Chapter 8

EXERGY: A MEASURE OFEXERGY: A MEASURE OF WORK POTENTIAL

Dr Ali JawarnehDr Ali Jawarneh

Department of Mechanical EngineeringHashemite University

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Objectives• Examine the performance of engineering devices in light

of the second law of thermodynamics.

D fi hi h i th i f l k th t• Define exergy, which is the maximum useful work that could be obtained from the system at a given state in a specified environment.

• Define reversible work, which is the maximum useful work that can be obtained as a system undergoes a process between two specified states.process between two specified states.

• Define the exergy destruction, which is the wasted work potential during a process as a result of irreversibilities.

• Define the second-law efficiency.

• Develop the exergy balance relation.

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• Apply exergy balance to closed systems and control volumes.

8-1 EXERGY: WORK POTENTIAL OF ENERGYThe useful work potential of a given amount of energy at someThe useful work potential of a given amount of energy at some specified state is called exergy, which is also called the availability or available energy.A system is said to be in the dead state when it is in thermodynamicA system is said to be in the dead state when it is in thermodynamic equilibrium with the environment it is in. The work

potential of the energy containedenergy contained in a system at a specified stateis simply the

i f l

At the dead state, the useful

maximum useful work that can be obtained from the system.

A system that is in equilibrium with its environment is said to be at the dead state.

At the dead state, the useful work potential (exergy) of a system is zero.

y

The property exergy is the work potential of a system in a specified

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The property exergy is the work potential of a system in a specified environment and represents the maximum amount of useful work that can be obtained as the system is brought to equilibrium with the environment.

A system delivers the maximum possible work as it undergoes a reversible process from the specified initial state to the state of its environment that is the dead state This represents the useful workenvironment, that is, the dead state. This represents the useful work potential of the system at the specified state and is called exergy. It is important to realize that exergy does not represent the amount of work that a work-producing device will actually deliver upon installation. Rather, itrepresents the upper limit on the amount of work a device can deliver withoutviolating any thermodynamic laws

The immediate surroundings of a hot potato are simply the temperature

The atmosphere contains a tremendous amount of energy, but no exergy

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gradient zone of the air next to the potato.

no exergy.Unless specified otherwise, the dead-state temp and pressure are taken to be T0 =25°C and P0= 1 atm (101.325) kPa . A system has zero exergy at the dead state

Distinction should be made between the surroundings, immediate surroundings, and the environment. By definition, surroundings are everything outside the system boundaries. The immediate surroundings refer to the portion of the surroundings that is affected by the process, and environment refers to the region beyond the immediate surroundings whose properties are not affected by the process at any point. Therefore, any irreversibilities during a process occur within the system and its immediate surroundings, and the environment is free of any irreversibilities.

I l i th i iti l t t i ifi d d th it i t i blIn an exergy analysis, the initial state is specified, and thus it is not a variable.The work output is maximized when the process between two specified

states is executed in a reversible manner, as shown in Chap. 7. Therefore, all the irreversibilities are disregarded in determining the work potential.

Finally the system must be in the dead state at the end of the process to maximize the workFinally, the system must be in the dead state at the end of the process to maximize the work output.

The notion that a system must go to the dead state at the end of the process to maximize the work output can be explained as follows: If the system temperature at the final state is greater than (or less than) the temperature of the environment it is in, we can always produce additional ( ) p , y pwork by running a heat engine between these two temperature levels. If the final pressure is greater than (or less than) the pressure of the environment, we can still obtain work by letting the system expand to the pressure of the environment. If the final velocity of the system is not zero, we can catch that extra kinetic energy by a turbine and convert it to rotating shaft work, and so on. No work can be produced from a system that is initially at the dead state. The atmosphere around us contains a tremendous amount of energy. However, the atmosphere is in the dead state, and the energy it contains has no work potential

Note that the exergy of a system at a specified state depends on the conditionsf th i t (th d d t t ) ll th ti f th t

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of the environment (the dead state) as well as the properties of the system.Therefore, exergy is a property of the system–environment combinationand not of the system alone.

Exergy (Work Potential) Associated with Kinetic and Potential Energy Kinetic energy is a form of mechanical

d th it b t d t kExergy of kinetic energy:

energy, and thus it can be converted to work entirely. Therefore, the work potential or exergy of the kinetic energy of a system is equal to the kinetic energy itself regardless

Exergy of potential energy:

q gy gof the temp and pressure of the environment

Potential energy is also a form of mechanical energy, and thus it can be converted to work entirely. Therefore, the exergy of the potential energy of a system is equal to the potential energy itself regardless of the temperatureand pressure of the environment

The work potential or exergy of potential energy is equal to the potential energy itself.

