Exergy

Exergy refers to the portion of the total energy of a system , the work can perform when in the thermodynamic (thermal, mechanical and chemical) equilibrium is brought with its surroundings. Exergy is a potential between at least two states, one of which is usually the environmental state. In contrast to energy, exergy is not a conserved quantity because it is degraded by irreversible processes , i.e. H. it is converted into anergy .

The term exergy goes back to a suggestion by Zoran Rant in the 1950s.

example

Exergy losses occur z. B. in heat transfers . If energy flows into the environment in the form of heat from a poorly insulated hot water pipe , it can no longer be used to do work. However, the principle of conservation of energy applies : the pipe and the environment together have the same amount of energy as before the start of heat transfer.

The second law of thermodynamics ( entropy ) restricts the first law with regard to the possible energy conversions . If, for example, two substances of different temperatures are mixed with one another in a thermally insulated ( adiabatic ) mixing chamber, no losses can be identified in the energy balance equation, the total energy in the system remains the same.

Nevertheless, there are thermodynamic losses, since entropy is generated by this process. Before that, the system that contains the two substances has exergy that a heat engine could use to equalize the temperature between the substances . Afterwards this is no longer possible because of the general validity of the second law, this exergy was destroyed; all that remains is the exergy that the entire system possesses in relation to its surroundings .

Examples of exergetic losses are:

• Heat transport at a finite temperature difference
• friction
• Mixing at constant volume ( closed system ) or at constant pressure (flow processes), with the exception of an ideal gas mixture .
• chemical reactions .

See also : Carnotization to keep exergy losses low.

application

The exergy concept provides a tool with which, on the one hand, the maximum useful work of a system or material flow can be calculated. On the other hand, actual losses can be calculated. It can be of help for engineering problems , especially if the exergy concept is linked to economic variables, cf. using thermo-economic methods. Together with the concept of anergy, the two main principles of thermodynamics can also be summarized as follows:

The 1st law of thermodynamics (energy law) says:

• In a closed system , with reversible and irreversible processes, the sum of exergy and anergy, i.e. the energy, remains constant (energy conservation).

The 2nd law of thermodynamics (entropy theorem) provides several conclusions:

• In a closed system, exergy and anergy remain constant in reversible processes.
• In irreversible processes, exergy is converted into anergy.
• Anergy cannot be converted into exergy.

In the literature one often reads the general connection:

${\displaystyle {\text{Exergie}}+{\text{Anergie}}={\text{Energie}}}$

where anergy denotes the unusable part of the energy.

This relationship apparently leads to a contradiction when processes take place below ambient temperature, e.g. B. in chillers . Below the ambient temperature , the exergy of a system increases with decreasing temperature, since the temperature difference to the environment can be used to operate a heat engine and thus gain useful work; however, the internal energy of the system decreases with decreasing temperature. With a corresponding system pressure, the exergy of a system below the ambient temperature can be greater than its (internal) energy, which would mean negative anergy.

The contradiction is resolved if the direction of energy flow is taken into account: in this case, energy, consisting of the two components exergy and anergy, flows differently from the environment into the system . The exergy is a potential that in the case under consideration has the environment compared to the system and not vice versa. Nevertheless, it is reasonable and customary to assign the exergy component to the system under consideration.

calculation

The exergy  E _{sys of} a system or material flow is made up of

• the internal exergy  E _in
• the chemical exergy  E _chem
• the kinetic exergy  E _kin (corresponds to the kinetic energy )
• the potential exergy  E _pot (corresponds to the potential energy ):

${\displaystyle E_{\mathrm {sys} }=E_{\mathrm {in} }+E_{\mathrm {chem} }+E_{\mathrm {kin} }+E_{\mathrm {pot} }}$

or mass-specific :

${\displaystyle e_{\mathrm {sys} }=e_{\mathrm {in} }+e_{\mathrm {chem} }+e_{\mathrm {kin} }+e_{\mathrm {pot} }}$

Internal exergy of a closed system

The mass-specific internal exergy of a closed system can be determined as follows:

${\displaystyle e_{\mathrm {in} }=(u-u_{0})+p_{0}(v-v_{0})-T_{0}(s-s_{0})}$

With

• mass-specific internal energy  u
• mass-specific volume  v
• mass-specific entropy  s
• Pressure  p
• Temperature  T .

The index  ₀ characterizes the state of the system or material flow at ambient pressure and ambient temperature in thermodynamic equilibrium.

