Exams 60% Final Exam 20% 3 Machine Problems 15% Classwork 5%
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 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
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 MICROSCOPIC MOLECULAR
Use of macroscopic balances
Overall assessment of a system
MICROSCOPIC
MOLECULAR

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
MICROSCOPIC
MOLECULAR

15. Levels of Analysis MACROSCOPIC MICROSCOPIC MOLECULAR
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
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
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. Material Balance
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
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

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

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

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

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)
NH3
air
W (water)
1.0 H2O
SCRUBBER
D (dry gas) = lbmol
0.75 air
0.25 NH3
S (water + ammonia)
x H2O
y NH3
Air balance: 0.75(42.35) = G G = amount of dry gas = lbmol dry gas

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

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

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