Tutorial ChE333 : Mass Transfer

1. Tutorial ChE333 : Mass Transfer

2. 24.12 : Highly purified tetrachlorosilane (SiCl4) gas is reacted with hydrogen gas (H2) to produce electronic grade polycrystalline silicon at 800 oC and 1.5x105 Pa according to the following equation: SiCl4 (g) + 2H2 (g)  Si(s) + 4HCl(g) There are concerns that the reaction experience diffusional limitation at the growing Si solid surface. Estimate the molecular diffusion of ----- (1) SiCl4 in H2 (2) SiCl4 in a gas phase mixture of 40 mol% SiCl4 40% mol H2 and 20 mol% HCl Lennard-Jones parameters for SiCl4 are εA/k = 358 and σA = 5.08 Å

4. Nomenclature : A = SiCl4 B = H2 C= HCl Table K2. page 695

6. Temperature = 800 oC + 273.15 = 1073.15 K Section (1) DSiCl4-H2 or DAB

7. Temperature = 800 oC + 273.15 = 1073.15 K Section (1) DSiCl4-H2 or DAB Use appendix K.1 ::: page 694 Find out the collision integral, ΩD.

8. P = 1.48 atm T1.5 = 35,155.3

10. Part 2 : Mixture diffusivity. P = 1.48 atm T1.5 = 35,155.3

11. Part 2 : Mixture diffusivity. Corrected mole fraction Sum add to 1

12. Question : Calculate the gas phase diffusivity of CS2 in air at 35°C at 1 atm using Fuller, Schettler and Giddings correlation? Nomenclature : A = CS2 B = Air Molecular wt of CS2 , MA= ? Diffusion volume of CS2 , VA= ? Molecular wt of Air , MB= ? Diffusion volume of air VB= ?

13. Question : Calculate the diffusivity of CS2 in air at 35°C at 1 atm using Fuller correlation? Nomenclature : A = CS2 B = Air Molecular wt of CS2 = 72 gm/mol Diffusion volume of CS2 = 50.5 Molecular wt of Air = 29 gm/mol Diffusion volume of air = 20.1

14. Calculate the diffusivity of CS2 in air at 35°C at 1 atm Solution For air: • VA = 20.1. • Molecular weight = 29.0. For CS2, from Table 24.3: • VCS2 = 50.5 • Molecular weight = 76.

15. Problem 24.18: The aeration of water is an important industrial operation. Determine the liquid diffusion coefficient of oxygen (O2) in an infinitely dilute solution of water at 288K using Wilke-Chang equation The Hayduk-Laudie equation Aeration is an optional part of the water treatment process which uses scrubbing action and oxidation to remove or modify constituents of the water.   Substances affected by aeration include volatile organic chemicals, carbon dioxide, hydrogen sulfide, methane, iron, and manganese. Aerators work by increasing the amount of surface area of air coming in contact with water.  This may be achieved by passing air through water, as in an air diffusion aerator.  In contrast, many aerators pass water through air, as in spray nozzle, cone tray, cascade, and coke tray aerators.  Finally, forced draft aerators both pass air through water and water through air.

16. Liquid – Liquid diffusivity calculation Nomenclature : Solute = Oxygen (O2) = A Solvent = Water (H2O) = B System Temperature T = 288K φB = association parameter of water =?? [Table 24.5 page 417 WWWR] MB = Molecular weight of water =?? [periodic table] VA = molar volume of solute (O2) at its normal boiling point from Table 24.4, page 416. μB = Viscosity of water @ 288K = ???? cP [use property table in appendix I , page 686]. 1 cP = 10−2 P = 10−3 Pa·s = 1 mPa·s

17. Nomenclature : Solute = Oxygen (O2) = A Solvent = Water (H2O) = B 1 cP = 10−2 P = 10−3 Pa·s = 1 mPa·s φB = association parameter of water =2.26 MB = Molecular weight of water =18 gm/mol. VA = molar volume of solute (O2) at its normal boiling point =25.6 cm3/mol. μB = Viscosity of water @ 288K = ???? cP Interpolation

18. Nomenclature : Solute = Oxygen (O2) = A Solvent = Water (H2O) = B 1000 cP = 1 Pa·s φB = association parameter of water =2.26 MB = Molecular weight of water =18 gm/mol. VA = molar volume of solute (O2) at its normal boiling point =25.6 cm3/g.mol. μB = Viscosity of water @ 288K = ???? cP 1.148 cP

19. Wilke Chang equation, DAB = 1.692 x 10-5 cm2/s • For infinite dilution of non-electrolytes in water, W-C is simplified toHayduk-Laudieeq. μB = Viscosity of water @ 288K = 1.148 cP VA = molar volume of solute (O2) = 25.6 cm3/g.mol DAB = 1.68 x 10-5 cm2/s

20. Molar volume of benzene at its normal boiling point from the atomic volume of the elements?? Benzene : Aromatic compound Chemical formula ( C6H6) Contains an aromatic ring

21. Table 24.5 : page 417 : atomic volumes table C = 14.8 cm3/mol H = 3.7 cm3/mol Benzene ring correction = deduct 15 Molar volume of benzene = 6 x 14.8 + 6 x 3.7 -15 = 96 cm3/mol

22. Molar volume of ‘ethanol’ at its normal boiling point ??

23. Nomenclature : Silane ( SiH4) = A [ molecular weight = 32] and Helium =B Temperature , T = 900K TotalPressure , P = 100Pa DA-B = 0.571 cm2/s @ 298 K and 1.013x105 Pa Mole fraction, ySiH4 =0.01 Mole franction, yHe = 0.99

26. 24.19: A silicon wafer is doped with phosphorous(P, Mwt =35). The nominal value for diffusion coefficient for phosphorous in silicon(Si, Mwt =28) at 1316 K is 1x10-13 cm2/s and at 1408 K is 1x 10-12 cm2/s . Determine the value of diffusion coefficient at 1373 K? In electronics, a wafer (also called a slice or substrate) is a thin slice of semiconductor material, such as a silicon crystal, used in the fabrication of integrated circuits and other micro-devices. The wafer serves as the substrate for microelectronic devices built in and over the wafer and undergoes many micro-fabrication process steps such as doping or ion implantation, etching, deposition of various materials, and photolithographic patterning.

27. Solid diffusion is an activated process and diffusivity relation with temperature is given by Arrhenius equation ::: Do = Proportionality constant Q = Activation energy , kcal/mol, J/mol etc. R = Gas constant = 8.314 J/mol/K Rearrange the Arrhenius equation , at two different temperature DAB|T1 and DAB|T2. T1 = 1316 K DAB|T1 =1x10-13 cm2/s T2 = 1408 K DAB|T2 = 1x10-12 cm2/s Unknown parameters to calculate the DAB value at T=1373 K are Q, activation energy. Do, Proportionality constant. CAN YOU CALCULATE THESE VALUES????

28. T1 = 1316 K DAB|T1 =1x10-13 cm2/s T2 = 1408 K DAB|T2 = 1x10-12 cm2/s Straight line equation on a semi log plot Slope Intercept Subtract and rearrange

29. Q/R = 46375.26325 Do = 201.53 DAB| T = 4.318 x 10-13 cm2/s

30. Partial pressure of a component in a gas mixture is 2x104 Pa. The system pressure is 1 atm and temperature is 25 oC. what is the concentration of the gas component in the mixture?