Is the author integrating constants?

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The discussion focuses on the integration of constants in the context of calculating a velocity profile for annular flow. Peter questions the relocation of the R constant into the logarithmic expression during integration. Participants clarify that the integration process involves a change of variables, specifically u = r/R, which leads to the correct formulation of the logarithmic term as ln(r/R). They emphasize that R is treated as a constant throughout the integration, and the integration constant C2 is necessary for boundary conditions.

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Peter Schles
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Dear Sirs,

I am currently calculating a velocity profile of an annular flow. Unfortunatelly I am not understanding the following step:

http://[url=https://postimg.org/image/vl256ffhj/][ATTACH=full]200119[/ATTACH]

That seems the author had integrated the R constant. And remains the question: why had R been realocated into the ln´s parenthesys?

Thanks,
Peter
 

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The first term R*r/R is obvious so I assume that your question regards the second term. Make a change of variables u = r/R, so that dr = R du. This R is pulled out in front making R^2, while int{du/u} becomes ln(u) = ln(r/R). There's no integration of constants.
 
No, he integrates rightly over ##r##. How do you come to a different conclusion. Note that
$$\frac{\mathrm{d}}{\mathrm{d} r} \ln(r/R)=\frac{1}{r}$$.
Rightly he avoids a dimensionful logarithm by introducing an arbitrary constant. You need initial/boundary conditions anyway to fix the integration constant ##C_2##. So that's the correct general solution of the ODE (2.4-5).
 
Hi, before the integration you can multiply and divide by ##R##, one of this remain outside the parentesis so the ##R^2##, after observe that ##\frac{1}{R}\left(\frac{R}{r}\right)## is ##\frac{d}{dr}\ln{\frac{r}{R}}##. Yes, here ##R## is trated as constant and ##C_{2}## is the constant of integration that remains multiplied by the constant term in front of the bracket.
 

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