Exercise 2 Solution

David B. McWhorter

Exercise 2 Solution

Start with Equation 5, which defines the non-equimolar flux.

$\displaystyle N^{D}\equiv N_{A}^{D}+N_{B}^{D}=(1+N_{B}^{D}/N_{A}^{D})N_{A}^{D}$

Use Graham’s Law to obtain the following expression.

$\displaystyle N^{D}=(1-M_{AB}^{0.5})N_{A}^{D}$

Then replace the total diffusion flux of component A using Equation 23 to obtain the desired constitutive equation (Equation Exercise Solution 2-1).

$\displaystyle N^{D}=-\frac{\left ( 1-M_{AB}^{0.5}DCdx_{A}/dl \right )}{1-\left ( 1-M_{AB}^{0.5} \right )x_{A}}$ $\displaystyle =-\frac{\left ( 1-M_{BA}^{0.5}DCdx_{B}/dl \right )}{1-\left ( 1-M_{BA}^{0.5} \right )x_{B}}$

(Exercise Solution 2-1)

Underpinning this development is the prescription of uniform pressure. However, Section 5.2 illustrates that Equation Exercise Solution 2-1 is a satisfactory approximation under non-isobaric conditions when diffusion that is driven by variable pressure is negligible relative to that driven by gradients of mole fraction.

Return to Exercise 2

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