The exergies of kinetic and potential energies are equal to

and pressure of the environment

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The exergies of kinetic and potential energies are equal to themselves, and they are entirely available for work.

Example: The electric power needs of a community are to be met by windmills with 10-m-diameter rotors. The windmills are to be located

h th i d i bl i t dil t l it f 8 /where the wind is blowing steadily at an average velocity of 8 m/s. Determine the minimum number of windmills that need to be installed if the required power output is 600 kW.

Solution: Air is at standard conditions of 1 atm and 25°CThe gas constant of air is 0.287 kPa.m3/kg.K (Table A-1).

8In practice, the actual efficiency ranges between 20 and 40 percent and is about 35 percent for many wind turbines.

Example: Consider a thermal energy reservoir at 1500 K thatcan supply heat at a rate of 150,000 kJ/h. Determine the exergy pp y , gyof this supplied energy, assuming an environmental temperature of 25°C.

Solution:

Unavailable energy isUnavailable energy is the portion of energy

that cannot be converted to work by even a

reversible heat engine

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reversible heat engine.

8-2 REVERSIBLE WORK AND IRREVERSIBILITY

Surrounding work(Wsurr): Work done by or against the surroundings during a process.

As a closed

When a system is expanding and doing work, part of the work done is used to overcome the atmospheric pressure, and thus Wsurr represents a l Wh t i d

W:actual work(i.e Wb)

Wu:useful work

As a closed system expands, some work needs to be done to push the atmospheric

loss. When a system is compressed, however, the atmospheric pressure helps the compression process, and thus Wsurr represents a gain.Note that the work done by or againstthe atmospheric

air out of the way (Wsurr).

Note that the work done by or against the atmospheric pressure has significance only for systems whose volume changes during the process (i.e., systems that involve moving

For constant-volume systems, the total actual and useful works are identical

( , y gboundary work). It has no significance for cyclic devices and systems whose boundaries remain fixed during a process such as rigid tanks and

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(Wu = W). steady-flow devices (turbines, compressors, nozzles, heatexchangers, etc.)

8-2 REVERSIBLE WORK AND IRREVERSIBILITYReversible work W : The maximum amount of useful work that can be When evaluating Reversible work Wrev: 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 (or expended) when the process between the initial and final states is executed in a totally reversible manner

gexergy, the final state is always assumed to be the dead state, which is hardly ever the case for actual

i i tbetween the initial and final states is executed in a totally reversible manner. When the final state is the dead state, the reversible work equals exergy.

The irreversibility is

engineering systems.

Any difference between the reversiblework W and the useful work W

Irreversibility I:

e e e s b y sequivalent to the exergy destroyed (or exergy destruction).

work Wrev and the useful work Wuis due to the irreversibilities

Irreversibility is a positive

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 The

The difference between

Irreversibility is a positivequantity for all actual (irreversible) processes since Wrev ≥Wu for work-producing

converted to work but was not. The smaller the irreversibility associated with a process, the greater the work that is produced (or the smaller the work that is consumed). The performance of a system can be

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The difference between reversible work and

actual useful work is the irreversibility.

devices and Wrev ≤ Wu for work-consuming devices.

performance of a system can be improved by minimizing the irreversibility associated with it.

8-3 SECOND-LAW EFFICIENCY, ηIIThermal efficiency and the coefficient of performance are sometimes referred to as the first-law efficiencies

Two heat engines that have the

Second-law efficiency is a measure of the performance of a device relative to its have the

same thermal efficiency, but different

i

performance under reversible conditions.

Now it is becoming apparent that engine B has a greater work potential available to it (70% of the heat supplied as compared to 50%for engine A), and thus h ld d l t b tt th i A Th f

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maximum thermal efficiencies.

should do a lot better than engine A. Therefore, we can say that engine B is performing poorly relative to engine A even though both have the same thermal efficiency.

The definitions above for the second-law efficiency do not apply to devices that are not intended to produce or consume work.

General definition of exergy efficiency

The second-law efficiency is intended to serve as a measure of approximation toapproximation to reversible operation, and thus its value should range from zero in the worst case (complete

The second-law efficiency of t ll i i

( pdestruction of exergy) toone in the best case (no destruction of exergy).

