1. R. Shanthini 31 July 2009 PM3125 Lectures 7 to 9 Lecture Content of Lectures 7 to 9: Mathematical problems on heat transfer Mass transfer: concept and theory

2. R. Shanthini 31 July 2009 Mathematical Problems on Heat Exchanger T c,in T c,out T h,in T h,out.. Q = m c c c (T c,out – T c,in ) = m h c h (T h,in – T h,out ).

3. R. Shanthini 31 July 2009 Mathematical Problems on Heat Exchanger T c,in T h,out T h,in T c,out T c,in T c,out T h,in T h,out Parallel-flow heat exchanger high heat transfer low heat transfer

4. R. Shanthini 31 July 2009 Mathematical Problems on Heat Exchanger T c,in T h,out T h,in T c,out ab Q = U A ΔT. where ΔT = ΔT a - ΔT b ln(ΔT a / ΔT b ) is the log mean temperature difference (LMTD) ΔTaΔTa ΔTb ΔTb Parallel-flow heat exchanger

5. R. Shanthini 31 July 2009 Mathematical Problems on Heat Exchanger T c,out T c,in T h,in T h,out.. Q = m c c c (T c,out – T c,in ) = m h c h (T h,in – T h,out ). Counter-flow heat exchanger

6. R. Shanthini 31 July 2009 Mathematical Problems on Heat Exchanger T c,in T c,out T h,in T h,out T h,in T h,out Counter-flow heat exchanger T c,in T c,out

7. R. Shanthini 31 July 2009 Mathematical Problems on Heat Exchanger ab Q = U A ΔT. where ΔT = ΔT a - ΔT b ln(ΔT a / ΔT b ) is the log mean temperature difference (LMTD) ΔTaΔTa ΔTb ΔTb Counter-flow heat exchanger T c,in T c,out T h,in T h,out

8. R. Shanthini 31 July 2009 An exhaust pipe, 75 mm outside diameter, is cooled by surrounding it by an annular space containing water. The hot gases enters the exhaust pipe at 350 o C, gas flow rate being 200 kg/h, mean specific heat capacity at constant pressure 1.13 kJ/kg K, and comes out at 100 o C. Water enters from the mains at 25 o C, flow rate 1400 kg/h, mean specific heat capacity 4.19 kJ/kg K. The heat transfer coefficient for gases and water may be taken as 0.3 and 1.5 kW/m 2 K and pipe thickness may be taken as negligible. Calculate the required pipe length for (i) parallel flow, (ii) counter flow. Example in heat Exchanger Design

9. R. Shanthini 31 July 2009 Solution: Example in heat Exchanger Design.. Q = m c c c (T c,out – T c,in ) = m h c h (T h,in – T h,out ). (1400 kg/hr) (4.19 kJ/kg K) (T c,out – 25) o C = (200 kg/hr) (1.13 kJ/kg K) (350 – 100) o C The temperature of water at the outlet = T c,out = 34.63 o C.

10. R. Shanthini 31 July 2009 Solution continued: (i)Parallel flow: Example in heat Exchanger Design ΔT a = 350 – 25 = 325 o C ΔT b = 100 – 34.63 = 65.37 o C ΔT = ΔT a - ΔT b ln(ΔT a / ΔT b ) 325 – 65.37 ln(325 / 65.37) = = 162 o C Q = U A ΔT. = (UA) 162 o C What is UA?

11. R. Shanthini 31 July 2009 Solution continued: 1/U = 1/h water + 1/h gases = 1/1.5 + 1/0.3 = 4 (kW/m 2 K) -1 Therefore, U = 0.25 kW/m 2 K A = π (outer diameter) (L) = π (0.075 m) (L m) Example in heat Exchanger Design Q. = (UA) 162 o C What is Q? = (0.25) π (0.075) L (162) kW.

12. R. Shanthini 31 July 2009 Solution continued: Example in heat Exchanger Design Q. = (UA) 162 o C = (0.25) π (0.075) L (162) kW.. Q = m c c c (T c,out – T c,in ) = m h c h (T h,in – T h,out ). = (200 kg/h) (1.13 kJ/kg K) (350 – 100) o C = 15.69 kW Substituting the above in we get L = 1.64 m

13. R. Shanthini 31 July 2009 Solution continued: (ii) Counter flow: Example in heat Exchanger Design ΔT a = 350 – 34.63 = 315.37 o C ΔT b = 100 – 25 = 75 o C ΔT = ΔT a - ΔT b ln(ΔT a / ΔT b ) 315.37 – 75 ln(315.37 / 75) = = 167.35 o C Q = U A ΔT. = (UA) 167.35 o C Q = 15.69 kW; U = 0.25 kW/m 2 K ;. A = π (0.075) L m 2 Therefore, L = 1.59 m

14. R. Shanthini 31 July 2009 Other Heat Exchanger Types Cross-flow heat exchanger with both fluids unmixed The direction of fluids are perpendicular to each other. The required surface area for this heat exchanger is usually calculated by using tables. It is between the required surface area for counter-flow and parallel-flow heat exchangers.

