First of all, what is exergy?

Exergy, or availability, is the maximum amount of work that can be extracted from a system as it comes into equilibrium—temperature, pressure, chemical potential—with its environmental surroundings. Exergy analysis is frequently used in industry to support the efficient use of energy. 

While kinetic energy (flowing water) and potential energy (height of water fall) can be used to generate work, I am ignoring these contributions for this post.

Total exergy = physical exergy + chemical exergy

Physical Exergy

Physical exergy is the productive work generated by an ideal reversible process that brings the temperature and pressure of a system to ambient conditions, e.g., 298K and 1 atm.  For example, one can envision a high pressure, high temperature gas stream that generates maximum work in a sequence of two reversible steps:  isentropic expansion in a turbine to bring pressure down to 1 atm and then constant pressure heat exchange with an infinite number of Carnot heat engines to bring the resulting temperature to 298K. 

As outlined by Kotas (1),

Exergy _physical = b₁  –  b_o  =  (H₁ – T_oS₁)  –  (H_o  –  T_oS_o)                                                     [1]

where the subscript 1 refers to the conditions of the stream and subscript 0 refers to environmental conditions, typically 298K and 1 atm.

For an ideal gas, the above equation becomes

Exergy _physical = c_p (T – T_o)  –  T_o [c_p ln(T/T_o) – R ln (P/P_o)]                                              [2]

Chemical Exergy

Chemical exergy is the negative of the work required to reversibly prepare a specific mixture starting from what is available in the environment, given that such mixtures don’t naturally exist. (Operating this process in reverse would generate the maximum productive work.)  For example, while oxygen is indeed available in the environment, pure oxygen at ambient conditions of temperature (298K) and pressure (1 atm) is not.  So one must use a select membrane to separate oxygen from air and then isothermally compress the oxygen up to 1 atm pressure.  The same goes other molecules present in air such as nitrogen, carbon dioxide, and water vapor.

As for molecules that aren’t present in air, these must be prepared by chemical reactions between those molecules that are present in air.  For example, consider methane.  To prepare methane at ambient temperature and pressure, on must first use selective membranes to separate, isothermally compress, and mix carbon dioxide and water (vapor).  These two molecules are then reversibly reacted at constant temperature and pressure to form methane and oxygen: (CO2 + 2H20) –> (CH4 + 2O2).  Methane product is then available for use while the oxygen is isolated by a selective membrane and isothermally expanded into the environment to the concentration, i.e., partial pressure, that naturally exists.

While the chemical exergy calculations involved can be complicated, tables of data have been prepared to make them less so.  The tables provide exergy values (e^o) of pure materials at standard conditions of 1 atm and 298K.

As outlined by Kotas (1),

Exergy _chemical =  b_total  =  b_{chem, pure}  +  b_mixing = sum [n_ie_i]  +  RT_o sum [n_i ln x_i] [3]

where e is equal to the tabulated standard exergy values and x_i are the mole fractions of the species in the mixture.

Does combustion destroy exergy?

I have read in multiple publications that combustion, being an irreversible, entropy-generating process, destroys exergy. For example, in S. Can Gulen’s Gas Turbines for Electric Power Generation (Cambridge Univ. Press, 2019, p. 151) is written: “There are two loss mechanisms [in a gas turbine], about which nothing can be done, [the first being] combustion irreversibility or exergy destruction.” For another example, in John B. Heywood’s Internal Combustion Engine Fundamentals (McGraw Hill, 1988, p. 190) is written: “This loss in availability [for a specific example he was considering] results from the increase in entropy associated with the irreversibilities of the combustion process.” In neither of this references, nor in any other references I have seen, do the authors explain exactly why, on a molecular basis, exergy is destroyed. This made me question the premise that it was destroyed. And so I sought to test the premise by comparing the exergy of a mixture of methane and oxygen to the exergy of the combustion products, i.e., high temperature/pressure carbon dioxide and water vapor.

