From my long time experience in teaching thermodynamics for engineers I come to the conclusion that "enthalpy" as combination of inner energy and the product of pressure and specific volume should better not be used in energy conversion analysis like the analysis of power cycles. The two main reasons are:
In this kind of analysis the inner energy is a state quantity, the product of pressure and specific volume, however, appears as a work term, i.e. as a process quantity, Mixing both catagories in one quantity is misleading.
When the enthalpy is used one automatically refers to the total thermodynamic energy and not to the two partial equations of mechanical energy (in which the pressure times specific volume occurs) and the thermal energy (in which the inner energy occours), Then, however, the dissipation term does not appear explicitely and thus dissipation (losses) cannot be determined.
For more details see the paper
The Misleading Use of “Enthalpy” in an Energy Conversion Analysis
which can be found on my research gate page.
I must say that I have difficulty understanding where the problem is. We have the thermodynamic laws (1st and 2nd) which we cannot alter. We might write them with enthalpy (h) or with internal energy (u), but if we apply the laws correctly, we will find correct and unique results for the processes we tackle. For instance, if we ignore kinetic and potential energies, and look at open systems, then in a
adiabatic process: work = - enthalpy difference
isobaric process: heat = enthalpy difference
other processes: heat - work = enthalpy difference
The given reference on misleading use of enthalpy considers, it seems, only incompressible fluids, and thus misses out on compressibility effects. But these are important, in general. In isobaric processes compressibility effects lead to the heat being related to enthalpy, not internal energy difference: some energy is required for the expansion (or contraction) against the constant pressure. In incompressible liquids, however, this effect vanishes, and specifc heat at constant pressure equals specific heat at constant volume.
Hence, it's not that straightforward to split the first law into work and heat contributions--to do this properly, one needs to consider first and second law together. If one does that, by means of the Gibbs equation, the mechanical work in open systems is not the difference in pressure (p) times volume (v), but rather
work = - Integral [ v dp] (+ other terms, incl. dissipation)
Only for incompressible fluids, where v=const. this becomes
work = difference in (p v) (+ other)
So, I am happy to have enthalpy in my equations, alone for convenience: why should I write (u+pv) in all expressions for a steam engine (say), where all heat and work contributions are determined as enthalpy differences?
I might also add the following: I never had problems with enthalpy in teaching thermodynamics and energy conversion. It's just the quantity that appears in the proper form of the thermodynamic laws, if these are written in a convenient way, and hence shows up in expressions of work and heat for a number of standard processes (isobaric, adiabatic,...). One could do without, but that would just made things rather inconvenient--there is a reason that steam tables have a column for enthalpy
Enthalpy also is one of the thermodynamic potentials (when variables are entropy and pressures), so it has additional meaning in other parts of thermodynamics.
Reading the question again, I see the statement about the pv term appearing from work (it's the work per unit mass to push mass acrross a system boundary), hence referring to a process quantity. In general, as the name states, process quantities depend on the process (for instance the above Integral [v dp] depends on the relation v(p) ). But for the work to push the mass over the system boundary, there is no process dependency: it's just what it is (or zero, if no mass is transferred). And pv is a property as well. Maybe I just don't understand why this should be a problem?
In the end, what we need is the proper equations to describe the system at hand, and the proper procedure to solve them for what we need to know for particular systems (heat, work, etc). Whether enthalpy shows up for convenience or not should not alter the equations, nor the result.
I add a short file, where I show how 1st and 2nd law can be split between work and heat by means of the Gibbs equation (adapted from Chapter 9 of my new textbook with Springer--sorry, I could not resist to mention it)
I must say that I have difficulty understanding where the problem is. We have the thermodynamic laws (1st and 2nd) which we cannot alter. We might write them with enthalpy (h) or with internal energy (u), but if we apply the laws correctly, we will find correct and unique results for the processes we tackle. For instance, if we ignore kinetic and potential energies, and look at open systems, then in a
adiabatic process: work = - enthalpy difference
isobaric process: heat = enthalpy difference
other processes: heat - work = enthalpy difference
The given reference on misleading use of enthalpy considers, it seems, only incompressible fluids, and thus misses out on compressibility effects. But these are important, in general. In isobaric processes compressibility effects lead to the heat being related to enthalpy, not internal energy difference: some energy is required for the expansion (or contraction) against the constant pressure. In incompressible liquids, however, this effect vanishes, and specifc heat at constant pressure equals specific heat at constant volume.
Hence, it's not that straightforward to split the first law into work and heat contributions--to do this properly, one needs to consider first and second law together. If one does that, by means of the Gibbs equation, the mechanical work in open systems is not the difference in pressure (p) times volume (v), but rather
work = - Integral [ v dp] (+ other terms, incl. dissipation)
Only for incompressible fluids, where v=const. this becomes
work = difference in (p v) (+ other)
So, I am happy to have enthalpy in my equations, alone for convenience: why should I write (u+pv) in all expressions for a steam engine (say), where all heat and work contributions are determined as enthalpy differences?
I might also add the following: I never had problems with enthalpy in teaching thermodynamics and energy conversion. It's just the quantity that appears in the proper form of the thermodynamic laws, if these are written in a convenient way, and hence shows up in expressions of work and heat for a number of standard processes (isobaric, adiabatic,...). One could do without, but that would just made things rather inconvenient--there is a reason that steam tables have a column for enthalpy
Enthalpy also is one of the thermodynamic potentials (when variables are entropy and pressures), so it has additional meaning in other parts of thermodynamics.
