A catalyst particle is modeled as a symmetrical double cone solid as shown in the figure. For each conical sub-part, the radius of the base is \( r \) and the height is \( h \). The sphericity of the particle is given by:
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Sphericity is important for understanding how particles behave in processes such as catalytic reactions, filtration, and fluidization. It gives a measure of how "spherical" a particle is compared to an ideal sphere with the same volume.
The sphericity of a particle is a measure of how closely the shape of the particle approximates that of a sphere. It is given by the formula:
\[
\phi = \frac{\text{Surface area of a sphere with same volume}}{\text{Surface area of the particle}}
\]
For a double-cone shape:
Volume of a single cone:
\[
V_{\text{cone}} = \frac{1}{3} \pi r^2 h
\]
Surface area of a single cone:
\[
A_{\text{cone}} = \pi r \sqrt{r^2 + h^2}
\]
Since the particle consists of two identical cones, the total volume \( V_{\text{total}} \) and total surface area \( A_{\text{total}} \) are:
\[
V_{\text{total}} = 2 \times \frac{1}{3} \pi r^2 h = \frac{2}{3} \pi r^2 h
\]
\[
A_{\text{total}} = 2 \times \pi r \sqrt{r^2 + h^2} = 2 \pi r \sqrt{r^2 + h^2}
\]
Now, the volume of a sphere with the same volume as the double cone is:
\[
V_{\text{sphere}} = \frac{4}{3} \pi R^3
\]
Equating the volumes of the sphere and the double cone:
\[
\frac{4}{3} \pi R^3 = \frac{2}{3} \pi r^2 h
\]
Solving for \( R \):
\[
R = \left( \frac{r^2 h}{2} \right)^{1/3}
\]
The surface area of the sphere with radius \( R \) is:
\[
A_{\text{sphere}} = 4 \pi R^2 = 4 \pi \left( \frac{r^2 h}{2} \right)^{2/3}
\]
Finally, the sphericity \( \phi \) is the ratio of the surface area of the sphere to the surface area of the double cone:
\[
\phi = \frac{4 \pi \left( \frac{r^2 h}{2} \right)^{2/3}}{2 \pi r \sqrt{r^2 + h^2}} = \frac{2 \left( \frac{r^2 h}{2} \right)^{2/3}}{r \sqrt{r^2 + h^2}}
\]
