Thermo-mechanical exergy of a substance below environmental pressure

In this work, we demonstrate why classical thermo-mechanical exergy functions are incorrect below the environmental reference pressure, since they result in too-small or even negative values. We present new derivations for thermo-mechanical exergy for fluids at any temperature or pressure above or below the environmental reference temperature and pressure. We then present new pressure-enthalpy-exergy diagrams of real chemicals to visually demonstrate the results


INTRODUCTION
Exergy is a thermodynamic metric of matter or energy which is useful for describing its energetic value in terms of both quantity (according to the 1st law of thermodynamics) and quality (according to the 2nd law of thermodynamics).Although several definitions of exergy have been proposed, the definition of Moran et al. (p372) [1] is the most general (emphasis ours):

Definition 1. (Definition of Exergy)
Exergy is the maximum theoretical work obtainable from an overall system consisting of a system and the environment as the system comes into equilibrium with the environment (passes to the dead state).
Since there are many kinds of exergy, it is quite useful to break down exergy into various kinds and classifications.Many taxonomies for this have been proposed, but for this work we use the taxonomy of Deng et al. [2], illustrated in simple form in Figure 1.
This paper only considers the thermo-mechanical exergy of fluids.Thermo-mechanical exergy is the exergy associated with pressure, temperature, and phase of a substance1 .It does not include chemical exergy (associated with chemical bonds and phase), exergy of height (potential) or motion (kinetic), electrostatic / electrodynamic exergy, or atomic/nuclear exergy.Some works decompose thermo-mechanical exergy into a temperature contribution and a pressure contribution, however, this is done for convenience; it has no fundamental basis.In addition, such a decomposition is not strictly correct in general because it neglects phase change and mixedphases (e.g.vapor-liquid), but the sentiment is easy to follow.Instead, thermo-mechanical exergy (  ) is better generalized in terms of how a material's enthalpy and entropy differ from its enthalpy and entropy at the environmental state.The environmental reference state 0 is defined as ( 0 ,  0 ), which is the environmental reference temperature and pressure.Many classical texts have derived the equation for thermo-mechanical exergy ( tm ) of a substance as [3]: where ℎ is the specific enthalpy (we use kJ/kg in this work),  is the specific entropy (kJ/kg-K), and  is temperature (K).Subscript 0 indicates the temperature, enthalpy, or entropy of the substance at the environmental reference state, and subscript 1 indicates the enthalpy or entropy of the substance in its actual state.In this work, we will show that Eq. ( 1) is invalid below the environmental pressure.For this paper, we use environmental reference conditions ( 0 ,  0 ) of (298.15K, 1.01325 bar).
Eq. 1 is used widely in the literature for thermomechanical exergy in industrial, chemical, and mechanical systems analysis.It is often referred to as the specific flow exergy because it is the most appropriate form for flowing systems and arises naturally from general flow balances.However, McGovern noted that what has been labeled flow exergy in the literature is perhaps better described as "the sum of the exergy…of the substance…and the exergy transfer corresponding to a work interaction at a boundary" (emphasis ours) [4].Thus it is not strictly limited to flowing systems.
However, as we will show, if  1 <  0 , Eq. 1 will result in too-small values.In fact, it will result in a negative value for exergy when (ℎ 1 − ℎ 0 ) <  0 ( 1 −  0 ).Very little has been published on sub-environmental pressures for Eq. 1 in the open literature.The few works that address it sometimes incorrectly conclude that any negative value resulting from Eq. 1 implies that   can sometimes be negative.
In this paper, we demonstrate that (1) Eq. 1 is incorrect when  1 <  0 ; (2) we use an optimization approach to rederive   for cases when  1 <  0 and show that this new equation is always positive; (3) we illustrate these findings visually with real chemicals by introducing a new pressure-enthalpyexergy diagram.The resulting general form of the equation is very useful for practicing engineers.

Fundamental Principles of Exergy
We can rewrite Definition 1 as an optimization problem, as follows: Here,  ∈ ℙ is a process within the set of all possible processes ℙ that bring the substance into equilibrium with its environment while producing work  (here expressed in specific form, or kJ per kg of substance).As currently defined, there is no restriction on the number of  that are solutions to Eq. 2, and in fact there may be an infinite number of solutions as will be discussed later.

