Acces violation error #15489
Replies: 11 comments 47 replies
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We would need the entire input file and we would have to run it ourselves to see what the problem is. The error is some undefined internal error. |
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Hello, Here is a simplified version of my case given that i can't share the file for confidentiality reasons. My aim is to visualise flame spread across the walls of this chamber made out of composite. I tried firstly to test it with wet wood and polyester but i didn't notice the flame spread. I also removed the ramps of the burners to get instant results and the ceiling because the O2 was being consumed too quickly leading to phantom flames. Regarding the phantom flames i also have a side question: what way is usually used to get rid of these especially if we have a really high HRRUA applied inside a small confined chamber like this case for example. Thanks again for your help! |
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This case does not run with the current release. The ACTIVE reaction lacks HEAT_OF_REACTION inputs. When trying to model pyrolysis you need to make sure your inputs are sensible. Look at the .out file for this case and look at how FDS computed the A and E values for inputs with REFERENCE_TEMPERATURE. There are nonsensical values resulting from overflowing a 64 bit real (which takes a value >1x10^300). You should be reviewing the MATL parameter input echo in the .out file to make sure everything looks OK and you should also be doing virtual TGA tests for each unique layer with a pyrolysis reaction and virtual cone tests for each SURF that pyrolyzes to make sure the behavior is sensible (see the User's Guide). |
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Hello, Thank you for the clarification. I took a look at the .out file and i found out effectively unreasonable values for the kinetic parameters. 2- When i try to run the case using the wood properties that i extracted from the TGA_analysis example file from FDS, the simulation runs but still no flame spread is being done to the different wood surfaces. Are there any additional parameters that i must define? |
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Your input file is overly complicated and messy. Why are you using multiple reactions for the materials? I suggest you use a single |
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Sorry for the confusion. I erased all the extra stuff in the file and i restarted the case using wood properties that were given in the tga analysis case example. In order to simplify the problem i'm trying to set a table on fire by applying underneath it a fire of 100 KW/m2 HRRPUA . The simulation works but the flame doesn't spread and burn the table and i can't figure out the reason so any helpwould be greatly appreciated. |
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If you are predicting pyrolysis you are doing research and need to take the time to look at all the information available to you. Is the wood heating up to a temperature where you expect pyrolysis to occur? |
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Thank you for the advice. I just have a follow up question. Supposing that i defined a fuel in my reaction Like Polyester for example because the fire occurs on a bedsheet made of polyester. Knowing after a certain time pyrolysis will occur on a wall made from balsa wood for example, what should i define its SPEC_ID? Is it the cellulose which in that case i must define as a specie? Or is it the fuel defined in REAC? In the old version of FDS it used to be the latter i believe but i don't know if this has changed. |
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You can define multiple fuel gases now, but if you are doing a large scale fire simulation, there is not much value in it. First, it adds to the cost of the simulation to track multiple fuel gases, and second, you really don't know exactly what those fuel gases are because something made of polyester is probably going to pyrolyze to form an alphabet soup of different gases. Then you have to specify soot and CO yields too. So I would recommend that you choose an effective fuel molecule that is some average of what is burning. If the molecular weight of the molecule is large, say over 100 g/mol, make it lighter but retain the right C to H to O ratios. |
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Thank you all for your help and guidance. After some time i managed to get a reasonable flame spread on my object and apply the burn_away feature. However i'm stuck on some certain results that are bugging me and i am unable to justify them: 1-I put surface density sensors for the wood birch material and the char residue in order to see if the mass of the wood is converted to the specified amount of residue noted in the code (0.172) and i don't obtain the same value (1.43 max value in devc.csv whereas it should be 10.890.172=1.87). Is it possibly due to the fact that i need to define a finer mesh (even though i doubt it because i calculated the ratio of the D and the cell size and it is of a value that is between 10 and 20)? I also defined a density sensor for the cellulose fuel and it doesn't give me anywhere near the expected mass from the pyrolysis of the wood. 2-Secondly after a certain time the mass of char and cellulose start to decrease reaching 0 for the char and remaining constant at 0.02 for the cellulose. Is this reasonable given that i didn't specify that the char must be all consumed as well. Also the solid density devices are not being registered in the device file and i can't figure out why since the simulation runs normaly. 3-Finally i have a question about the burn away feature. At first i notice the flame spreads from the sides of the table to the center, and so cubes little by little start disappearing from the side to the center( i attached a couple of figures at certain time periods). Is this physically logical, especially since i have a very high temperature at the surface of the table around 25 secs , whereas the table disappears entirely around 250 sec, as well as having cubes that are not attached to anything floating in the air before disappearing? Is it also possibly a problem of refining the mesh ? I have attached a much cleaner version of the code, as well as the devc and hrr csv files and a couple of results at certain times of the simulation. The wood properties were taken from the work of Ana Matala - Estimation of solid phase reaction parameters for fire simulation. Any sort of feedback and advices would be appreciated since i am fairly new to pyrolysis studies on FDS and i believe that processing these results for this "fairly simple case" is essential in order to achieve my main goal which is modeling the pyrolysis of composite sandwiches. pyrolysis_birch_devc.csv |