Reading the question again, I see the statement about the pv term appearing from work (it's the work per unit mass to push mass acrross a system boundary), hence referring to a process quantity. In general, as the name states, process quantities depend on the process (for instance the above Integral [v dp] depends on the relation v(p) ). But for the work to push the mass over the system boundary, there is no process dependency: it's just what it is (or zero, if no mass is transferred). And pv is a property as well. Maybe I just don't understand why this should be a problem?
In the end, what we need is the proper equations to describe the system at hand, and the proper procedure to solve them for what we need to know for particular systems (heat, work, etc). Whether enthalpy shows up for convenience or not should not alter the equations, nor the result.
I add a short file, where I show how 1st and 2nd law can be split between work and heat by means of the Gibbs equation (adapted from Chapter 9 of my new textbook with Springer--sorry, I could not resist to mention it)
Indeed, enthalpy is one of the thermodynamic potentials, but H is not an independent potential, in my opinion, enthalpy is an math equation, denotes the sum of the some independent potentials.
Internal energy: U= TS+Yx-pV+ΣμN.
Enthalpy: H=U-Yx+pV=TS+ΣμN.
Where Y is the generalized force, dx is the generalized displacement, and H is the sum of the thermodynamic potentials TS and ΣμN.
Y, x, p and V are the state variables, W is “a process quantity”, but pdV is the change of “a state quantity".
I have myself taught Thermodynamics for mor than 40 years. I agree with the answer of prof. Henning Struchtrup.
I agree with prof. Henning Struchtrup.
Enthalpy is used when we analyze the energy conversion (or 1st thermodynamics law) of an open system (hence a control volume) instead of closed system. Some examples of open systems are turbine, heat exchanger, compressor, pump, nozzle, diffuser.
When we deal with a control volume, there is energy carried by the flowing fluid. This energy is called "Pv". It shows energy of a flowing fluid either entering or leaving the control volume. So, a flowing fluid has internal energy (u) + Pv . This combination is defined as enthalpy.
When there is a change phase from liquid to vapor in a system, the energy needed is also called 'enthalpy of vaporization'.
Therefore, we can not stop using enthalpy in energy conversion analysis :)
I also agree with prof. Henning particularly when energy transitions are examined in the scope of climatology
Enthalpy is my favorite thermodynamic entity particularly when I am trying to understand the behavior of a 800 MW combined cycle under variable operating conditions in a real power plant and discuss with the enduser what we have to do to improve the efficiency (first law obviously).
Enthalpy is very useful in the blade cooling area where we have to consider systems that exchange simultaneously heat and work (i.e. in classical thermodynamic usually we have separate effects: adiabatic expander - compressor or heat exchangers where no work is present).
Moreover, the use of the separate terms present in the enthalpy function (internal energy, pressure and volume) may be useful for theoretical speculation but quite complex to be used in the turbomachinery field where also total conditions must be considered.
I agree with Ekadewi Handoyo . Closed system is not useful but in open system ( a control volume) is very useful.
This sounds like shades of "The New Heat Transfer" by Eugene F. Adiutori. I remember years ago he offered to give a copy to anyone who could solve a simple heat transfer problem the "old" way. I did and sent it in, but added, "please don't send me one of your books!" Without enthalpy, how would we know that T=dh/ds)p=const? I couldn't sleep at night if g didn't equal h-Ts. I think I just heard James Clerk Maxwell and Willard Gibbs groan.
The question to address is about determining dissipation (losses) from the enthalpy.
h(Enthalpy )=u(internal to system) + Pv (boundary phenomena, mechanical).
Now dissipation occurs through entropy rise only due to heat transfer or by work done on an adiabatic system or through a mixing process. Which is represent by the first term on left hand side. 'u' and be expressed in terms of entropy rise using laws of thermodynamics.
Even in isenthalpic process if dissipation occurs then un-availability of the energy increases. Although the total energy(enthalpy) is conserved, still the entropy rises.
So it better to use entropy generated and express the enthalpy in terms of entropy.
I am sorry that I was not aware of this question so I later asked the following:
Do we have to consider enthalpy, or not, in an energy conversion analysis?
Before trashing tens of books of my library which make use of enthalpy, including my own:
G.M. Carlomagno, Elementi di Gasdinamica, 1-475, 2009, Liguori, Napoli, Italy,
I would like to know other people comments on the sentence reported in prof. H. Herwig’s paper:
Herwig, H. (2014) The Misleading Use of “Enthalpy” in an Energy Conversion Analysis. Natural Science, 6, 878-885. http://dx.doi.org/10.4236/ns.2014.611085
also present in Research Gate.
This paper declaims at its end:
With all these points in mind the final recommendation is: try to avoid enthalpy in an energy conversion analysis.
Please see the attached file.
I don't agree with prof. Henning Struchtrup
Enthalpy difference dH doesn't represent energy conversion, below is an example.
For Joule-Kelvin effect, throttling of a gas through a porous plug and cause cooling or heating, it is known as a constant enthalpy process, we have
dH=0
Enthalpy difference dH is unable to describe the cooling or heating effects (or the energy conversion).
The problem can be solved by recognizing enthalpy as a proper form of energy. See:
Article Energy-balance of the Joule-Thomson experiment: Enthalpy cha...
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