TAMU MEEN 344 - HW8 (3 pages)

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HW8



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HW8

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School:
Texas A&M University
Course:
Meen 344 - Fluid Mechanics
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ChE 541 Mass Transfer Homework 8 Due 11 14 12 Fall 2012 M Sahimi 1 Early Times Solution for Diffusion in a Membrane In homework 2 problem 2 we analyzed diffusion in a membrane under pseudo steadystate condition In the present problem we wish to analyze the same problem for early times when the PSS assumption is not valid We also assume that similar to the practice the concentration of the species at z 0 is a time dependent function C1 t If we introduce the dimensionless concentration CA C1 t then the governing equation for CA is the standard one dimensional unsteady state diffusion equation with the initial condition z 0 0 In the previous problem the thickness of the membrane was finite But since we seek the early times solution the opposite side of the membrane is not felt by the species diffusing into the membrane As a result we assume that t 0 a Introduce a similarity variable z 4Dt and rewrite the governing equation in terms of b Show that the similarity technique works only if C1 t at where a and are constant c With such a choice for C1 t derive the governing equation for Specify its boundary conditions d Introduce a new variable 1 2 exp 2 2 w 2 where w 2 is a function to be determined and show that the governing equation for w is given by d2 w 1 1 4 2 w 0 d 2 4 2 where 2 1 4 and 1 4 This equation is called the Whittaker s equation named in honor of Edward Taylor Whittaker 1873 1956 the great English mathematician Consult mathworld wolfram com WhittakerDifferentialEquation html and write down the solution of the equation for w see also homework 6 problem 1 e Derive an expression for the flux Nz of the species at z 0 2 More on Unsteady State Evaporation As discussed in the class solving the problem of unsteady state evaporation reduces to solving the differential equation d2 f df 0 2 2 d d df xA0 1 0 2 1 xA0 d when we use the similarity variable The solution of the equation was derived in the class and as a bonus problem Show how for a given value of xA0 one determines Compute numerically for xA0 0 5 3 Increasing Production of a Catalyst by Fluid Flow One way of increasing the rate of reaction in a catalyst particle is by using flow through its pore space so that the delivery of the reactants to the active sites can be more efficient and faster Consider a catalyst particle with a slab shape with a large surface so that diffusion and flow are important only in one direction say x The concentrations of the reactant at the upstream x L and downstream x 0 are C0 and 0 respectively There is fluid flow in the slab in the x direction with a constant velocity v The reaction is second order RA kC 2 and the phenomenon is at steady state a Write down the governing equation for the concentration b Let C C0 and x L Rewrite the governing equation in terms of and If done correctly two dimensionless groups the Damko hler number Da and the Pe clet number Pe should emerge Provide the expressions for both c Let us assume that the reaction is fast relative to diffusion but not so compared with convection so that Da 1 but Da Pe O 1 O 1 means order of 1 Then Da 1 is a parameter in terms of which the governing equation may be solved Rewrite the equation in terms of and d First set 0 i e Da and solve the simplified equation with the boundary condition at 1 Show that the solution does not satisfy the boundary condition at 0 Why Such problems are called singular perturbation problems There is a boundary layer near 0 i e a region in which the solution with 0 is incorrect which means that no matter how fast is the reaction there is always a region the boundarylayer region in which it is slower than diffusion recall boundary layer problems in fluid mechanics Here we learn how to solve such problems e Next let Y b Substitute this in the governing equation in c and select b in such a way that the coefficients of the first and second derivatives have the same power of Then give the final differention equation for Y f Thus we divide the domain of the problem into two regions i An outer region outside the boundary layer and an inner or boundary layer region For the outer region we write 0 1 O 2 Substitute this into the governing equation in c to derive the governing equations for the two functions 0 and 1 Solve for 0 What does the solution mean g For the boundary layer region we write Y 0 Y 1 Y O 2 Substitute this in the governing equation in e to derive the differential equations for 0 Y and 1 Y Solve the equation for 0 Y h The solution for 0 Y should contain a constant Determine the constant by using the matching condition namely lim Y lim Y 0 Then the composite solution for is given by 0 Y 0 limY 0 Y O 2 Give the solution The solution is valid for small For larger values of smaller Da one must also compute the 2 terms


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