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For simulations like this, the first thing I do is to create a very small test case in which I have a single obstruction ( |
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Each reaction occurs at its own rate based on the A and E you define and each reaction makes its own products based on the NU inputs for that reaction. Reaction 1 does not pay any attention to any inputs for Reaction 2 and vice versa. The two reactions are independent. |
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so the reactant of the first reaction is the same as the second, i.e. the material initially defined using &MATL? |
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On any given &MATL input, all the reactions defined on that single input will consume that MATL. |
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I would think that if there are two simultaneous reactions assigned to a single |
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Let me put it another way. If you have a TGA curve for this material, you can use multiple materials, each with a single reaction, to mimic the TGA curve. Sure, there is much more going on within the solid, but from a modeling point of view, you are capturing the macroscopic phenomena. |
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Well noted. I was just having a look at one of the box burn away examples (box burn away 2D residue) and wanted to verify if the same initial mass i had for the box is conserved for the produced fuel and residue. It states in the file that the box mass is 200.40.4*.4=1.28 kg, however when you measure it in the devc file it gives a mass of 12.8 ,what is the reason behind this? Also i tried to measure the residue mass formed and i don't get the logic behind the values i get ( why does the mass after a certain while drop to 0 and shouldn't the maximum mass reached be also 0.64?). I attached the excel results and the line of command i added to measure the mass of the residue to the code. box_burn_away_2D_residue_modified.txt |
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The reason why integrating surface density is not appropriate is that this is a homogenous solid. I think the DEVC line was left over from another case where we have an empty box and we integrate the mass per unit area of the thin box liner. |
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Adding to the confusion, I just noticed that this case is 2D. It is an odd test case, and I do not remember what the purpose was. In any case, pick another example to work with. The results here are confusing for several reasons. |
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So if i want to measure the total mass of a homogenous solid, is there a specific quantity or statistic i can use because i can't seem to find one for the solid phase. |
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One way to do it is to use the 'SURFACE INTEGRAL' for just one face of the solid. That measures the mass per unit area all the way through. |
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I figured out why this case is so confusing, and I will fix the input file to be more clear. Here is why the initial solid mass is reported as 12.8 kg. First, the surface density of the solid is its density, 20 kg/m3, times its thickness, 0.4 m, or 8 kg/m2. This value is integrated over the surface of the solid, which is 4 times 0.4 m x 1.0 m = 1.6 m2. The factor of 4 refers to the fact that only 4 faces of the solid are exposed---the other two are flush with the y boundaries. The factor of 1 m refers to the fact that the original solid dimensions are 0.4 m by 1.0 m by 0.4 m. FDS has adjusted for the fact that this 1 m wide solid is being squeezed into a 0.4 wide domain. As I said before, this case is not really "well posed" and is used to test certain features. I will remove some of the output that makes no sense. |
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I tried running a more simplified case in an inert environment with 700 C applied at the walls just like Mr McGratta recommended for the birch wood. It is only one reaction as well with 0.173 mass fraction of char residue and the rest is the fuel. I put the surface density devices and applied the surface integral on one face of the geometry in order to calculate the decreasing mass of the birch(5500.0051.82.2=10.89 kg), and to verify its conversion to char(0.17310.89=1.884)and fuel(10.89*0.828=9.006). My initial measured mass of birch is equal to what i expected which leads me to believe that i have correctly defined my surface density devices. However in my case i get a quantity of residue that not only is smaller than specified in the NU_MATL(a max of 1.5327) , but is also decreaseing to 0 leading me to have 100% of the birch wood being converted to fuel at the end. I also integrated the mass loss ratio of the fuel in function of time and i found out that the total quantity produced is also equal to the inital mass of the birch wood. I'm trying to figure out the reasoning of these results. Does the produced residue also pyrolyzes thus justifying a peak different than expected, its decreasing mass and its addition to the fuel's? Any help or explanations would be appreciated. The birch pyrolysis parameters are from the article of Anna Matala: estimation of pyrolysis model parameters for solid materials using TGA Data. I attached the fds script and the device csv file. 