Theorem 1. Reversibility at Solution
The  that produces the maximum  must be a reversible process.This is a well-known thermodynamic result shown in other sources [1] and taken as true in this work.A reversible process has by its definition no thermodynamic losses (irreversibilities).This obviously means maximum work production or minimum work consumption for processes involving mechanical energy (work).In fact, other definitions of exergy go so far as to incorporate this result into the definition itself, such as that of Szargut [5] (emphasis ours):

Definition 2. (Alternative Definition of Exergy)
Exergy is the amount of work obtainable when some matter is brought to a state of thermodynamic equilibrium with the common components of its surrounding nature by means of reversible processes, involving interaction only with the above mentioned components of nature.
However, it should be noted that Definition 1 is more general because it does not assume reversibility of , and so we use Definition 1 in this work.Definition 2 is still useful, but has lead to some misinterpretations and misapplications which will be noted later.Both definitions rely on the assumption that the "environment" or "surrounding nature" is an infinite reservoir of mass and energy at ( 0 ,  0 ).
Although inelegant, it is technically possible to solve the optimization problem in Eq. 2 through guess-and-check on : propose a process , 2 Except for the trivial case where the substance already is at environmental conditions, in which case its thermo-mechanical exergy is 0 and it is reversible because there is no process to reverse.
compute the work that results, and keep guessing new  until it has somehow been determined that no  that produces a larger  could exist.With this approach, each valid  results in a valid lower bound on  tm .

Exergy can Never be Negative
It is clear then that the thermo-mechanical exergy of a substance can never be negative, as noted by Moran et al. (p375) [1].This is because it is always possible to propose a trivial process  that does zero work but brings the substance into equilibrium with its environment by simply exposing the substance (at any temperature and pressure) to the environment and leaving it for infinite time.Because the environment is infinite, the substance will necessarily equilibrate to ( 0 ,  0 ) and the environmental temperature and pressure remain unchanged.Obviously this process does zero work and is not reversible 2 .However, because of Eq. 2, zero is therefore a lower bound on  tm .

Reversibility Implications
Because  is reversible at the solution to Eq. 2., the exergy is not only the maximum amount of work that can be obtained while bringing the substance into equilibrium with its environment, it is also the minimum theoretical work required to take the substance from its environment to the state of interest.For thermo-mechanical exergy, this means we are concerned only with taking the substance at state 0 (ℎ 0 ,  0 ) to new conditions of interest at state 1 (ℎ 1 ,  1 )3 .Eq. 2 can be rewritten equivalently: where  ∈ ℚ is a process within the set of all possible processes ℚ that bring the substance from state 0 to state 1 while consuming work .Since all   that are solutions of Eq. 2 are reversible, we can easily state:

Corollary 1
For each process   that solves Eq. 2, let the reverse process be called   .Then all   are solutions to Eq. 3, and all solutions to Eq. 2 and all solutions to Eq. 3 result in the same  tm .This is simple but important.It means that  tm can never be zero except at state 0, it must always be positive everywhere away from state 0. If this were not true, it means it would be theoretically possible to spontaneously bring the substance from state 0 to state 1 without the consumption of work.This, however, cannot be true because then we would be able to construct a perpetual motion machine, violating the 2 nd Law of Thermodynamics, as will be discussed later.

