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Dept of Chemical and Biomolecular Engineering CN2125E Heat and Mass Transfer Dr. Tong Yen Wah, E5-03-15, 6516-8467 chetyw@nus.edu.sg (Mass Transfer, Radiation)

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Course Outline Week 9-12: Mass Transfer –Week 9: Steady-state Diffusion (WWWR Ch 26) –Week 10: Unsteady-state Diffusion (WWWR Ch 27) –Week 11: Convective Mass Transfer (WWWR Ch 28) Week 13: Radiation Heat Transfer (WWWR Ch 23, ID Ch 12-13)

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HW/Tutorial Week #9 WWWR Chapters 26, ID Chapter 14 Tutorial #9 WWWR # 26.17 & 26.27 To be discussed on March 24, 2015. By either volunteer or class list.

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Molecular Diffusion General differential equation One-dimensional mass transfer without reaction

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Unimolecular Diffusion Diffusivity of gas can be measured in an Arnold diffusion cell

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Assuming –Steady state, no reaction, and diffusion in z- direction only We get And since B is a stagnant gas,

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Thus, for constant molar flux of A, when N B,z = 0, –with boundary conditions: at z = z 1, y A = y A1 at z = z 2, y A = y A2 Integrating and solving for N A,z

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since the log-mean average of B is we get This is a steady-state diffusion of one gas through a second stagnant gas;

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For film theory, we assume laminar film of constant thickness , then, z 2 – z 1 = and But we know So, the film coefficient is then

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To determine concentration profile, if isothermal and isobaric, integrated twice, we get

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–with boundary conditions: at z = z 1, y A = y A1 at z = z 2, y A = y A2 So, the concentration profile is:

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Example 1

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Pseudo-Steady-State Diffusion When there is a slow depletion of source or sink for mass transfer Consider the Arnold diffusion cell, when liquid is evaporated, the surface moves, at any instant, molar flux is

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Molar flux is also the amount of A leaving Under pseudo-steady-state conditions, which integrated from t=0 to t=t, z=z t0 to z=z t becomes

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Example 2

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Equimolar Counterdiffusion Flux of one gaseous component is equal to but in the opposite direction of the second gaseous component Again, for steady-state, no reaction, in the z-direction, the molar flux is

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In equimolar counterdiffusion, N A,z = -N B,z Integrated at z = z 1, c A = c A1 and at z = z 2, c A = c A2 to: Or in terms of partial pressure,

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The concentration profile is described by Integrated twice to With boundary conditions at z = z 1, c A = c A1 and at z = z 2, c A = c A2 becomes a linear concentration profile:

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Systems with Reaction When there is diffusion of a species together with its disappearance/appearance through a chemical reaction Homogeneous reaction occurs throughout a phase uniformly Heterogeneous reaction occurs at the boundary or in a restricted region of a phase

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Diffusion with heterogeneous first order reaction with varying area: –With both diffusion and reaction, the process can be diffusion controlled or reaction controlled. –Example: burning of coal particles –steady state, one-dimensional, heterogeneous

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–3C (s) + 2.5 O 2 (g) 2 CO 2 (g) + CO (g) –Along diffusion path, R O2 = 0, then the general mass transfer equation reduces from –to –For oxygen,

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–From the stoichiometry of the reaction, –We simplify Fick’s equation in terms of oxygen only, –which reduces to

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–The boundary conditions are: at r = R, y O2 = 0 and at r = , y O2 = 0.21, –Integrating the equation to: –The oxygen transferred across the cross- sectional area is then:

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–Using a pseudo-steady-state approach to calculate carbon mass-transfer output rate of carbon: accumulation rate of carbon: input rate of carbon = 0 Thus, the carbon balance is

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–Rearranging and integrating from t = 0 to t = , R = R i to R = R f, we get –For heterogeneous reactions, the reaction rate is

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–If the reaction is only C (s) + O 2 (g) CO 2 (g) and if the reaction is not instantaneous, then for a first-order reaction, at the surface, then,

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–Combining diffusion with reaction process, we get

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Example 3

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Diffusion with homogeneous first-order reaction: –Example: a layer of absorbing liquid, with surface film of composition A and thickness , assume concentration of A is small in the film, and the reaction of A is

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–Assuming one-direction, steady-state, the mass transfer equation reduces from to with the general solution

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–With the boundary conditions: at z = 0, c A = c A0 and at z = , c A = 0, –At the liquid surface, flux is calculated by differentiating the above and evaluating at z=0,

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–Thus, –Comparing to absorption without reaction, the second term is called the Hatta number. –As reaction rate increases, the bottom term approaches 1.0, thus

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–Comparing with we see that k c is proportional to D AB to ½ power. This is the Penetration Theory model, where a molecule will disappear by reaction after absorption of a short distance.

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Example 4

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