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Introduction to Mass Transfer CHAPTER 6. Introduction Three fundamental transfer processes: i) Momentum transfer ii) Heat transfer iii) Mass transfer.

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Presentation on theme: "Introduction to Mass Transfer CHAPTER 6. Introduction Three fundamental transfer processes: i) Momentum transfer ii) Heat transfer iii) Mass transfer."— Presentation transcript:

1 Introduction to Mass Transfer CHAPTER 6

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

3 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.

4 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 )

5 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.

6 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.

7 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

8 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

9 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.

10 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

11 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)

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

13 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

14 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.

15 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)

16 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)

17 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.

18 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

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

20 CONTENTS Mass Transfer Molecular Diffusion GasesLiquidSolid Convective Mass Transfer

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22 Molecular Diffusion in Gases Equimolar Counterdiffussion in Gases

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

26 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

27 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.

28 Diffusion of Gases A and B Plus Convection

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

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33 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.

34 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)

35 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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37 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.

38 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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41 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.

42 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

43 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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