4 replies to this topic

Hello newsgroup,

I am struggling with a conversion of HHV to LHV based on the amount of water in a gas mixture. This is what its about:

Gas composition mole frac:

H2O: 0.0395
N2: 0.3682
CO: 0.3311
CO2: 0.043
H2: 0.2183

I can now calculate the higher heating on a dry mole basis (i.e. excluding water) based on the enthalpy of combustion for CO (283 kJ/mol) and H2 (608 kJ/mol), giving me 235 kJ/mol, for a molar volume of 24.789l/mol this results in a higher heating value of 9510 kJ/Nm3.

However, I am baffled on how to converting this into a LHV now. All formula I have found are based on estimations for fuels, but the composition used there varies way too much that I would like to use them...

Thanks a lot and best regards,

MatL

[ankur2061]

Hi,

LHV and HHV are the same when the combustion of the fuel gives no water as a combustion product. For CO the HHV and LHV are the same. Please see the following link to download a spreadsheet I have posted regarding combustion calculations.

http://www.cheresour...;hl=spreadsheet

Regards,
Ankur.
Ankur.

Hi Ankur,

and thanks a lot for your message.

LHV and HHV are the same when the combustion of the fuel gives no water as a combustion product.

This i see, but that doesn't hold for the hydrogen in the gas. And there's also the fact that some share of the gas itself is water vapor, so when calculating the heating value of this, then the heat of evaporisation of this water must be included as well, or am i on the wrong path here?

Regards,

MatL

[djack77494]

then the heat of evaporisation of this water must be included as well, or am i on the wrong path here?

MatL,
I'm going with Option 2 here (you're on the wrong path). You're dealing with a gaseous fuel that has a bit of humidity (the water vapor). Ignore the water vapor - it does not contribute to the fuel's heating values.

HHV and LHV are defined values that begin with the compounds in the fuel that will be combusted. That is basically carbon, hydrogen, and sulfur bearing compounds that are oxidized (combusted) in the process. Nitrogen, carbon dioxide, water vapor, and sulfur oxides can be considered inert. You may use ankur's generously provided spreadsheet to calculate your heating values which apply the standard definitions -
C + O2 ----> CO2
CO + 1/2*O2 ----> CO2
H2 + 1/2*O2 ----> H2O
S + O2 ----> SO2
For HHV, the water product is assumed to be in vapor form and for LHV it is assumed to be liquid.

Hi djack,

MatL,
I'm going with Option 2 here (you're on the wrong path). You're dealing with a gaseous fuel that has a bit of humidity (the water vapor). Ignore the water vapor - it does not contribute to the fuel's heating values.

thanks a lot - now I got it! It took a while, but now I see - it's only the water that is formed by the reaction that impacts the LHV, not the water present... that makes sense

Thanks and best regards to you both,

MatL