Physics Asked by Łukasz on May 22, 2021
I have a problem with understanding basic concept about enthalpy.
I know there is an enthalpy (heat) of formation which tells how much energy is going to be released or consumed during chemical reaction. I found information that specific heat of formation for one of the hydrocarbon jet fuel is equal to -200 BTU/lbm. How to interpret that? If all the elemental ‘building blocks’ of the fuel undergoes reaction of mixing of whatever there will be 200 BTU/lbm heat released?
There must be second ‘meaning’ of enthalpy because I also found example problem of burning jet fuel with air, where in the given data is information that fuel enthalpy is equal to +189 BTU/lbm (there was no information if that’s heat of formation). What does it mean that fuel has enthalpy? It sits in the tank, and before burning process is no chemical reaction
How to interpret both cases? What’s the difference here. Generally we talk about enthalpy of fluid. Like compressor inlet flow enthalpy is equal to XXX, compressor exit flow enthalpy = YYY. There is no chemical reaction between inlet/outlet plane so …
Could you give me a ‘dummy’ type explanation?
Thank you,
Lukasz from Poland
Let me paste one of my comments here:
This is how book describes combustion in jet engine:
*specific enthalpy of mix air and fuel
hIn = ( mAir * hAir + mFuel * hFuel ) / ( mAir + mFuel)
*specific enthalpy at the end of combustion process
hOut = hIn + (LHV * (mFuel/mAir) * efficiency)
Looks like that I can’t understand what hFuel, and hAir mean in that case.
Consider a generic hydrocarbon (as a model for a fossil fuel), $text{C}_xtext{H}_{2y}$:
$$xtext{C}(s)+ytext{H}_2 (g)to text{C}_xtext{H}_{2y}(g,l)tag{1}$$
$$text{C}_xtext{H}_{2y}(g,l)+(x+frac{y}{2})text{O}_2(g)to xtext{CO}_2(g)+ytext{H}_2text{O}(l)tag{2}$$
$(1)$ is its formation reaction, with it is associated an Enthalpy of Formation.
$(2)$ is a typical combustion reaction, with it is associated an Enthalpy of Combustion (aka Enthalpy of Oxydation).
Both are very distinct and should not be confounded.
Answered by Gert on May 22, 2021
The enthalpy of formation of a compound is found relative to some specified reference state (of temperature and pressure) for the elements comprising the compound. The enthalpy of the elements comprising the compound is taken as zero in this reference state. So the enthalpy of formation of the compound is the change in enthalpy in going by chemical reaction from an initial state of the pure elements in the reference state to the compound in the reference state; it is equal to the amount of heat that has to be added to the system such that the temperature of the product is the same as that of the elementary reactants. This heat addition (or removal in the case of a negative heat of formation) is the result of the energy required to make and break chemical bonds.
Once you know the enthalpies of formation of a wide range of elements and compounds, you can determine the heat of reaction (enthalpy of reaction, equal to heat that has to be added to convert reactants into products) of any reactions involving these elements and compounds by applying Hess' law. One example of such a reaction is combustion.
Answered by Chet Miller on May 22, 2021
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