Second-law efficiency of all ibl d i i 100%

naturally occurring processes is zero if none of the work potential is recovered.

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reversible devices is 100%. Note that the second-law efficiency cannot exceed 100 percent

Example: A freezer is maintained at -7°C by removing heat from it at a rate of 80 kJ/min The power input to the freezer is 0 5 kW and therate of 80 kJ/min. The power input to the freezer is 0.5 kW, and the surrounding air is at 25°C. Determine (a) the reversible power, (b) the irreversibility, and (c) the second-law efficiency of this freezer.

Solution:

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EXERGY CHANGE OF A SYSTEMExergy of a Fixed Mass: NonflowExergy of a Fixed Mass: Nonflow (or Closed System) Exergy

In general, internal energy consists of sensible, latent, chemical, and nuclearenergies. However, in the absence of any chemical or nuclear reactions, thechemical and nuclear energies can be disregarded and the internal energy canchemical and nuclear energies can be disregarded and the internal energy canbe considered to consist of only sensible and latent energies that can betransferred to or from a system as heat whenever there is a temperature differenceacross the system boundary. The second law of thermodynamics states that heat cannot be converted to work entirely, and thus the work potential of internal energy must be less than the internal energy itself. But how much less?

Th f ifi d t

gy gy

The exergy of a specified mass at a specified state is the useful work that can be produced as the mass undergoes a reversible process to the state of the environment.

A closed system, in general, may possess kinetic and potential energies, and the total energy of a l d t i l t th f it i t l

15Exergy of a closed system

closed system is equal to the sum of its internal, kinetic, and potential energies.

Closed system exergy per unit mass

Exergy change of a closed system: for isothermalprocesses

φ:phi

or, on a unit mass basis:

The exergy of a closed system is either positive or zero. It is never negative.

The exergy of a When the properties of a t t if

: are the properties of the system evaluated at the dead state

Even a medium at low temperature (T <T0) and/or low pressure (P <P0) contains exergy since a cold medium can serve as

cold medium is also a positive quantity since

work can be

system are not uniform, the exergy of the system is

cold medium can serve as the heat sink to a heat engine that absorbs heat from the environment at T0, and an evacuated space

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work can be produced by

transferring heat to it.

makes it possible for the atmospheric pressure to move a piston and do useful work

Exergy of a Flow Stream: Flow (or Stream) Exergy

E f flExergy of flow energy

Flow exergy:Flow exergy:

Exergy change of flow:ψ:psi

The exergy associated with flow energy is the

The flow work is essentially theflow energy is the

useful work that would be delivered by an

yboundary work

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imaginary piston in the flow section.

The energy and t t fexergy contents of

(a) a fixed mass(b) a fluid stream.( )

Note that the exergy change of a closed system or a fluid stream representsthe maximum amount of useful work that can be done (or the minimum amount of useful work that needs to be supplied if it is negative) as the system changes from state 1 to state 2 in a specified environment, and represents the reversible work Wrev. It isrepresents the reversible work Wrev. It is independent of the type of process executed, the kind of system used, and the nature of energy interactions with the surroundings. Also note that the exergy of a closed system cannot be negati e b t the e erg of a flocannot be negative, but the exergy of a flow stream can at pressures below the environment pressure P0.

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8-5 EXERGY TRANSFER BY HEAT, WORK, AND MASS

Exergy, like energy, can be transferred to or from a system in three forms: heat, work, and mass flow. Exergy transfer is recognized at the system boundary, ,

Exergy by Heat Transfer, QExergy

Exergy transfer is recognized at the system boundary as exergy crosses it, and it represents the exergy gained or lost by a system during a process. The only two forms of exergy interactions associated with a fixed mass or closed system are heat transfer and

ktransfer by heat

When temperature is not constant

work.

constant

The transfer and destruction of

exergy during aexergy during a heat transfer

process through a finite

The Carnot efficiency ηc=1−T0 /T represents the fraction of the energy transferred from a heat source

temperature difference.

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fraction of the energy transferred from a heat source at temperature T that can be converted to work in an environment at temperature T0.