The absolute value results from the multiplication of the specific value with the mass  m of the system:

${\displaystyle E_{\mathrm {in} }=m_{\mathrm {sys} }\cdot e_{\mathrm {in} }}$

Exergy of a material flow

For the exergy of a material flow can be written:

mass-specific value

${\displaystyle e_{\mathrm {str} }=(h-h_{0})-T_{0}(s-s_{0})}$

With

• mass-specific enthalpy  h

absolute value

${\displaystyle {\dot {E}}_{\mathrm {str} }={\dot {m}}\cdot e_{\mathrm {str} }}$

The point above the respective size denotes a current or a time derivative , e.g. B. the mass flow

${\displaystyle {\dot {m}}=\lim _{\Delta t\rightarrow 0}{\frac {\Delta m}{\Delta t}}}$

Exergy of a heat flow

${\displaystyle {\dot {E}}_{q}=\left(1-{\frac {T_{0}}{T_{m}}}\right){\dot {Q}}<{\dot {Q}}\qquad \mathrm {mit} \,T_{0}<T_{m}}$

With

• transferred thermal power ( heat flow ) ${\displaystyle {\dot {Q}}}$
• Mean temperature of heat transfer . ${\displaystyle T_{m}}$

Exergy from work

${\displaystyle E_{w}=W+p_{0}\cdot V}$

With

• job ${\displaystyle W}$
• Volume change work performed by the system on the environment or by the environment on the system. ${\displaystyle p_{0}\cdot V}$

Exergy balance equations

The exergy of a system can change through the exergy destruction in the system and in the open system additionally through exergy flows that are connected with material and energy flows across the system boundary. ${\displaystyle {\dot {E}}_{D}}$ ${\displaystyle \sum _{\mathrm {ein} }{\dot {E}}_{\mathrm {ein} }-\sum _{\mathrm {aus} }{\dot {E}}_{\mathrm {aus} }}$

Therefore the exergy balance equation for a closed system reads:

${\displaystyle {\dot {E}}_{sys}=\sum _{j}{\dot {E}}_{q,j}+{\dot {E}}_{w}-{\dot {E}}_{D}}$

and for an open system:

${\displaystyle {\dot {E}}_{sys}=\sum _{j}{\dot {E}}_{q,j}+{\dot {E}}_{w}+\sum _{\mathrm {ein} }{\dot {E}}_{\mathrm {ein} }-\sum _{\mathrm {aus} }{\dot {E}}_{\mathrm {aus} }-{\dot {E}}_{D}}$

The exergy destruction is caused by irreversibilities during the process. The connection between it and the generation of entropy is

${\displaystyle {\dot {E}}_{D}=T_{0}\cdot {\dot {S}}_{\text{Erzeugung}}}$

Difference between exergy and free enthalpy

Exergy is not to be confused with the free enthalpy  G.

The free enthalpy is only a function of state that describes the state of a substance with a certain composition at a given temperature and pressure. However, it depends not on the parameters of the environment such as ambient temperature, pressure and - wet from.

Exergy, on the other hand, also depends on the ambient temperature, pressure and composition, since it represents mechanical work that can be obtained in a suitable machine if this substance is cooled / warmed / relaxed from a given temperature and pressure down to ambient temperature and pressure / condensed etc. Exergy is a relative quantity between two states and therefore not a function of a single state.

The exergy of a material flow can be understood as the difference between the free enthalpy in a given state and the free enthalpy at ambient temperature and pressure.

Exergy of compressed gases

Exergy flow diagram ( Sankey diagram ) of a compressed air system, example: The exergetic power of the compressed air flow after compression ( p ₂ = 7.3 bar ; T ₂ = 80 ° C = 353 K ; = 0.167 m ³ / s ; = 1.185 kg / m³ ) is 39.3 kW . Downstream of the aftercooler, pressure and temperature drop ( p ₃ = 7.0 bar ; T ₃ = 25 ° C = 298 K ) while the mass flow rate remains the same. Therefore the exergy flow drops to 36.4 kW , which corresponds to an exergy loss in the aftercooler of 4.6% based on the input power. ${\displaystyle {\dot {V}}_{2N}}$ ${\displaystyle \rho _{\text{N}}}$

In compressed air technology and pneumatics - just as in other technical disciplines - there is a need to qualitatively assess system parts and components by z. B. Energy losses and efficiencies are given.

To describe the current state of the compressed air at a certain point in the system, it initially seems plausible to use the thermodynamic values ​​of the internal energy  U (closed system) or the enthalpy  H (open system). Although both quantities correctly map the energy content, no statement can be made about the usability of this energy, since the energy gradient against the environment is not taken into account in both quantities. This can also be seen in the fact that both  U and  H are only functions of temperature; however, the pressure in the current state has no effect.