15. R. Shanthini 31 July 2009 Other Heat Exchanger Types Two shell passes and two tube passes T h,in T h,out T c,in T c,out The required surface area for this heat exchanger is calculated by using tables.

16. R. Shanthini 31 July 2009 Other Heat Exchanger Types One shell pass and two tube passes T h,in T c,in The required surface area for this heat exchanger is calculated by using tables. T h,out T c,out

17. R. Shanthini 31 July 2009 Batch Sterilization (method of heating): Steam heating Electrical heating Direct steam sparging

18. R. Shanthini 31 July 2009 Direct steam sparging T = T 0 + H m s t c (M + m s t) T - final temperature (in kelvin) T 0 - initial temperature (in kelvin) c - specific heat of medium M - initial mass of medium m s - steam mass flow rate t - time required H - enthalpy of steam relative to raw medium temperature For batch heating by direct steam sparging:

19. R. Shanthini 31 July 2009 For batch heating with constant rate heat flow: T = T 0 + q t c M T - final temperature (in kelvin) T 0 - initial temperature (in kelvin) c - specific heat of medium M - initial mass of medium t - time required q - rate of heat transfer Electrical heating Could you prove the above?

20. R. Shanthini 31 July 2009 For batch heating with isothermal heat source: T - final temperature (in kelvin) T H - temperature of heat source (in kelvin) T 0 - initial temperature (in kelvin) c - specific heat of medium M - initial mass of medium t - time required U - overall heat transfer coefficient A - heat transfer area Steam heating T = T H + (T 0 - T H ) exp - U A t c M ( ) Could you prove the above?

21. R. Shanthini 31 July 2009 Mass transfer occurs when a component in a mixture goes from one point to another driven by a concentration gradient of the component. Under a concentration gradient, mass transfer can occur by either diffusion or by convection. Diffusion refers to the mass transfer that occurs in a stationary solid or fluid in which a concentration gradient exists. Convection refers to mass transfer that occurs across a moving fluid in which a concentration gradient exists. Mass Transfer

22. R. Shanthini 31 July 2009 By stirring the water with a spoon to create forced convection, sugar molecules are transferred to the bulk water much faster. Example of Mass Transfer Consider the dissolution of a sugar cube in water.

23. R. Shanthini 31 July 2009 Example of Mass Transfer In a fermentation process, nutrients and oxygen dissolved in the solution diffuse to the microorganisms by mass transfer.

24. R. Shanthini 31 July 2009 Oxygen transfer from gas bubble to cell 1.Transfer from the interior of the bubble to the gas-liquid interface 2.Movement across the gas film at the gas-liquid interface 3.Diffusion through the relatively stagnant liquid film surrounding the bubble 4.Transport through the bulk liquid 5.Diffusion through the relatively stagnant liquid film surrounding the cells 6.Movement across the liquid-cell interface 7.If the cells are in floc, clump or solid particle, diffusion through the solid of the individual cell 8.Transport through the cytoplasm to the site of reaction.

25. R. Shanthini 31 July 2009 Diffusion (also known as molecular diffusion) is a net transport of molecules from a region of higher concentration to a region of lower concentration by random molecular motion. Diffusion

26. R. Shanthini 31 July 2009 AB AB Liquids A and B are separated from each other. Separation removed. A goes from high concentration of A to low concentration of A. B goes from high concentration of B to low concentration of B. Molecules of A and B are uniformly distributed everywhere in the vessel purely due to the DIFFUSION. Diffusion

27. R. Shanthini 31 July 2009 Scale of mixing: Mixing on a molecular scale relies on diffusion as the final step in mixing process because of the smallest eddy size Solid-phase reaction: The only mechanism for intra particle mass transfer is molecular diffusion Mass transfer across a phase boundary: Oxygen transfer from gas bubble to fermentation broth; Penicillin recovery from aqueous to organic liquid; Glucose transfer liquid medium into mould pellets Examples of Diffusion

28. R. Shanthini 31 July 2009 Fick’s Law of Diffusion NANA CACA C A + ΔC A ΔxΔx N A = D AB ΔCAΔCA ΔxΔx A & B N A = D AB ΔCAΔCA ΔxΔx

29. R. Shanthini 31 July 2009 N A = D AB ΔCAΔCA ΔxΔx Fick’s Law of Diffusion diffusion coefficient or diffusivity What is the unit of diffusivity? concentration gradient (mass per volume per distance) diffusion flux (mass per area per time)

30. R. Shanthini 31 July 2009 For dissolved matter in water: D ≈ 10 -5 cm 2 /s For gases in air at 1 atm and at room temperature: D ≈ 0.1 to 0.01 cm 2 /s Diffusivity

