Hydro carbon fuel - H2 % value of total heat

walterjw
:rofl: Hi I did this simple calculation as a mature student in 1965 & now I need to prove my statement. 'In most hydrocarbon fuels the combustion of Hydrogen H2 gives about 90% of the toal heat'
Eg Propane C3 H8 gives 50,350 KJ/Kg, H2 gives 141,790 KJ/Kg I know Carbon is a lump of black stuff like coal which is a hydrocarbon fuel & gives 15 - 17,000 KJ/Kg.
How true use it or loose it - more on H2 on my web page
http://myweb.tiscali.co.uk/h2energy4all" [Broken]
- looking for an elderly puzzled smiley.:rofl:

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Gold Member
Showing your data like that is just plain confusing though, because you can easily carry a worthwhile amount of a hydrocarbon fuel around in your car; it's stored as a liquid. It's no engineering problem to put petrol or Diesel in a tank. That's not true for hydrogen, which is far more difficult to store. Yes, hydrogen contains a large amount of energy per kilogram, but how about per litre?

Find us an economical way of producing hydrogen on a scale large enough for all automotive use. Then find us a way of storing it such that it's viable for cars.

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walterjw
Are you into Politics? standard procedure for them is - if you don't understand a question ask another

Homework Helper
now I need to prove my statement. 'In most hydrocarbon fuels the combustion of Hydrogen H2 gives about 90% of the toal heat'

That's a meaningless statement, unless you have a chemical reaction that can turn a hydrocarbon fuel into water plus unburnt carbon.

I have never been able to get a good source for the % energy per element in a fuel. I can say that weight % of H2 in jet fuel is only about 13%. Granted, carbon is a much heavier molecule. The only thing I have ever been able to find is this graphic:

http://www.elmhurst.edu/~chm/vchembook/images/512energycombust.JPEG [Broken]

This gives the energy input to break bonds, the energy released forming new bonds and the overall net energy released for the combustion of methane. All of the energy comes from the formation of bonds. I can't see how this would support the statement that all of the energy comes from the hydrogen alone.

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walterjw
From my calculation in the 1960's the 90% was for H2 and the other 10% for the heat provided by the Carbon - many thanks for the link. Calculations were much easier in the 60's we only knew a little about a lot - now?

Gold Member
Urm, what's your question? The energy from the combustion of hydrocarbon fuels does not come from the separate combustion of hydrogen and carbon, it's essentially from the oxidation of the hydrocarbons themselves (breaking of carbon/hydrogen bonds, and reforming carbon dioxide and water, depending on whether the combustion is complete).

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walterjw
The question is eg take Propane as an example of a Hydrocarbon fuel (HCF), formulae is C3H8 & when burnt +5O2 + 18N2 it gives say 50,350 KJ/Kg. What is the % of that 50,350KJ that came from the H2 - the remainder must be carbons contribution as O only supports combustion and N is an inert gas.
I remember dimly doing the 'simple' calculation that involved using the formulae ( & ? valency of the elements) the heat value (energy) of the elements H is 141,790 KJ/Kg & I can't find out C carbon but coal is 15 - 17,000 so ? for carbon say ?? 2,000 KJ/Kg ??
I have seen the answer is of the order of 90% for H but I need to be able to prove it - unless somebody invents an ice engine.

Mentor
Fred's link gives you exactly the procedure required to find your answer in an example for methane. Just repeat it for propane. The only piece of information not in the link is the bond energy in the c-c bond and you can google that.

Caveat: you can't really attribute the energy in the c-h bond to one or the other, so the question is kinda flawed to begin with (which is what alphazero was getting at).

Specific heats (the numbers you just posted) have nothing to do with chemical reactions, they are the the energy required to raise the temperature of one kg of the material one degree C.

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Cyrus
Then find us a way of storing it such that it's viable for cars.

I believe they are looking into ways of diffusing it into a (substrate is the correct term?) so that its not in liquid form. When you use it, it diffuses back out.

Gold Member
I believe they are looking into ways of diffusing it into a (substrate is the correct term?) so that its not in liquid form. When you use it, it diffuses back out.

