Hydrogen Flame temperature

[SamofTexas]

Can anyone please tell, why is the adiabatic flame temperature of Hydrogen greater than Methane?

Ideally, I would think that since methane can release a higher BTU/SCF, it would have a higher flame temperature than Hydrogen, but it's not so. I am not able to reason out this fact.

 

[Feric]

Hi there:

Here a few plots for an ideal and complete combustion of carbon, sulfur, hydrogen, coal, oil and gas (methane -- CH4) when oxidant is standard air (79% N2 and 21% O2 on volume/mole basis).

I do hope that you will find the above plots useful ...

Thanks,

Gordan Feric, PE
Engineering Software

http://members.aol.com/engware

 

[25362]

An adiabatic flame temperature involves keeping all the developed heat in the products of combustion. Thus, the answer rests on the amount of heat taken up by the moles of gaseous products obtained per mol of fuel.

The stoichiometry of combustion with air tells us that the products are:

for hydrogen 2.88 moles/mol
for methane 10.53 moles/mol

I hope the above answers your query.

 

[SamofTexas]

Dear 25362:

I did not understand the 2.88 moes/mol of H2
and 10.53 moles/mole of CH4(methane) in your reply.

Can you please explain it in a bit of detail?

 

[jistre]

Without going into too much detail, think of it this way. You have one liter of CH3 and one liter of H2. Now, because of the nature of gases, each liter contains exactly the same number of molecules. However, each molecule of CH3 is much heavier than each molecule of H2. So, even though you have equal volumes and equal number of molecules, you have a much greater mass in the liter of CH3.

When you burn the CH3, a greater total amount of energy may indeed be released in the reaction that that released by burning the H2 (I don't know one way or the other off the top of my head). However, you're talking about the adiabatic flame temperature, and you're not looking solely at the energy liberated. What you're considering when you talk about the adiabatic flame temperature is how that liberated energy is absorbed by the products of combustion and figuring out what the highest flame temperature would be if everything went perfectly during the combustion. So, when you consider that the CH3 will produce a much larger mass in its products of combustion than the H2 will, the CH3 can burn with a lower flame temperature than the H2 will even though the combustion reaction itself may liberate more energy because it has much more "stuff" to heat up.

 

[Feric]

To All:

Here are the basic H2 and CH4 combustion reactions balanced on mole (volume) basis:

1 H2 + (1/2) O2 + (1/2)*3.76 N2 = 1 H2O + 1.88 N2

Therefore, for 1 mol of H2, one gets 2.88 moles of combustion products.

1 CH4 + 2 O2 + 2*3.76 N2 = 1 CO2 + 2 H20 + 7.52 N2

Therefore, for 1 mol of CH4, one gets 10.52 moles of combustion products.

Note, for standard air the mole composition is as follows:

79% N2 and 21% O2

or for 1 mol of O2, there come 3.76 moles of N2.

Even though this is just basic engineering combustion reaction balancing, it is not quite simple to get it right because of mole/volume and mass basis that get thrown into the subject matter.

I do hope that this helps connect the dots provided by other Eng-Tips.com contributors ...

Thanks,

Gordan Feric, PE
Engineering Software

http://members.aol.com/engware