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Advanced Thermodynamics Exergy / Availability:
A Measure of Work Potential
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Definitions Exergy (also called Availability or Work Potential): the maximum useful work that can be obtained from a system at a given state in a given environment; in other words, the most work you can get out of a system Surroundings: outside the system boundaries Environment: the area of the surroundings not affected by the process at any point (For example, if you have a hot turbine, the air next to the turbine is warm. The environment is the area of the surroundings far enough away that the temperature isn’t affected.) Dead State: when a system is in thermodynamic equilibrium with the environment, denoted by a subscript zero; at this point no more work can be done
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The energy content of the universe is constant, just as its mass content is. Yet at times of crisis we are bombarded with speeches and articles on how to “conserve” energy. As engineers, we know that energy is already conserved. What is not conserved is exergy, which is the useful work potential of the energy. Once the exergy is wasted, it can never be recovered. When we use energy (to heat our homes, for example), we are not destroying any energy; we are merely converting it to a less useful form, a form of less exergy. Exergy and the Dead State The useful work potential of a system is the amount of energy we extract as useful work. The useful work potential of a system at the specified state is called exergy. Exergy is a property and is associated with the state of the system and the environment. A system that is in equilibrium with its surroundings has zero exergy and is said to be at the dead state. The exergy of the thermal energy of thermal reservoirs is equivalent to the work output of a Carnot heat engine operating between the reservoir and the environment.
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Example A coal-fired furnace is used in a power plant. It delivers kW at 1000 K. The environment is at 300 K. What is the exergy of the added heat? You can use two steps to solve this problem. Determine the maximum percentage of the heat that can be converted to work. Using your answer from the first part, determine the maximum work possible. This is the maximum work output possible between the given state and the dead state, i.e., the heat’s exergy. In this case, 30% of the 5000 kW is unavailable energy—it can’t be converted to work.
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Why Study Exergy? In the last several decades, exergy analysis has begun to be used for system optimization. By analyzing the exergy destroyed by each component in a process, we can see where we should be focusing our efforts to improve system efficiency. It can also be used to compare components or systems to help make informed design decisions.
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Reversible Work Wrev (reversible work): the maximum amount of work it’s possible to produce (or minimum necessary to input) in a process between given initial and final states. Note that this is different from an isentropic process where we were given an inlet state and solved for the exit state using s2=s1. Since the exit and inlet states are both fixed, the process is not necessarily isentropic. What two conditions will cause a process to be isentropic?
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Irreversibilities, I Irreversibility, I: exergy destroyed; wasted work potential. It represents energy that could have been converted into work but was instead wasted What are some sources of I? To have high system efficiency, we want I to be as small as possible.
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I, cont. I=Wrev, out –Wu, out (work output device, like a turbine) OR
I=Wu, in –Wrev, in (work input device, like a pump) Wu: useful work; the amount of work done that can actually be used for something desirable Wu=W-Wsurr where W=actual work done
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Surroundings Work, Wsurr
Here some work is used to push the atmospheric air (the surroundings) out of the way; that work can’t be used for other purposes.
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Surroundings Work, Wsurr, cont.
Here Patm helps push the piston in; this is gained work. In a process where the piston goes in and out continually, the surrounding work values cancel out. What is Wsurr for a control volume?
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Second Law Efficiency, ηII
Thermal efficiency tells us what we get out compared to what we put in. The second law efficiency tells us how much we get out compared to the maximum possible we could get out, given the inlet and exit conditions.
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Second Law Efficiency, cont.
hth,max=1-TL/TH=1-300/800=0.635 Say hth=0.45 hII=0.45/0.625=0.72 We want a high hth and hII Another way to look at this: for a work output device hII=Wu/Wrev
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Second Law Efficiency, cont.
A general definition:
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Three Efficiency Definitions
The second two are defined for work OUTPUT devices
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Example A freezer is maintained at -10 oC by removing heat from it at a rate of 3 kW. The power input to the freezer is 2.5 kW, and the surrounding air is 30 oC. Determine a) the reversible power, b) the irreversibility, an c) the second-law efficiency of this freezer. Ref: Cengel & Boles, Thermodynamics, An Engineering Approach, 4th edition, Mc-Graw Hill, 2002.):
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Exergy, X We can calculate the exergy, X (work potential) at a given state. The work potential is a function of the total energy of the system. (remember that in a control mass, there will be no flow work) XKE (exergy due to kinetic energy): V2/2 (on a per unit mass basis XPE: gZ Xinternal energy: u-uo+Po(v-vo)-To(s-s0) To see a derivation of this last equation, see the appendices on the web site. The “o” stands for the dead state (atmospheric conditions). If a piston is at atmospheric pressure and temperature (the dead state), it can’t do any work.