However, since pressure is relevant as the driving factor for performing mechanical work, especially in pneumatics, it is hardly possible to make any statements about the benefit of the energy content with  U and  H. This is where exergy comes into play, which describes the usable portion of the energy content.

Pressure-dependent exergy calculation

The use of exergy as a balance parameter provides a remedy. To calculate the exergy content in state  a from measurable quantities, the following three values ​​are required:

• The absolute pressure  p _a
• the temperature  T _a
• the associated volume flow . ${\displaystyle {\dot {V}}_{\text{a}}}$

With this information (index  0 denotes the environmental condition) the exergetic power of the compressed air flow is calculated as:

${\displaystyle {\dot {E}}_{\text{a}}={\dot {V}}_{\text{a}}\cdot \rho _{\text{a}}\cdot c_{\text{p}}\cdot (T_{\text{a}}-T_{\text{0}})+{\dot {V}}_{\text{a}}\cdot \rho _{\text{a}}\cdot T_{\text{0}}\cdot \left(R_{\text{s}}\cdot \ln {\frac {p_{\text{a}}}{p_{\text{0}}}}-c_{\text{p}}\cdot \ln {\frac {T_{\text{a}}}{T_{\text{0}}}}\right)}$

With

• the mass flow ${\displaystyle {\dot {V}}_{\text{a}}\cdot \rho _{\text{a}}={\dot {m}}_{\text{a}}={\dot {V}}_{\text{N}}\cdot \rho _{\text{N}}}$
  • the density ${\displaystyle \rho _{\text{a}}}$
  • the standard volume flow ${\displaystyle {\dot {V}}_{\text{N}}}$
  • the standard density ${\displaystyle \rho _{\text{N}}}$
• the specific heat capacity  c _p
• the specific gas constant  R _s .

The exergy analysis enables all important events in the chain of effects to be recorded:

• occurring pressure changes
• Temperature changes
• Changes in the mass flow (e.g. due to leakage ).

A state can be a specific point in the chain of effects, e.g. B. the final state of the compressed air after compression . The comparison of two states allows the exergy loss between two states to be calculated. If you put this in relation to the output energy as a percentage, you get the percentage exergy loss at each station in the chain of transformation.

A graphical representation of the losses can be done, for example, in the form of a Sankey diagram (figure on the right). The exergetic analysis of energy flows offers a clear and comprehensible method to evaluate compressed air systems qualitatively, to show losses and to create a comparison basis for the evaluation of systems and system parts.

Example: Calculating the exergy in a bicycle tire

Exergy of compressed air: inflating a bicycle tire

A bicycle tire should be inflated with a hand pump according to the sketch opposite, starting from an external pressure of 1 bar to 4 bar. The minimum work required for this must be determined. This minimum work corresponds to the exergy contained in the tire after inflation. In a reversible process, the losses are zero and the work to be done is minimal. This means that isothermal compression must be assumed, i.e. a process that is theoretically frictionless and takes an infinitely long time in order to avoid heating. To calculate the pumping work, a piston volume of three times the hose volume is assumed, which can perform the compression process with a single stroke. Then the entire mass of the gas in ambient conditions is already contained in the tire plus pump system at the beginning . With the piston stroke, the total volume is now compressed to the volume of the tire. An isothermal process can take place without friction and with (infinitely) much time. H. without losses. ${\displaystyle m_{2}}$

Calculation of the volume or the mass

Tube volume:

${\displaystyle V_{\mathrm {S} }=\pi ^{2}\cdot D_{\mathrm {S} }\cdot {\frac {d_{\mathrm {S} }^{2}}{4}}=4{,}205\,\mathrm {l} }$

Output volume:

${\displaystyle V_{\mathrm {A} }=V_{\mathrm {S} }+3\cdot V_{\mathrm {S} }}$

Hose content 1 in the initial state:

${\displaystyle m_{1}={\frac {p_{\mathrm {a} }\cdot V_{\mathrm {S} }}{R_{\mathrm {s} }\cdot T_{\mathrm {a} }}}=0{,}005\,\mathrm {kg} \quad {\text{mit}}\quad R_{\mathrm {s} }=287\mathrm {\frac {J}{kg\cdot K}} }$

Hose content 2 in the final state:

${\displaystyle m_{2}={\frac {p_{2}\cdot V_{\mathrm {S} }}{R_{\mathrm {s} }\cdot T_{\mathrm {a} }}}=0{,}020\,\mathrm {kg} }$

Calculation of work

Volume change work along the isotherm:

