Introduction to Mass Transfer

Outline Mass Transfer Mechanisms 2. Fick’s Law for Molecular Diffusion Convective Mass Transfer 2. Fick’s Law for Molecular Diffusion 3. Molecular Diffusion in Gases Equimolar Counterdiffusion Combined Diffusion and Convection Uni-component Diffusion

Mass Transfer Mechanisms 1. Convective Mass Transfer 2. Diffusion Analogous to heat convection and conduction http://www.timedomaincvd.com/CVD_Fundamentals/xprt/xprt_conv_diff.html

Mass Transfer Mechanisms 3. Convective and Diffusion http://www.timedomaincvd.com/CVD_Fundamentals/xprt/xprt_conv_diff.html

Outline Mass Transfer Mechanisms 2. Fick’s Law for Molecular Diffusion Convective Mass Transfer 2. Fick’s Law for Molecular Diffusion 3. Molecular Diffusion in Gases Equimolar Counterdiffusion Combined Diffusion and Convection Uni-component Diffusion

Fick’s Law for Molecular Diffusion We’ll first consider diffusion of molecules when the bulk fluid is not moving… 𝐽 𝐴𝑧 ∗ =− 𝐷 𝐴𝐵 𝑑 𝑐 𝐴 𝑑𝑧 𝑑 𝑐 𝐴 =𝑑 𝑐 𝑥 𝐴 =𝑐𝑑 𝑥 𝐴 𝐽 𝐴𝑧 ∗ =− 𝑐𝐷 𝐴𝐵 𝑑 𝑥 𝐴 𝑑𝑧 For a binary mixture of A and B J = molar diffusivity C = total concentration Xa = mole fraction of A in the mixture Other driving forces for diffusion: temperature, pressure, electrical potential, etc. (Will not be discussed in 131)

Molecular Transport Equations RECALL: MOMENTUM HEAT MASS

Fick’s Law for Molecular Diffusion Example A mixture of He and N2 gas is collected 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 pA1 of He is 0.60 atm and at the other end 0.2 m pA2 = 0.20 atm. Calculate the flux of He at steady state if DAB of the He-N2 mixture is 0.687 x 10-4 m2/s. Steady state diffusion Concentration at the boundaries are constant Diffusion is limited to molecular motion Diffusivity is independent from concentration No temperature gradient Assume ideal gas, calculate c values from the P given. Answer: J = 5.63 x 10-6 kg mol of A/s-m2

Convective Mass Transfer Coefficient For fluids in convective flow… 𝑘 𝑐 is very similar to h, 𝑁 𝐴 = 𝑘 𝑐 ( 𝐶 𝐿1 − 𝐶 𝐿𝑖 ) 𝑁 𝐴 = 𝑘 𝐺 ( 𝑝 𝐴1 − 𝑝 𝐴𝑖 ) 𝑁 𝐴 = 𝑘 𝑦 ( 𝑦 𝐴1 − 𝑦 𝐴𝑖 ) To have a fluid in convective flow usually requires the fluid to be flowing past another immiscible fluid or a solid surface. Na = convective molar flux, k = MTC with different units depending on the units of the driving force *MTC is a function of system geometry, fluid properties, and flow velocity. We will discuss convective mass transfer in more detail LATER… What factors influence 𝑘 𝑐 ?

Outline Mass Transfer Mechanisms 2. Fick’s Law for Molecular Diffusion Convective Mass Transfer 2. Fick’s Law for Molecular Diffusion 3. Molecular Diffusion in Gases Equimolar Counterdiffusion Combined Diffusion and Convection Uni-component Diffusion

Molecular Diffusion in Gases Equimolar Counterdiffusion Flux of one gaseous component is equal to but in the opposite direction of the second gaseous component A B Consider 2 gases A and B in a chamber, initially separated by a partition. At some instant in time, the partition is removed, and A and B diffuse in opposite direction as a result of the concentration gradients. Examples: distillation, alcohol/H2O, HOAc/H2O A B 𝐽 𝐴𝑧 ∗ =− 𝐽 𝐵𝑧 ∗

Molecular Diffusion in Gases Equimolar Counterdiffusion At constant pressure, 𝑃= 𝑝 𝐴 + 𝑝 𝐵 Then, 𝑐= 𝑐 𝐴 + 𝑐 𝐵 and 𝑑 𝑐 𝐴 =−𝑑 𝑐 𝐵 Fick’s law for B, 𝐽 𝐵𝑧 ∗ =− 𝐷 𝐵𝐴 𝑑 𝑐 𝐵 𝑑𝑧 A B Consider 2 gases A and B in a chamber, initially separated by a partition. At some instant in time, the partition is removed, and A and B diffuse in opposite direction as a result of the concentration gradients. A B