Derivation of Eq. 1
The derivation of Eq. 1 is well-known and summarized here briefly.Consider Figure 2A, which shows a generic hypothetical reversible engine  that can produce work  while bringing a substance from state 1 to state 0. Figure 2A is valid for situations in which the substance at state 1 is hot enough such that there is sufficient heat above  0 to produce work and reject heat to the environment at  0 .
The steady-state energy balance for Figure 2A is shown below, written in the form Accumulation equals Input minus Output plus Generation: where  0 is the heat transferred to the environment (in terms of kJ per kg of the substance of interest).By noting that the entropy of heat  0 at temperature  0 is  0  0 � and that the entropy of work is zero, the corresponding steady-state entropy balance in the form Accumulation equals Input minus Output plus Generation using the model in Figure 2A is: where   is the entropy generated by the process.When the process is reversible,   = 0, since reversible processes do not generate entropy by definition.Therefore, one can solve Eq. 5 for  0 and substitute that into Eq.4, and arrive at: However, because we are only considering a reversible process, by Theorem 1,  tm =  and therefore we arrive at Eq. 1.
The same procedure can be performed for "cold" exergy, considering Figure 2B.Other examples include cases in which  1 might be somewhat above  0 , but  1 is well above  0 .The energy and entropy balances for Figure 2B become: Repeating the procedure by solving for  0 in Eq. 8 and substituting into Eq.7 again results in Eq. 1 for reversible processes where   = 0. Note that alternatively, one can simply use only Figure 2A and use a sign convention on  0 instead of using arrows.All approaches result in Eq. 1 and are valid for the model shown in

Non-uniqueness of 𝒫𝒫
The derivation for Eq. 1 does not require actually knowing what kind of reversible engine that is used.However, many have preferred to think about breaking the process  down into two subprocesses in series,  1 * followed by  * 0 , where  1 * is a process where the substance moves from state 1 to some intermediate state *, and  * 0 is a process where the substance moves from state * to state 0. In addition, it is obvious and commonly understood that if  is reversible then the subprocesses are also reversible, and vice versa:

Corollary 2
Let   be a process going from state A to state B such that   can be broken down into two processes in series   * →  *  through some intermediate state *.  is reversible ⟺   * and  *  are both reversible.
Corollary 2 can be applied recursively to further break down the subprocesses.
This corollary is commonly used in the literature to help in the computation of  tm using Eq 2. For example, one common approach is to propose a  consisting of two steps, a thermal step  1 * followed by a pressure change step  * 0 .Very commonly, the  1 * chosen is a Reversible Heat Engine (RHE), also called a Carnot Engine, which moves the substance from state 1 ( 1 ,  1 ) to state * at ( 0 ,  1 ) while producing work.Then  * 0 moves the substance from state * to state 0 ( 0 ,  0 ) through isothermal expansion.This has led to a commonly used conceptual framework where  tm can be fundamentally decomposed into thermal and mechan- ical (pressure) based components [6][7][8][9][10].However, Marmolejo-Correa [11,12] demonstrated 4 that this is misleading, because one can simply choose * to be ( 1 ,  0 ) instead, giving different processes for  1 * and  * 0 , each producing a different value of work for each subprocess, but yet still arriving at the same total  tm .In fact, 4 See p34 in reference [12] Marmolejo-Correa showed that for any * that does not cause subprocesses to cross  0 , and even for  1 <  0 : 1 tm = −∆ 1 * tm −∆ * 0 tm (11) such that ∆ 1 * tm and ∆ * 0 tm are the change in thermomechanical exergy of the substance during  1 * and  * 0 respectively.By substituting Eq. 9 and Eq. 10 into Eq.11, the * terms cancel, simply resulting in Eq. 1 for any reversible .Therefore, the thermal component and the pressure component are ambiguously defined.This shows that thermo-mechanical exergy can certainly be split into thermal and mechanical components as a conceptual or computational aid, but this decomposition is not fundamental.

Remark 1
Marmolejo-Correa's result [11,12] is only valid if Eq. 1 is valid since Eq. 9 and Eq. 10 assume that Eq. 1 is a valid way of computing ∆ tm between any two states.We will challenge this later.

