Leading order solution

To obtain a unique solution of (5-79) for the pressure, the values of p(0> must be specified at the ends of the lubrication layer. The necessary boundary values come from the requirement that the solution within this inner region must match the solution in the outer domain, at the ends of the thin film. Note, however, that this would seem to suggest that the solution within the lubrication layer cannot be determined completely without simultaneously obtaining the leading-order solution in the outer region, including specifically the pressure. 

The right-hand side is the first two terms of the inner solution 6q + Pexl20, expressed in terms of outer variables. The first two terms on the two sides of (9-189) match exactly. However, there is a mismatch at O(Pe). In particular, on the right-hand side, the term T/2v/7r(—Te/p) is not yet matched with a corresponding term in the outer solution. Of course, this is to be expected because we have so far considered only the leading-order solution of 0(Pe1,12) in the outer region, likewise, the term (()(/>Pei/2) in the outer solution has not yet been matched with any term in the inner expansion, but this is, again, to be expected because a term of this form could derive only from a term O(rPe) in the inner solution, and we have not yet considered the 0(Pe) approximation in the inner expansion (9-164). 

The most important point to note is that the leading-order solution = 1 does not imply that the heat flux at the body surface is zero, even though this might at first appear to be the case. In general, the condition of matching between the inner and outer approximations within the thermal boundary layer requires not only (11—67c) but also that the spatial derivatives of 9 and should match at each level of approximation for Pr -> 0. Thus, in particular,... 

Thus, the leading order solutions in the subdomain (0,A ) (Eqs. 12.85 and 12.88) suggest that in the inner core of the particle, there is no solute of any form, either in the free or adsorbed state. 

Therefore the leading-order solution is of potential flow. Moreover the solutions can take a separable form, e.g.,... 

This indicates that the velocity field can be described by a potential function (p(x,y,t) in such a way that u = V(/). As usual, this inviscid solution must satisfy the botmdary condition for the normal component of the velocity at each of the walls but not the tangential component. This is the most important part of the analysis, because otherwise the leading-order solution is not of the progressive-wave type. Applying the conditions given in Eqs. 30b and 30d to the Laplace equation = 0 and solving for (p, we obtain 4> = (cosh y/sinh fi)sm(x t) and... 

It is notorious that analytical solutions for the case of nonlinear wall reactions are substantially more difficult to obtain. Lopes et al. [40] attempted to achieve some insight on this problem by considering the limits of a slow and fast reaction. In this case, the conversion profile is perturbed fi om the leading-order solutions for the cases when Da—yQ andDa->-oo, respectively. The calculation of this perturbation resembles the problem of... 

In the boundary layers, Eqs. (154,156) yield poz = woz = Poz = Oz = 0, hence these functions are constants. Thus BC (162,164,164) can be directly applied to po and Wq. Finally, the leading order solution in the bulk coincides with that in the inviscid liquid case ... 

In the problem discussed, e = 2.5 10 ", and, hence, in Equation 5.201, the second derivative 9 l>/9z is multiplied by a very small factor = 6.25 10 . The smallness of e allows obtaining the leading-order solution to Equation 5.216, neglecting the left-hand side of this equation. Setting in Equation 5.216 e = 0, one arrives at... 

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