Molecular Diffusion in Gases Equimolar Counterdiffusion Substitution of Fick’s law into the equation for equimolar counter diffusion, A B 𝐽 𝐴𝑧 ∗ =− 𝐽 𝐵𝑧 ∗ Consider 2 gases A and B in a chamber, initially separated by a partition. At some instant in time, the partition is removed, and A and B diffuse in opposite direction as a result of the concentration gradients. A B − 𝐷 𝐴𝐵 𝑑 𝑐 𝐴 𝑑𝑧 =−(− 𝐷 𝐵𝐴 𝑑 𝑐 𝐵 𝑑𝑧 )

Molecular Diffusion in Gases Equimolar Counterdiffusion − 𝐷 𝐴𝐵 𝑑 𝑐 𝐴 𝑑𝑧 =−(− 𝐷 𝐵𝐴 𝑑 𝑐 𝐵 𝑑𝑧 ) A B − 𝐷 𝐴𝐵 𝑑 𝑐 𝐴 𝑑𝑧 =−(− 𝐷 𝐵𝐴 (− 𝑑 𝑐 𝐴 𝑑𝑧 )) For a binary mixture of A and B the diffusivity coefficient DAB for A diffusing in B is the same as DBA for B diffusing into A. A B 𝐷 𝐴𝐵 = 𝐷 𝐵𝐴

Molecular Diffusion in Gases Equimolar Counterdiffusion For gases, A B 𝑐 𝐴1 = 𝑝 𝐴1 𝑅𝑇 = 𝑛 𝐴 𝑉 𝐽 𝐴𝑧 ∗ =− 𝐷 𝐴𝐵 𝑅𝑇 𝑑 𝑝 𝐴 𝑑𝑧 A B

Molecular Diffusion in Gases Equimolar Counterdiffusion In terms of mole fraction, A B 𝑐 𝐴 =𝑐 𝑥 𝐴 𝐽 𝐴𝑧 ∗ =− 𝑐𝐷 𝐴𝐵 𝑑 𝑥 𝐴 𝑑𝑧 A B

Molecular Diffusion in Gases Example A large tank filled with a mixture of methane and air is connected to a second tank filled with a different composition of methane and air. Both tanks are at 100 kN/m2 and 0°C. The connection between the tanks is a tube of 2 mm inside diameter and 150 mm long. Calculate the steady state rate of transport of methane through the tube when the concentration of methane is 90 mole percent in one tank and 5 mole percent in the other. Assume that transport between the tanks is by molecular diffusion. The mass diffusivity of methane in air at 0°C and 100 kN/m2 is 1.57 x 10-5 m2/s.

Molecular Diffusion in Gases Diffusion plus Convection 𝑣 𝐴 = 𝑣 𝐴𝑑 + 𝑣 𝑀 Multiplying by 𝑐 𝐴 , 𝑐 𝐴 𝑣 𝐴 = 𝑐 𝐴 𝑣 𝐴𝑑 + 𝑐 𝐴 𝑣 𝑀 𝑁 𝐴 = 𝐽 𝐴 ∗ + 𝑐 𝐴 𝑣 𝑀 𝑣 𝑀 𝑣 𝐴 Va= velocity of A relative to a stationary point, sum of diffusion velocity and the average or convective velocity Vad = diffusion velocity of A, measured relative to the moving fluid Vm = molar average velocity of the whole fluid relative to a stationary point 𝑣 𝐴 = 𝑣 𝐴𝑑 + 𝑣 𝑀 𝐽 𝐴 ∗ = 𝑣 𝐴𝑑 𝑐 𝐴

Molecular Diffusion in Gases Diffusion plus Convection 𝑁 𝐴 = 𝐽 𝐴 ∗ + 𝑐 𝐴 𝑣 𝑀 𝑣 𝑀 Total convective flux of A wrt stationary pt 𝑁 𝐴 𝑣 𝐴 Va= velocity of A relative to a stationary point, sum of diffusion velocity and the average or convective velocity Vad = diffusion velocity of A, measured relative to the moving fluid 𝐽 𝐴 ∗ 𝑣 𝐴 = 𝑣 𝐴𝑑 + 𝑣 𝑀 Diffusion flux wrt moving fluid 𝐽 𝐴 ∗ = 𝑣 𝐴𝑑 𝑐 𝐴 Convective flux wrt to stationary point 𝑐 𝐴 𝑣 𝑀

Molecular Diffusion in Gases Diffusion plus Convection 𝑁= 𝑐𝑣 𝑀 = 𝑁 𝐴 + 𝑁 𝐵 𝑁 𝐴 = 𝐽 𝐴 ∗ + 𝑐 𝐴 𝑣 𝑀 Replacing 𝐽 𝐴 ∗ and 𝑣 𝑀 , 𝑁 𝐴 =− 𝑐𝐷 𝐴𝐵 𝑑 𝑥 𝐴 𝑑𝑧 + 𝑐 𝐴 ( 𝑁 𝐴 + 𝑁 𝐵 𝑐 ) Solving for 𝑣 𝑀 , 𝑣 𝑀 = 𝑁 𝐴 + 𝑁 𝐵 𝑐 Va= velocity of A relative to a stationary point, sum of diffusion velocity and the average or convective velocity Vad = diffusion velocity of A, measured relative to the moving fluid

Molecular Diffusion in Gases Diffusion plus Convection 𝑁 𝐴 =− 𝑐𝐷 𝐴𝐵 𝑑 𝑥 𝐴 𝑑𝑧 + 𝑐 𝐴 ( 𝑁 𝐴 + 𝑁 𝐵 𝑐 ) General equation for diffusion plus convection For equimolar counterdiffusion, Na = -Nb and the convective term becomes zero.

