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Author Topic: ‏ how many times does carbon equal hydrogen for combustion ?  (Read 4367 times)

Offline taregg

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 for exsample c=h+h+h  ......and explain why?
« Last Edit: 05/01/2012 11:43:40 by taregg »


 

Offline rosy

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I'm afraid this question is completely incomprehensible. Please be more specific about what you'd like to know!
 

Offline Geezer

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Hydrogen is about 4.5 times carbon by mass.
 

Offline damocles

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Hydrogen is about 4.5 times carbon by mass.


I'm afraid this questionreply is completely incomprehensible. Please be more specific about what you'd like to knowsay!

 ;D
 

Offline CZARCAR

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c+o2=co2= xBTU
h+o2=h2o=yBTU
????????
 

Offline imatfaal

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c+o2=co2= xBTU
h+o2=h2o=yBTU
????????

For starters your equations don't balance. 

Damocles will rap my knuckles for this - but assuming limitless supply of oxygen and perfect conditions; high pressure hydrogen will provide about 123 MJ/kg and coal will provide about 24MJ/kg. 

A kilo of hydrogen has twelve(ish) times more atoms than a kilo of carbon.  carbon wins - but i know what they use to drive rockets and it ain't coal
 

Offline CZARCAR

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c+o2=co2= xBTU
h+o2=h2o=yBTU
????????

For starters your equations don't balance. 

Damocles will rap my knuckles for this - but assuming limitless supply of oxygen and perfect conditions; high pressure hydrogen will provide about 123 MJ/kg and coal will provide about 24MJ/kg. 

A kilo of hydrogen has twelve(ish) times more atoms than a kilo of carbon.  carbon wins - but i know what they use to drive rockets and it ain't coal
thanx, just trying to help OP. Q= H floats vs C & might this assist in "rising" the rocket?
 

Offline imatfaal

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c+o2=co2= xBTU
h+o2=h2o=yBTU
????????

For starters your equations don't balance. 

Damocles will rap my knuckles for this - but assuming limitless supply of oxygen and perfect conditions; high pressure hydrogen will provide about 123 MJ/kg and coal will provide about 24MJ/kg. 

A kilo of hydrogen has twelve(ish) times more atoms than a kilo of carbon.  carbon wins - but i know what they use to drive rockets and it ain't coal
thanx, just trying to help OP. Q= H floats vs C & might this assist in "rising" the rocket?

I think in rockets its more the higher energy density per kilo that matters rather than per mole.  And ease of combustion and control.  Note that rockets use liquid hydrogen which (even though it is very undense for a liquid) is far denser than air and would fall rather than rise if left by itself.
 

Offline damocles

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for exsample c=h+h+h  ......and explain why?

So we seem to have decided, on behalf of taregg:

(1) that the question does not really have anything to do with temperature, but with heat energy output
(2) that we are specifically referring to an oxidation reaction with oxygen gas
(3) that we are discussing things on a mass basis rather than an atomic basis (per kilogram rather than per mole)

This is probably the best set of assumptions we can make, in spite of the fact that the subject line specifically mentions "temperature" and the notation "c=h+h+h" suggests an atomic rather than a mass basis.

So, to make things perfectly clear:
- actual flame temperature depends on a lot of factors, but can be tabulated as "adiabatic flame temperature".
- flame temperatures in °C do not scale as multiples. To say that a temperature is "twice as high" as another is quite meaningless. Absolute temperatures, kelvin temperatures, do scale for some purposes, but this area is not one of them.

As Amatfaal has said, the heat output from hydrogen combustion is about 5 times that from carbon combustion on a mass basis, given a plentiful supply of oxygen gas (or normal air, of course).
The heat output from carbon combustion is about 1.3 times that from hydrogen combustion on a molar basis (heat per molecule), or 2.6 times that of hydrogen on a heat per atom basis, since hydrogen gas is H2

 

Offline flr

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 The combustion reactions of CH4 and H2 are:

 CH4 + 2O2 ---> CO2 + 2H2O

 H2 + 1/2 O2 --->  H2O

 A first (and quick) estimation of the heat released by combustion can be obtained from standard enthalpies of combustion.
 In general these enthalpies are available:
 -for methane DH_combustion(CH4,standard) = –882.0 kJ/mol  [ http://en.wikipedia.org/wiki/Methane_%28data_page%29 ]
 -for H2 is: DH_combustion(H2,standard = -286 kJ/mol [ http://en.wikipedia.org/wiki/Hydrogen]

So, per mol, you would get more energy from methane. For practical purposes however, more useful would be to report these energies per unit mass (because an object is as heavy as its mass is). To do so, divide the enthalpies in kJ/mol to the molar mass of the compound. One will easily get:

 -for methane DH_combustion(CH4,standard) = –882.0 kJ/mol/(16g/mol)= -55.1kJ/g
 -for methane DH_combustion(H2,standard) = –286.0 kJ/mol/(2g/mol)= -143kJ/g
 
   Per unit mass, H2 is provide a lot more energy than CH4.
 

