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Coordinate Transformation of the equation of continuity for a vaporizing droplet

  1. Oct 28, 2014 #1
    Hey there,

    I trying to understand the following coordinate transformation of the equation of continuity (spherical coordinates) for a vaporizing liquid droplet[tex]\frac{\partial \rho}{\partial t} + \frac{1}{r^2} \frac{\partial}{\partial r} (r^2 \rho v) = 0[/tex] into [tex]\epsilon \sigma \frac{\partial \gamma}{\partial \sigma} + \frac{\partial \gamma}{\partial \tau} + \frac{1}{\sigma^2} \frac{\partial \mu}{\partial \sigma} = -\gamma \frac{\partial ln\rho_0}{\partial \tau}[/tex] with [tex]\sigma=r/r_0,\\ \tau=\int \rho D/\rho_0 r_0^2\, dt,\\ \gamma=\rho/\rho_0,\\ \mu=r^2\rho v/r_0\rho D \\ \epsilon=-(r_0\rho_0/\rho D)dr_0/dt[/tex]where [tex]\rho D=const.[/tex] and subscript 0 means condition at the droplet surface.
    This transformation comes from F.A. Williams "On the Assumptions Underlying Droplet Vaporization and Combustion Theories" J. Chem. Phys. 33, 133 (1960); doi: 10.1063/1.1731068

    Actually I only have problems reproducing the first term, so I appreciate if someone can guide me through this transformation.
  2. jcsd
  3. Oct 29, 2014 #2
    Some questions:

    What is D, the diffusion coefficient?
    Why is the density varying with time (and position?)?
  4. Oct 30, 2014 #3
    Yes, D is the diffusion coefficient. The density is vary in time and space since it represents the mass at the point r in every instance t.

    [tex]\frac{\partial \rho}{\partial t}=\frac{\partial}{\partial t} \gamma\rho_0=\gamma \frac{\partial \rho_0}{\partial t}+\rho_0\frac{\partial \gamma}{\partial t}[/tex]which may give the second term on the left and the one on the right by using [tex]\frac{d\tau}{dt}=\frac{\rho D}{\rho_0 r_0^2}[/tex] But I'm not sure if I could only divide [tex]\tau \ \text{by}\ dt \text{, since}\ \rho_0 r_0^2\ \text{are both time dependent}[/tex]However, I'm assumed that [tex]\tau=\int_0^\tau d\tau[/tex]so that the former division should be valid, right?
    For the [tex]\frac{1}{r^2}\frac{\partial}{\partial r}(r^2\rho v)[/tex] I got [tex]\frac{1}{r_0 \sigma^2}\frac{\partial}{\partial \sigma}(\frac{r^2\rho v}{\rho D})[/tex] taking into account [tex]\frac{\rho D}{\rho_0 r_0^2}[/tex] from the time derivative and ρD=const.. Thus leading to [tex]\frac{1}{r_0 \sigma^2}\frac{\partial}{\partial \sigma}(\frac{r^2\rho v}{\rho D})=\frac{1}{\sigma^2}[\frac{\partial}{\partial \sigma}\mu-\frac{r^2\rho v}{\rho D}\frac{\partial}{\partial \sigma}\frac{1}{r_0}][/tex]. The first term is in accordance with the third term on the left of the solution. However, I got headache getting the first term on the left of the solution from [tex]-\frac{r^2\rho v}{\rho D}\frac{\partial}{\partial \sigma}\frac{1}{r_0}[/tex]
    Last edited: Oct 30, 2014
  5. Oct 30, 2014 #4
    You still haven't adequately articulated for me what is actually going on here physically. I'm beginning to get the idea that you are making a change of variable to remap the drop radial coordinate, by scaling it by the instantaneous radius of the drop. I still don't know why the density is varying with time and position unless (a) the temperature is also changing or (b) you have more than one component in the liquid phase (only one of which is volatile). I'm also wondering why the diffusion coefficient is present in the overall mass balance equation, unless D represents the diffusion coefficient in the gas phase outside the drop.

    You need to provide more details on this problem: fewer equations and more words.

  6. Oct 30, 2014 #5
    I think I may be able to help. If f is a function of t and r, then

    [tex]df=\left(\frac{\partial f}{\partial t}\right)_{\sigma}dt+\left(\frac{\partial f}{\partial \sigma}\right)_{t}d\sigma[/tex]
    So, [tex]\left(\frac{\partial f}{\partial t}\right)_r=\left(\frac{\partial f}{\partial t}\right)_{\sigma}+\left(\frac{\partial f}{\partial \sigma}\right)_{t}\left(\frac{\partial \sigma}{\partial t}\right)_r[/tex]
    [tex]\left(\frac{\partial f}{\partial t}\right)_r=
    \left(\frac{\partial f}{\partial t}\right)_{\sigma}
    -\left(\frac{\partial f}{\partial \sigma}\right)_{t}
    \frac{\sigma}{r_0} \frac{dr_0}{dt}[/tex]

    The second term here corresponds to your missing term.

    Last edited: Oct 30, 2014
  7. Oct 30, 2014 #6
    Ok I'm sorry for that.
    The continuity equation is only the simplest one of the transformed equations. The hole set of equations comprise in addition the "species conservation equation" and "energy conservation equation". These equations form the differential equations for a droplet vaporizing or burning in quiescent air.
    The density is changing due to temperature changes (in the droplet as well as outside) and due to different species (only outside in the gas phase - the liquid contains only one component). The diffusion coefficient represents the value in the gas phase outside the droplet due to the different species (air for vaporizing case and products for burning case).
  8. Oct 31, 2014 #7
    First of all, thank you very much dealing with this problem.
    Of course, the total derivative will give the missing (first) term of the solution if I put [tex]df=d\gamma[/tex]. But this leaves me with the residual partial derivatives, which are not in accordance with my former route of transformation. Maybe I did mistakes already before.....
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