ChE 541 Homework 6 Fall 2012Mass Transfer Due 10/31/12 M. Sahimi1. Modeling of Intercellular Calcium Response of Endothelial Cells. Part II: TheSolutionIn this part of the analysis we wish to derive the solution of the problem that we formulatedin Part I. Recall that we used the method of separation of variables to write the solutionas, C = X(ξ)Y (γ). The solution for the function X(ξ) is obtained straightforwardly andis given by, X(ξ) = B exp(−λ2ξ/6), where λ2is the separation of the variables constant,and B is a constant.(a) If Part I is solved correctly, the governing equation for Y (γ) should be,d2Ydγ2+ λ2(γ − γ2)Y = 0 . (1)Equation (1) can be transformed to the following equation,zd2ωdz2+ (b − z)dωdz− aω = 0 , (2)which is called the confl uent hypergeometric equation, the solution of which is known.Let, a =14−λ16, b = 1/2, ω = γAexp[f(γ)]Y (γ), and z = KH(γ), where A and K are twoconstants, and f (γ) and H(γ) two fun ctions to be determined. Substitute these in Eq.(2) to convert it to an equation for Y . Then, compare the resulting equation with Eq. (1).Show that one obtains, A = 0, K = λ, H(γ) = (γ −12)2, and f(γ) = z/2 =12λ(γ −12)2.(b) The solution of Eq. (2) is given byω = C1M(a, b, z) + C2√zM(1 + a − b, 2 − b, z) , (3)where C1and C2are constant, and M(a, b, z) is called the confluent hypergeometric func-tion (recall ChE 501!), given byM(a, b, z) =∞Xn=0(a)n(b)nn!zn,with (a)n= a(a + 1)(a + 2) ······(a + n −1) (and similarly for (b)n), and (a)0= (b)0= 1.Use the reactive boundary condition to obtain a relation between C1and C2. Then, usethe second boundary condition at the non-reactive, impermeable wall to derive a secondrelations between the two constants. Combine the two equations to obtain the governingequation for λ and solve one of the two constants in terms of the other one.(c) The solution is now written asC =∞Xn=0Cnexp(−λ2nξ/6 − zn/2)[M(an, 1/2, zn) −√znαn/βnM(an+ 1/2, 3/2, zn)] ,where, zn= λn(γ −12)2, an=14−λn16, and αnand βnare two constants that arise whenyou solve for C1or C2in terms of the other; identify αnand βn. We must then determineCn. For this purpose we use the last boundary condition, and the orthogonal propertiesof the function Yn. Then,Cn=Z10(γ − γ2)YndγZ10(γ − γ2)Y2ndγ. (4)Consider, first, the numerator of Eq. (4). Equation (1) is the governing equation for Ynwith λ replaced by λnand Y by Yn. Integrate Eq. (1) from 0 to 1. This would yield thenumerator. Then, multiply both sides of Eq. (1) by Ynand integrate it from 0 to 1. Thiswould give an expression for the denominator. Hence, give the expression for Cn, whichcompletes the solution.2. Drug Delivery and Release. Part II: Spherical Drug-Delivery Material with noSurface Coating(a) Repeat the analysis of Part I, but assume that the spherical drug-delivery material isnot coated on its surface.(b) Compare the solutions of the two parts. What conclusion can you draw?3. Stretching a Polymeric Sheet to Remove Volatile Monomers. Part II: Analysisof Their TransportIn this part of the problem we wish to study the transport of the volatile monomers inthe polymer sheet. In order to do that, the results of Part I must be utilized.(a) Write down th e governing equation for the concentration C(x, y, t) of the monomers inthe two-dimensional polymer sheet. Keep in mind that convection cannot be neglected,but the fluid velocities vxand vywere determined in Part I. Assume that the initialconcentration is uniform and equal to C0everywhere. Write down all the boundaryconditions.(b) We now assume that the polymer sheet’s length is large enough that the edge effectat x = L(t) can be neglected. This means that diffusion and convection in the x directioncan be neglected. Then, suppose that, η = y/δ(t), and τ = t/f(t), where f(t) is a functionto be determined, δ(t) is the half-thickness of the sheet defined and determined in PartI, and let, θ = C/C0. Rewrite the governing equation in terms of the new dimensionlessvariables.(c) Derive an equation for f(t) in such a way that the equation in (b) reduces to a diffusionequation.(d) Use the equation for f(t) to show that, f(0) = δ20/D.(e) Solve the equation for