Alternative Thermo-Mechanical Exergy Equation
An alternative expression for thermo-mechanical exergy has been derived for closed systems (see for example Moran et al. [1] p372, while neglecting the kinetic and potential energy terms not related to thermo-mechanical exergy, or McGovern [4]): where  is specific internal energy (in kJ/kg) and  is the specific volume (in daL/kg, noting that 1 daLbar = 1 kJ).Since ℎ =  +  by definition, this can be rearranged: Thus, Eq. 13 is equivalent to Eq. 1 if and only if  1 =  0 .When  1 =  0 , thermo-mechanical exergy is restricted to thermal exergy only, and can be computed using an RHE model that evaluates to the same value as computed by either Eq. 1 or Eq. 13 along  0 .This is a well known result.Therefore we take the following as true: Theorem 2.   at   Eq. 1 accurately computes  tm of a substance at any ℎ 1 if  1 =  0 .
At other pressures, Eq. 1 and Eq. 13 produce very different quantities.Their difference (the right hand side of Eq. 1 minus the right hand side of Eq. 13) is  1 ( 1 −  0 ).This term is often called either the "flow work" when used in the context of flows [1] or "work interaction at a boundary" [4] more generally or in other contexts, such as boundary expansion or contraction (even for non-flowing systems).
The question considered in this paper therefore is which form (Eq. 1 or 13) should be chosen to describe the specific thermo-mechanical exergy of a substance itself.Both forms compute  tm as a function of independent intensive state variables (ℎ 1 ,  1 ,  1 ,  1 ) and the reference state (ℎ 0 ,  0 ,  0 ,  0 ,  0 ) which is constant.Either could be applied to a substance in a general sense as a function of its state variables without further context as to how that substance exists in a part of some larger system.
The choice between the two depends on the definition of thermo-mechanical exergy itself.For example McGovern uses the following definition (emphasis ours) [4]:

Definition 3 (Alternate definition of thermo-mechanical exergy)
The thermomechanical exergy of any system is defined as the maximum useful work (assuming that any exergy associated with potential energy or kinetic energy remains unchanged) that could be produced by any interaction of the system with the specified reference environment under the constraint that the system remains closed during any such interaction.
The constraint in this definition is a key element: forcing the system to remain closed to the environment results in Eq. 13, and not having the constraint results in Eq. 1 as it is traditionally used.
As noted in Moran p.387, the Eq. 1 form is "most useful for engineering analyses" [1], and indeed our observation is that the large majority of the systems exergy analysis literature uses the Eq. 1 form.This is because most real systems of interest are not closed, and so the constraint in Definition 3 is not suitable.
That begs the question: which definition is the right one to use for computing the thermo-mechanical exergy of substances as a function of state, independent of application?We argue that Definition 3 is the wrong form because constraining a substance to closed systems is too limiting and impractical.Definition 1 is the preferred, more general form.According to Definition 1, exergy is the maximum amount of work a system could produce, which is stated as an optimization problem in Eq. 2. When seen in this context, the "system" is the substance plus any hypothetical process  that brings the substance into equilibrium and is an optimal solution to Eq. 2, whether flowing or not.The  tm computed is, for all intents and purposes, the specific thermo-mechanical exergy of the substance itself.
As will be discussed later, for substances at pressures other than  0 , it is trivial to find practical processes  that produce more work from a substance than the value of  tm computed by Eq. 13 if the process were allowed to interact with the environment through boundary work interactions such as compression, expansion, or mass flowing to/from the environment.Compression, expansion, or mass flowing to/from the environment are present in many real industrial applications, and so excluding them by using Eq. 13 does not seem useful.However, Eq. 1 is still incorrect below the environmental reference pressure, as discussed next.

