How to freeze the reactions in the Quench zone without affecting the combustion the Lean zone?

[Francisco M. Guimarães]

Dear all,

In the CRN model I developed of a Rich-Quench-Lean combustor I am currently having a problem. I want to simulate the Quench stage (QZ) as a zone where no reactions take place, but only in the Rich (RZ) and Lean (LZ) stages. To do this I set "QZ.chemistry_enabled = False" for the Quench zone.

 

However, the problem is that putting "QZ.chemistry_enabled = False" will not lead to complete combustion in Lean, as you might expect. On the contrary, by writing "QZ.chemistry_enabled = True" combustion starts in the Quench, which is not supposed to, but there is complete combustion in the LZ, which is supposed to.

Below you can see some relevant lines of code.

# -------------------------------------------------------------------------------------------------------

# high temperature inside Rich stage to ignite mixture (NH3/H2)

gas = ct.Solution('mechanismsRQL\\{}'.format(mech))
gas.TPX = 1500, P, 'O2:0.21, N2:0.79'
rhoGas = gas.density

# ambient temperature in Quench-Lean stages
gas_0 = ct.Solution('mechanismsRQL\\{}'.format(mech))
gas_0.TPX = 298, P, ' O2:0.21, N2:0.79  '

# --- Rich----

(...)

PFR = ct.IdealGasReactor(gas, energy='on')
PFR.volume = dVol_PFZ

# --- Quench ----

QZ = ct.IdealGasReactor(gas_0, energy='on')
QZ.volume = vol_QZ
QZ.chemistry_enabled = False

# --- Lean ----

LZ = ct.IdealGasReactor(gas_0, energy='on')
LZ.volume = dVol_LZ
LZ.chemistry_enabled = True

# -------------------------------------------------------------------------------------------------------

Results using "QZ.chemistry_enabled = False":

It would be expected that NOx would increase considerably from 310 mm and NH3 and H2 would go to zero from that point as well.

Does anyone have any idea how I can overcome the problem of being able to cancel the reactions in Quench without it affecting Lean? Do I make myself clear?

Thanks in advance!

Francisco

[Bryan Weber]

Hi Francisco,

I'm not sure I understood you. It seems clear to me that if QZ has reactions or not, the input conditions to the Lean reactor will be quite different. As such, I don't see how you can cancel the reactions in QZ with affecting Lean, since it is not a problem of an interaction between the code. Can you try to clarify your objective?

Best,

Bryan

[Francisco M. Guimarães]

Hi Bryan,

Yes, I definitely didn't explain myself well.... What I wanted to understand is the reason that NOx does not increase from Lean, as is expected. Just as NH3 and H2 are also expected to drop to zero, which was not happening.

However, I think I have figured out the error. I model the Lean stage by a PFR, for which I use the example from Cantera's site (pfr.py) as a reference. My error was that I was initializing the Reservoir upstream of this PFR with gas_0 (see previous post) which due to the low temperature did not allow the ignition of the fuel present in the effluent coming from the Rich stage.   

However, now a new problem has arisen! What happens now is that the NOx is increasing more than it should be in the Lean stage, resulting in values that I think are not reasonable.

Furthermore, comparing the cases of considering reactions in the Quench (on the left in the figures) and not considering reactions in the Quench (on the right in the figures), we can see that all the profiles, except those relative to NOx, are practically the same for both cases, except for the jump that exists from Quench to Lean in the figures on the right side.

Only in the NOx prediction are the final results completely different! I don't understand why there is such a discrepancy for NOx, unlike the others.  

Does anyone have any idea what might be the basis for this discrepancy in NOx, when comparing results for an RQL system considering Quench with or without chemical reactions?

Many thanks,

Francisco

[Francisco M. Guimarães]

I would really appreciate some commentary on this problem (see the previous post), if possible, of course. I have some urgency to solve it...

Sincerely grateful!

Francisco

[Tomm]

Maybe you could share the code? I find it quite hard to reply on this topic with around 10 lines of code.