birch inert pyrolysis test.txt |
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If you want solid material to remain, then you cannot use BURN_AWAY. |
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I figured that was case. But still i don't seem t understand the case of the fuel mass being almost equal to the initial birch mass when i turn off the burn away. |
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Every wall cell does its own independent calculation. You have 22x18x2 grid cells top and bottom and 2x(22+18)*1 grid cells along the edge. The top bottom grid cells use LAYER_DIVIDE to inject the fuel gas. Even though each wall cell top and bottom does its on calculation the top wall cells only inject mass from pyrolysis of the first half of the SURF and vice versa on the bottom. This means you effective have 22x18x1 + 2x(22+18)*1 wall cells each of which is 0.1 m x 0.1 m x 0.005 m. Those edge cells add 20 % to the total of the top cells. You are only integrating surface density over the top face and not all faces. You actually have 20 % more mass were you to include the edges. With a roughly 20 % char yield ~20 % more mass with a ~20 % residue gives you back the mass of the just the top/bottom cells. |
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If i don't want to take into acount the added mass from the edges, can i simply define the faces as INERT? After this trial i got what i was trying to reach with my mass balance output for the char and fuel being exactly as i specified. |
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Yes, for the sake of this exercise, you are correct. In a more practical case, however, you need to decide if it makes sense physically. |
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Thank you all for your help. I was satisfied with the results i got by simulating the pyrolysis of the birch wood in inert atmosphere . At first for a mesh of 0.10.10.1 (D*/δx)=11.7), i get thIs total HRR curve. 
As you can see there is a significant difference in terms of not only the HRR peak, but the time to reach it as well. I thought that by using the 0.1 mesh i was in satisfying fine resolution simulation but by refining it, it doesn't seem that is the case even though i have a respectable ratio of D*/δx. I tried to look at the validation guide for adequate ratios that match my case but i can't seem to find any. Any help or explanation would be appreciated as always. The fds file is attached below here as well. |
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You go from a peak of 7500 at ~25 s to a peak of ~9000 at 35 s. These are not unexpected results. Prediction of pyrolysis has very strong feedback between the burning rate and heat feedback to the surface. As you chage resolution both convective and radiative fluxes are changing. Prediciton of pyrolysis is still an active area of research. All issues related to grid convergence, what grid size is needed, the best approach for getting convective heat transfer, blowing effects or heat transfer, etc. have by no means been solved. |
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After obtaining satisfying results for the the pyrolysis model of birch, i decided to try a multi layered material made of 2 polyester layers and a core of balsa wood. The pyrolysis models for these materials were taken from an experiment performed in 2008 by the LNE in France. However in the older version of FDS (v5.3), there wasn't a way to define the SPEC_ID for each material when defining the pyrolysis model, instead it was just the fuel defined in the reaction. Is it always the same case, or can i define for example cellulose as the fuel produced by pyrolysis of the balsa, and another fuel that is produced by the pyrolysis of the polyester? I tried this and it seems to work fine. I just want to make sure that this definition is coherent and if it requires any modification (i.e. defining additionaly the heat of combustion of the cellulose to make sure i have the proper amount of energy and consequently define a HOC_COMPLETE to adjust the effective mass loss rate in order to obtain a mass balance between the produced fuel and residue and the original material). Any advice or guidance is always appreciated. |
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I only see a |
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I didn't define a REAC for the cellulose initially because i wanted to measure the produced mass (by integrating its the MLR in the .hrr file) and verify if i have a mass conservation equivalent to that defined in the model. Eventually i would like it to burn and so define a REAC for it as well. Although i would like to know the dfference between defining the heat of combustion in the MATL line and in the REAC for the same fuel. Like if i define propane as a SPEC_ID with a heat of combustin of 40k KJ/kg and then i define it in REAC with the same value, does this indicate that i am completly burning the propane fuel that is being produced by pyrolysis? |