Eq. 1 Fails Below the Reference Pressure
Figure 3 shows a pressure-enthalpy-exergy diagram (PHE) for 1-butene using Eq. 1.The figure shows the phase envelope and isometric lines for temperature, entropy, and thermo-mechanical exergy, with exergy computed using Eq. 1. State functions relating ℎ, , , and  were computed using the CoolProp package in Python [13] and the equation of state for 1-butene given by Lemmon and Ihmels [14].
In Figure 3, point 0 is the reference state (ℎ 0 = 439.7 kJ/kg,  0 = 1.64 kJ/kg-K).Consider point A. Marmolejo-Correa's work [12] showed that one can compute the  tm at A by first computing the exergy change at constant pressure by moving from A to B reversibly, and then compute the exergy change at constant temperature by moving to point 0 reversibly (process  0 =   →  0 ).Marmolejo-Correa showed that this is equivalent to  0 =   →  0 by doing the reversible isothermal pressure change first, followed by the reversible isobaric temperature change.Similarly, she showed that this applies to point D which has a lower temperature than  0 .Again, Marmolejo-Correa showed that one could choose two paths, DB0 or DE0 to get the same exergy value.However, her work did not consider pressures below the reference pressure.
Figure 3 is based on Eq. 1, which clearly fails for  <  0 .First, there is a large region where  tm is zero or negative, which is impossible as noted in Section 2.2.Everything on or below the  tm = 0 line, which occurs whenever (ℎ 1 − ℎ 0 ) ≤  0 ( 1 −  0 ), violates this principle and must be wrong.
Second, even the positive values below  0 are too small.Consider point G (  = 2.2 kJ/kg-K, ℎ  = 627.3kJ/kg), which on this plot has  tm = 20.0 kJ/kg.It is quite easy to show that this value is not the true solution to Eq. 2 because one can propose a process  0 =   →  0 that generates more than 20 kJ/kg of work.This proposed process consists of isentropic compression of 1-butene from point G to F (  =   ,   =  0 , and ℎ  = 638.6 kJ/kg), using the environment as the source of work, followed by isobaric cooling from F to 0. One simple device to accomplish this might be to place the substance at state G inside of a movable pistoncylinder apparatus with a turbine attached such that it can produce shaft work.The atmospheric pressure  0 is higher than the pressure   , so the environment will push the piston down, compressing the substance isentropically (adiabatically and reversibly) until it reaches  0 , producing work  as it does.The work is easily computed by the enthalpy difference between G and F ( = ℎ  − ℎ  ) which is 11.3 kJ/kg.Now at point F, the substance in the pistoncylinder is at  0 but at a higher temperature   .More work could be produced using the substance as a heat source to power an RHE in process  0 .The reversible work for that step would be computed by Theorem 2, which is (ℎ  − ℎ 0 ) −  0 (  −  0 ) = 31.3kJ/kg.Therefore, the proposed machine should be able to produce 11.3 + 31.3 = 42.6 kJ/kg of work total.Even an irreversible, imperfect machine should be able to produce more than 20 kJ/kg.So clearly, the exergy numbers calculated by Eq. 1 are incorrect below the reference pressure by using arguments from Eq. 2.
Kotas [3] considered a similar situation, but for a point in the low-pressure region where Eq. 1 is negative.There, Kotas proposed a very similar machine to our previous example (the GF0 process), with isentropic compression of a substance at low pressure followed by an RHE.However, he erred by determining that the compression step would consume external work, offset by work produced in the RHE step, resulting in a negative exergy value since the isentropic compression step had a higher  than the  of the RHE in his example.This approach is essentially to apply Eqs. 9 and 10 naively, resulting in the direct application of Eq. 1.The text incorrectly concluded that negative exergies therefore can exist.It is easy to see why though; this seems at first to be in line with Definition 2 of exergy, since after all, both proposed steps in his machine are in fact reversible.Though technically correct, the particular language of Definition 2 is likely the source of confusion because it defines exergy as the "work obtainable" and thus only implies that it is the maximum work attainable, rather than using a stronger statement like "maximum theoretical work" in Definition 1.In the end, the Kotas example clearly does not solve the more basic Definition 1 as expressed explicitly in Eq. 2.
If Eq. 1 is derived from the abstract fundamental energy and entropy balances shown in Figure 2, while applying principles of reversibility, why is it wrong below the reference pressure?The explanation is that the abstract balance shown in Figure 2 does not allow energy in the form of pressure to move from the environment into the reversible engine box.The model in Figure 2B allows heat to move from the environment into the system (e.g., the substance acts as a heat sink) producing work, and this is an important point that made it possible to compute "cold exergy".In the same way, we need a new model that is generic enough to allow us to compute "vacuum exergy".This is discussed in Section 3.

