## HW6

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## HW6

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- School:
- Texas A&M University
- Course:
- Meen 344 - Fluid Mechanics

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ChE 541 Mass Transfer Homework 6 Due 10 31 12 Fall 2012 M Sahimi 1 Modeling of Intercellular Calcium Response of Endothelial Cells Part II The Solution In this part of the analysis we wish to derive the solution of the problem that we formulated in Part I Recall that we used the method of separation of variables to write the solution as C X Y The solution for the function X is obtained straightforwardly and is given by X B exp 2 6 where 2 is the separation of the variables constant and B is a constant a If Part I is solved correctly the governing equation for Y should be d2 Y 2 2 Y 0 d 2 1 Equation 1 can be transformed to the following equation z d2 d b z a 0 2 dz dz 2 which is called the confluent hypergeometric equation the solution of which is known Let a 14 16 b 1 2 A exp f Y and z KH where A and K are two constants and f and H two functions 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 H 21 2 and f z 2 21 21 2 b The solution of Eq 2 is given by C1 M a b z C2 zM 1 a b 2 b z 3 where C1 and C2 are constant and M a b z is called the confluent hypergeometric function recall ChE 501 given by M a b z X a n n z n 0 b n n 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 C1 and C2 Then use the second boundary condition at the non reactive impermeable wall to derive a second relations between the two constants Combine the two equations to obtain the governing equation for and solve one of the two constants in terms of the other one c The solution is now written as C X n 0 Cn exp 2n 6 zn 2 M an 1 2 zn zn n n M an 1 2 3 2 zn where zn n 21 2 an 41 16n and n and n are two constants that arise when you solve for C1 or C2 in terms of the other identify n and n We must then determine Cn For this purpose we use the last boundary condition and the orthogonal properties of the function Yn Then Z 1 Cn Z 01 0 2 Yn d 2 4 Yn2 d Consider first the numerator of Eq 4 Equation 1 is the governing equation for Yn with replaced by n and Y by Yn Integrate Eq 1 from 0 to 1 This would yield the numerator Then multiply both sides of Eq 1 by Yn and integrate it from 0 to 1 This would give an expression for the denominator Hence give the expression for Cn which completes the solution 2 Drug Delivery and Release Part II Spherical Drug Delivery Material with no Surface Coating a Repeat the analysis of Part I but assume that the spherical drug delivery material is not 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 Analysis of Their Transport In this part of the problem we wish to study the transport of the volatile monomers in the polymer sheet In order to do that the results of Part I must be utilized a Write down the governing equation for the concentration C x y t of the monomers in the two dimensional polymer sheet Keep in mind that convection cannot be neglected but the fluid velocities vx and vy were determined in Part I Assume that the initial concentration is uniform and equal to C0 everywhere Write down all the boundary conditions b We now assume that the polymer sheet s length is large enough that the edge effect at x L t can be neglected This means that diffusion and convection in the x direction can be neglected Then suppose that y t and t f t where f t is a function to be determined t is the half thickness of the sheet defined and determined in Part I and let C C0 Rewrite the governing equation in terms of the new dimensionless variables c Derive an equation for f t in such a way that the equation in b reduces to a diffusion equation d Use the equation for f t to show that f 0 02 D e Solve the equation for f t

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