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Exergy Transfer by Work, WDuring expansion, this work cannot be used for any useful purpose since it is just enough to push the atmospheric air aside. During compression, when the gas is cooled the piston moves down

such as shaft work and electrical work

when the gas is cooled, the piston moves down, compressing the gas. Again, no work is needed from an external source to accomplish this compression process. Thus we conclude that the work done by or against the atmosphere is not available for any useful purpose and should be

Exergy Transfer by Mass, m

available for any useful purpose, and should be excluded from available work.

when the fluid properties are variable

There is no useful work transfer associated with

Mass contains energy, entropy, and exergy, and thus mass flow into or out of associated with

boundary work when the pressure of the system is maintained constant

thus mass flow into or out of a system is accompanied by energy, entropy, and exergy transfer.

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maintained constant at atmospheric pressure.

Note that exergy transfer by heat X heat is zero for adiabatic systems, and the exergy transfer by mass X mass is zero for systems that involve no mass flow across their boundaries (i.e., closed systems). The total exergy transfer is zero for isolated systems since they involve no heat, work, or mass transfer.

8-6 THE DECREASE OF EXERGY PRINCIPLE AND EXERGY DESTRUCTION

The isolated system considered in the development of the decrease of exergydecrease of exergy principle.

The exergy of an isolated system during a process always decreases or, in the limiting case of a reversible process, remains constant. In other words, it never

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limiting case of a reversible process, remains constant. In other words, it never increases and exergy is destroyed during an actual process. This is known as the decrease of exergy principle. For an isolated system, the decrease in exergy equals exergy destroyed.

Exergy DestructionAnything that generates entropy always destroys exergy

F E 8 31From Eq. 8–31:

Exergy destroyed is a positive quantity for

Exergy destroyed represents the lost work potential and is also called the wasted work or lost work.

Exergy destroyed is a positive quantity for any actual process and becomes zero for a reversible process. Exergy destroyed represents the lost work

The exergy change of a system can be negative, but the exergy destruction

potential and is also called the irreversibility or lost work.The exergy change of a

t b iti exergy destruction cannot.

system can be positive or negative during a process

Equations 8–32 and 8–33 for the decrease of exergy and the exergy destruction are li bl t ki d f t d i ki d f i t d

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applicable to any kind of system undergoing any kind of process since any system and its surroundings can be enclosed by a sufficiently large arbitrary boundary across which there is no heat, work, and mass transfer, and thus any system and its surroundings constitute an isolated system.

GENERAL EXERGY BALANCE The exergy change of a system during a process is equal to the difference qbetween the net exergy transfer through the system boundary and the exergy destroyed within the ysystem boundaries as a result of irreversibilities.Q)

TT(Xheat

01−=usefulwork WX = mψXmass =

112212 φ−φ=−= mmXXΔX

=

Mechanisms of exergy transfer

112212 φφ mmXXΔX

= transfer.

steady-flow process:

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∑∑∑∑∫••••

−=−=outin

cv

outincv

mmdtdmormmdρ

dtd V

Non flow exergy

2 W: actual work that]gzV)vv(P)ss(T)uu[(mX ooooo ++−+−−−=

2

2

φ= mX

W: actual work that appears in 1st law (such as Wb, Wshaft, WelecWu : useful work that appers in Exergy balance

112212 φ−φ=−= mmXXΔX

Closed system:

appers in Exergy balance equation such as Wshaft,Welec, surrb WW −surrW +:expansion

-ve:compr

Flow exergyfor isothermal processes

Reversible work Wrev: The maximum (or the minimum ) t f f l k

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amount of useful work

Comment on Boundary Work

usefulwork WX =

Expansion:

)VV(PdVPWX ousefulwork 12

2

1

−−== ∫Expansion:

1

∫

bWCompression:surrW

)VV(PdVPWX ousefulwork 212

−−== ∫Example:The boundary work for polytropic process during expansion of piston cylinderp y p y p p g p p y

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8-7 EXERGY BALANCE:CLOSED SYSTEMS

A closed system does not involve any mass flow and thus any exergy transfer associated with

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mass flow.

0.0

The heat transfer to t d ka system and work

done by the system are taken to be positive quantities.

Qk is the heat transfer through the boundary at temperature Tk at location k.

Exergy destroyed

outside system b d i

Exergy b l f boundaries can

be accounted for by writing an

exergy balance

balance for a closed system when heat

on the extended system that includes the

system and its

transfer is to the system and the work is

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yimmediate

surroundings.from the system.