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Exergy of a Closed System
Exergy of a closed system, per unit mass j, can be found be adding all the terms This gives us the maximum work we could possibly get out of a system. Usually we will be more interested in the change in exergy from the beginning to end of a process. For a closed system,
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Example 8-1 Oxygen gas is compressed in a piston-cylinder device from an initial state of 0.8 m3/kg and 25oC to a final state of 0.1 m3/kg and 287oC. Determine the reversible work input and the increase in the exergy of the oxygen during this process. Assume the surroundings to be at 25oC and 100 kPa. We assume that oxygen is an ideal gas with constant specific heats. From Table A-2, R = kJ/kgK. The specific heat is determined at the average temperature Table A-2(b) gives Cv, ave = kJ/kgK.
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The entropy change of oxygen is
We calculate the reversible work input, which represents the minimum work input Wrev,in in this case, from the exergy balance by setting the exergy destruction equal to zero.
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Therefore, the change in exergy and the reversible work are identical in this case. Substituting the closed system exergy relation, the reversible work input during this process is determined to be The increase in exergy of the oxygen is
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For a control volume Xcv=Xclosed+Xflow work
y=Xcv/m (exergy per unit mass) Xflow work=Wflow-Wagainst atmosphere=Pv-Pov Now combine terms: u+Pv=h; uo+Povo=ho
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Change in exergy If we only have one fluid stream
If we have multiple streams
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Exergy Balance We will use these equations in an exergy balance to solve for such quantities as reversible work or exergy destroyed. Xin-Xout-Xdestroyed=DXsys Xdestroyed is potential work that was destroyed due to irreversibilities like friction. Exergy can be transferred (Xin-Xout) by heat, work, and mass flow
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Exergy Transfer by Heat Transfer
As we add heat to a system, we increase its ability to do work.
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Exergy Transfer by Work and Mass Flow
If we do work on a system, we increase its ability to do work. Xwork=W-Wsurr for boundary work Xwork=W for all other kinds of work Remember Xmass=my
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Xdestroyed Xdestroyed=I=ToSgen Recall DSsys=Sin-Sout+Sgen
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Entropy Generated, Sgen
For a steady-state control volume, this leads us to For a control mass, this becomes Here Tk is the temperature of the heat source or heat sink (not the system temperature).
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Final Equation for DXsys for control mass
Terms in [ ] are W-Wsurr=Wu If we want to find Wrev, then ToSgen=0 and Wu=Wrev Note that if heat transfer is to/from the surroundings, the Q term drops out.
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Example A 12-ft3 rigid tank contains R-134a at 30 psia and 40% quality. Heat is transferred now to the refrigerant from a source at 120°F until the pressure rises to 60 psia. Assuming the surroundings to be at 75°F, determine a) the amount of heat transfer between the source and the refrigerant and b) the exergy destroyed during the process. Ref: Cengel and Boles
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Final Equation for DXsys for control volume
For multiple fluid streams, unsteady flow: For one fluid streams, steady flow: To find Wrev, set Sgen=0. If adiabatic, Q=0.
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Example 8-2 Steam enters an adiabatic turbine at 6 MPa, 600C, and 80 m/s and leaves at 50 kPa, 100C, and 140 m/s. The surroundings to the turbine are at 25C. If the power output of the turbine is 5MW, determine (a)the power potential of the steam at its inlet conditions, in MW. (b) the reversible power, in MW. (c)the second law efficiency. We assume steady-flow and neglect changes in potential energy.
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Conservation of mass for the steady flow gives
The mass flow rate of the steam is determined from the steady-flow energy equation applied to the actual process, 0 (steady) Conservation of mass for the steady flow gives The work done by the turbine and the mass flow rate are
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where From the steam tables:
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The power potential of the steam at the inlet conditions is equivalent to its exergy at the inlet state. Recall that we neglect the potential energy of the flow.
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The power output of the turbine if there are no irreversibilities is the reversible power and is determined from the rate form of the exergy balance applied on the turbine and setting the exergy destruction term equal to zero. 0 (steady)
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The second-law efficiency is determined from
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Set up the following problems.
