Is the author integrating constants?

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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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  • Two textbook equations: dv_z/dr = -((P_0 - P_L)R/(2μL))[(r/R) - λ²(R/r)], labelled (2.4-5), and the integrated result v_z = -((P_0 - P_L)R²/(4μL))[(r/R)² - 2λ² ln(r/R) + C_2], labelled (2.4-6).
    Integration.png
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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.