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8-8 EXERGY BALANCE: CONTROL VOLUMES

The rate of exergy change within the control volume during a process is equal to the rate of net exergy transfer through the control volume boundary by heat, work, and mass flow minus the rate of exergy destruction within therate of exergy destruction within the boundaries of the control volume.

Exergy is transferred into or out of a control volume by mass as

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o a co t o o u e by ass aswell as heat and work transfer.

Exergy Balance for Steady-Flow SystemsMost control volumes encountered in practice such as turbines, compressors, nozzles, diffusers, heat exchangers, pipes, and ducts operate steadily, and thus they experience no changes in their mass, energy, entropy, and exergy contents as well as their volumes. Therefore, dVCV/dt = 0 and dXCV/dt = 0 for such systems.

The exergy transfer to a steady-flow system is equal to the exergy transfer from it plus the exergytransfer from it plus the exergy destruction within the system.

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Reversible Work, WrevThe exergy balance relations presented above can be used to determine the reversible work Wrev by setting the exergy destroyed equal to zero. The work W in that case becomes the reversible work.

The exergy destroyed is zero only for a reversible process, and reversible work represents the maximum work output for work-producing devices such as turbines and the minimum work input for p g pwork-consuming devices such as compressors.

The wasted work potential is equivalent to the exergy destroyed,

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p q gy y

The difference between the reversible work and the actual useful work is the exergy destroyed:

Second-Law Efficiency of Steady-Flow Devices, ηII

Th d l ffi i f i t d fl d i b d t i d f it lThe second-law efficiency of various steady-flow devices can be determined from its general definition, ηII = (Exergy recovered)/(Exergy supplied). When the changes in kinetic and potential energies are negligible and the devices are adiabatic:

Turbine

Compressor

Heat exchanger

Mixing chamber A heat exchanger with two unmixed

fluid streams

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fluid streams.

EXAMPLE 8-10:Consider steady heat transfer through a 5-m 6-m brick wall of a house Co s de steady eat t a s e t oug a 5 6 b c a o a ouseof thickness 30 cm. On a day when the temperature of the outdoors is 0°C, the house is maintained at 27°C. The temperatures of the inner and outer surfaces of the brick wall are measured to be 20°C and 5°C, respectively, and the rate of heat transfer through the wall is 1035 W. Determine the rate of exergy destruction in the wall, and the rate of total exergy destruction associated with this heat transfer process.

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Solution: We first take the wall as closed system

Applying the rate form of the exergy balance to the wall givespp y g gy g

the direction of exergy transfer is the same as the direction of heat transfer

The exergy balance for this extended system (system + immediate surroundings) is the same as that given above except the two boundary temperatures are 300is the same as that given above, except the two boundary temperatures are 300and 273 K instead of 293 and 278 K, respectively

The difference between the two exergy destructions is 41.2 W and represents the exergy destroyed in the air layers on both sides of the wall.

Discussion This problem was solved in Chap. 7 for entropy generation. Weld h d t i d th d t d b i l lti l i th

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could have determined the exergy destroyed by simply multiplying theentropy generations by the environment temperature of T0 =273 K.

=0.341 x 273=93.1

Example: A piston cylinder device contains 5 kg of refrigerant 134a at 0 7 MPa and 60 0C The refrigerant isrefrigerant 134a at 0.7 MPa and 60 0C. The refrigerant is now cooled at constant pressure until it exists as a liquid at 24 0C. If the surroundings are at 100 kPa and 24 0C, determine (a) the exergy of the refrigerant at the initial and final states and (b) the exergy destroyed during this process. p

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Solution:From the refrigerant tables (Tables A-11 through A-13),

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(b) The reversible work input, which represents the minimum work input W rev, in, this case can be determined from the exergy balance (extended system) by setting the exergy destruction equal to zeroequal to zero,

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Alternative solution (Extended system): sysdesoutin XXXX Δ=−−

