Visibility vs mechanical ventilation using FDS
Jason
shopping mall to maintain a 10m-visibility at 3m above the floor.
Since the fuel is not known, I used 0.05 as the assumed soot
production rate. 0.01 is for wood, which can not be accepted by the
NSW fire brigade, and some times they even require 0.10 for
polyurethane.
In this scenario, the Sprinkler controlled fire size is 2.56MW, the
shop trading area has a ceiling height of 3.6m (floor area is ~3000m2,
the only opening is a 16m wide x 3m high door), and require at least
10m-visibility at 3.0m above the floor for this assessment.
My FDS modeling can not meet the above criteria with fan capacity of 4
x 9m3/s, which activate at 230 sec (even for fan activation from
Time=0).
However, Building Code Australia (BCA) Deem-to-Satisfy (DtS) Clause
only requires 15~20m3/s to keep a layer height of 3m for a 2.56MW
fire. So there is an inconsistency between the FDS results and the
Standard.
My feeling is that FDS over-predicted the smoke in this case. Can FDS
be used as a capable tool for choosing the mechanical ventilation
capacity for fire-smoke scenarios? Is there any validation for similar
cases?
Thanks.
clauten
smoke control used to be based primarily on 4-6 air changes per hour
before they moved to plume entrainment equations) so I wouldn't always
expect a deemed to satisfy condition to actually meet all tenability
criteria. Also, you have picked a scenario that FDS is very "good" at
(tracking bulk smoke and heat transport) so FDS probably isn't
overpredicting the smoke development.
A few things to think about:
How do you get the makeup air in? Mechanically (through vents) or
naturally (through the open door)? The particularities of the makeup
air can sometimes have a big effect on the visibility.
What is the heat of combustion of your reaction? The lower the heat of
combustion the worse the visibility gets because FDS has to "pump"
more mass into the domain to maintain the same (specified) HRR.
In the FDS model, are sprinklers spraying on the fire? This can have a
significant effect on the smoke buoyancy.
Are you using the default mass_extinction_coefficient (7600 m2/kg) and
visibility_factor (3)?
Is your visibility criterion exceeded locally at a height of 3 m or
everywhere? FDS calculates the visibility locally, but really
visibility is a quantity that should be integrated along a path length
in the direction an observer is looking.
Do you see any plugholing in the simulations (sucking of clean air
into the exhaust vents)?
What are the thermal properties of your wall lining materials,
particularly the ceiling? If they're the default (INERT) then they
stay cold and the smoke loses buoyancy and makes it harder to
extract.
What is the egress time? How much after (or before) the required safe
egress time does the space lose tenability?
What is the grid size? You're trying to maintain a smoke layer 0.6 m
under the ceiling, and this length is probably only slightly larger
than your grid size (guessing 0.2-0.3 m?) so using more cells may
help.
Do you have any draft curtains/downstands to confine the smoke to a
particular smoke zone? This can also help.
Chris
Kevin
sprinklers, 3000 m2 of floor area, a 3.6 m high ceiling and keep the
smoke within a zone 0.6 m from the ceiling. Smoke just doesn't behave
so nicely. It could be that the calculation method used to get the
15-20 m3/s exhaust requirement assumes that the fans work perfectly
and exhaust pure "smoke", when in reality they must suck out alot of
fresh air too.
Maybe I'm biased (of course!) but I think I'd believe the FDS results
in this case, unless I am reading something wrong here.
Jason
I have uploaded the input file (Supermarket.fds).
Sorry I mis-type the ceiling height in the previous post, ceiling
height should be 4.2m (not 3.6m) above the floor. Mesh number in Z
directrion is 45, so the mesh size in the vertical direction is below
0.1m.
I only specified the HRR=2.56MW, the heat of combustion is the default
value as discussed in the FDS Users Guide 5.
No calculation of sprinklers sprays is involved in FDS. 2.56MW is used
as design fire because of the Mall has sprinklers designed, othewrwise
the fire size of 2.56MW can not be justified.
Makeup air is supplied by natural ventilation through the open door,
the door size is 16m wide x 3m high.
The default mass_extinction_coefficient (7600 m2/kg) and
visibility_factor is used.
Plugholing can be observed in the simulations but the designer want to
have 4 fans only, in total we have a exhaust capacity of 4 x 9m3/s.
All the walls are set as "concrete", there are no smoke curtains/
downstands to define zones, and the designer want to gain infinite
tenable time, once the fan is avtivated at 230 sec.
Jason
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myy
> What is the heat of combustion of your reaction? The lower the heat of
> combustion the worse the visibility gets because FDS has to "pump"
> more mass into the domain to maintain the same (specified) HRR.
This bring me a question.
As far as I know, FDS calculate the heat release rate based on the
amount of oxygen depleted. Then, the heat of combustion affects the
heat release rate and subsequently the visibility results? Somebody
can make this clear? Thank you.
dr_jfloyd
combustion, and the fixed product yields (soot yield). The fuel
chemistry defines how much O_2 is needed to fully combust the fuel.
The default fuel is propane which has a large heat of combustion as
compared to typical fuels in an atrium. The default soot yield for
propane is also lower than typical fuels (especially if there is
upholstered furniture.).
The heat of combustion defines how much fuel flows from the burner for
a given HRRPUA. The soot yield says how much smoke is made when that
fuel burns. The fuel chemisty defines how much O_2 is needed to burn
the fuel, and if not enough is available then the heat release will be
lower.