In the free flame propagation solution it is required that one of the nodes has to be at a particular temperature. I was trying to go through the code to find exactly how that temperature was set and failed to find it. I would be looking there first. If that temperature makes no sense for the problem that you are trying to solve that would screw up any numerical solution that you seek. From what you report this could be the difficulty. If you find it and that is the problem then you can modify the C++ code and compile your own amended version of Cantera and proceed.
Hope this helps.
Dave Mikolaitis
________________________________________
From: canter...@googlegroups.com <canter...@googlegroups.com> on behalf of Fabio Di Lenarda <fdl....@gmail.com>
Sent: Monday, December 11, 2017 5:31 PM
To: Cantera Users' Group
Subject: [cantera-users] Re: Laminar Premixed Flame Simulation Problems
Nobody could help me? :( It's a very important point for my thesis.
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Fabio,
It appears to me that you have some bad thermo data for the species named “CH2CHCHCHC.H2”. The first clue is the huge discrepancy at T=1000 which you are warned about when importing the mechanism:
**** WARNING ****
For species CH2CHCHCHC.H2, discontinuity in s/R detected at Tmid = 1000
Value computed using low-temperature polynomial: 18.0695
Value computed using high-temperature polynomial: 60.9667
While small discrepancies (on the order of a few percent) can be potentially ignored, this is a sign of bigger problems. The second sign is that the heat capacity for this species is actually negative at high temperatures:
>>> S = gas.species('CH2CHCHCHC.H2')
>>> S.thermo.cp(2500)
-111885.1767752868
which is clearly non-physical. The incorrect thermo data ends up giving reverse rate coefficients that are impossibly high, which is what ends up giving the 1D solver trouble:
>>> gas.TPX = 700, None, f.inlet.X
>>> rr = gas.reverse_rate_constants
>>> for i in range(gas.n_reactions):
>>> if rr[i] > 1e16:
>>> print gas.reaction_equation(i), rr[i]
CH2CHC.HCHCH2 <=> CH2CHCHCHC.H2 8.58725587371e+27
CH2CHCHCHCH3 + H <=> CH2CHCHCHC.H2 + H2 9.81104809369e+18
CH2CHCHCHCH3 + OH <=> CH2CHCHCHC.H2 + H2O 3.03424875448e+16
If you fix this thermo data, I think you will find that the 1D solutions will converge without any further troubles.
Regards,
Ray
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Fabio
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