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Guosheng Kang October 10, 1999 19:08

Velocity Field in Combustion
 
Hello all, I met a probelm which is described below, I hope experts on gas combustion simulation give me a clue. DESCRIPTION OF MY QUESTION I am simulate a premixed natural gas-pure oxygen combustion, the stoichiometri mixture is introduced at a speed of 80 m/s, the simualtion result shows the highest velocity is around 200 m/s, I doubt it even it is a convergent simulation result. Is it possible that the gas flow may be accelerated due to combustion?

Jin Wook LEE October 11, 1999 02:57

Re: Velocity Field in Combustion
 
Premixed combustion with natural gas as fuel and pure oxygen as oxidant ? It is very dangerous situation in the real situation because of the possibility of very high temperature and very high velocity. Anyway, let us focus on simulation only now.

In the computational domain, the highest temperature may be very high, say 5000 Kelvin or higher, for stoichiometric mixture(without heat loss and without dissocation). The highest velocity depends both on your inlet TEMPERATURE and GEOMETRY. Of course, 200m/s can be possible IN THE COMPUTATIONAL DOMAIN if inlet temperature is low and expansion ratio(Area_reactor/Area_nozzle) is samll(say order of unity). You can simply calculate APPROXIMATE outlet velocity by following equation.

V_out = V_in * (T_out/T_in) * (Area_in/Area_out)

According to my order analysis, T_out is above 5000 Kelvin (without heat loss and without dissociation) with 300 Kelvin inlet temperature, so that local highest velocity may exceeds 200 m/sec in your case. Check your geometry, especially ratio of A_out/A_in. In addition, if the highest velocity exceeds 200m/sec, compressibile effect might be dominant(check Mach number) included. I would like to say that natural gas combustion with pure oxygen is physically unreallistic.

Hoping it will help you, Jinwook.


John C. Chien October 11, 1999 10:47

Re: Velocity Field in Combustion
 
(1). The heat release in a closed system will cause the temperature to increase. As a result, the pressure will also increase. It is like a bomb. (2). The heat release in a constant pressure system will increase the temperature and reduce the density. To satisfy the continuity, decrease in density will cause the velocity to increase.

Guosheng Kang October 11, 1999 11:55

Re: Velocity Field in Combustion
 
Thank you, Jin Wook and John, here is another question, based on theoretical calculations, the peak temperature of stoichiometric methane-oxygen combustion is no more than 4000K (3080K by STANJA method, 3800K by my calculation), why my simulation and also Jin Wook's simulation resulted in a peak temperarture of 5000K? Because I don't know the source code, I am not sure what kind of mechanism is used in temperature prediction. Hope you two and other peers refresh my mind again, thanks a lot.

Rainer Kurz October 11, 1999 16:11

Re: Velocity Field in Combustion
 
Two possible reasons:(1) flame temperature is pressure dependent. Example: CH4-O2 at 1 bar: 3030K; at 20 bar: 3460K (2) Correct implementation of dissociation.

Jin Wook LEE October 11, 1999 20:55

Re: Velocity Field in Combustion
 
1) Today, I have had a calculation for methane-O2 mixture temperature by fairly well known process simulation S/W ASPEN PLUS(which is based on zero dimensional analysis) with its default chemical databank, PURECOMP. Resulting flame temperature was 5230K under complete combustion(without heat loss and without dissociation).

My simple calculation for order analysis was :

1 kmole CH4 + 2 kmole O2 (80kg) --> 1 kmole CO2 + 2 Kmole H2O (80kg)

Q_in = 5 * 10^4 *10^3 * 16 (50000KJ/kg x 16kg x1000Kj/kg)

Let Cp_mix = 2000 (about 1450 J/Kg-K at room temp. and 2250 J/Kg-K at 5200 Kelvin)

Where, mixture means 44kg(1 Kmole) CO2 + 36kg(2 Kmole) H2O

Delta T = 8*10^8/(80*2000) = 5000 Kelvin

T = T_room + Delta T = 5300 Kelvin

Again, emphasizing that my calculation was based on 1 bar pressure, without heat loss and without dissociation.

Sincerely, Jinwook.


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