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I took a look at the guide and other discussions on the forum and i hope i correctly understood the difference between the heat of combustion defined in the matl and reac lines. Basically if i understood correctly, it is used to represent the actual gases being produced by pyrolysis when defining a single gas in the reac line, and it scales the mass in order to have a conservation of energy such as mass_matlhoc_matl=mass_reachoc_reac. When i studied previously the pyrolysis+combustion of the birch wood, i defined a single fuel which happened to be the one used in the burner and at the same time produced by pyrolysis hence the energy balance was already established. But supposing i want to define a different burner fuel such as propane with an H_O_C of 43058 kJ/kg, then i should specify the H_O_C of the birch wood gas produced in the pyrolysis model (14988 kJ/Kg) while defining the SPEC_ID=PROPANE, in order to signal that i get a different gas specie produced than the one defined in REAC. Therefore the mass of the formed pyrolysed gas will be equal to the value defined in the NU_SPEC (10.890.828=9.016), and so the mass_reac = mass_gas from matl hoc matl/hoc_reac=9.016* 14988/43058 and so this value+ the mass in the burner is equal to the value integrated from the MLR column in the hrr column. Is my analysis correct? An additional question i would like to ask as well is why the HRR peak is smaller than the one calculated with MLR* heat of combustion of propane? I attached the modified fds file and the hrr csv files in order to clarify my explanation a little bit. Any calrification or response would be greatly appreciated. |
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You are correct on how the mass from the material gets scaled as a result of differences in the HoC. The MLR in the hrr file is the gas phase source term and accounts for the scaling. There are a few of reasons you might see a lower HRR than expected based on the MLR: |
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Thank you for the clarification. I have a follow up question. Is it possible to define multiple species like for example the cellulose gas that is formed from the pyrolysis of the birch wood and propane in the burner, and therefore define 2 reactions in order to avoid the scaling done when defining one reaction? Will this give different results? |
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Yes you can define multiple reactions. This is discussed in the user's guide. If you were to add a species that has a different molecular weight and requires a different amount of air to be entrained to burn, for the same HRR do you think you would see the exact same reults or similar but different results? A good way to get more familiar with how FDS works is to get in the habit of making simple test cases. Here you could have two cases where you have a burner each with one of the two fuels and the same HRRPUA and look at things like centerline plume temperature and velocity to see what types of differences you see. |
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Thank you for your recommendation. That is exactly what i thought of doing. I decided to increase the height of domain and add 2 additional OPEN vents in order to ensure enough oxygen circulates and mixes with the fuel.Then i conducted 2 simulations, method 1 was by defining a single reaction of propane, the second by defining 2 reactions for the propane burner and the cellulose gas produced. In terms of HRR i got very similar results in terms of distributions and peak, the key difference was that the combustion occurs earlier when defining 2 reactions in comparison to the first method as seen below . Now i tried to replicate this to my Balsa wood material with a predefined pyrolysis model in an older FDS file (v 5.3),based on experiment conduct by the LNE, and unfortunately i do not have experimental data and i'm struggling to find some published articles in order to validate my TGA and cone calorimeter tests. Nonetheless parting from that i ran the same 2 cases and the results of the HRR curve below and even the total energy produced are extremely different which make me question the validity of the pyrolysis model, since for for the version 5.3 only a single reaction fuel was defined and applied to all the materials of the composite sandwich i'm trying to study, which makes me question the validity of assuming cellulose gas (the same one used for birch) as the pyrolysed gas formed from thermal decomposition of Balsa.Could this be the main reason why such key differences are observed between the 2 methods? I additionaly ran this case for a defined Polyester model which gave very similar results between the methods just like the birch wood tests . Any insight would be appreciated as always. 
