An overview of exergy calculation
Two important points in exergy calculation
Total exergy
If the system is adiabatic with no heat losses to the surroundings, all energy is contained within the system. The exergy change for such a system is the sum of the free energy change of cold and hot streams. Free energy takes into account the entropy losses. The sum total of two free energy is the available energy for the maximum useful work
Remember work does not generate entropy. Therefore, work transfer by the adiabatic process does not contribute to exergy.
Do your own calculation of a process. Identify the process boundary for calculation. It is easy. You would know where your process stands with respect to energy losses.
Exergy efficiency
Exergy efficiency
Just remember, in thermodynamics, exergy is a measure of the maximum useful work that can be obtained from a system as it reaches equilibrium with its surroundings. It is often defined as the difference between the actual free energy of a system and the free energy it would have if at the same temperature as its surroundings. Therefore, the ratio of the free energy of a system to the free energy of a hot system can be considered indicative of the exergy efficiency.
To calculate the exergy efficiency, you can use the following formula:
Exergy Efficiency = (Free Energy of the System) / (Free Energy of the Hot System)
This ratio represents the portion of the available exergy that is successfully converted into useful work.
What is exergy
The exergy of a system is the maximum useful work possible during a process before the system reaches equilibrium with a heat reservoir, reaching maximum entropy. In real-life processes which are irreversible, there is always some entropy generation and correspondingly, loss of work potential of energy. ?All real processes imply inevitably a loss of exergy as anergy.
Problem:?Exergy change of a heat recovery system and Exergy efficiency
Data given
As the image shows air at 200 kg/sec enters the heat recovery system at T1 = 60 degc and exits at T2=35 degc. Coldwater enters at T3 = 25 degc and leaves at T4 =50 degc. Specific heat of water CW = 4.18 kj/kg-k, the specific heat of air CA= 1.005 kj/kg-k. The heat recovery system is well insulated and the heat exchange process is adiabatic. No work interaction is involved. Neglect the kinetic and potential energy changes. The heat exchange process is an isobaric steady process. Determine [1] the Mass flow rate of water [2] Exergy change and [3] Exergy efficiency.
Approach
How will you approach step by step
The first point is read the problem couple of times and evaluate the given process conditions. ?[1] The process says that the heat recovery system is well insulated and the heat exchange process is adiabatic. This means there is no heat transfer outside the system, dQ=0.
[2] No work interaction is involved. This means there is no work involved in the process
[3] The process is isobaric
[4] Potential and kinetic energy changes are negligible
Solution
Let us focus on the solution now
Find out the mass flow rate of water
Consider the heat recovery system as a control volume.
What is control volume definition [Wikipedia]: At steady state, a control volume can be
thought of as an arbitrary volume in which the mass of the continuum remains constant.
As a continuum moves through the control volume, the mass entering the control volume
is equal to the mass leaving the control volume. At steady-state, and in the absence of
work and heat transfer, the energy within the control volume remains constant.
According to the assumptions,
MW[h4-h3] = MA[h1-h2] MW and MA are mass of water and mass of air.
领英推荐
MW = MA [h1-h2]/[h4-h3] [ h is Enthalpy = Mass x Specific heat]
= 200[1.005] [333-308]/4.18[323-298] = 48.1 kg / s
Exergy analysis
Exergy: In thermodynamics, the exergy of a system is the maximum useful work possible
during a process that brings the system into equilibrium with a heat reservoir, reaching
maximum entropy
Why exergy changes:
When the surroundings are the reservoir, exergy is the potential of a system to cause a
change as it achieves equilibrium with its environment. Exergy is the energy that is
available to be used. After the system and surroundings reach equilibrium, the exergy is
zero.
The specific exergy change dS between two states (e.g., inlet and outlet) in a control
volume can be expressed as [ this is the fundamental equation]
= (hb-ha) – T0 (Sb- Sa), a, and b are two states. T0 is standard temperature. [S is entropy]
Some exergy is lost during the process due to entropy generation.
Therefore, the exergy change = [Exergy change of cold stream + Exergy change of hot
stream]
= MA [ (h1-h2) – T0 (S1-S2)] + MW [(h3-h4) – T0 (S3-S4)]
Explanation
This is Gibbs free energy equation. The free energy after accounting for entropy losses is approximated as exergy. The sum of free energy changes dG for the cold and hot steam is the total exergy change. ?
Since the process is isobaric. ?
= MA [ CA (T1-T2) – T0 CA ln T1/T2] + MW [ CW (T3-T4) - T0 CW ln T3/T4]
[ CA and CW are specific heat of water and air].
= 200 [ 1.005 (333-308) – 298 (1.005) ln (333/308)] + 48.1 [ 4.18 (298-323) – 298 ln(298
/323)]
Exergy change = 150.7 kj/s
Exergy efficiency
Exergy efficiency (also known as the second-law efficiency) computes the effectiveness of
a system relative to its performance in reversible conditions.
= Exergy change in the cold stream / Exergy change in the hot stream. When exergy change in
both cold and hot streams are equal, thermodynamic efficiency = 1. This is only possible if
the system is reversible.
Exergy efficiency
= MW [ CW (T4-T3) - T0 CW ln T4/T3] / MA [ CA (T1-T2) – T0 CA ln T1/T2]
= 48.1 [ 4.18 (323-298) – 298 ln (323 /298)] / 200 [ 1.005 (333-308) – 298 (1.005) ln [333/308] = 0.57
--
10 个月I am not a very theoretical person. But in the real world, understand that 3 tonnes of woodchips with a moisture content of 56% and a LHV (lower heat value) of 7.44GJ/tonne has less ability to do useful work/heating than 1 tonne of torrefied wood briquettes with a moisture content of 2% and a LHV of 20GJ/tonne. But how would this difference be explained and quantified with exergy calculations. For example the fuel is to create steam for a milk powder plant
Associate Director, Judicial Expert at Ministry of Justice UAE.- BSc, MSc (Hons) Eng. Mech. CEM, LEED AP O&M. ???? ?????????? ?????????????? ???? ??????????, ?????? ?????????????????????? those of ???? ????????????????.
1 年While I appreciate your efforts in outlining the concepts of total exergy and exergy efficiency, I believe there are a few areas that might require clarification. For example, when you're defining exergy, it might be more accurate to say that it's the maximum useful work that can be extracted from a system at a given state while the system moves to equilibrium with its surroundings, including the reference environment. It's not merely the difference between the actual free energy of a system and the free energy it would have if at the same temperature as its surroundings, but rather a measure of potential work which considers both the energy and entropy changes in the system and its surroundings. Also, while the ratio of the free energy of a system to the free energy of a hot system can be indicative of the exergy efficiency, it's a simplification and might not hold true in all scenarios. For a more precise assessment, we should consider the exergy destruction due to irreversibilities, and hence, the exergy efficiency would be better defined as the ratio of the useful work output to the exergy input. That being said, your emphasis on practical application and urging readers to do their own calculations is certainly commendable.