12d XXXW −=−0.0Q)TT1(X o

heat =−=)XX(WX12desu XXXWT )XX(WX 12udes −−=

bW T=T0

5.831.1256.208XX 12 =−=− 6.18)1.1256.208(1.102)XX(WX 12udes =−−=−−=

N t th W i ΔX d WNote: the Wsurr appears in ΔX system and Wu

Alternative solution (Extended system):)uu(mQWEEE 12bsysoutin −=−→Δ=−

Wb appears in 1st law not Wu as in Exergy Balance

Q

1067Q)01.27444.84(5Q)0008261.0034875.0(7005)uu(mQ)vv(mP

)(Q

1221

12bsysoutin

=→−=−−×→−=−−→

)0256.131958.0(5S10670

)ss(mSTQ0SSSS

gen

12gensurr

sysgenoutin

−=+−→

−=+−→Δ=+−

390625.0S

)(27324

gen

gen

=→+

6.180625.0)27324(STX genodes =×+==

Example: An insulated piston–cylinder device contains 2 L ofsaturated liquid water at a constant pressure of 150 kPa. Anelectric resistance heater inside the cylinder is turned on, andelectrical work is done on the water in the amount of 2200 kJ.Assuming the surroundings to be at 25°C and 100 kPa, determine (a) the minimum work with which this process could be accomplished and (b) the exergy destroyed during thisprocess.

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Solution: From the steam tables (Tables A-4 through A-6),

The reversible work input which represents the minimum work input Wrev in

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The reversible work input, which represents the minimum work input Wrev,in in this case can be determined from the exergy balance by setting the exergy destruction equal to zero:

The exergy destruction (or irreversibility) associated with this process can be determined from its definition Xdes= T0Sgen where the entropy generation is determined from an entropy balance on thedefinition Xdes T0Sgen where the entropy generation is determined from an entropy balance on the cylinder, which is an insulated closed system,

Alternative solution: XXXX Δ=−−Alternative solution:sysdesoutin XXXX Δ=−−

XX)WW(W sysdessurrbe Δ=−−−

)]vv(P)vv(P[m)VV(P)VV(PWW 12o1212o12surrb −−−=−−−=−

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1705XXX)VV(PW(W dessysdes12obe =⇒Δ=−−−−→

7.2142Wu =17057.4377.2142WW7.2142XI revudes =−=−===

Example: An iron block of unknown mass at 85°C is d d i t i l t d t k th t t i 100 L f tdropped into an insulated tank that contains 100 L of water at 20°C. At the same time, a paddle wheel driven by a 200-W motor is activated to stir the water. It is observed that thermal equilibrium is established after 20 min with a final temperature of 24°C. Assuming the surroundings to be at 20°C determine (a) the mass of the iron block and (b) the20 C, determine (a) the mass of the iron block and (b) the exergy destroyed during this process.

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Solution:

44

Alternative solution:

sysdesoutin XXXX Δ=−−

waterirondesin,pw

sysdesoutin

XXXW Δ+Δ=−

7.146)}SS(T)TT(mc{X iron12o12piron −=−−−=Δ

237.11)}SS(T)TT(mc{X water12o12pwater =−−−=Δ

XXXW Δ+Δ=−

375X237.117.146X240

XXXW

desdes

waterirondesin,pw

=⇒+−=−

Δ+Δ=−

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Can you find the reversible work?

Example: A 0.35 m3 rigid tank contains refrigerant-134a at280 kPa and 55 percent quality Heat is transferred now to the280 kPa and 55 percent quality. Heat is transferred now to therefrigerant from a source at 50°C until the pressure rises to400 kPa. Assuming the surroundings to be at 25°C, determine(a) the amount of heat transfer between the source and the(a) the amount of heat transfer between the source and therefrigerant and (b) the exergy destroyed during this process.

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Solution: From the refrigerant tables (Tables A-11 through A-13),

(a) The mass of the refrigerant is

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Alternative solution: use the extended system for exergy balanceAlternative solution: use the extended system for exergy balance

XXXX Δ=−−

sysdesin0

sysdesoutin

XXQ)TT1(

XXXX

Δ=−−

Δ=−−

32)]ss(T)vv(0)uu[(mX 12o2112sys −=−+=+−=ΔXXQ)

TT1( sysdesin

0 Δ=−−

48

)]()()[( 12o2112sys

65X32X425)27350273251( desdes =⇒−=−

++

−

Example: Steam enters an adiabatic turbine at 6 MPa, 600°C and 80 m/s and leaves at 50 kPa 100°C and 140600°C, and 80 m/s and leaves at 50 kPa, 100°C, and 140m/s. If the power output of the turbine is 5 MW, determine (a) the reversible power output and (b) the second-law efficiency f th t bi A th di t b t 25°Cof the turbine. Assume the surroundings to be at 25°C.