${\displaystyle W_{1,2,\mathrm {V} }=m_{2}\cdot R_{\mathrm {s} }\cdot T_{\mathrm {0} }\cdot \ln {\left({\frac {p_{\mathrm {a} }}{p_{\mathrm {0} }}}\right)}=2{,}332\,\mathrm {kJ} }$

Shift work through the atmosphere:

${\displaystyle W_{\mathrm {VA} }=3\cdot V_{\mathrm {S} }\cdot p_{\mathrm {0} }=1{,}262\,\mathrm {kJ} }$

Compressor work by the pump:

${\displaystyle W_{\mathrm {H} }=W_{1,2,\mathrm {V} }-W_{\mathrm {VA} }=1{,}07\,\mathrm {kJ} }$

Calculating the exergy E _x using the equation for the closed system leads to the same result:

${\displaystyle {\begin{aligned}E_{\mathrm {x} }&=(U-U_{\mathrm {0} })+p_{\mathrm {0} }\cdot (V-V_{\mathrm {0} })-T_{\mathrm {0} }\cdot (S-S_{\mathrm {0} })\quad {\text{mit}}\quad T_{\mathrm {0} }=T_{\mathrm {a} }=T_{\mathrm {1} }=T_{\mathrm {2} }{\text{;}}\qquad V_{\mathrm {0} }={\frac {m_{2}\cdot R_{\mathrm {s} }\cdot T_{\mathrm {0} }}{p_{\mathrm {0} }}}\\E_{\mathrm {x} }&=m_{2}\cdot \left[c_{\mathrm {p} }\cdot (T_{2}-T_{\mathrm {0} })+T_{\mathrm {0} }\cdot \left(R_{\mathrm {s} }\cdot \ln {\left({\frac {p_{\mathrm {2} }}{p_{\mathrm {0} }}}\right)}-c_{\mathrm {p} }\cdot \ln {\left({\frac {T_{2}}{T_{\mathrm {0} }}}\right)}\right)\right]-p_{\mathrm {0} }\cdot (V_{\mathrm {0} }-V_{\mathrm {S} })\\E_{\mathrm {x} }&=m_{2}\cdot \left[T_{\mathrm {0} }\cdot R_{\mathrm {s} }\cdot \ln \left({\frac {p_{\mathrm {2} }}{p_{\mathrm {0} }}}\right)\right]-p_{\mathrm {0} }\cdot V_{\mathrm {P} }=1{,}07\,\mathrm {kJ} \end{aligned}}}$

The real work to be done is because of the finite time to compress, the air heats up and consequently a higher back pressure has to be overcome, and due to friction losses in the valve and on the piston, especially due to the dead space in the pump; it can be twice as much.

See also

• Calculation of the maximum work
• Carnot method

literature

• Hans Dieter Baehr, Stephan Kabelac: Thermodynamics. Basics and technical applications. 13th, revised and expanded edition. Springer, Berlin et al. 2006, ISBN 3-540-32513-1 ( Springer textbook ).
• Jochen Fricke, Walter L. Borst: Energy. A textbook on the basics of physics . Oldenbourg Verlag, Munich / Vienna 1981, chap. 2: exergy and energy .
• Adrian Bejan, George Tsatsaronis, Michael Moran: Thermal Design and Optimization. Wiley, New York NY et al. 1996, ISBN 0-471-58467-3 .
• Zoran Rant: Exergy, a new word for technical work ability. In: Research in the field of engineering. 22, 1956, ZDB -ID 212959-0 , pp. 36-37.
• Jan Szargut: Exergy Method. Technical and Ecological Applications. WIT Press, Southampton et al. 2005, ISBN 1-85312-753-1 ( Developments in heat transfer 18).

Web links

• Exergy calculator of the Exergoecology portal
• Brief introduction to the subject of exergy

Individual evidence

1. ↑ Presented at the VDI heat conference in Lindau, 1953, quoted from Fran Bošnjaković, Karl-Friedrich Knoche: Technical Thermodynamics Part . 8th edition. Steinkopff Verlag, Darmstadt 1998, ISBN 978-3-642-63818-3 . ; Another source Bošnjaković refers to research in the field of engineering. 22, (1956), p. 36.
2. ↑ For reversible mixing see section 7.6 “Entropy of ideal gas mixtures” in Bošnjaković / Knoche: Technical Thermodynamics Part 1. 8th edition. Steinkopff-Verlag, Darmstadt 1998, ISBN 978-3-642-63818-3 .
3. ↑ Christoph Kail: Analysis of power plant processes with gas turbines under energetic, exergetic and economic aspects. In: Dissertation at the Technical University of Munich. Chair for Thermal Power Plants with Cogeneration Plants, March 1998, accessed on June 25, 2018 .  P. 11f.