tailieunhanh - Computational Fluid Mechanics and Heat Transfer Third Edition_18

Tham khảo tài liệu 'computational fluid mechanics and heat transfer third edition_18', kỹ thuật - công nghệ, điện - điện tử phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | 652 An introduction to mass transfer which yields NsL I x1 e N1 s Ns c D12 X1 s Ni s Ns or Ns c D12 ln 1 x1 e x1 s L x1 s N1 s Ns The last boundary condition is the value of N1s Ns. Since we have allowed for the possiblity that species 2 passes through the bottom of the layer N1s Ns may not equal unity. The ratio depends on the specific problem at hand as shown in the two following examples. Example Find an equation for the evaporation rate of the liquid in the Stefan tube described at the beginning of this section. Solution. Species 1 is the evaporating vapor and species 2 be the stationary gas. Only vapor is transferred through the s-surface since the gas is not significantly absorbed into the already gas-saturated liquid. Thus N2 s 0 and Ns N1 s Nvapor s is simply the evaporation rate of the liquid. The s-surface is just above the surface of the liquid. The mole fraction of the evaporating liquid can be determined from solubility data for example if the gas is more-or-less insoluble in the liquid Raoult s law eqn. maybe used. The e-surface is at the mouth of the tube. The gas flow over the top may contain some concentration of the vapor although it should generally be near zero. The ratio N1s Ns is unity and the rate of evaporation is Ns Nvapor s D In 1 X1 e xj Example What is the evaporation rate in the Stefan tube if the gas is bubbled up to the liquid surface at some fixed rate Ngas Solution. Again N1 s Nvapor s is the evaporation rate. However the total mole flux is Ns Ngas N1 s Steady mass transfer with counterdiffusion 653 Thus Ngas N1 s cD12 ln 1 x1 e - x1 s L _ x1 s N1 s N1 s Ngas This equation fixes N1s but it must be solved iteratively. Once we have found the mole fluxes we may compute the concentration distribution x1 y using eqn. N1 s I X1 y N X1 s N1 s Ns exp Nsy cD12 Alternatively we may eliminate Ns between eqns. and to obtain the concentration distribution in a form that