1. ChE 149: TRANSPORT PHENOMENA
Engr. Denise Ester O. Santiago
Engr. Michael Vincent O. Laurio
2nd Semester A.Y – 2012

2. METHOD OF ANALYSIS

3. Objective
To have a systematic approach in analyzing transfer systems involving molecular transport

4. Basic balance equation
Rate of Input+ Rate of generation=
Rate of output + Rate of accumulation

5. Input, Output
flow of transferent property across the boundary of the system, the flux is drawn into and out of the element in a direction parallel to the chosen arbitrary axis

6. Generation
source or sink within the system, these phenomena must occur uniformly throughout the physical
because of uniformity, the generation term can be expressed on a per volume basis

7. Accumulation
The term refers only to accumulation with time, it accounts for the decrease or increase of the conserved quantity within the system as time increases.

8. Step-by-step procedure for differential balance technique
1. A mathematical model is developed to the physical system
determine what is being transported
decide on the direction of transport
choose the location and orientation of an appropriate coordinate system
construct differential volume element (or a small finite volume element) by choosing a regular geometry appropriate to the physical problem. (Sketch)
draw fluxes into and out of the differential element. Note that the transfer surface is always normal or perpendicular to the flux.
determine whether there is generation or consumption term. Is the generation term constant or varying?
determine whether there is accumulation term. Is it a steady or unsteady-state problem?

9. Step-by-step procedure for differential balance technique
2. Differential balance written for the pertinent transferent property
Generally, it is necessary to obtain the balance for an infinitesimal volume element because concentration and mechanism may vary within the volume of the system.
To obtain the differential equation, the limits of the balance on a small finite volume are taken as the dimensions of the volume element approach zero.
If the transferent property is a function of more than one independent variable, then a partial differential equation is formulated.
Be careful of signs
For some problems with varying transfer area, the variation of the transfer area along the transport direction of flux is expressed in mathematical form

10. Step-by-step procedure for differential balance technique
3. Use the phenomenological equations
the flux terms in the differential equations can be substituted by concentration gradient expression
note that in some problems, it may be convenient to integrate before substituting the phenomenological equations

11. Step-by-step procedure for differential balance technique
4. The differential equations are integrated using the appropriate boundary conditions
5. Assess the adequacy of the model by comparing its results to empirical (experimental) data.

12. Differential volume based on geometry
Flat slab
Cylindrical pellet
Cylindrical shell
Sphere

13. SIMPLE TRANSFER

14. SIMPLE TRANSFER Transfer without internal generation
Assuming a steady state system
In general, input and output may be by both molecular transfer and fluid flow.

15. For flat-slab geometry
But since we are considering molecular transport only. -x y x z Δx

16. Step 1 Assume: SS molecular transport in the x- direction
Simple transfer ( no internal generation)

17. Step 2
Balance equation

18. Step 3 Balance is taken on a finite volume element
Rearranging and dividing by Δx gives;
means: rate is constant for infinitesimal thickness dx

19. Example
A chimney brick is shown in the sketch. Heat is transferred from the 3 in. end to the 5 in end. No heat is lost through the other sides, because these sides adjoin other bricks. The 3 in. side is at 800◦F because it is exposed to hot gases inside the chimney. The 5 in. side is exposed to the atmosphere and is at 200◦F. The thermal conductivity of the brick is 0.4 Btu/hr-ft-◦F. Calculate the heat loss per hour through the brick.
3in
4in
6in
5in

20. Example
Determine the mass transport rate on the conical section shown in the figure. The concentration of CO₂ in air is 30 mol% at the 10 cm. opening and 3 mol% at the 5 cm. opening. For this mixture, the diffusivity of CO₂ in air is cm2/s. The system is at 1 atm and and 25oC at every point. The section is 30 cm long.
30 cm
10 cm
5 cm

21. ExAmplE
A spherical fishbowl which is open to the atmosphere is half full of water. The air at the mouth of the bowl is dry, the total pressure is 1 atm and the temperature is at 75oF. The radius of the fishbowl is 20cm. (D of water in air is cm2/s).
Derive an expression for the transport area as the function of distance from the water surface. (Hint: The transfer area is a circle)
At what rate in mol/s is the water evaporating?

22. TOP SCORERS 1st Exam Section B and C
CANTONG, JOHN ANTHONY L 92 RAMOS, ROSEMARIE R 90 BAYANI, VERONICA B 87 SAN PEDRO, DYAN ERICKA G 82.5 CARO, MARK ANTHONY D 79 DELA CRUZ, REYSON DAVE G 78 MANGUBAT, JUNE IRIS AIDYL A 78 CIAR, LOWEN B 74.5 JOAQUIN, KENNETH KEVIN DC 74.5

23. Momentum Transfer
Consider, two parallel moving planes containing a fluid -x y x z
Fy2
Fy1 Δx

24. Step 1 Assume: SS molecular transport in the x- direction
Simple transfer ( no internal generation)

25. Step 2
Balance equation

26. Step 3

27. Example 1
Two parallel flat plates are spaced 2 in apart. One plate is moving at a velocity of 10 ft/min and the other is moving in the opposite direction of 35 ft/min. The viscosity of the fluid between the plates is constant at 363 lb/ft-hr.
Calculate the stress on each plate
Calculate the fluid velocity at ½ intervals from plate to plate

28. Answer 1

29. Example 2
Determine the shear stress and velocity profiles for the conditions of Example 1 where the fluid between the plates is a Bingham-plastic non- Newtonian liquid described by
with μB = 363 lb/ft-hr τ0 = 0.01 lbf/ft2

30. Answer 2

31. Non-Newtonian fluids
Bingham plastics

32. Non-Newtonian fluids
Dilatant Fluids

33. Non-Newtonian fluids
Pseudoplastics Fluids

34. Heat Transfer
Step 1
Assume: SS molecular transport in the x- direction
Simple transfer ( no internal generation)

35. Heat Transfer
Flat-slab
Step 2
Step 3

36. Step 4

37. Step 4

38. Step 4

39. Step 4

40. Example 3
A steam pipe 2 in in outside diameter has an outside surface temperature 350oF. The pipe is covered with a coating material 2 in thick. The k of coating varies with temperature such that k=0.5+5x10-4T where T is in oF and k in Btu/hr ft2(oF/ft). The outside surface of the coating is at 100oF. Calculate the heat loss per foot of pipe length.

41. Answer 3
q=877 Btu/hr

42. Mass Transport B1 B2 Ct Diffusion of b CB2 CB1 Bulk flow of a & b CA1
Diffusion of a

43. Two Mechanism 1. Molecular transport
- diffusion of &b due to concentration gradient
- Equimolar Counter Diffusion (ECD)
Na=-Nb
2. Bulk flow of a &b
-counteracts the diffusion of b from B2 to B1
-in effect comp b becomes stationary
-due to depletion of components at the boundary

44. Step 1 Assume: SS molecular transport in the x- direction
Simple transfer ( no internal generation)

45. Flat-slab
Step 2

46. ECD

47. DTSM

48. DTSM

49. Example 4
An open cylindrical tank is filled to within 2 ft of the top with pure methanol. The tank is tapered, as shown in the figure. The air within the tank is stationary but circulation of air immediately above the tank is adequate to a assure negligible concentration of methanol at this point. The tank and air space are at 77oC & 1atm. The D of CH3OH in air at 77oC & 1atm is 0.62 ft2/hr. Calculate the rate of loss of methanol from the tank at SS.

50. Answer 4
(Na)t= 2.914x10-3 lbmol/hr

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