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Introduction to Mass Transfer CHAPTER 6

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Introduction Three fundamental transfer processes: i) Momentum transfer ii) Heat transfer iii) Mass transfer

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Mass transfer may occur in a gas mixture, a liquid solution or solid. Mass transfer occurs whenever there is a gradient in the concentration of a species. The basic mechanisms are the same whether the phase is a gas, liquid, or solid.

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Definition of Concentration i) Number of molecules of each species present per unit volume (molecules/m 3 ) ii) Molar concentration of species i = Number of moles of i per unit volume (kmol/m 3 ) iii) Mass concentration = Mass of i per unit volume (kg/m 3 )

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Diffusion phenomena Fick’s law: linear relation between the rate of diffusion of chemical species and the concentration gradient of that species. Thermal diffusion: Diffusion due to a temperature gradient. Usually negligible unless the temperature gradient is very large. Pressure diffusion: Diffusion due to a pressure gradient. Usually negligible unless the pressure gradient is very large.

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Forced diffusion: Diffusion due to external force field acting on a molecule. Forced diffusion occurs when an electrical field is imposed on an electrolyte ( for example, in charging an automobile battery) Knudsen diffusion: Diffusion phenomena occur in porous solids.

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Whenever there is concentration difference in a medium, nature tends to equalize things by forcing a flow from the high to the low concentration region. The molecular transport process of mass is characterized by the general equation: Rate of transfer process = driving force resistance Before After

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Consider a tank that is divided into two equal parts by a partition. Initially, the left half of the tank contains nitrogen N 2 gas while the right half contains O 2 at the same temperature and pressure. When the partition is removed the N 2 molecules will start diffusing into the air while the O 2 molecules diffuse into the N 2. If we wait long enough, we will have a homogeneous mixture of N 2 and O 2 in the tank. Example of Mass Transfer Processes

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Liquid in open pail of water evaporates into air because of the difference in concentration of water vapor at the water surface and the surrounding air. A drop of blue liquid dye is added to a cup of water. The dye molecules will diffuse slowly by molecular diffusion to all parts of the water.

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Molecular Diffusion Equation Fick’s Law is the molar flux of component A in the z direction in kg mol A/s.m 2. is the molecular diffusivity of the molecule A in B in m 2 /s is the concentration of A in kg mol/m 3. z is the distance of diffusion in m

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Fick’s Law of Diffusion Molecular diffusion or molecular transport can be defined as the transfer or movement of individual molecules through a fluid by mean of the random, individual movements of the molecules. If there are greater number of A molecules near point (1) than at (2), then since molecules diffuse randomly in both direction, more A molecules will diffuse from (1) to (2) than from (2) to (1). The net diffusion of A is from high to low concentration regions. B B BB B B B B B B B A A Figure 3: Schematic diagram of molecular diffusion process (2) (1)

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The two modes of mass transfer: - Molecular diffusion - Convective mass transfer

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Molecular diffusion The diffusion of molecules when the whole bulk fluid is not moving but stationary. Diffusion of molecules is due to a concentration gradient. The general Fick’s Law Equation for binary mixture of A and B c = total concentration of A and B [kgmol (A + B)/m 3 ] x A = mole fraction of A in the mixture of A and B

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Example A mixture of He and N 2 gas is contained in a pipe at 298 K and 1 atm total pressure which is constant throughout. At one end of the pipe at point 1 the partial pressure p A1 of He is 0.6 atm and at the other end 0.2 m p A2 = 0.2 atm. Calculate the flux of He at steady state if D AB of the He-N 2 mixture is x m 2 /s.

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Solution Since a total pressure P is constant, the c is constant, where c is as follows for a gas according to the perfect gas law: Where n is kg mol A plus B, V is volume in m 3, T is temperature in K, R is m 3.Pa/kg mol.K or R is x cm 3. atm/g. mol. K, and c is kg mol A plus B/m3. For steady state the flux J* Az in Eq.(6.1-3) is constant. Also D AB for gas is constant. Rearranging Eq. (6.1-3) and integrating. (6.1-11)

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Also, from the perfect gas law, p A V=n A RT, and Substituting Eq. (6.1-12) into (6.1-11), This is the final equation to use, which is in a form eqsily used for gases. Partial pressures are p A1 = 0.6 atm = 0.6 x x 10 5 = 6.04 x 10 4 Pa and p A2 = 0.2 atm = 0.2 x x 10 5 = x 10 4 Pa. Then, using SI units, (6.1-13)

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If pressures in atm are used with SI unit, Other driving forces (besides concentration differences) for diffusion also occur because of temperature, pressure, electrical potential, and other gradients.

