Discrepancy between visibility and optical density limits

[D D]

Hello all,

while examining visibility and optical densities for a given situation, I came across a discrepancy, or at least I think it is.

Below we see two result slices on the same given time, one superimposed to the other. The warm (reddish) colors are the visibility slice of 5 m. All values (i.e. colors) above that value are truncated. Similarly, the blueish hues is from the optical density slice with all values below 0.2 truncated. So basically at the blue border we have practically the FED of smoke  = 1.

It is clearly seen that these two slices do not match exactly, so using the theoretical background about smoke obscuration (also in FDS UG, 21.10.4), using a C=3, with a OD=0.2, S comes to equal 6.51 m, not 5 m. Similarly, looking at slice with visibility set to 6.51 m makes it look much better (although not perfect at some areas, especially near the fire).

Tons and tons of handbooks around the globe all talk for a visibility criterion of 5 m which corresponds to an OD=0.2. Obviously I miss something here but what?

Any comments appreciated.

[dr_jfloyd]

From the FDS User's Guide 21.10.4

Visibility = C/K

where C is 3 or 8 for light reflecting or light emitting surfaces and K= K_m rho Y_soot with K_m=8700 kg/m2

Optical density = K log_10 (e) = K / 2.3

If you were to set C to 2.3, then Visibility would equal 1/OD.

[DRAKE JONES]

In relation to the above query, I want to calculate the light extinction coefficient K at a point where K= K_m rho Y_soot  with K_m=8700 kg/m2. I have the MASS FRACTION of the SOOT (in kg/kg). How to determine the density of the smoke particulate, rho (in kg/m^3) at that particular point? 

[D D]

OK so  basically I realize that my understanding about a default C value of 3 does not align fully with the recommendations of the handbooks and that I need to set it to 2.3 in order to follow the recommendations for non self-illuminating elements.

[D D]

The experts should confirm or prove me wrong but since K=2.3*D and since you can measure optical density, just multiply all values with 2.3.

[D D]

@ DRAKE

I just realized that there is an output device called EXTINCTION COEFFICIENT, so wouldn't that be enough for you?

[DRAKE JONES]

@D D 

Thank you for your input. I did record the  EXTINCTION COEFFICIENT and VISIBILITY at a point. Visibility =C/ EXTINCTION COEFFICIENT, where C is 3 by default. But the visibility values by FDS and by C/ EXTINCTION COEFFICIENT are not being the same. Could anyone of you tell me why?. Thank you

[dr_jfloyd]

DEVC values are time averages over the output interval.  

Visibility(t) = C/Extinction Coefficient(t)

Average of (C/ Extinction Coefficient(t)) does not equal  C/Average(Extinction Coefficient(t))

[DRAKE JONES]

Thank you @dr_jfloyd. May I know how in the density of smoke particulate (rho*Y_s) in kg/m^3 at a point is calculated for K?

[dr_jfloyd]

Equation 2.1 through M.16 in the FDS Technical Reference Guide Volume 1: Mathematical Model.

[DRAKE JONES]

 I did record the visibility for two soot yields at the same point separately 0.191 kg/kg, 0.103 kg/kg separately. However the times to reach 3 m visibility at the point for two cases are 11.13 and 11.11 seconds respectively. Similarly, the times at other points also do not vary much for the two different soot yields. I would like to know why the times do not show much difference even though the soot yield values are significantly different. TIA

[Kevin McGrattan]

11 s? That's too fast to make a meaningful comparison. Try a scenario where it takes 10s of minutes to slowly reach a visibility of 3 m at a given point.