Doubts about the result of my first simulation - window temperature during simulation

[Loredana Lo Monaco]

Good morning everyone, I'm attaching my first simulation (600 seconds), because despite not having had any error messages (apart from the usual one about density clipping which I "solved" by lowering the minimum limit) the result doesn't convince me. The objective was to verify the spread of the fire from the closed mezzanine (used as an office, with three PC stations, an archive cabinet and a table) and the timing of opening of the thermo-melting skylights and possibly of the windows (double glazing) and the effect that this supply of air would have had on the spread of the fire. What doesn't convince me is the temperature of the window on the yellow block (stacked empty wooden pallets). For the mezzanine and the pallets (wood) I set a characteristic speed of 300 seconds, while for the other blocks (covered with exposed plastic film) 150 seconds and a thickness of one millimeter (precisely the plastic film). I have not set the particulate, having considered all the materials in the gaseous phase, for simplicity. I assume that the error message that initially appeared (the density one) was due to the position of the mezzanine (set as full obstruction) too close to the pillar and the perimeter wall. What is your point of view? Does it seem realistic as a simulation to you? What might I have failed to implement? Thank you for your patience and availability. I attach the input file

[carlop...@gmail.com]

While waiting for other more authoritative opinions I will try to tell you my first impressions.

You have assigned to the loft walls the area F_SPC to which you have assigned a HRRPUA of 4,000 kW. This means that during full combustion it will give off about 475 mW (considering only the front wall, floor and ceiling of the loft). Just a little too much for a fire of three PC workstations, an archive cabinet, and a table....
You set that the fire grows linearly, not necessarily, but it is usually schematized as t^2 fire. The mezzanine will have window openings and at least one door, the fire will have to "break" these elements before spreading to the ground floor.

What HRR curve were you planning to simulate?  Why did you assign each loft face a value of 4,000 kW/m^2

The grid size in my opinion is not appropriate for your fire scenario, which I think needs to be revised. The number of cells should be suitable to avoid numerical instability problems and be factorizable and traceable to 2^l * 3^m * 5^n.
Obstructions should happen exactly inside the cells and not overlap as it seems to be in your model. I would leave more space between the ceiling of your warehouse and the end of the computational domain.

[Loredana Lo Monaco]

Thank you for your answer, in reality the heat release rate also includes what is inside the mezzanine, not just the partitions. I created it as a closed block to simplify the simulation, but if you look at the species it also includes the air included in the volume and the materials inside, in percentages compared to the total volume. Otherwise I would have had to insert the individual furnishings and equipment, for which I do not have sufficient data for the individual configuration, burdening a simulation which has already required 3 days of calculation. I had tried to insert them separately, creating the individual partitions of the mezzanine, but by assigning the SURF with the curve, the floor detached from the walls. This is probably due to the grill. Regarding the grid, can you explain the sizing criterion better, which I didn't understand well? Now it is 0.5x0.5x0.5 and obviously the various elements do not fill the cells perfectly. But as already said, I'm a beginner and I really want to improve

[Loredana Lo Monaco]

regarding the values, from the calculation of the fire load divided by the surfaces I get an hrr of 248 Mj/m2 (considering internal furnishings and individual internal faces of the mezzanine closing partitions). If 1 MJ corresponds to 1 MW/s, but HRRPUA is required in KW = 1/000 of MW, I should have 248000 HRRPUA achieved in TA =4724s with a rate of 300s. 1000KW at 300s is 4000KW at 600s calculated as t^2, but it would be 4 MW/m2. Can you explain to me where that 475 MW comes from?

Il giorno venerdì 20 ottobre 2023 alle 11:03:46 UTC+2 carlop...@gmail.com ha scritto:

[dr_jfloyd]

MJ/m2 is the amount of fuel that is present. MJ/m2 is independent from the HRRPUA. There are an inifinte number of HRRPUA values that could result in a specific MJ/m2.  For example you can get 100 MJ/m2 if you have a 1 MW/m2 fire for 100 s, a 10 MW/m2 fire for 10 s, an 0.1 MW/m2 fire for 1000 s.   In your case assuming you want the fire burning all the fuel in 600 s,  248 MJ/m2 / 600 s = 413 kW/m2

&MATL ID='SPC' SPEC_ID='SPC' CONDUCTIVITY=0.2 SPECIFIC_HEAT=2.5 DENSITY=120/
&SURF ID='F_SPC' MATL_ID='SPC' COLOR='SILVER' THICKNESS=0.005 HRRPUA=4000 RAMP_Q='F_SPC'/
&RAMP ID='F_SPC' T=0 F=0/
&RAMP ID='F_SPC' T=600 F=1/
&OBST XB=22.05,26.45,0.85,11.1,2.05,4.85 SURF_ID='F_SPC'/
&HOLE XB=25.45,26.45,9.45,11.1,2.05,4.85/

The SURF of F_SPC is being applied to every exposed wall cell on the OBST.  For this OBST it looks like the two ends, one side, and the bottom are exposed. That gives about 100 m2 of surface area.  100 m2 * 4 MW/m2 HRRPUA = 400 MW.

50 cm is a very coarse grid.  It is fine if your intent is just developing inputs and using a coarse grid so things run fast while you test inputs. For a production run you would want to do a grid study.

[carlop...@gmail.com]

Hi, 

A surface (F_SPC) has been applied to all the surfaces of your mezzanine that releases linearly and up to 600 s a power per unit air of 4,000 kW/m2. Carrying out the calculations comes out with a very high peak HRR.

Assimilating the mezzanine to a parallelepiped we have:

Top side and bottom side= (4.40 m x 10.25 m) x 2 = 90.20 sq. m.
Short side= (4.40 m x 2.80 m) x 2= 24.64 sq m
Long side= (10.25 m x 2.80 m) = 57.40 sq m

Total Area= 172.24 sqm

Total HRR= 172.24 sq m x 4,000 kW/sq m = 688,960.0 kW

This represents a very large fire, and not comparable to your office fire in any way.

What you are calling the fuel load is the total amount of energy released when the fuel is burned. If you had a HRR curve, then the area under the curve would be the fuel load. In terms of units,

HRR (kJ/s) x Time (s) = Fuel Load  (kJ)

So, if you have calculated the fuel load for a given room, you then have to decide how long it will take for it to burn.

Fuel Load (kJ) / Time (s) = HRR (kJ/s = kW)

The cell size is usually a function of the HRR or according to the sensitivity you want to achieve. There are relationships that on the basis of the power released allow you to estimate the mesh size.
If your structural elements (beams, pillars, walls, etc...) do not fill the whole cell it does nothing, via surface you can imost the thickness value and tell FDS to consider a smaller size.

Finally, I think you need to model the loft openings, perhaps as holes, simulating the worst-case situation, i.e., open doors and windows. Evaluate whether as a result of the fire your fumes are hot enough to propagate the fire to the ground floor.

dr_jfloyd please correct me if I said something wrong.

Have a nice day