IntroReactionDiffusion [read]

Nb 5 it follows from bc1 that k2 c0 a log 1 34 n1 c

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Unformatted text preview: rom BC2 we have c0 A n1 ∆ Log 1 K1 C n1 (35) 0 C AB Solving for K1 gives C AB K1 c0 A Log 1 n1∆ (36) n1 C So finally we have the following expression for the concentration profile cA Log 1 C n1 AB c0 A Log 1 n1∆ C n1 z Log 1 C c0 A n1 (37) C Simplifying gives cA Log 1 n1 Log 1 c0 A C C n1 1 AB z (38) n1∆ C which can also be written as C C 1 AB z c0 A Log 1 C n1 ∆ cA z n1 (39) 1 n1 Note that this expression does satisfies the BCs. given in (29) In this limiting case we can also determine the flux C AB c0 A Log 1 A n1∆ A which is given by K1 (see Eq. 21 and Eq. 30 ) (40) n1 C Concentration Profile when Φ= (1) When Φ 1 we cannot assume that cA ∆ cA AB 1 0. Instead we must solve the following system A cA C z n1 (41) BC1 : cA 0 BC2 : A c0 A k cA ∆ We can proceed as before to get Log 1 z cA n1 n1 (42) A C C AB Integrating with respect to z gives and applying BC 1 gives Log 1 cA C n1 n1 z Log 1 A C AB c0 A C n1 (43) 6 IntroReactionDiffusion.nb Now we can apply BC2 to find an equation for determining c0 A n1 A Log 1 A n1 ∆ Log 1 A kC C This is a transcendental equation for A (44) n1 C AB in which the only unknown is A. Example Calculation Let us specify parameters so that we can determine a concentration profile for a given reaction. We will introduce a set of dimensionless variables. If we let Φ A Log 1 A n1 kC k∆ n 1 Φ2 AB Log 1 then we can write (43) as c0 A kC This suggests we define a dimensionless flux (45) n1 C A A k C and a dimensionless concentration cA C. Our equation for the flux becomes ΧA Log 1 A n1 n 1 Φ2 A Log 1 Χ0 n 1 A (46) while the equation for the dimensionless concentration profile becomes ΧA n 1 Log 1 where Ξ n 1 Φ2 Ξ Log 1 A Χ0 n 1 A (47) z ∆. In terms of the dimensionless variables the BC at the catalytic surface becomes: A k cA ∆ A kC k cA Ξ 1 (48) A xA 1 Solving for ΧA gives 1 ΧA 1 n1 n 1 Φ2 Ξ Χ0 n 1 A A 5 Solving for A gives 1 (49) Sample Calculation Let us suppose that Χ0 A In[16]:= n 0.6, n 3; Φ 5; sol Out[18]= 3, Φ2 FindRoot Log 1 A A n 1 A n 1 Φ2 Log 1 0.06638 Using this value for A we get the following concentration profile 0.6 n 1 , A, 0.2...
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This document was uploaded on 03/14/2014.

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