Verification case WUI/hot_rods.fds

[lihy213]

Dear team,
In the Verification case WUI/hot_rods.fds, you said that the radius of rods were uniformly distributed in the range of 0.002 and 0.003 m. Meanwhile, it also refered that $f(r)=1000$, $\bar{r^2} = \int_{0.002}^{0.003}r^2 f(r) dr \approx 6.333\times 10^{-6}$. I'm confused about the meaning of $f(r)$ and how to get the distributed ranges.
LI

[mcgratta]

The expression $\overline{r^2}$ is known as the expected value of $r^2$. The expected value of the radius, $r$, denoted by $\overline{r}$ or $E(r)$, is 0.0025, but the expected value of $r^2$ is a little more than $\overline{r}^2$. Another way to think about it is to say that the mean area of the circles is not $\pi$ times the mean radius squared.

[lihy213]

My doubt is that there's no information about the range of radius in the input file hot_rod.fds, so how to determine its range between 0.002 and 0.003 m? Besides, I still don't know the meaning of $f(r)$.

[drjfloyd]

The CNF_RAMP on PART points to a RAMP that defines the number distrubution as a function of the diamter in $\mu m$.

This RAMP says the CNF is uniform between diameters of 4000 and 6000 $\mu m$ or a radius of 0.002 to 0.003 m

&RAMP ID='dist', T=4000., F=0. /
&RAMP ID='dist', T=6000., F=1. /

[lihy213]

Ohhh! I got it!