# Homework_3_2008_upload - _PaulGonzales _W35 BME 314, Fall...

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____________Paul Gonzales _________________W 3-5  BME 314, Fall 2010 HW 3 Due November 23, 2010 Total points = 75 Always work problems in the units provided (if provided with English units, work in English, if provided with metric units, then work in metric). Define all equations used and provide explanatory comments to allow one to easily follow your work. Box final solutions. In problem A, B and C show all calculations and use the given values only to derive/obtain related parameters. A. SOLVE the following problems from the EXERCISES given at the end of CHAPTER 11 of your textbook (Introduction to Biomedical Engineering, Enderle, Blanchard, Bronzino, SECOND EDITION). 3. Assume that a membrane is permeable to Ca ++ and Cl _ . The initial concentrations on the inside are different from the outside, and these are the only ions in the solution. (a) Write an equation for J Ca and J Cl . J Ca = J Ca (drift) + J Ca (diffusion) J Ca = -2μ[Ca 2+ ] dv dx - KTm q d [ 2+ Ca ] dx Where the valence, Z, for Ca is 2+ J Cl = J Cl (drift) + J Cl (diffusion) J Cl = μ[Cl -1 ] dv dx - KTm q d [ - Cl ] dx Where the valence, Z, for Cl is -1 J x = flow of ions due to drift in an Electric Field μ = mobility in m 2 /sV Z = ionic valence dv dx = voltage gradient across membrane [I] = ion concentration (b) Write an expression for the relationship between J Ca and J Cl . From space charge neutrality, 2J Ca = J Cl By plugging in values from part 1a we obtain: 2(-2μ[Ca 2+ ] dv dx - KTm q d [ 2+ Ca ] dx )= μ[Cl -1 ] dv dx - KTm q d [ - Cl ] dx -4μ[Ca 2+ ] dv dx - 2 KTm q d [ 2+ Ca ] dx = μ[Cl -1 ] dv dx - KTm q d [ - Cl ] dx

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BME 314 HW #3 Fall 2010 (Roy) (c) Find the equilibrium voltage. J Ca = -2μ[Ca 2+ ] dv dx - KTm q d [ 2+ Ca ] dx ; begin with the flow rate for Calcium Under a constant electric field, dv dx = V d ; Substitute into original equation J Ca = -2μ[Ca 2+ ] V d - KTm q d [ 2+ Ca ] dx Now lets obtain a relationship for the permeability of Calcium: P Ca = KTm dq = D Ca d Using this relationship in our equation for J Ca we obtain: J Ca = - P Ca q KT V [ 2+ Ca ] - 2 P Ca d d [ 2+ Ca ] dx Rearranging the terms yields: dx = d [ 2+ Ca ] - J Ca 2 P Ca + qV [ 2+ Ca ] KTd Taking the integral of both sides, while assuming J Ca is independent of x, gives: dx 0 d ò = d [ 2+ Ca ] - J Ca 2 P Ca + qV [ 2+ Ca ] KTd [ 2+ Ca ] i [ 2+
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## This note was uploaded on 11/29/2010 for the course BME 314 taught by Professor Frey during the Fall '08 term at University of Texas.

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Homework_3_2008_upload - _PaulGonzales _W35 BME 314, Fall...

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