B. Generating a Distribution for the Droplet Radius

The random distribution for the droplet radius for the ESD spray pyrolysis model in (6.1) is derived in this section. The volume fraction is evenly distributed along the droplets whose radii range from $ r_{min}=2.5\mu m$ to $ r_{max}=55\mu m$. Therefore, the first step is to relate the radius distribution linearly to a value $ x\in\left[0,1\right]$ so that as $ x$ goes from 0 to 1, $ r_{d}$ goes from $ r_{min}$ to $ r_{max}$:

$\displaystyle r_{d}\left(x\right)=\left(r_{max}-r_{min}\right)x+r_{min},\quad\textrm{where }x\in\left[0,1\right].$ (249)

Next, the assertion is made that $ x$ represents the evenly distributed volume number fraction, or normalized volume $ \xi =x$. Using the equation for the volume of a sphere, the relationship between volume and radius is established as

$\displaystyle V_{sphere}=\cfrac{4\pi}{3}\,r_d^{3}.$ (250)

Therefore, when the volume is evenly distributed, the effect on the radius will be $ \xi_V\propto\cfrac{1}{r_d^3}$. Initially, it might be counter-intuitive to note the inverse relationship since $ V\propto r^3$. However, when a volume of $ V=100m^3$ is distributed for a radius $ r_1=10$m, then

$\displaystyle 100=N(r=10)\cdot\cfrac{4\pi}{3}(10)^{3},$ (251)

where $ N$ is the number fraction, resulting in $ N=0.0238$. When the same volume is distributed for droplets of a radius $ r_2=20$m, then the calculation above leads to the number fraction $ N(r=20)=0.00298$, which is 8 times less, or $ \left(\cfrac{r_1}{r_2}\right)^3$ times less.

Now we know that the radius distribution should follow the equation

$\displaystyle r_{\xi}=\cfrac{C}{r_d^3},$ (252)

where $ C$ is a normalization constant which must be found, $ r_{\xi}$ is the randomly distributed radius, and $ r_d$ is the radius relating $ r_{min}$ and $ r_{max}$ to an even volume distribution $ \xi_V\in\left[0,1\right]$ from (B.1)

$\displaystyle r_{\xi}=\cfrac{C}{\left[\left(r_{max}-r_{min}\right)\xi_V+r_{min}\right]^{3}}.$ (253)

Inverting (B.5) to solve for $ \xi_V$ allows to find the CPD function $ \Phi\left(r_{\xi}\right)$

$\displaystyle \Phi\left(r_{\xi}\right)=\xi_V=\cfrac{1}{r_{max}-r_{min}}\left[\left(\cfrac{C}{r_{\xi}}\right)^{1\slash3}-r_{min}\right].$ (254)

The derivative of (B.6) gives the PDF

$\displaystyle f\left(r_{\xi}\right)=\cfrac{d\Phi\left(r_{\xi}\right)}{dr_{\xi}}=-\cfrac{C^{1\slash3}}{3}\cdot\cfrac{1}{r_{max}-r_{min}}\cdot r_{\xi}^{-4\slash3},$ (255)

where it can be noted that the last $ r_{min}$ term from (B.6) has disappeared. The only non-constant term in (B.7) is $ r_{\xi}^{-4\slash 3}$, therefore a replacement constant, which will be the new normalization constant is introduced for simplicity

$\displaystyle A\equiv-\cfrac{C^{1\slash3}}{3}\cdot\cfrac{1}{r_{max}-r_{min}}$ (256)

and (B.7) can be rewritten to

$\displaystyle f\left(r_{\xi}\right)=A\cdot r_{\xi}^{-4\slash3}.$ (257)

Using the PDF from (B.9), we can now proceed to find the normalized distribution $ r_{\xi}$, but first the normalization constant $ A$ must be found by integrating (B.9) with respect to $ r_{\xi}$ from $ r_{min}$ to $ r_{max}$ and equating the integral to 1

$\displaystyle \intop_{r_{min}}^{r_{max}}f\left(r_{\xi}\right)dr_{\xi}=\intop_{r_{min}}^{r_{max}}A\cdot r_{\xi}^{-4\slash3}dr_{\xi}=1,$ (258)

which can be solved to

$\displaystyle -3A\left[\left(\cfrac{1}{r_{max}}\right)^{1\slash3}-\left(\cfrac{1}{r_{min}}\right)^{1\slash3}\right]=1,$ (259)

giving the normalization constant $ A$

$\displaystyle A=-\cfrac{1}{3\left[r_{max}^{-1\slash3}-r_{min}^{-1\slash3}\right]}$ (260)

and the normalized PDF

$\displaystyle f\left(r_{d}\right)=-\cfrac{1}{3\left[r_{max}^{-1\slash3}-r_{min}^{-1\slash3}\right]}\cdot r_{d}^{-4\slash3}.$ (261)

Now one can integrate the normalized PDF from $ r_{min}$ to $ r_{\xi}$ to find the CPD

$\displaystyle \intop_{r_{min}}^{r_{\xi}}f\left(r\right)dr=\cfrac{r_{\xi}^{-1\slash3}-r_{min}^{-1\slash3}}{r_{max}^{-1\slash3}-r_{min}^{-1\slash3}}=\xi_{V}$ (262)

and invert the CPD to find the quantile function and solve for $ r_{\xi}$

$\displaystyle r_{\xi}=\left\{ \xi_V\cdot\left[\left(r_{max}\right)^{-1\slash3}-...
...{min}\right)^{-1\slash3}\right]+\left(r_{min}\right)^{-1\slash3}\right\} ^{-3},$ (263)

which gives the equation for the radius distribution $ r_{\xi}$ between $ r_{min}$ and $ r_{max}$ when the volume number fraction $ \xi_V$ is evenly distributed and $ \xi_V\in\left[0,1\right]$.

L. Filipovic: Topography Simulation of Novel Processing Techniques