Material Science > QUESTIONS & ANSWERS > CBE 120C – Mass Transfer Spring 2020, Homework 2 Solution University of California, Irvine - CBEMS (All)

CBE 120C – Mass Transfer Spring 2020, Homework 2 Solution University of California, Irvine - CBEMS 120CCBE120C-HW2-Soln

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CBE 120C – Mass Transfer Spring 2020, Homework 2 Solution PROBLEM 1 !N A = −CDAB !∇ yA + yA !N A + !N ( B) Assuming C = constant, we get: !N A = −DAB !∇ C A + yA !N A + !N... ( B) To convert to mass form, multiply by MA (molecular weight of A): !N AM A = −M ADAB !∇ C A + yAM A !N A + !N ( B) Note that: N AM A = nA and ω A = yAM A yAM A + yBM B , which gives: !n A = −DAB !∇ M ( ACA) +ω A(yAM A + yBM B)(N! A + N! B) !n A = −DAB !∇ ρ A +ω A y AM A + yBM B ( )(N! A + N! B) !n A = −DAB !∇ ρ A +ω A M avg !N A + M avg !N ( B) Now only if M A ≈ M B = M avg , we have: !n A = −DAB !∇ ρ A +ω A !n A + !n ( B) Therefore, the assumptions made are: C = const and M A ≈ M B Note that, alternatively, you can derive the same final equation by assuming that the system is incompressible. But you have to start with the mass form of the flux equation. PROBLEM 2 The mixture is dilute in both A and B. So the mass transfer of each of those species can be considered to be mediated by DA,H2O and DB,H2O, respectively, instead of a full mixture diffusivity. This is the pseudo-binary approximation, and based on the premise that, as molecules of A or B are diffusing through the mixture, they primarily only see water molecules around them (because of dilute conditions). So their diffusivity is not really affected by the presence of the other species, and they can be considered to be diffusing through water only [Show More]

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