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With HRR > 20 MW it appears you are running a somewhat complicated/large case. This can make it difficult to develop a detailed understanding of what is happening. I suggest you make a very very simple case. Start with a solid phase only case and one wall cell with your material and an external flux. Instrument it heavily so you understand what is happening. Make sure that you are getting results that make sense and that your inputs are doing what you want. Then move to a small sealed box. Set the ambient oxygen to near zero to turn off combusiton. One wall cell with your material and make the other box walls hot. Again look to see if you get the amount of fuel gas you expect and the correct behavior of the solid phase. Then OPEN a boundary and have typical oxygen make sure you get the HRR you expect. |
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Thank you for the suggestion. I did the solid phase only simulation and it gives coherent results in terms of conservation of mass. For the sealed box, where method 1 consisting of defining 2 reactions ( in this case a single reaction with celulose as fuel since there is no burner) gives accurate results of the mass produced of the gas species (cellulose and char masses same as the one found in the solid phase simulation). Method 2 on the other hand where i define a single reaction fuel which is propane in this case, and a heat of combustion equivalent to the cellulose at the Matl Line, gives the mass of propane + mass char= mass balsa. The scaling of the propane mass following the equation HOC matl * spec_matl( propane)=HOC_reac*spec_reac is not done in this case and i assume it is because there is no combustion being done. The problem appears when i open the box, where for method 1, when integrating the value of the cellulose's MLR in order to find the total quantity of burnt cellulose gas which is equal to the cellulos gas produced by pyrolysis, this mass is lesser than the expected amount and therefore the sum of char and cellulose is lesser than the original balsa mass. Method 2 also gives incoherence in results and a noticeable difference in the HRR curve behavior as well as the total energy produced (calculated by integration of the HRR). In addition when calculating the propane mass,and scaling it following the equation mentioned above, i found a spec_matl mass that is also larger than the expected value found in the previous 2 cases. It is worth mentioning that the balsa in this case is being pyrolyzed following 2 parallel reactions with different kinetic parameters as well, and when testing this with my polyeser material, which has its own proper defined fuel formula formed by pyrolysis, and 2 parallel pyrolysis reactions as well, i get accurate results. The same goes for the birch wood material that only has 1 pyrolysis reaction. 
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Please provide the inputs for your closed and open box simple cases. |
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Here are the input files for the 2 different cases. I just had to erase the balsa's properties due to confidentialy reasons and i can guarantee that the sum of nu_spec and nu_matl is equal to 1 for each reaction. |
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I cannot help you without the input files. |
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If you cannot provide the actual inputs you need to provide me with some input that I can run that demonstrates the issues. |
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Hello everyone ,
I am a fairly new user of FDS (v.6.10.1)and i am encountering an error when trying to simulate pyrolysis of sandwich composite with its core being made out of balsa and the layers are from fiber glass and polyester. Fire occurs at a bed made out of polyester, and the fire spreads to the narby walls, ceiling and ground that are all made of this composite material. I defined them first as steel and i ran the case and it worked perfectly, but when i change the material to composite in order to simulate the pyrolysis, it gives me the following error leading me to assume that i certainly made an error in definiing the pyrolysis model and materials but i can't quite figure out what it is.
