0
$\begingroup$

I'm working on some code that computes adiabatic flame temperature given a balanced equation and the relevant thermodynamic properties and I'm starting with the simplest combustion reaction I can think of: stoichiometric combustion of pure hydrogen with pure oxygen.

$2H_2+O_2 \Rightarrow 2H_2O$

I'm using these thermodynamic values:

$h_{f,H_2} = 0.0 \text{ kJ/kmol}$

$h_{f,O_2} = 0.0 \text{ kJ/kmol}$

$h_{f,H_2O} = -241820 \text{ kJ/kmol}$

$c_p = 1.864$ kJ/kg-K at 300K for $H_2O_{(g)}$

$c_p = 3.217$ kJ/kg-K at 4000K for $H_2O_{(g)}$

If the combustion equation I wrote is on a kmol basis, I get a total change in enthalpy of 483640 kJ with the combustion. If that heat goes into heating the product (water vapor), I get temperature increases ranging from 4173K to 7201K (final temperatures of 4471K and 7499K) depending on the specific heat value you use.

$\frac{483640\text{ kJ}}{(3.217 \text{ kJ/kg-K})(36.03\text{ kg})} = 4173\text{ K}$

$\frac{483640\text{ kJ}}{(1.864 \text{ kJ/kg-K})(36.03\text{ kg})} = 7201\text{ K}$

When I run the code that actually considers changes in the specific heat throughout the whole heating process (interpolating based on a table at each step), I get 5024K. Even my lowest value, assuming constant specific heat at the highest value, gives a temperature increase higher than the number listed on Wikipedia, ~3500K. Am I making some kind of error or is there some kind of factor that makes the real thing different from the idealization? Thanks.

$\endgroup$
  • $\begingroup$ You assume complete combustion but even in ideal conditions Le Chatelier's principle prohibits that. en.wikipedia.org/wiki/… .High temperatures affect the reaction equilibrium constants. $\endgroup$ – Gert Jul 10 '16 at 1:43
0
$\begingroup$

I thought it is related to dissociation and confirmed that from the book by Michael Liberman.

The assumed reaction $2H_2+O_2 \Rightarrow 2H_2O$ is not true at high temperature. The molecules will dissociate to $H_2, H, O_2, O, OH, HO_2$. The dissociation reactions (e.g. $H_2O \rightarrow OH + H + O$) are endothermic and thus reduce the flame temperature.

To calculate the adiabatic flame temperature, you may need a program such as Stanjan to include more species.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.