Team:Guanajuato Mx/Difussion

iGEM Guanajuato Mx

MODEL

Difussion through the nitrocellulose membrane

Mass balance of A through the nitrocellulose membrane

\(A\) = Bactericide
\(B\) = Infecting agent

Assumptions:

Steady state

Constant temperature

\(C_A\) is unidimensional
\(C_A = C_A(Z)\)

Low concentration
(Fick's Law)

\(m \) = polymer permeability

Mass balance, flux of A times the area (input minus output): $$\pi r^2J_A\rvert_{_Z} - \pi r^2J_A\rvert_{_{Z+\Delta Z}} = 0$$ Divided by \(\pi r^2\Delta Z\) (control volume) and limit as \(\Delta Z\) approaches to zero: $$\lim_{\Delta Z \to 0} \frac{J_A\rvert_{_Z} - J_A\rvert_{_{Z+\Delta Z}}}{\Delta Z} = 0$$ $$-\frac{d J_A}{dZ}=0$$ Fick's Law Equation: $$J_A = -\partial_{AB}\frac{d C_A}{dZ}$$ Replacing Fick's Law into the previous equation: $$\partial_{AB}\frac{d^2 C_A}{dZ^2}=0$$ Solving the differential equation: $$C_A = c_1Z + c_2$$ Boundary conditions: $$Z = 0 \hspace{1.3cm} C_A = mC_{A_0}$$ $$Z = L \hspace{1.3cm} C_A = mC_{A_L}$$ B.C. 1: $$C_A = mC_{A_0} = c_2$$ $$C_A = c_1Z + mC_{A_0}$$ B.C. 2: $$C_A = mC_{A_L} = c_1L + mC_{A_0}$$ $$c_1 = \frac{m(C_{A_L}-C_{A_0})}{L}$$ Final solution: $$C_A = m(C_{A_L}-C_{A_0})\frac{Z}{L} + mC_{A_0}$$ With this equation it is possible to know the concentration of the bactericide (\(C_A\)) at any point within the nitrocellulose membrane (\(Z\)).

Non-equimolar counter-difussion of A and B

\(A\) = Bactericide
\(B\) = Infecting agent

Assumptions:

Steady state

Constant temperature

\(C_A\) is unidimensional
\(C_A = C_A(Z)\)

Low concentration
(Fick's Law)

\(m \) = polymer permeability

\(C\) = Medium concentration Total flux of A (diffusion plus convection) $$N_A = -C\partial_{AB}\frac{d Y_A}{dZ} + Y_A(N_A + N_B)$$ Mass balance, flux of A times the area (input minus output): $$\pi r^2N_A\rvert_{_Z} - \pi r^2N_A\rvert_{_{Z+\Delta Z}} = 0$$ Divided by \(\pi r^2 \Delta Z\) (control volume) and limit as \(\Delta Z \to 0\) $$-\frac{dN_A}{dZ} = 0 \rightarrow N_A = constant$$ $$N_A = -C\partial_{AB}\frac{d Y_A}{dZ} + Y_A(N_A + N_B)$$ $$N_B = -nN_A$$ Combining the two equations: $$N_A = -C\partial_{AB}\frac{dY_A}{dZ} + Y_A(N_A-nN_A)$$ $$N_A - Y_AN_A(1-n) = -C\partial_{AB}\frac{dY_A}{dZ}$$ $$N_A [1 - Y_A(1-n)] = -C\partial_{AB}\frac{dY_A}{dZ}$$ Boundary conditions: $$Z = 0 \hspace{1.3cm} C_A = mC_{A_0} = mCY_{A_0}$$ $$Z = L \hspace{1.3cm} C_A = mC_{A_L} = mCY_{A_L}$$ $$C_A = Y_AC$$ $$Y_A = \frac{C_A}{C}$$ $$N_A = -\frac{C\partial_{AB}}{1-Y_A(1-n)}\frac{dY_A}{dZ}$$ $$N_AdZ = -\frac{C\partial_{AB}}{1-Y_A(1-n)}dY_A = -\frac{C\partial_{AB}}{1-Y_A(1-n)}d\left(\frac{C_A}{C}\right)$$ $$1-Y_A(1-n) = 1 - \frac{C_A}{C}(1-n) = \frac{C-C_A(1-n)}{C}$$ $$\int_0^L N_A dZ = - \frac{\partial_{AB}}{\frac{C-C_A(1-n)}{C}}dC_A = \int_{mC_{A_0}}^{mC_{A_L}} -\frac{C\partial_{AB}}{C-C_A(1-n)} dC_A$$ $$N_A L = -C\partial_{AB} \int_{mC_{A_0}}^{mC_{A_L}} \frac{dC_A}{C-C_A(1-n)}$$ Integration by substitution:
\(u = C-C_A(1-n)\)
\(du = -(1-n)dC_A\)

