ChE 541 Homework 8 Fall 2012Mass Transfer Due 11/14/12 M. Sahimi1. Early-Times Solution for Diffusion in a MembraneIn homework 2, problem 2, we analyzed diffusion in a membrane under pseudo-steady-state condition. In the present problem we wish to analyze the same problem for earlytimes, when the PSS assumption is not valid. We also assume that, similar to th e practice,the concentration of the species at z = 0 is a time-dependent function C1(t). If weintroduce the dimensionless concentration, θ = CA/C1(t), then the governing equationfor CAis the standard one-dimensional unsteady-state diffusion equation, with the initialcondition, θ (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 memb rane is not “felt”by the species diffusing into the membrane. As a result, we assume that, θ(∞, t) = 0.(a) Introduce a similarity variable, η = z/√4Dt, and rewrite the governing equation interms of η.(b) Show that the similarity technique works only if, C1(t) = atγ, where a and γ areconstant.(c) With such a choice for C1(t), derive the governing equation for θ(η). Sp ecify itsboundary conditions.(d) Introduce a new variableθ = η−1/2exp(−η2/2)w(η2) ,where w(η2) is a function to be determined, and show that the governing equation for wis given byd2wdζ2+ −14+κζ+1/4 − µ2ζ2!w = 0 ,where, ζ = η2, κ = −(γ + 1/4), and µ = 1/4. This equation is called the Whittaker’sequation, named in honor of Edward Taylor Whittaker (1873-1956), the great Englishmathematician. Consultmathworld.wolfram.com/WhittakerDifferentialEquation.htmland write down the solution of the equation for w(ζ); see also homework 6, problem 1.(e) Derive an expression for the flux Nzof the species at z = 0.2. More on Unsteady-State EvaporationAs discussed in the class, solving the problem of unsteady-state evaporation reduces tosolving the differential equation,d2fdη2+ 2(η − φ)dfdη= 0 ,φ ≡ −12xA01 − xA0dfdη|η=0.when we use the similarity variable η. The solution of the equation was derived in theclass (and as a bonus problem). Show how for a given value of xA0one determines φ.Compute φ (numerically) for xA0= 0.5.3. Increasing Production of a Catalyst by Fluid FlowOne way of increasing the rate of reaction in a catalyst particle is by using flow throughits pore space, so that the delivery of the reactants to the active sites can be more efficientand faster. Consider a catalyst particle with a slab shape with a large surface, so thatdiffusion and flow are important only in one direction, say x. The concentrations of thereactant at the upstream, x = L, and downstream, x = 0, are C0and 0, respectively.There is fluid flow in the slab in the −x direction with a constant velocity v. The reactionis second order, RA= −kC2, and the phenomenon is at steady state.(a) Write down the governing equation for the concentration.(b) Let, θ = C/C0, and η = x/L. Rewrite the governing equation in terms of η and θ.If done correctly, two dimensionless groups, the Damk¨ohler number Da and the P´ecletnumber Pe, should emerge. Provide the expressions for both.(c) Let us assume that the reaction is fast relative to diffusion, but not so comparedwith convection, so that Da ≫ 1, but α = Da/Pe ∼ O(1) [O(1) means order of 1]. Then,ǫ = Da−1is a parameter in terms of which the governing equation may be solved. Rewritethe equation in terms of ǫ and α.(d) First, set ǫ = 0 (i.e. Da = ∞), and solve the simplified equation with the boundarycondition at η = 1. Show that the solution does not satisfy the boundary conditionat η = 0. Why? Such problems are called singular perturbation problems. There is aboundary layer near η = 0, i.e. a region in which the solution with ǫ = 0 is incorrect, whichmeans that no matter how fast is the reaction, there is always a region - the boundary-layer region - in which it is slower than diffusion (recall boundary-layer problems in fluidmechanics). Here, we learn how to solve such problems.(e) Next, let, Y = ηǫb. Substitute this in the governing equation in (c) and select b insuch a way that the coefficients of the first and second derivatives have the same powerof ǫ. Then, give the final differention equation for θ(Y ).(f) Thus, we divide the domain of the problem into two regions: (i) An outer regionoutside the boundary layer, and an inner or boundary-layer region. For the outer regionwe write, θ(η) = θ0(η) + θ1(η)ǫ + O(ǫ2). Substitute this into the governing equation in (c)to derive the governing equations for the two functions θ0and θ1. Solve for θ0(η). Whatdoes the solution mean?(g) For the boundary-layer region, we write, θ(Y ) =˜θ0(Y ) +˜θ1(Y )ǫ + O(ǫ2). Substitutethis in the governing equation in (e) to derive the differential equations for˜θ0(Y ) and˜θ1(Y ). Solve the equation for˜θ0(Y ).(h) The solution for˜θ0(Y ) should contain a constant. Determine t he constant by usingthe matching condition, namely,limY →∞θ(Y ) = limη→0θ(η) .Then, the composite solution for θ is given by, θ =˜θ0(Y ) + θ0(η) + limY →∞˜θ0(Y ) + O(ǫ2).Give the solution. The solution is valid for small ǫ. For larger values of ǫ (smaller Da)one must also compute the