31. R. Shanthini 31 July 2009 Prediction of Binary Gas Diffusivity D AB - diffusivity in cm 2 /s P - absolute pressure in atm M i - molecular weight T - temperature in K V i - sum of the diffusion volume for component i D AB is proportional to 1/P and T 1.75

32. R. Shanthini 31 July 2009 Prediction of Binary Gas Diffusivity

33. R. Shanthini 31 July 2009 Prediction of Diffusivity in Liquids D AB - diffusivity in cm 2 /s T - temperature in K μ - viscosity of solution in kg/m s V A - solute molar volume at its normal boiling point in m 3 /kmol D AB is proportional to 1/μ and T D AB = 9.96 x 10 -12 T μ V A 1/3 For very large spherical molecules (A) of 1000 molecular weight or greater diffusing in a liquid solvent (B) of small molecules: applicable for biological solutes such as proteins

34. R. Shanthini 31 July 2009 Prediction of Diffusivity in Liquids D AB - diffusivity in cm 2 /s M B - molecular weight of solvent B T - temperature in K μ - viscosity of solvent B in kg/m s V A - solute molar volume at its normal boiling point in m 3 /kmol Φ - association parameter of the solvent, which 2.6 for water, 1.9 for methanol, 1.5 for ethanol, and so on D AB is proportional to 1/μ B and T D AB = 1.173 x 10 -12 ( Φ M B ) 1/2 T μ B V A 0.6 For smaller molecules (A) diffusing in a dilute liquid solution of solvent (B): applicable for biological solutes

35. R. Shanthini 31 July 2009 Prediction of Diffusivity of Electrolytes in Liquids D o AB is diffusivity in cm 2 /s n + is the valence of cation n - is the valence of anion λ + and λ - are the limiting ionic conductances in very dilute solutions T is 298.2 when using the above at 25 o C D AB is proportional to T D o AB = 8.928 x 10 -10 T (1/n + + 1/n - ) For smaller molecules (A) diffusing in a dilute liquid solution of solvent (B): (1/λ + + 1/ λ - )

36. R. Shanthini 31 July 2009 N A = D AB ΔCAΔCA ΔxΔx Diffusion and Convection Velocity of the fluid flow + v C A

37. R. Shanthini 31 July 2009 N A = k ΔCAΔCA Mass Transfer Coefficient is used when the mass transfer is caused by molecular diffusion plus other mechanisms such as convection. k is known as the mass transfer coefficient What is the unit of k?

38. R. Shanthini 31 July 2009 NANA Mass Transfer Coefficient k ΔCAΔCA = NANA C A1 C A2 A & B k (C A1 – C A2 ) =

39. R. Shanthini 31 July 2009 NANA Mass Transfer Coefficient k ΔCAΔCA = NANA C A1 C A2 A & B k (C A1 – C A2 ) = C A1 = P A1 / RT C A2 = P A2 / RT

40. R. Shanthini 31 July 2009 NANA Mass Transfer Coefficient k (P A1 – P A2 ) / R T = NANA P A1 P A2 A & B

41. R. Shanthini 31 July 2009 Other Driving Forces Mass transfer is driven by concentration gradient as well as by pressure gradient as we have just seen. In pharmaceutical sciences, we also must consider mass transfer driven by electric potential gradient (as in the transport of ions) and temperature gradient. Transport Processes in Pharmaceutical Systems (Drugs and the Pharmaceutical Sciences, vol. 102), edited by G.L. Amidon, P.I. Lee, and E.M. Topp (Nov 1999)

42. R. Shanthini 31 July 2009 Oxygen transfer from gas bubble to cell 1.Transfer from the interior of the bubble to the gas-liquid interface 2.Movement across the gas film at the gas-liquid interface 3.Diffusion through the relatively stagnant liquid film surrounding the bubble 4.Transport through the bulk liquid 5.Diffusion through the relatively stagnant liquid film surrounding the cells 6.Movement across the liquid-cell interface 7.If the cells are in floc, clump or solid particle, diffusion through the solid of the individual cell 8.Transport through the cytoplasm to the site of reaction.