Offline damocles

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(...snip...)

So, per mol, you would get more energy from methane. For practical purposes however, more useful would be to report these energies per unit mass (because an object is as heavy as its mass is). To do so, divide the enthalpies in kJ/mol to the molar mass of the compound.
(...snip...)


I am a little puzzled -- for which particular practical purpose is the energy value per unit mass more useful (for a gaseous fuel)?

-- The weight of a gaseous fuel is very much smaller than that of its container (or pipeline infrastructure)
-- The amount of a gaseous fuel contained in a particular container with a particular pressure rating is an equal number of mol, but a very different mass for different gases (ideal gas law).
-- It is also the case that flow of gas down a pipeline depends on gas viscosity (varies for different gases) and volume flux, not mass flux.
 

Offline flr

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I am a little puzzled -- for which particular practical purpose is the energy value per unit mass more useful (for a gaseous fuel)?

Which particular purpose? Well, for example would one use a gaseous propellant to fuel a car/ship/plane rather than liquid/solid (which carry more energy per volume)? And would not be useful to know how much extra mass the fuel carry?

See also wikipedia: energy densities are reported in energy / unit mass rather than energy /moles.
http://en.wikipedia.org/wiki/Energy_density
 

Offline damocles

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If you take a compressed gas cylinder of a particular size, and fill it with hydrogen or methane to the same pressure, the heat energy that you can get from burning the methane will be more than 3 times that available from burning the hydrogen, because both cylinders will contain an equal number of mole (but quite a different mass) of gas. Because of the requirements for a compressed gas cylinder, the mass of gas will only be a very small fraction of the mass of the cylinder. So for PRACTICAL purposes it is the number of mole of a gaseous fuel that counts, and not the mass.
« Last Edit: 26/11/2011 07:31:41 by damocles »
 

Offline Geezer

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If you take a compressed gas cylinder of a particular size, and fill it with hydrogen or methane to the same pressure, the heat energy that you can get from burning the methane will be more than 3 times that available from burning the hydrogen, because both cylinders will contain an equal number of mole (but quite a different mass) of gas. Because of the requirements for a compressed gas cylinder, the mass of gas will only be a very small fraction of the mass of the cylinder. So for PRACTICAL purposes it is the number of mole of a gaseous fuel that counts, and not the mass.


Which is why gaseous hydrogen (or any other gaseous fuel?) is a lousy fuel for a vehicle :D
 

Offline damocles

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If you take a compressed gas cylinder of a particular size, and fill it with hydrogen or methane to the same pressure, the heat energy that you can get from burning the methane will be more than 3 times that available from burning the hydrogen, because both cylinders will contain an equal number of mole (but quite a different mass) of gas. Because of the requirements for a compressed gas cylinder, the mass of gas will only be a very small fraction of the mass of the cylinder. So for PRACTICAL purposes it is the number of mole of a gaseous fuel that counts, and not the mass.


Which is why gaseous hydrogen (or any other gaseous fuel?) is a lousy fuel for a vehicle :D

Yes gaseous hydrogen is a lousy fuel for a vehicle, and but for economic considerations, so is methane. But methane = CNG (compressed/combustible natural gas, with some ambiguity about what the C stands for) is a suitable fuel and much used for heavy vehicles with short haul duty. Many urban bus fleets and some heavy goods suburban scale delivery transports run quite successfully and very cheaply on methane fuel (here in the antipodes, both Aus and NZ). Liquid petroleum gas (mostly propane) is a different matter, because although it is a gaseous fuel it is stored in the tank as a liquid under rather lower pressure than that used for CNG. It is quite competitive with petrol/gasoline, and actually more economical here because of more favourable tax treatment.

A 50 litre tank would contain 400 mol of hydrogen or methane gas if compressed to 200 atmosphere, meaning 0.8 or 6.5 kg of fuel respectively. It would contain 25 kg of liquid petroleum gas (propane) or 560 mol as (mostly) liquid compressed to 20 atm, or 35 kg of petrol/gasoline, amounting to roughly 350 mol at normal atmospheric pressure.
« Last Edit: 26/11/2011 11:36:48 by damocles »
 

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