Sundials library version: v6.7.0
Job TITLE :
Job ID string : pyrolyse
Time Step: 1, Simulation Time: 0.08926 s
forrtl: severe (157): Program Exception - access violation
Image PC Routine Line Source
fds_openmp.exe 00007FF7CEBCE203 Unknown Unknown Unknown
libiomp5md.dll 00007FFFEA850653 Unknown Unknown Unknown
libiomp5md.dll 00007FFFEA7B0A97 Unknown Unknown Unknown
libiomp5md.dll 00007FFFEA7B1E4E Unknown Unknown Unknown
libiomp5md.dll 00007FFFEA768D21 Unknown Unknown Unknown
fds_openmp.exe 00007FF7CEBBA0B6 Unknown Unknown Unknown
fds_openmp.exe 00007FF7CEC17D4D Unknown Unknown Unknown
fds_openmp.exe 00007FF7CE37FB5B Unknown Unknown Unknown
fds_openmp.exe 00007FF7CEE53C2C Unknown Unknown Unknown
KERNEL32.DLL 00007FF865BE7374 Unknown Unknown Unknown
ntdll.dll 00007FF867BFCC91 Unknown Unknown Unknown
i ran the case using 5 mpi processors given i have 5 mesh domains, with 2 openmp threads,. The total number of cells is relatively small of 240k cells, the majority being in the mesh domain where my pyrolysis occurs. Below are the materials i defined for the composite model :
&SURF ID = 'COMPOSITE'
MATL_ID(1,1:2)='Polyester','GLASS'
MATL_MASS_FRACTION(1,1:2)= 0.3,0.7
MATL_ID(2,1)='Polyester2'
MATL_ID(3,1:4)='WOOD','LIGNIN','WATER','Polyester2'
MATL_MASS_FRACTION(3,1:4)= 0.8,0.05,0.05,0.1
MATL_ID(4,1)='Polyester2'
MATL_ID(5,1:2)='Polyester','VERRE'
MATL_MASS_FRACTION(5,1:2)= 0.3,0.7
THICKNESS(1:5)= 0.003,0.0015,0.037,0.0015,0.003
BACKING= 'EXPOSED'/
& MATL ID = 'Polyester'
EMISSIVITY = 0.89
SPECIFIC_HEAT_RAMP = 'c_polyester'
CONDUCTIVITY_RAMP = 'k_polyester'
DENSITY = 1189
N_REACTIONS = 1
HEAT_OF_REACTION(1:2) = 1050., 1550.
MATL_ID(1,1)='CHAR'
MATL_ID(2,1)='CHAR'
NU_MATL(1,1)=0.05
NU_MATL(1,2)=0.0
SPEC_ID(1,1)='Polyester_fuel'
SPEC_ID(1,2)='Polyester_fuel'
NU_SPEC(1,1)=0.95
NU_SPEC(1,2)=1
HEAT_OF_COMBUSTION = 21000
REFERENCE_TEMPERATURE(1:2)= 395.,455. /
MATL_ID(1,1)='CHAR'
MATL_ID(1,2)='CHAR'
NU_MATL(1,1)=0.05
NU_MATL(1,2)=0.0
SPEC_ID(1,1)='Polyester_fuel'
SPEC_ID(1,2)='Polyester_fuel'
NU_SPEC(1,1)=0.95
NU_SPEC(1,2)=1
HEAT_OF_COMBUSTION = 19000
REFERENCE_TEMPERATURE(1:2)= 355.,405. /
&MATL ID = 'CHAR'
EMISSIVITY = 0.89
SPECIFIC_HEAT_RAMP = 'c_polyester'
CONDUCTIVITY_RAMP = 'k_polyester2'
DENSITY = 1109./
MATL_ID='FIBRE'
NU_MATL=1/
& MATL ID= 'WOOD'
CONDUCTIVITY_RAMP= 'k_balsa'
SPECIFIC_HEAT_RAMP= 'c_balsa'
DENSITY= 140.
A = 2.5E5
E = 8.8E4
HEAT_OF_REACTION = 0.
N_REACTIONS = 1
MATL_ID='ACTIVE'
NU_MATL=1/
HEAT_OF_COMBUSTION = 13408.
A(1:2) = 2.5E5, 3.23E14
E(1:2) = 8.8E4, 1.965E5
MATL_ID(1,1)='CHARCOAL'
MATL_ID(1,2)='CHARCOAL'
NU_MATL(1,1)=0.25
NU_MATL(1,2)=0.0
SPEC_ID(1,1)='Polyester_fuel'
SPEC_ID(1,2)='Polyester_fuel'
NU_SPEC(1,1)=0.75
NU_SPEC(1,2)=1/
&MATL ID = 'LIGNIN'
EMISSIVITY= 1.0
DENSITY = 150.