Eq. 13 Underestimates Exergy in Practice
Figure 4 shows the PHE plot for 1-butene using Eq. 13 instead.Compared to Eq. 1 (Figure 3), exergy values obtained by Eq. 13 are generally lower than values from Eq. 1 above  0 and generally higher than values from Eq. 1 below  0 , because the work interaction at the boundary term is positive below  0 and negative above it.The net effect is that  tm computed by Eq. 13 is always positive except at the reference state 0.
However, even though it is never negative,  tm is greatly underestimated with this equation compared to the work that could be produced with various theoretical and even real processes.Consider point A in Figure 4 (  = 2.0 kJ/kg-K, ℎ  = 587.6 kJ/kg), which is at the same location as point A in Figure 3. On Figure 4, the  tm is 18.2 kJ/kg, but it is easy to propose a process through path AH0 that could produce much more work.The isentropic expansion from A to H alone (  =   , ℎ  = 560.5 kJ/kg) should produce 27.2 kJ/kg of work (based on the enthalpy difference ℎ  − ℎ  ) 5 .Then, from H to 0 we should be able to employ an RHE for another (ℎ  − ℎ 0 ) −  0 (  −  0 ) = 12.8 kJ/kg (which we know to be correct by Theorem 2), bringing the total work to 40.0 kJ/kg for this machine, far greater than the about 17 kJ/kg predicted by Eq. 13.In fact, 40.0 kJ/kg is the same exergy predicted by Eq. 1, implying that Eq. 1 is likely correct for point A given Theorem 2.
Although not shown for brevity, a similar check of the low-pressure region shows Eqs. 1 and 13 to be under-estimates as well.This means that both Eq. 1 and Eq. 13 are incorrect below  0 .
Therefore, the closed form of thermo-mechanical exergy (Eq.13) is not an appropriate choice for computing the thermo-mechanical exergy of substances at any pressure except at the reference pressure.Similarly, the specific-flow exergy form (Eq. 1) is also not appropriate for pressures below the reference pressure.This is addressed next.

Derivation
We now derive a new equation for  tm based on Definition 1 expressed in the optimization form of Eq. 3. We could use Eq. 2 equivalently, but our experience has shown that the Eq. 3 approach is easier to follow.To find  tm of a substance at point 1 ( 1 tm ) at any temperature, pressure, or phase, we propose the following process: where  01 is broken into two subprocesses through an intermediate point *.Point * is selected such that it is at state ( 0 ,  1 ).
The subprocess  0 * that requires the minimum work is very well known: an RHE run in reverse (i.e. a reversible heat pump, or RHP) that performs isobaric enthalpy change (including both temperature and phase change if necessary).The subprocess  * 1 that requires the minimum work is also very well known: an isentropic compression performed in a piston-cylinder arrangement by pushing the piston down or an isentropic expansion for lowering the pressure below  0 , by pulling the piston up.For example, to arrive at point A in either Figure 3 or 4, one could follow the path 0HA, where H is the intermediate point *.
If we select those two reversible pathways ( 0 * and  * 1 ), then by Corollary 2 we know that  01 is reversible since both of the subprocesses are reversible.Furthermore, because those two subprocesses require the minimum work and  01 is reversible, process  01 also requires the minimum work to get from point 0 to point 1.Then we know by Corollary 1 that the total work required for process  01 is a solution to Eq. 3 and therefore equal to  1 tm .To be sure, there are many other pathways and approaches that could be proposed instead.However, by Corollary 1, a single pathway is sufficient.
For pathway 0* (for example, paths 0H or 0E in Figures 3 and 4), the work required using subprocess  0 * is equal to its exergy change.Since this is an RHP, by Theorem 2, this is simply: Since we are choosing * to be at the state ( 0 ,  1 ), we can simplify Eq. 15: For pathway *1 (for example, paths HA or FG in Figures 3 and 4), the work required using subprocess  * 1 is equal to its exergy change.Since this is isentropic expansion or compression, this is simply: The absolute value is important to make it general to both expansion and compression away from  0 .For example, for the reversible pathway HA, the enthalpy difference ℎ A − ℎ H (or Noting that ∆ 0 * tm = −∆ * 0 tm and ∆ * 1 tm = −∆ 1 * tm , we can then combine Eqs.11, 16 and 17 for process  01 and compute the total thermo-mechanical exergy at state 1: where ℎ * = ℎ( 0 ,  1 ).

