Hi,
I have a problem calculating Adiabatic Flame Temperature of ethanol burned with air in fuel rich conditions. I cannot solve the mass balance and find the number of moles of each component. I have to solve the reaction with the following products: CO2, H2O,CO, NO and H2. I tried to use the equilibrium coefficients of the elementary reactions in addition to the mass balance equations but I cannot find any solution. In fact I have to examine the effect of changing the (fuel/oxygen)/(fuel/oxygen)stoichiometric ratio on the adiabatic flame temperature.
I will be very grateful, if anyone can help me with some advice.
Thank you!
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Adiabatic Flame Temperature Calulation -- Fuel Rich Conditions
Started by , Sep 29 2007 08:42 PM
5 replies to this topic
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#1
Posted 29 September 2007 - 08:42 PM
#2
Posted 01 October 2007 - 03:52 AM
For complete combustion the following method can be used for gas fuel;
Ht = 0,03457 x ( combustion air temp – 65,6 ) + Lower Heating Value fuel / ( 37,2540220635 x air / gas ratio )
theoretical maximum flame temp = 456 + 17,2 x Ht + 0,002 x combustion air temp x (Ht-1)
all values in metric units.
In your case the problem is a surplus of fuel, no idea how to apply this method in such a case nor how to calculate the amount of different components (the CO to NO ratio being the most difficult thing; NO and NOx formation depending on the flame temperature and exact burner conditions).
Ht = 0,03457 x ( combustion air temp – 65,6 ) + Lower Heating Value fuel / ( 37,2540220635 x air / gas ratio )
theoretical maximum flame temp = 456 + 17,2 x Ht + 0,002 x combustion air temp x (Ht-1)
all values in metric units.
In your case the problem is a surplus of fuel, no idea how to apply this method in such a case nor how to calculate the amount of different components (the CO to NO ratio being the most difficult thing; NO and NOx formation depending on the flame temperature and exact burner conditions).
#3
Posted 01 October 2007 - 10:17 AM
Simplistically, all your hydrogen contained in the fuel will burn to form water. Then you will burn your carbon to CO, and then you will start burning that to CO2. This will continue until you run out of oxygen.
This is, of course, a gross simplification. It does not take into account the fact that some nitrogen will combine with oxygen present to form NOx (though the amount should be quite small under fuel-rich conditions). You may also form some carbon (soot) at the expense of CO and CO2. Nonetheless, I feel this simple approach will be sufficiently adequate for many calculations. If you need better, I hope you have a process simulator available.
Doug
This is, of course, a gross simplification. It does not take into account the fact that some nitrogen will combine with oxygen present to form NOx (though the amount should be quite small under fuel-rich conditions). You may also form some carbon (soot) at the expense of CO and CO2. Nonetheless, I feel this simple approach will be sufficiently adequate for many calculations. If you need better, I hope you have a process simulator available.
Doug
#4
Posted 03 October 2007 - 07:43 AM
QUOTE (djack77494 @ Oct 1 2007, 11:17 AM) <{POST_SNAPBACK}>
Simplistically, all your hydrogen contained in the fuel will burn to form water. Then you will burn your carbon to CO, and then you will start burning that to CO2. This will continue until you run out of oxygen.
This is, of course, a gross simplification. It does not take into account the fact that some nitrogen will combine with oxygen present to form NOx (though the amount should be quite small under fuel-rich conditions). You may also form some carbon (soot) at the expense of CO and CO2. Nonetheless, I feel this simple approach will be sufficiently adequate for many calculations. If you need better, I hope you have a process simulator available.
Doug
This is, of course, a gross simplification. It does not take into account the fact that some nitrogen will combine with oxygen present to form NOx (though the amount should be quite small under fuel-rich conditions). You may also form some carbon (soot) at the expense of CO and CO2. Nonetheless, I feel this simple approach will be sufficiently adequate for many calculations. If you need better, I hope you have a process simulator available.
Doug
Thanks for the idea but I think that it cannot be applied since I have to calculate H2 also as a product . So the hydrogen goes to H2O + H2.
Thanks anyway.
#5
Posted 11 October 2007 - 06:05 AM
Dear ZZZ
You have to write down all the reaction which ever are possible, and you think that are possible at that conditions. You can assume that whatever coming out are in the at chemical equilibrium. First assume concentration of all the products (some variable) and then do the material balance. That will lead to give the equations.
Now do the enthaply balance to give the outlet temperature in terms of the variables that are used.
Now write chemical equilibrium equations (thermodynamics) at outlet temperatures. All this equations has to be solved simultaneously to give you the answers. My guess is that you wont get any NO as inorganic N2 which comes from Air wont burn.
Hope this will help you
Regard
Suhas
You have to write down all the reaction which ever are possible, and you think that are possible at that conditions. You can assume that whatever coming out are in the at chemical equilibrium. First assume concentration of all the products (some variable) and then do the material balance. That will lead to give the equations.
Now do the enthaply balance to give the outlet temperature in terms of the variables that are used.
Now write chemical equilibrium equations (thermodynamics) at outlet temperatures. All this equations has to be solved simultaneously to give you the answers. My guess is that you wont get any NO as inorganic N2 which comes from Air wont burn.
Hope this will help you
Regard
Suhas
#6
Posted 13 November 2007 - 06:36 PM
Hi there:
Here is an MS Excel that can help you with coal and oil fuel type stoichiometric and oxidant rich combustion cases.
http://engware.i-dentity.com/eng-tips/combustion.xls
For fuel rich combustion, one needs to work with already developed software packages that are based on the minimization of the Gibbs free energy. Even simple combustion cases are very difficult to be handled by hand calculations -- there is a problem of no convergence when the combustion products temperature gets very low.
In general, this is not an easy subject matter to deal with.
Thanks,
Gordan
Here is an MS Excel that can help you with coal and oil fuel type stoichiometric and oxidant rich combustion cases.
http://engware.i-dentity.com/eng-tips/combustion.xls
For fuel rich combustion, one needs to work with already developed software packages that are based on the minimization of the Gibbs free energy. Even simple combustion cases are very difficult to be handled by hand calculations -- there is a problem of no convergence when the combustion products temperature gets very low.
In general, this is not an easy subject matter to deal with.
Thanks,
Gordan
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