Refrigerant at T1 and P1 is throttled to a pressure of P2. Find the reversible work and exergy destroyed during this process. The atmosphere has a temperature of To. Air at T1 and P1 with a velocity of V1 enters a nozzle and exits at P2 and T2 with a velocity of V2. There is a heat loss Q from the nozzle to the surroundings at To. Find the exergy destroyed during this process. Air enters a compressor at ambient conditions (To and Po) and leaves at P2 and T2. The compressor is deliberately cooled, and there is a rate of heat loss of Q to the surroundings. The power input to the compressor is PWR. Find the rate of irreversibility, I, for this process.
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Exergy Analysis for a Cycle, 1 fluid stream, steady flow
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Second Law Efficiency for a Cycle
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Exergy of kinetic energy
Exergy Forms Exergy of kinetic energy Kinetic energy is a form of mechanical energy and can be converted directly into work. Kinetic energy itself is the work potential or exergy of kinetic energy independent of the temperature and pressure of the environment. Exergy of kinetic energy: Exergy of potential energy Potential energy is a form of mechanical energy and can be converted directly into work. Potential energy itself is the work potential or exergy of potential energy independent of the temperature and pressure of the environment. Exergy of potential energy:
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Useful Work The work done by work producing devices is not always entirely in a useable form. Consider the piston-cylinder device shown in the following figure. The work done by the gas expanding in the piston-cylinder device is the boundary work and can be written as The actual work done by the gas is
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The word done on the surroundings is
Any useful work delivered by a piston-cylinder device is due to the pressure above the atmospheric level. Reversible Work Reversible work Wrev is defined as the maximum amount of useful work that can be produced (or the minimum work that needs to be supplied) as a system undergoes a process between the specified initial and final states. This is the useful work output (or input) obtained when the process between the initial and final states is executed in a totally reversible manner. Irreversibility The difference between the reversible work Wrev and the useful work Wu is due to the irreversibilities present during the process and is called the irreversibility I. It is equivalent to the exergy destroyed and is expressed as
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where Sgen is the entropy generated during the process
where Sgen is the entropy generated during the process. For a totally reversible process, the useful and reversible work terms are identical and thus irreversibility is zero. Irreversibility can be viewed as the wasted work potential or the lost opportunity to do work. It represents the energy that could have been converted to work but was not. Exergy destroyed represents the lost work potential and is also called the wasted work or lost work. Second-Law Efficiency The second-law efficiency is a measure of the performance of a device relative to the performance under reversible conditions for the same end states and is given by for heat engines and other work-producing devices and
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for refrigerators, heat pumps, and other work-consuming devices.
In general, the second-law efficiency is expressed as Exergy of change of a system Consider heat transferred to or from a closed system whenever there is a temperature difference across the system boundary. The exergy for a system may be determined by considering how much of this heat transfer is converted to work entirely. Let’s take a second look at the following figure.
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Taking the heat transfer to be from the system to its surroundings, the conservation of energy is
The work is the boundary work and can be written as Any useful work delivered by a piston-cylinder device is due to the pressure above the atmospheric level. To assure the reversibility of the process, the heat transfer occurs through a reversible heat engine.
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Integrating from the given state (no subscript) to the dead state (0 subscript), we have
This is the total useful work due to a system undergoing a reversible process from a given state to the dead state, which is the definition of exergy. Including the kinetic energy and potential energy, the exergy of a closed system is on a unit mass basis, the closed system (or nonflow) exergy is
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Here, u0, v0, and s0 are the properties of the system evaluated at the dead state. Note that the exergy of the internal energy of a system is zero at the dead state is zero since u = u0, v = v0, and s = s0 at that state. The exergy change of a closed system during a process is simply the difference between the final and initial exergies of the system, On a unit mass basis the exergy change of a closed system is
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Exergy of flow The energy needed to force mass to flow into or out of a control volume is the flow work per unit mass given by (see Chapter 3). The exergy of flow work is the excess of flow work done against atmospheric air at P0 to displace it by volume v. According to the above figure, the useful work potential due to flow work is
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Thus, the exergy of flow energy is
Flow Exergy Since flow energy is the sum of nonflow energy and the flow energy, the exergy of flow is the sum of the exergies of nonflow exergy and flow exergy. The flow (or stream) exergy is given by
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Exergy Transfer by Heat, Work, and Mass
The exergy of flow can be negative if the pressure is lower than atmospheric pressure. The exergy change of a fluid stream as it undergoes a process from state 1 to state 2 is Exergy Transfer by Heat, Work, and Mass Exergy can be transferred by heat, work, and mass flow, and exergy transfer accompanied by heat, work, and mass transfer are given by the following. Exergy transfer by heat transfer By the second law we know that only a portion of heat transfer at a temperature above the environment temperature can be converted into work. The maximum useful work is produced from it by passing this heat transfer through a reversible heat engine. The exergy transfer by heat is Exergy transfer by heat:
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Note in the above figure that entropy generation is always by exergy destruction and that heat transfer Q at a location at temperature T is always accompanied by entropy transfer in the amount of Q/T and exergy transfer in the amount of (1-T0/T)Q. Note that exergy transfer by heat is zero for adiabatic systems.