It's definitely an area under moderate research at the moment. However, studies so far have not found the mass of hydrogen able to be stored by these devices to have been pretty impressive.

Further info:
http://www.inl.gov/hydrogenfuels/projects/storage.shtml [Broken]
http://gcep.stanford.edu/research/factsheets/nanocluster_composite.html
http://cat.inist.fr/?aModele=afficheN&cpsidt=13542763

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walterjw
Many thanks for all your posts - I am going to try & get a 1965 textbook on thermo & combustion & see if I can then post what was printed.

removing carbon from combustion

That's a meaningless statement, unless you have a chemical reaction that can turn a hydrocarbon fuel into water plus unburnt carbon.

That's more than just easy. Combustion has to be deliberately designed to prevent soot formation.
Hydrogen burns more easily than carbon, so combustion is normally arranged to provide enough oxygen and enough temperature to prevent the carbon from forming soot. Deliberately collecting most of the carbon from combustion would be very easy. It would require a relatively cool surface for the soot to collect on, a means of continually removing it, a reduction of O2 at the flame, and provision of adequate hot air further along the flame path for complete combustion of the hydrogen and carbon monoxide.

walterjw
many thanks - does a candle flame burning just under a saucer & it leaves a black (?carbon) deposit illustrate your point, vary the flame distance from the saucer & its a circle or a ring - all I need to do is to validate my statement in my one web page on H2 fuel (90%) as stated in http://myweb.tiscali.co.uk/h2energy4all [Broken] - things were much simpler in the 1960's

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Mentor
C'mon, walter- this is a simple problem, but I'm not going to do it for you. Have you googled for the bond energies in propane yet?

many thanks - does a candle flame burning just under a saucer & it leaves a black (?carbon) deposit illustrate your point, vary the flame distance from the saucer & its a circle or a ring - all I need to do is to validate my statement in my one web page on H2 fuel (90%) as stated in http://myweb.tiscali.co.uk/h2energy4all [Broken] - things were much simpler in the 1960's

Yes, that illustrates carbon deposition but it doesn't validate your point. The reason furnaces are not designed to avoid burning the carbon is that your 90% figure is way off. You're overlooking something.

Bill

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eaboujaoudeh
Here goes: i really had to dig up the past to get you the answers, lucky i still know where i have put them:)
In ideal combustion the combustion energy of hydrogen is 120000KJ/Kg, the 141800 is the higher heating value and cannot be used in engine yield calculations or combustion calculations. The combustion of pure carbon (coal) has 33800KJ/Kg, while C3H8 gives 46190KJ/Kg. The missing value you are searching for lies in your negligence to the energy needed to break the C3H8 internal bonds prior to combustion, breaking bonds consumes the energy u are looking for, so combustion of C is not 2000KJ/Kg, its that value + the value needed to break the bonds. C is actually coal and is rated like i stated above. Hope this answers your question

To come up with anything near Walter's percentage for hydrogen energy, I have to "forget" something more obvious than the C-H bond energy.

Using eaboujaoudeh's figures, I get 47% from H, 53% from C.

When I use Walter's numbers and ignore the energy used in breaking the bonds I get 54% for H, 46% for C.

eaboujaoudeh
actually its 56% C and 44% H, and a value nearly equal to 7% to the final yield is used for bond breaking. but how can u ignore the energy needed to break the bonds? u can't do that.

actually its 56% C and 44% H, and a value nearly equal to 7% to the final yield is used for bond breaking. but how can u ignore the energy needed to break the bonds? u can't do that.

I'm sorry, I guess I wasn't clear. Walter had calculated that more than 90% of the energy of combustion comes from hydrogen. My point was that the energy used to break the bonds prior to oxidation is much less than the difference between the actual contribution of hydrogen and the amount he had calculated. I didn't mean to imply that you can ignore the bond energies of a fuel, only that this isn't the main source of the error.

I'm no physicist, but it seems to me that what Walter has overlooked is very simple and very clear.

eaboujaoudeh
oh ok ! :)