My calculations

To determine whether or not exergy is destroyed during an irreversible reaction such as combustion, consider the following reaction, the combustion of methane in air with excess nitrogen.

CH4  +  2 O2  +  7.4 N2  –>  CO2  +  2 H2O  +  7.4  N2  (10.4 mols total)

Assume that this reaction takes place at constant internal energy in a constant volume, isolated vessel, like a bomb calorimeter, and compare the exergy of the feed to the exergy of the products.

Exergy of the Feed = Chemical Exergy (the feed has zero physical exergy since it’s at ambient T,P)

e _CH4^o = 836.5 kJ / mol

e _O2^o =  3.97 kJ / mol

e _N2^o = 0.72 kJ / mol

Per [3]

b_total  =  b_{chem, pure}  +  b_mixing

         =  (1 x 836.5  +  2 x 3.97  +  7.4 x 0.72)  +  RTo [(1)xln (1/10.4) + (2)xln(2/10.4) + (7.4)xln(7.4/10.4)]

         = 849.8  –  (0.0083 x 298) x (-10.8)

         =  858.8  –  26.7

= 832.1

Total exergy of feed       =  832.1 kJ

Exergy of Products = Physical Exergy + Chemical Exergy

Products Physical Exergy

To quantify the physical exergy of the products, one must first quantify the final temperature and pressure of the adiabatic reaction in the constant volume vessel.

The enthalpy change of reaction for methane combustion is −802.32 kJ/mol CH4.  The resulting adiabatic temperature rise in a fixed volume, isolated vessel is calculated from the equation (based on product properties):

C_{v, avg} (T_f  −  T₀)  =  -ΔH  =  802.32 kJ – – – based on 1 mol CH4 and 10.4 total product moles (1 mol CO2, 2 mol H2), 7.42 mol N2)

Assuming ideal gas and given constant volume,

P_f  =  T_f x (P_o/T_o)  =  2789 x (1/298)  =  9.4 atm

Again assuming ideal gas and a value of C_p = C_v  +  R  =  0.32  +  10.4 x .008314  =  0.41 kJ / K

Exergy _physical = c_p (T – T_o)  –  T_o [c_p ln(T/T_o) – R ln (P/P_o)] [2]

= 0.41 (2789 – 298) – 298 [0.41 ln (2789/298) – (10.4 x .008314) ln (9.4/1)]

                       =  1021 – 298 [0.917 – 0.194]

                       =   805 kJ

Products Chemical Exergy

e _CO2^o = 20.14 kJ / mol

e _H2O^o =  11.71 kJ / mol

e _N2^o = 0.72 kJ / mol

n = 10.4 moles

Per [3]

b_{chem, pure}  =  (1 x 20.14) + (2 x 11.71) + (7.4 x 0.72)  = 48.9

b_mixing  =  (0.0083)(298) x [1 x ln(1/10.4)  +  2 x ln(2/10.4)  +  7.4 x ln(7.4/10.4)]

            =  2.47 x (-8.15)

            = -20.1

b_total  =  48.9 – 20.1 = 29

Total exergy products = 805  +  29  =  834  kJ

Compare Total Exergy Feed vs.  Total Exergy Combustion Product

Total Exergy Feed = 832 kJ

Total Exergy Product = 834 kJ

The destruction of exergy is not observed in these calculations, despite the fact that the reaction is irreversible and that the entropy increases from xxx kJ/K (ideal gas at 298K, 1 atm) to xxx kJ/K (ideal gas at 2789K, 9.4 atm).

This begs the question: does the increase in entropy of an irreversible chemical reaction carry the same “exergy loss” meaning that an increase in entropy does in a process that changes only T and P but not chemical potential? What does the increase in entropy of a chemical reaction truly mean? This is something I want to further explore.

References

(1)  Kotas, T. J., The Exergy Method of Thermal Plant Analysis, Exergon Publishing Company, U.K., 2012, Chapter 2.