Question on the Cable Tray Fire in the FDS user's guide

[Fasteel]

In Section 22.2 of the FDS user's guide, a case was described to model
a fire spreading along a cable tray. The following settings are
specified for the cable tray. I have no problem with these settings.
But when I read the statement in the guide "The pile of cables is
assumed to be a solid slab, 28 cm wide and 2 cm deep", I have a
question regarding whether this statement is correct or not.

To me, the statement should be "The pile of cables is assumed to be a
solid slab, 28 cm wide and 4 cm deep" instead of 2 cm in the guide. My
argument is as follows:

Whist the 'loose cable' is specified as 2 cm thick in the SURF lines,
this surface property is applied to each surface of the OBST described
below which is 4 cm deep. To me, a 2 cm thick loose cable is placed on
the top of the OBST and another 2 cm thick loose cable is placed on
the bottom of the OBST. As a result, the cable tray is 4 cm thick.

I could be wrong. But would appreciate if anyone can provide some
clarifications. I guess this is important for the modellers with
respect to how to construct the model correctly.

&MATL ID = 'PLASTIC'
CONDUCTIVITY = 0.2
SPECIFIC_HEAT = 1.5
DENSITY = 1500.
N_REACTIONS = 1
HEAT_OF_REACTION = 3000.
HEAT_OF_COMBUSTION = 25000.
REFERENCE_TEMPERATURE = 400.
NU_FUEL = 1.0 /

&MATL ID = 'COPPER'
SPECIFIC_HEAT = 0.38
CONDUCTIVITY = 387.
DENSITY = 8940. /

&SURF ID = 'Loose Cable'
COLOR = 'IVORY BLACK'
MATL_ID(1,1:2) = 'PLASTIC','COPPER'
MATL_MASS_FRACTION(1,1:2) = 0.4,0.6
BACKING = 'EXPOSED'
THICKNESS = 0.02 /

&OBST XB=-2.00, 2.00,-0.14, 0.14, 0.51, 0.55, SURF_ID='Loose Cable' /

[shostikk]

The part of the surface thickness, from where the fuel vapors are
accounted for, is controlled by LAYER_DIVIDE keyword. By default,
LAYER_DIVIDE is 0.5 times the number of layers for surfaces with
EXPOSED backing, and
equal to the number of layers for other surfaces.

This means that both sides only release fuel from 1.0 cm of cabling,
but do the heat transfer calculation for the whole thickness of 2 cm.

[Jason]

Thanks for your reply. It helps a lot. I looked further at Section
8.3.5 of the guide about the use of LAYER_DIVIDE keyword. It is clear
to use this keyword to specify the certain number of layers which are
involved in the combustion.

In the Cable tray case, because the default value of the LAYER_DIVIDE
is used, it means only half the thickness of the layer is combusted
since the layer consists of plastic (combustible) and copper (non-
combustible). Does the default value 0.5 mean half of the mass of the
layer or purely half the volume? I noticed that the plastic only
occupies 40 percent of the mass of the layer. Is it more accurate to
specify the LAYER_DIVIDE as 0.4?

I agree that only half of the cable tray is combusted since the
default value is used. But am I correct if I say that the physical
size of the cable try is 4 cm thick?

If I change the OBST specification to the following (reduce the
thickness to 2cm), does that make any difference to the modelling?

&OBST XB=-2.00, 2.00,-0.14, 0.14, 0.51, 0.53, SURF_ID='Loose Cable' /

Thanks again, Simo. Look forward to your reply at your earliest time.

> > &OBST XB=-2.00, 2.00,-0.14, 0.14, 0.51, 0.55, SURF_ID='Loose Cable' /- Hide quoted text -
>
> - Show quoted text -

[shostikk]

LAYER_DIVIDE refers to volume (of thickness). 0.5 means that the fuel
mass flux is integrated only over the first 50 % of the surface
thickness, starting from the front surface. When this is made on both
sides of the obstacle, we get 100 % of the volume.

The physical size is 2 cm, because the surface solutions on top and
bottom sides solve the "same thing". That means the same heat
conduction equation is solved twice, with just front and back boundary
conditions switched. This is done for practical convenience of code
implementation.

The OBST size makes no difference. The SURF THICKNESS matters.