Discussion
Note that when  1 ≥  0 , |ℎ 1 − ℎ * | = ℎ 1 − ℎ * since ℎ 1 ≥ ℎ * for all states 1 at or above the reference pressure.Thus Eq. 18 reduces to Eq. 1 for all  1 ≥  0 .Eq. 18 was not derived from the general energy and entropy balances of Figure 2. It only relies on Theorem 2, which is the computation of exergy for a reversible heat engine at the reference pressure agreed upon by both Eq. 1 and Eq. 13.This is important because it means that Eq. 1 is correct for all  1 >  0 .Eq. 13 underestimates exergy in these cases because we can always propose a process  10 that is the reverse of process  01 of Eq. 14, which would produce at least as much work as Eq. 1 and is always higher than the value computed by Eq. 13.
Below atmospheric pressure, |ℎ 1 − ℎ * | = ℎ * − ℎ 1 , such as path FG in Figures 3 and 4.This is the work required to perform process  FG , which is to expand the fluid isentropically against the surrounding environment.For example, a substance could be in a piston-cylinder device at state F and then work would be required to pull the piston up to state G, such that the fluid is now at a lower pressure and temperature because of its expansion.This means Eq. 18 could be rewritten: By recognizing that process  FG requires the consumption of work to reduce the pressure of a substance below  0 , it is easy to see that its reverse,   , should produce work by the same amount.The mistake of Kotas [3] noted in Section 2.7 was, for all intents and purposes, equivalent to computing ∆ * 1 tm = ℎ 1 − ℎ * instead of ∆ * 1 tm = |ℎ 1 − ℎ * |.The erroneous result is Eq. 1 at low pressures, leading to negative exergies.
In another example, Szargut et al.'s classical book (p25) [15] provides an example using the closed form (Eq. 13) for ideal gas air at reduced pressure ( 1 = 0.3 bar,  1 =  0 = 290 K,  0 = 1 bar), resulting in a positive  tm of 94 kJ/kg.However, this value is too small, since from Eq. 18 (using a real equation of state [16]) the value should be 139 kJ/kg.The difference is not explained by Szargut's assumption of ideal gas equation of state (we determined that using the rigorous equation of state for air resulted in only a 0.01 kJ/kg differential in  tm ).It is ultimately explained by a difference in the boundary work term.
Figure 5 shows the final PHE diagram for 1-butene using Eq.18.Here, exergy is everywhere positive.All points above  0 have  tm equivalent to Eq.

An Example with a Normal Liquid
Figure 7 shows a PHE diagram of R113 (1,1,2-trichloro-1,2,2-trifluoroethane, a normal liquid) generated using Eq.18 and the equation of state of Marx et al. [17].For this normal liquid example, the reference state is in the liquid (left) side of the phase envelope, which results in a characteristically different shape from the normal gas example in Figure 5.
In Figure 7, the iso-exergy lines are much more convex in nature, especially in the lower enthalpy, high pressure regions of the diagram.However, the shapes formed by the iso-exergy lines are still not completely convex, as can be seen most easily in the area where iso-exergy lines cross the phase envelope above the reference pressure.The nonsmooth nature of the iso-exergy lines at the reference pressure is also evident in this example as well.Similarly, the nearly vertical iso-exergy lines in the low pressure, low temperature liquid region remain a characteristic, since the isentropic lines are also nearly vertical.

Further Examples
Although not shown for brevity, we also found that the characteristics shown in this work for 1-butene and R113 are the same for 25 other real chemicals, including water, air, and many normal liquids and normal gases.We did not find a single counter example for any real chemical.Pressure-enthalpy-exergy diagrams for these can be found in Deng et al. [2].

Other choices of 𝑝𝑝 0
Because Eq. 18 is general, it holds for any  0 regardless of its value.Whether the analyst has chosen a  0 that is correct for the analysis at hand is a different matter.For example, if analyzing substances on the surface of Mars, or in case-specific systems where there is a huge low-pressure reservoir that effectively serves as the environmental sink for analysis purposes and not the actual planetary environment on which it sits, the analyst should choose an appropriate  0 lower than 1 atmosphere.The selection of appropriate reference conditions is always an important part of any exergy analysis.