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Convection Mass Transfer When a fluid flowing outside a solid surface in forced convection motion, rate of convective mass transfer is given by: k c - mass transfer coefficient (m/s) c L1 - bulk fluid conc. c Li - conc of fluid near the solid surface Kc depend on: 1. system geometry 2. Fluid properties 3. Flow velocity

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CHAPTER 2 Molecular Diffusion in Gases Principles of Mass Transfer CHAPTER 2 Molecular Diffusion in Gases

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CONTENTS Mass Transfer Molecular Diffusion GasesLiquidSolid Convective Mass Transfer

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Molecular Diffusion in Gases Equimolar Counterdiffussion in Gases

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For a binary gas mixture of A and B, the diffusivity coefficient D AB =D BA

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Example Ammonia gas (A) is diffusing through a uniform tube 0.10 m long containing N 2 gas (B) at x 10 5 Pa pressure and 298 K. The diagram is similar to Fig At point 1, p A1 = x 10 4 Pa and at point 2, p A2 = x 10 4 Pa. The diffusivity D AB = x m 2 /s. (a) Calculate the flux J* A at steady state (b) Repeat for J* B

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Solution Equation (6.1-13) can be used, where P = x 10 5 Pa, z 2 -z 1 = 0.10 m, and T = 298 K. Substituting into Eq. (6.1-13) for part (a), Rewriting Eq. (6.1-13) for component B for part (b) and noting that p B1 = P – p A1 = x 10 5 – x 10 4 = x 104 Pa and pB 2 = P – p A2 = x 10 5 – x 10 4 = x 10 4 Pa. The negative for J* B means the flux goes from point 2 to point 1.

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Diffusion of Gases A and B Plus Convection

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For equimolar counterdiffussion, N A =-N B, then N A =J* A =-N B =-J* B

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Example Water in the bottom of a narrow metal tube is held at a constant temperature of 293 K. The total pressure of air (assumed dry) is x 10 5 Pa (1.0 atm) and the temperature is 293 K (20 °C). Water evaporates and diffuses through the air in the tube, and the diffusion path z 2 -z 1 is m (0.5 ft) long. The diagram is similar to Fig a. Calculate the rate of evaporation at steady state in lb mol/h.ft 2 and kg mol/s.m 2. The diffusivity of water vapor at 293 K and 1 am pressure is x m 2 /s. Assume that the system is isothermal. Use SI and English units.

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Solution The diffusivity is converted to ft 2 /h by using the conversion factor: From Appendix A.2 the vapor pressure of water at 20 °C is mm, or p A1 = 17.54/760 = atm = ( x 10 5 ) = x 10 3 Pa, p A2 = 0 (pure air). Since the temperature is 20 °C (68 °F), T = °R = 293 K. From Appendix A.1, R = ft 3.atm/lb mol.°R. To calculate the value of p BM from Eq. (6.2-21)

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Since p B1 is close to p B2, the linear mean (p B1 +p B2 )/2 could be used and would be very close to p BM. Substituting into Eq. (6.2-22) with z 2 -z 1 = 0.5 ft (0.1524m),

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Example A sphere of naphthalene having a radius of 2.0 mm is suspended in a large volume of still air at 318 K and x 10 5 Pa (1 atm). The surface temperature of the naphthalene can be assumed to be at 318 K is 6.92 x m 2 /s. Calculate the rate of evaporation of naphthalene from the surface.

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Solution The flow diagram is similar to Fig a. D AB = 6.92 x m 2 /s, p A1 = (0.555/760)( x 10 5 ) = 74.0 Pa, p A2 = 0, r 1 = 2/1000 m, R = 8314 m 3.Pa/kg mol.K, p B1 = P-p A1 = x 10 5 – 74.0 = x 10 5 Pa, p B2 = x 10 5 – 0. since the values of p B1 and p B2 are close to each other, Substituting into Eq. (6.2-32),

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Example Normal butanol (A) is diffusing through air (B) at 1 atm abs. Using the Fuller et al. method, estimate the diffusivity D AB for the following temperatures with the experimental data: (a) For 0 °C. (b) For 25.9 °C. (c) For 0 °C and 2.0 atm abs.

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Solution For part (a), P = 1.00 atm, T = = 273 K, M A (butanol) = 74.1, M B (air) = 29. From Table 6.2-2, Substituting into Eq. (6.2-45), This values deviates +10% from the experimental values of 7.03 x m 2 /s from Table 6.2-1

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For part (b), T = = Substituting into Eq. ( ), D AB = 9.05 x m 2 /s. This values deviates by +4% from the experimental value of 8.70 x m 2 /s For part (c), the total pressure P = 2.0 atm. Using the value predicted in part (a) and correcting for pressure, D AB = 7.73 x (1.0/2.0) = 3.865x10 -6 m 2 /s

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