Transport Phenomena: Mass Transfer Procedure 1) Mass Balance Over A

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Transport Phenomena: Mass Transfer Procedure 1) Mass balance over a thin shell 2) Obtain 1st ODE 3) Insert the relation between mass flux and concentration gradient 4) Result in 2nd ODE 5) Integration 6) Apply BCs to determine the constants (from integration) 7) Obtain concentration distribution The law of conservation of mass of species A in a binary system is written over the volume of the shell in the form Rate of Rate of rate of production of Mass of - Mass of - mass of A by =0 A in A out homogeneous reaction Mass flux = the number of moles of A that go through a unit area in unit time. Combined molecular convective flux flux flux (Diffusive flow) (Due to Bulk motion of A) Boundary conditions 1) The concentration at surface can be specified; for example . 2) The mass flux at surface can be specified; for example, . 3) Solid surface substance A is lost to a surrounding stream 4) Rate of chemical reaction at the surface can be specified; for example, Example 1 Diffusivity through a stagnant gas (gas-air) Air flow B S: Cross sectional area z2 z1 A Loop A z 0 Figure 1 Steady-state diffusion of A through stagnant B with the liquid vapor interface maintained at a fixed position. Assumption 1) B does not dissolve in liquid A 2) A does not react to B 3) Temperature and pressure is constant Mole A balance around shell Divided by and taking limit we obtain Mass diffusion in binary system From Fick’s law; * is the molar flux of component A in the z direction due to molecular diffusion. * is the molecular diffusivity of the molecule A in B Mass convection in binary system From Binary system: From molar relative fixed co-ordinate: Set equation (3) = equation (4) we get Since B is stagnant, We obtain the molar flux of A with Substitute equation (5) into equation (1) we get Divided through the equation by we obtain Integration with respect to z gives Second integration then gives If we replace by and by , the above equation becomes and may then be determined from the boundary conditions BC1: BC2: Apply BC1, Apply BC2, Divided equation (7) by equation (8) gives Substitute into equation (7) we obtain Substitute and into the equation (6) we get The profile of gas A are The profile of gas B are obtained by using .

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