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Exergy transfer by work
Exergy is the useful work potential, and the exergy transfer by work can simply be expressed as Exergy transfer by work: where , P0 is atmospheric pressure, and V1 and V2 are the initial and final volumes of the system. The exergy transfer for shaft work and electrical work is equal to the work W itself. Note that exergy transfer by work is zero for systems that have no work. Exergy transfer by mass Mass flow is a mechanism to transport exergy, entropy, and energy into or out of a system. As mass in the amount m enters or leaves a system the exergy transfer is given by Exergy transfer by mass:
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Note that exergy transfer by mass is zero for systems that involve no flow.
The Decrease of Exergy Principle and Exergy Destruction The exergy of an isolated system during a process always decreases or, in the limiting case of a reversible process, remains constant. This is known as the decrease of exergy principle and is expressed as Exergy Destruction Irreversibilities such as friction, mixing, chemical reactions, heat transfer through finite temperature difference, unrestrained expansion, non-quasi-equilibrium compression, or expansion always generate entropy, and anything that generates entropy always destroys exergy. The exergy destroyed is proportional to the entropy generated as expressed as
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The decrease of exergy principle does not imply that the exergy of a system cannot increase. The exergy change of a system can be positive or negative during a process, but exergy destroyed cannot be negative. The decrease of exergy principle can be summarized as follows: Exergy Balances Exergy balance for any system undergoing any process can be expressed as General: General, rate form:
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General, unit-mass basis:
where For a reversible process, the exergy destruction term, Xdestroyed, is zero. Considering the system to be a general control volume and taking the positive direction of heat transfer to be to the system and the positive direction of work transfer to be from the system, the general exergy balance relations can be expressed more explicitly as
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where the subscripts are i = inlet, e = exit, 1 = initial state, and 2 = final state of the system. For closed systems, no mass crosses the boundaries and we omit the terms containing the sum over the inlets and exits. Example 8-1 Oxygen gas is compressed in a piston-cylinder device from an initial state of 0.8 m3/kg and 25oC to a final state of 0.1 m3/kg and 287oC. Determine the reversible work input and the increase in the exergy of the oxygen during this process. Assume the surroundings to be at 25oC and 100 kPa. We assume that oxygen is an ideal gas with constant specific heats. From Table A-2, R = kJ/kgK. The specific heat is determined at the average temperature Table A-2(b) gives Cv, ave = kJ/kgK.
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The entropy change of oxygen is
We calculate the reversible work input, which represents the minimum work input Wrev,in in this case, from the exergy balance by setting the exergy destruction equal to zero.
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Therefore, the change in exergy and the reversible work are identical in this case. Substituting the closed system exergy relation, the reversible work input during this process is determined to be The increase in exergy of the oxygen is
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Example 8-2 Steam enters an adiabatic turbine at 6 MPa, 600C, and 80 m/s and leaves at 50 kPa, 100C, and 140 m/s. The surroundings to the turbine are at 25C. If the power output of the turbine is 5MW, determine (a)the power potential of the steam at its inlet conditions, in MW. (b) the reversible power, in MW. (c)the second law efficiency. We assume steady-flow and neglect changes in potential energy.
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Conservation of mass for the steady flow gives
The mass flow rate of the steam is determined from the steady-flow energy equation applied to the actual process, 0 (steady) Conservation of mass for the steady flow gives The work done by the turbine and the mass flow rate are
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where From the steam tables:
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The power potential of the steam at the inlet conditions is equivalent to its exergy at the inlet state. Recall that we neglect the potential energy of the flow.
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The power output of the turbine if there are no irreversibilities is the reversible power and is determined from the rate form of the exergy balance applied on the turbine and setting the exergy destruction term equal to zero. 0 (steady)
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The second-law efficiency is determined from
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