CONDUCTIVITY_RAMP = 'k_balsa'
SPECIFIC_HEAT_RAMP= 'c_balsa'
N_REACTIONS = 1
HEAT_OF_COMBUSTION = 1208.
A = 2.77E+7
E = 1.24E+05
HEAT_OF_REACTION = 146.
MATL_ID='CHARCOAL'
NU_MATL=0.15
SPEC_ID='Polyester_fuel'
NU_SPEC= 0.85/
&MATL ID = 'WATER'
DENSITY = 1000.
CONDUCTIVITY = 0.1
SPECIFIC_HEAT= 4.184
N_REACTIONS = 1
REFERENCE_TEMPERATURE = 100.
REFERENCE_RATE = 0.0016
HEATING_RATE = 10.
SPEC_ID = 'WATER VAPOR'
NU_SPEC = 1.0
HEAT_OF_REACTION= 2500./
&MATL ID = 'CHARCOAL'
EMISSIVITY = 1
DENSITY = 130.
CONDUCTIVITY_RAMP = 'k_charbon'
SPECIFIC_HEAT_RAMP = 'c_charbon'/
! Balsa Wood Thermal Properties
&RAMP ID='k_balsa', T=20., F=0.066 /
&RAMP ID='k_balsa', T=50., F=0.068 /
&RAMP ID='k_balsa', T=100., F=0.071 /
&RAMP ID='k_balsa', T=150., F=0.080 /
&RAMP ID='k_balsa', T=200., F=0.110 /
&RAMP ID='k_balsa', T=250., F=0.160 /
&RAMP ID='k_balsa', T=300., F=0.200 /
&RAMP ID='c_balsa', T=20., F=1.420 /
&RAMP ID='c_balsa', T=50., F=1.454 /
&RAMP ID='c_balsa', T=100., F=1.488 /
&RAMP ID='c_balsa', T=150., F=1.522 /
&RAMP ID='c_balsa', T=200., F=1.556 /
&RAMP ID='c_balsa', T=250., F=1.590 /
&RAMP ID='c_balsa', T=300., F=1.624 /
! Char Thermal Properties
&RAMP ID='k_charbon', T=20., F=0.008 /
&RAMP ID='k_charbon', T=50., F=0.011 /
&RAMP ID='k_charbon', T=100., F=0.021 /
&RAMP ID='k_charbon', T=150., F=0.036 /
&RAMP ID='k_charbon', T=200., F=0.057 /
&RAMP ID='k_charbon', T=250., F=0.083 /
&RAMP ID='k_charbon', T=300., F=0.114 /
&RAMP ID='k_charbon', T=350., F=0.151 /
&RAMP ID='k_charbon', T=400., F=0.194 /
RAMP ID='c_charbon', T=20., F=3.207 /
&RAMP ID='c_charbon', T=50., F=3.260 /
&RAMP ID='c_charbon', T=100., F=3.327 /
&RAMP ID='c_charbon', T=150., F=3.393 /
&RAMP ID='c_charbon', T=200., F=3.460 /
&RAMP ID='c_charbon', T=250., F=3.526 /
&RAMP ID='c_charbon', T=300., F=3.593 /
&RAMP ID='c_charbon', T=350., F=3.659 /
&RAMP ID='c_charbon', T=400., F=3.726 /
! Ramps for properties of CHAR
&RAMP ID='k_char', T=20., F=0.05 /
&RAMP ID='k_char', T=100., F=0.08 /
&RAMP ID='k_char', T=200., F=0.12 /
&RAMP ID='k_char', T=300., F=0.18 /
&RAMP ID='k_char', T=400., F=0.25 /
&RAMP ID='k_char', T=500., F=0.35 /
&RAMP ID='k_char', T=600., F=0.45 /
&RAMP ID='k_char', T=700., F=0.60 /
&RAMP ID='k_char', T=800., F=0.80 /
&RAMP ID='c_char', T=20., F=1.5 /
&RAMP ID='c_char', T=100., F=1.7 /
&RAMP ID='c_char', T=200., F=1.9 /
&RAMP ID='c_char', T=300., F=2.2 /
&RAMP ID='c_char', T=400., F=2.5 /
&RAMP ID='c_char', T=500., F=2.8 /
&RAMP ID='c_char', T=600., F=3.1 /
