FDS default REAC parameters

[Mich B]

I have recently encountered a number of projects where no fire chemistry has been specified in the REAC line; I understand that propane is the fire chemistry which FDS (FDS5) defaults to in this instance.

 

The projects where I have encountered this include CFD modelling of retail premises which sell foam furniture and of prison cells where, in both cases, a polyurethane foam chemical formula might be more appropriate.

 

I'm wondering whether there is a reason why FDS5 defaults to a propane fire chemistry, and if, using such chemistry would be appropriate in these circumstances?

 

I understand that chemical composition must be specified in the REAC line within FDS6. Is there a reason that this has changed?

Thanks,

Michael

[Kevin]

You answered your own question. We did not like propane to be the default fuel for all these apps.

That being said, if you are just looking at heat and smoke transport away from the fire, the HRR and soot yield are the key parameters. The actual fuel stoichiometry is less important. However, if you interested in near field effects, you have to look more carefully at the chemistry.

[Randy McDermott]

Michael,

The details of the pyrolyzate are very much a research question.  Do you have a reference for the gas phase fuel that should be use for polyurethane foam pyrolysis?

Regarding why we ask users for the specific fuel gas in v6, you have basically answered your own question: the user should know what they are asking the model to do.  You are correct to question whether propane is an appropriate surrogate fuel.

Randy

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[dr_jfloyd]

Propane as a default goes all the way back to FDS 2 when we first developed the mixture fraction combustion model that was used for FDS 2 - FDS 4 and then modified for FDS 5. At that time there were not many users of FDS outside of research. We had many defaults for inputs just to make it easy to quickly get test cases running with FDS. Propane was selected in part since at the time I was at VT developing the combustion model and doing reduced scale compartment fires with propane.        

On Thursday, October 29, 2015 at 12:36:58 PM UTC-4, Mich B wrote:

[Mich B]

Thanks Kevin,

As your response suggests, I was expecting these answers.

However, the reason I have asked is because my experience with varying the fuel chemistry suggests that it is quite important (and can have a significant effect) in terms of the outputs from FDS which we use for design purposes, such as visibility. For example the difference in results between propane and polyurethane fuels can determine whether or not a design falls within the specified acceptance criteria.

Thanks,

[Randy McDermott]

I still do not understand what exactly you mean by "polyurethane fuel".  Suppose you set FUEL=CxHyNzO, SOOT_YIELD=0.0.  Are you saying you will not pass visibility?

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[dr_jfloyd]

What is important for a smoke control simulation is the heat release rate and the smoke production rate. As long as those are the same, fuel chemistry isn't very important.  However, to get the same smoke production rate does not mean the same soot yield.  If your actual fuel is PU foam with say a 10 % soot yield, then if you were to use propane you would need ~ 20 % soot yield.  Propane has a heat of combustion of ~ 45,000 kJ/kg.  PU foams are ~25,000 kJ/kg.  For the same fire size you burn twice as much PU and if soot yields were equal the PU would have a higher hazard.

[Mich B]

The guidance in BS7974 suggests SOOT_YIELD=0.1 for plastics fuels.

What I have seen recently is that this can be applied with any (a seemingly unrelated) FUEL="CxHyNzO" parameter within the modelling.

For two simulations with the same specified soot yield, changing only the subscripts X, Y and Z can significantly alter the results.

Where it has been stated that the smoke production rate is important, my understanding is that this is dependent on the FUEL stoichiometry and the SOOT_YIELD parameters (as stated in section 9.1 of the FDS5 user guide). Is this the case?

Regards,

[dr_jfloyd]

For simple chemistry, the smoke production rate is given by the rate at which fuel is burned times the soot yield. 

For a given heat release rate, the amount of fuel that is burned will vary depending on the heat of combustion of the fuel.

If you have the same fire with two fuels having the same soot yield, but different heats of combustion, then there will be varying soot production rates 

[Kevin]

You cannot just randomly choose x, y, and z, because in FDS 6, there is an extinction model that is based on the basic fuel stoichiometry. However, assuming you have chosen reasonable fuel molecules, and the HRR and soot yield are the same, I would think that the results would not be significantly different. Maybe if you have O in the fuel molecule would that change things. If you just have C_x H_y, the molecular weight might affect the fire dynamics in the near-field, but in the far-field, your chosen HRR and soot yield are most important.

On Wednesday, November 4, 2015 at 11:03:40 AM UTC-5, Mich B wrote:

[dr_jfloyd]

Any significant amount of O or N in the fuel molecule will have a significant impact on the heat of combustion and thus the smoke production rate.

CH4: HoC ~50 MJ/kg

CH3OH: HoC ~23 MJ/kg

For the same fire size and soot yield, the smoke production rate would vary by a factor of two.

[Khalid Moinuddin]

A bit unclear.

 

If one chooses

(a) CH4:          HoC ~50 MJ/kg, SOOT_YIELD=0.1

(b) CH3OH:    HoC ~23 MJ/kg, SOOT_YIELD=0.1

 

What will vary by a factor of two? Or is it a unrealistic proposition?

 

Regards

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[Randy McDermott]

To get the same HRR in each case the methanol flame would need twice the fuel flow rate.  The SOOT_YIELD has units of kg soot/kg fuel reacted.  Thus, you produce twice as much soot for the same HRR.

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[Khalid Moinuddin]

Thank you. That makes it clear.

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