Accuracy of Calculations
Like Eqs. 1 or 13, the accuracy of the exergy value computed in Eq. 18 is limited to the accuracy of the model or method used to compute the underlying enthalpies and entropies of a substance at any given state.For this work, the accuracy of Figures 5 and 7 are therefore limited to the accuracy of the equation of state models developed in references [14] and [17] respectively as implemented by CoolProps [13].For the 1-butene case, Lemmon et al. [14] compared the equation of state to 52 differ- ent experimental data sets measuring densities, heat capacities, vapour pressures, heats of vaporization, and other properties, and found an average absolute deviation between model and experiment of less than 1% in most cases for the applicable pressure ranges used in this work.Span and Wagner [18] compared the equation of state in [17] to 16 different experimental data sets in a variety of categories and found average absolute deviations between model and experiment to be less than 0.6% in all cases, with most being below 0.2%.

CONCLUSIONS
The new equation for thermo-mechanical exergy  tm derived in Eq. 18 should be used for fluids in general, flowing or not, at all temperatures and pressures.It reduces to the classical Eq. 1 when  1 ≥  0 .Since the vast majority of exergy analysis applications occur at or above environmental pressure, this means that most previous work using Eq. 1 is correct.However, any previous calculations using Eq. 1 below  0 are underestimates of  tm and should be recalculated by adding 2(ℎ * − ℎ 1 ) to the previously computed value to get the new exergy value using Eq.18. 2(ℎ * − ℎ 1 ) is simply the difference between Eq. 1 and Eq.18 when  1 <  0 .Note that this is only applicable if the reference conditions are the same.
The results here are significant because they resolve the inconsistencies in formulas and interpretations found in the literature, many of them in widely used textbooks.Our more general equation relies only on the most basic form of the exergy definition and the portions of exergy calculations in which the literature is in complete agreement.This is in contrast to the classical "flow exergy" form which is derived from an incomplete model that allows potential energy (pressure) from the substance to be released to the surroundings (when  >  0 ), but not the reverse direction.In other words, it fails below the reference pressure because it then fails to satisfy Definition 1.
Because Eq. 2 is an optimization problem, the only way Eq.18 cannot be the correct solution at any  is if one could find a process  ∈ ℙ that produces more work to go from state 1 to 0 than the value computed by Eq. 18 (without violating the laws of thermodynamics).Conversely, one could disprove Eq. 18 by finding a process  ∈ ℚ that requires less work to go from state 0 to state 1 than the value computed by Eq. 18.However, as explained in Section 3.1, Eq. 18 should be the global optimal solution to Eq. 2 or Eq. 3 and thus Eq. 18 computes the true thermo-mechanical exergy for a substance.Even if someone could find a process  that produces more work than the value computed by Eq. 18, the conclusion that the "flow exergy" form of thermo-mechanical exergy (Eq. 1) below the reference pressure is incorrect would remain unchanged.
For future work, it would be interesting to confirm Eq. 18 below the reference pressure by using a generalized optimal process synthesis approach in which Eq. 2 is solved using mathematical programming algorithms to provable global optimality (within convergence tolerances).For example, it should be possible to use formal superstructure optimization approaches in which a large number of possible processes (but still a subset of ℙ) is considered that would arrive at the same values as Eq. 18, but not higher.A more difficult but more powerful approach would be to somehow include all ℙ in the decision space (which is infinite) into the mathematical programming formulation.

Figure 1 :
Figure 1: A simplified form of the exergy taxonomy of Deng et al. [2].This paper focuses on thermo-mechanical exergy only.

Figure 2 .
Figure 2. Process flow diagrams for computing Eq 1. (A) when the reversible engine is such that it has waste heat to reject to the environment; (B) when the reversible engine consumes heat from the environment.

Figure 2 .
This derivation in various forms is well known and further explanations can be found in textbooks like Moran et al. (p388) [1] and Deng et al. (p13) [2].

Figure 3 :
Figure 3: Pressure-enthalpy-exergy diagram of 1-butene using Eq. 1. Point 0 is the reference state.A-H are described in the text.It is important to note that in this diagram,  tm is incorrect below the reference pressure.

Figure 4 :
Figure 4: Pressure-enthalpy-exergy diagram of 1-butene using Eq. 13.States A-H are the same as in Figure 3.It is important to note that in this diagram  tm is incorrect almost everywhere.