43. R. Shanthini 31 July 2009 1.Transfer through the bulk phase in the bubble is relatively fast 2.The gas-liquid interface itself contributes negligible resistance 3.The liquid film around the bubble is a major resistance to oxygen transfer 4.In a well mixed fermenter, concentration gradients in the bulk liquid are minimized and mass transfer resistance in this region is small, except for viscous liquid. 5.The size of single cell <<< gas bubble, thus the liquid film around cell is thinner than that around the bubble. The mass transfer resistance is negligible, except the cells form large clumps. 6.Resistance at the cell-liquid interface is generally neglected 7.The mass transfer resistance is small, except the cells form large clumps or flocs. 8.Intracellular oxygen transfer resistance is negligible because of the small distance involved

44. R. Shanthini 31 July 2009 Interfacial Mass Transfer air-water interface air water C w = concentration of solute in water C a = concentration of solute in air P a = partial pressure of solute in air volatilization absorption Transport of a volatile chemical across the air/water interface. P a = C a RT

45. R. Shanthini 31 July 2009 Interfacial Mass Transfer air-water interface air water δ a and δ w are boundary layer zones offering much resistance to mass transfer. δaδa δwδw C w,i P a,i P a = partial pressure of solute in air C w = concentration of solute in water P a,i vs C w,i ?

46. R. Shanthini 31 July 2009 Interfacial Mass Transfer air-water interface air water δaδa δwδw C w,i P a,i PaPa CwCw Henry’s Law: P a,i = H C w,i at equilibrium, where H is Henry’s constant δ a and δ w are boundary layer zones offering much resistance to mass transfer.

47. R. Shanthini 31 July 2009 Henry’s Law P a,i = H C w,i at equilibrium, where H is Henry’s constant Unit of H = [Pressure]/[concentration] = bar / (kg.m 3 ) P a,i = C a,i RT is the ideal gas equation Therefore, C a,i = (H/RT) C w,i at equilibrium, where (H/RT) is known as the dimensionless Henry’s constant H depends on the solute, solvent and temperature

48. R. Shanthini 31 July 2009 Gas-Liquid Equilibrium Partitioning Curve C w,i P a,i PaPa CwCw P*aP*a C*wC*w CwCw PaPa P a,i = H C w,i P* a = H’ C w P a = H’’ C* w H = H’ = H’’ if the partitioning curve is linear

49. R. Shanthini 31 July 2009 Interfacial Mass Transfer air-water interface air water δaδa δwδw C w,i P a,i PaPa CwCw P*aP*a C*wC*w N A = K G (P a – P a,i ) K G = gas phase mass transfer coefficient N A = K L (C w,i – C w ) K L = liquid phase mass transfer coefficient

50. R. Shanthini 31 July 2009 Interfacial Mass Transfer air-water interface air water δaδa δwδw C w,i P a,i CwCw C*wC*w K OG = overall gas phase mass transfer coefficient K OL = overall liquid phase mass transfer coefficient N A = K G (P a – P a,i ) N A = K L (C w,i – C w ) N A = K OG (P a – P* a ) N A = K OL (C* w – C w ) P*aP*a PaPa

51. R. Shanthini 31 July 2009 Interfacial Mass Transfer C w,i P a,i K OG = overall gas phase mass transfer coefficient K OL = overall liquid phase mass transfer coefficient N A = K G (P a – P a,i ) N A = K L (C w,i – C w ) = K OG (P a – P* a ) = K OL (C* w – C w ) K G = gas phase mass transfer coefficient K L = liquid phase mass transfer coefficient C*wC*w PaPa CwCw P*aP*a

52. R. Shanthini 31 July 2009 Relating K OL to K L P a - P a,i = H (C* w - C w,i ) C* w – C w = C* w – C w,i + C w,i – C w N A / K OL = C* w – C w,i + N A /K L If the equilibrium partitioning curve is linear over the concentration range C* w to C w,i, then N A / K G = H (C* w – C w,i ) (1) (2) Combining (1) and (2), we get 1 K OL 1 H K G =+ 1 KLKL

53. R. Shanthini 31 July 2009 Relating K OG to K G P a,i – P* a = H (C w,i – C w ) P a - P* a = P a – P a,i + P a,i – P* a N A / K OG = N A /K G + P a,i – P* a If the equilibrium partitioning curve is linear over the concentration range P a,i to P* a then (3) (4) Combining (3) and (4), we get 1 K OG 1 KGKG =+ H KLKL P a,i – P* a = H N A / K L

54. R. Shanthini 31 July 2009 Summary: Interfacial Mass Transfer N A = K G (P a – P a,i ) N A = K L (C w,i – C w ) = K OG (P a – P* a ) = K OL (C* w – C w ) 1 K OL 1 H K G =+ 1 KLKL 1 K OG 1 KGKG =+ H KLKL H = P* a / C w = P a,i / C w,i = P a / C* w Two-film Theory

55. R. Shanthini 31 July 2009 Encyclopedia of Pharmaceutical Technology (Hardcover) by James Swarbrick (Author) Transport Processes in Pharmaceutical Systems (Drugs and the Pharmaceutical Sciences, vol. 102), edited by G.L. Amidon, P.I. Lee, and E.M. Topp, Nov 1999 Visit the site for books: http://www.freemedicalebooks.net/?p=11332 57 MB ebook downloadable from http://rapidshare.com/files/79551667/Encyclopedia_Pharma_Tech_3Ed.zip