\( = \int \frac{du}{u} = -\ln u\) $$N_A L = C\partial_{AB} \ln \left[ \frac{C-mC_{A_L}(1-n)}{C-mC_{A_0}(1-n)} \right]$$ $$N_A = \frac{C\partial_{AB}}{L} \ln \left[ \frac{C-mC_{A_L}(1-n)}{C-mC_{A_0}(1-n)} \right]$$ $$N_A = \frac{-C\partial_{AB}}{\frac{C-C_A(1-n)}{C}} \frac{d\left( \frac{C_A}{C} \right)}{dZ}$$ $$N_A = \frac{-C\partial_{AB}}{C-C_A(1-n)} \frac{dC_A}{dZ} = c_1 $$ $$\frac{-C\partial_{AB}}{C-C_A(1-n)}dC_A = c_1 dZ$$ $$C\partial_{AB} \ln [C-C_A(1-n)] = c_1Z + c_2$$ B. C. $$Z = 0 \hspace{1.3cm} C_A = mC_{A_0}$$ $$Z = L \hspace{1.3cm} C_A = mC_{A_L}$$ B. C. 1: $$C\partial_{AB} \ln [C-mC_{A_0}(1-n)] = c_2$$ $$C\partial_{AB} \ln [C-mC_A(1-n)] = c_1Z + C\partial_{AB} \ln [C-\ln C_{A_0}(1-n)]$$ B. C. 2: $$C\partial_{AB} \ln [C-mC_{A_L}(1-n)] = c_1L + C\partial_{AB} \ln [C-\ln C_{A_0}(1-n)]$$ $$c_1 = \frac{C\partial_{AB}}{L}\ln \left[ \frac{C-mC_{A_L}(1-n)}{C-mC_{A_0}(1-n)} \right]$$ Replacing: $$C\partial_{AB} \ln [C-C_A(1-n)] = C\partial_{AB}\left( \frac{Z}{L}\right) \ln \left[ \frac{C-mC_{A_L}(1-n)}{C-mC_{A_0}(1-n)} \right] + C\partial_{AB} \ln [C-mC_{A_0}(1-n)]$$ Doing the respective mathematical analysis: $$\ln \left[ \frac{C-C_A(1-n)}{C-mC_{A_0}(1-n)} \right] = \ln \left[ \frac{C-mC_{A_L}(1-n)}{C-mC_{A_0}(1-n)} \right]^{Z/L}$$ Then: $$\frac{C-C_A(1-n)}{C-mC_{A_0}(1-n)} = \left[ \frac{C-mC_{A_L}(1-n)}{C-mC_{A_0}(1-n)} \right]^{Z/L}$$ Later on: $$C-C_A(1-n) = [C-mC_{A_0}(1-n)]\left[ \frac{C-mC_{A_L}(1-n)}{C-mC_{A_0}(1-n)} \right]^{Z/L}$$ Finally: $$C_A = \frac{C - [C-mC_{A_0}(1-n)]\left[ \frac{C-mC_{A_L}(1-n)}{C-mC_{A_0}(1-n)} \right]^{Z/L}}{1-n}$$ With this equation we can calculate the concentration of bactericine (\(C_A\)) at any point within the nitrocellulose membrane (\(Z\))

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