Molecular Diffusion in Gases Uni-component Diffusion One component (A)diffuses, while the other (B) remains stagnant A special case wherein, one boundary at the end of the diffusion path is impermeable to component B, so it cannot pass through. Example: Gas Absorption, evaporation of organic solvent in air Since B cannot diffuse, 𝑁 𝐵 =0 http://sst-web.tees.ac.uk/external/U0000504/Notes/ProcessPrinciples/Diffusion/Default.htm

Molecular Diffusion in Gases Uni-component Diffusion Since B cannot diffuse, 𝑁 𝐵 =0 A special case wherein, one boundary at the end of the diffusion path is impermeable to component B, so it cannot pass through. Example: Gas Absorption, evaporation of organic solvent in air 𝑁 𝐴 =− 𝑐𝐷 𝐴𝐵 𝑑 𝑥 𝐴 𝑑𝑧 + 𝑐 𝐴 ( 𝑁 𝐴 +0 𝑐 ) http://sst-web.tees.ac.uk/external/U0000504/Notes/ProcessPrinciples/Diffusion/Default.htm

Molecular Diffusion in Gases Uni-component Diffusion 𝑁 𝐴 =− 𝑐𝐷 𝐴𝐵 𝑑 𝑥 𝐴 𝑑𝑧 + 𝑐 𝐴 𝑁 𝐴 𝑐 A special case wherein, one boundary at the end of the diffusion path is impermeable to component B, so it cannot pass through. Example: Gas Absorption, evaporation of organic solvent in air 𝑁 𝐴 =− 𝑐𝐷 𝐴𝐵 𝑥 𝐵 𝑑 𝑥 𝐴 𝑑𝑧 http://sst-web.tees.ac.uk/external/U0000504/Notes/ProcessPrinciples/Diffusion/Default.htm

Molecular Diffusion in Gases Uni-component Diffusion When P is constant, 𝑁 𝐴 =− 𝐷 𝐴𝐵 𝑃 𝑅𝑇 𝑑 𝑝 𝐴 𝑑𝑧 + 𝑁 𝐴 𝑝 𝐴 𝑃 A special case wherein, one boundary at the end of the diffusion path is impermeable to component B, so it cannot pass through. Example: Gas Absorption, evaporation of organic solvent in air 𝑁 𝐴 =− 𝐷 𝐴𝐵 𝑃 𝑅𝑇 𝑃 𝐵 𝑑 𝑝 𝐴 𝑑𝑧 http://sst-web.tees.ac.uk/external/U0000504/Notes/ProcessPrinciples/Diffusion/Default.htm

Molecular Diffusion in Gases Example Water in the bottom of a narrow metal tune is held a t a constant temperature of 293 K. The total pressure of air (assumed dry) is 1.01325  105 Pa and the temperature is 293 K. Water evaporates and diffuses through the air in the tube, and the diffusion path z2-z1 is 0.1524m long. Calculate the rate of evaporation of water vapor at 293 K and 1 atm pressure. The diffusivity of water in air is 0.250 x 10-4 m2/s. Assume that the system is isothermal.

Long Exam Results LE 1 LE 2 Mean 33.32 36.55 Median 32.00 30.75 Mode 39.00 25.50 Passing Rate 0.00 9.09

Quizzes Machine Problems Target Average Scores Student No. Total Q 5/5   Quizzes Machine Problems Target Average Scores Student No. Total Q 5/5 Total* M 15/15 LE3 L 60/60 Final F 20/20 2011-18077 30 1.50 240 12 82 30.4 16.4 2011-57319 46 2.30 90 27.8 18.0 2010-04141 36 1.80 31 2010-01283 26 1.30 89 29.1 17.8 2010-31873 47 2.35 85 29 17.0 2011-07217 67 33.3 13.4 2011-03676 50 2.50 93 26.9 18.6 2010-36588 23 1.15 92 28.7 18.4 2011-18143 1.55 73 31.9 14.6 2011-18147 77 31.3 15.4 2011-09522 33 1.65 102 20.4 2011-30507 88 29.6 17.6 2011-09270 19 0.95 81 30.9 16.2 2010-53270 74 31.6 14.8 2011-14930 61 3.05 51 35 10.2 2009-21119 8 0.40 99 28.1 19.8 2011-21884 48 2.40 70 32.1 14.0 2011-19280 9 0.45 29.5 2011-26790 104 5.20 45 34.1 9.0 2010-21409 14 0.70 100 27.3 20.0 2011-01530 57 2.85 65 32.5 13.0 2011-30255 21 1.05 91 28.9 18.2