&RAMP ID='c_char', T=700., F=3.4 /
&RAMP ID='c_char', T=800., F=3.7 /
&RAMP ID = 'k_polyester', T = 20., F = 0.16 /
&RAMP ID = 'k_polyester', T = 100., F = 0.18 /
&RAMP ID = 'k_polyester2', T = 150., F = 0.16 /
&RAMP ID = 'k_polyester2', T = 200., F = 0.25 /
&RAMP ID = 'k_polyester2', T = 250., F = 0.3 /
&RAMP ID = 'k_polyester2', T = 300., F = 0.4 /
&RAMP ID = 'k_polyester2', T = 350., F = 0.50 /
&RAMP ID = 'k_polyester2', T = 400., F = 0.60 /
&RAMP ID = 'k_polyester2', T = 450., F = 0.80 /
&RAMP ID = 'k_polyester2', T = 500., F = 1.0 /
&RAMP ID = 'k_polyester2', T = 550., F = 2.0 /
&RAMP ID = 'c_polyester', T = 20., F = 1.247 /
&RAMP ID = 'c_polyester', T = 100., F = 1.680 /
&RAMP ID = 'c_polyester', T = 180., F = 2.180 /
!glass fiber properties
&RAMP ID = 'k_verre1', T = 20., F = 0.53 /
&RAMP ID = 'k_verre1', T = 200., F = 0.65 /
&RAMP ID = 'k_verre1', T = 300., F = 0.75 /
&RAMP ID = 'k_verre1', T = 500., F = 1.04 /
&RAMP ID = 'k_verre', T = 20., F = 1.1 /
&RAMP ID = 'k_verre', T = 200., F = 1.3 /
&RAMP ID = 'k_verre', T = 300., F = 1.47 /
&RAMP ID = 'k_verre', T = 500., F = 2.09 /
&RAMP ID = 'k_verre', T = 700., F = 5.02 /
&RAMP ID = 'k_verre', T = 900., F = 10.9 /
&RAMP ID = 'c_verre', T = 20., F = 0.87 /
&RAMP ID = 'c_verre', T = 200., F = 0.87 /
&RAMP ID = 'c_verre', T = 300., F = 0.96 /
&RAMP ID = 'c_verre', T = 500., F = 1.005/
&RAMP ID = 'c_verre', T = 700., F = 1.071/
!fiber properties
&RAMP ID = 'k_fibre', T = 0., F = 0.05 /
&RAMP ID = 'k_fibre', T = 100., F = 0.06 /
&RAMP ID = 'k_fibre', T = 200., F = 0.08 /
&RAMP ID = 'k_fibre', T = 300., F = 0.1 /
&RAMP ID = 'k_fibre', T = 400., F = 0.12 /
&RAMP ID = 'k_fibre', T = 500., F = 0.15 /
&RAMP ID = 'k_fibre', T = 600., F = 0.18 /
&RAMP ID = 'k_fibre', T = 700., F = 0.22 /
&RAMP ID = 'k_fibre', T = 800., F = 0.27 /
&RAMP ID = 'k_fibre', T = 900., F = 0.33 /
&SPEC ID='Polyester_fuel', FORMULA='C1.0H1.039O0.2N0.0035'/
&SPEC ID='WATER VAPOR'/
&SPEC ID='CARBON DIOXIDE'/
&REAC ID = 'Polyester_burn'
FUEL='Polyester_fuel'
CO_YIELD=0.047
SOOT_YIELD = 0.07 /
I tried to run the case with dry wood and so i defined and changed the spec_id to cellulose for the dry wood instead of polyester_fuel and it also works. I also don't understand what kind of species to put during the pyrolysis of certain materials like ACTIVE,LIGNIN,is it always the fuel used in REAC? Any help with pointing out the problem with the materials definition would be greatly appreciated!
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