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Solution: From the steam tables (Tables A-4 through A-6)

Th ibl ( i ) i d i d f h f f hThe reversible (or maximum) power output is determined from the rate form of the exergy balance applied on the turbine and setting the exergy destruction term equal to zero, You can add

the ΔKE effectthe ΔKE effect

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Example: Liquid water at 200 kPa and 20°C is heated in achamber by mixing it with superheated steam at 200 kPa andchamber by mixing it with superheated steam at 200 kPa and 300°C. Liquid water enters the mixing chamber at a rate of2.5 kg/s, and the chamber is estimated to lose heat to the

di i t 25°C t t f 600 kJ/ i If th i tsurrounding air at 25°C at a rate of 600 kJ/min. If the mixtureleaves the mixing chamber at 200 kPa and 60°C, determine(a) the mass flow rate of the superheated steam and (b) the

t d k t ti l d i thi i iwasted work potential during this mixing process.

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Solution:

52

Alternative solution: use the extended system for exergy balance

0XQ)T1(mmm

XXXX

desin0

33221

sysdesoutin

1 =−−−ψ−ψ+ψ

Δ=−−•••••

••••0.0

53

0XQ)T

1(mmm desin332211 ψψ+ψ

h f@25=104.83

)36720296490)(27325()831049183()ss(T)hh( o1oo11

−+−−=

−−−=ψ

152.0)3672.029649.0)(27325()83.10491.83(

=−+−−=

)3672089417)(27325()8310413072()ss(T)hh( o2oo22 −−−=ψ

25.724)3672.08941.7)(27325()83.1041.3072(

=−+−−=

)ss(T)hh( o3oo33 −−−=ψ

05.8)3672.08313.0)(27325()83.10418.251(

)()( o3oo33

=−+−−=

ψ

mmmX0Xmmm desdes ψ−ψ+ψ=⇒=−ψ−ψ+ψ••••••••

3.8605.8648.225.724148.0152.05.2X

mmmX0Xmmm

des

33221desdes33221 11

=×−×+×=

ψψ+ψ=⇒=ψψ+ψ•

Note: the two solutions do not meet, this mean the physical

54

p yinputs for the question are wrong, but the two solutions methodology are Ok.

Example: Liquid water at 15°C is heated in a chamber by mixing it with saturated steam Liquid water enters the chambermixing it with saturated steam. Liquid water enters the chamber at the steam pressure at a rate of 4.6 kg/s and the saturated steam enters at a rate of 0.23 kg/s. The mixture leaves the

i i h b li id t 45°C If th di tmixing chamber as a liquid at 45°C. If the surroundings are at 15°C, determine (a) the temperature of saturated steam entering the chamber, (b) the exergy destruction during this

i i d ( ) th d l ffi i f th i imixing process, and (c) the second-law efficiency of the mixing chamber.

55

Solution:

energy balance on the chamber gives:

The exergy destruction is determined from an exergy balance on the chamber to begy gy

56

Example: A well-insulated shell-and-tube heat exchanger is used to heat water (cp = 4.18 kJ/kg · °C) in the tubes from 20to 70°C at a rate of 4.5 kg/s. Heat is supplied by hot oil (cp =2.30 kJ/kg · °C) that enters the shell side at 170°C at a rate of10 kg/s. Disregarding any heat loss from the heat exchanger, determine (a) the exit temperature of oil and (b) the rate ofexergy destruction in the heat exchanger. Take T0 = 25°C.gy g 0

57

Solution:We take the cold water tubes as the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as

N ti th t h t i b th t i l t th h t l b th il th tl tNoting that heat gain by the water is equal to the heat loss by the oil, the outlet temperature of the hot water is determined from

The rate of entropy generation within the heat exchanger is determined by applying the rate form of the entropy balance on the entire heat exchanger:

58

Noting that both fluid streams are liquids (incompressible substances), the rate of entropy generation is determined to be

Alternative solution: use exergy balance

sysdesoutin XXXX•••••

••••

Δ=−−

424321des

des442231

mm,mmcesin)(m)(mX

0Xmmmm

3131

31

•••••••

•••••

==ψ−ψ+ψ−ψ=

=−ψ−ψ−ψ+ψ

59

PvΔTcdPvΔuΔh avg Δ+≅+= ΔTcΔuΔh avg≅= ΔP=0PvΔh Δ= ΔT=0

For liquids:

)Tln(cT)TT(c

)ss(T)hh(

1

21o2121

−−=

−−−=ψ−ψ

9.12)2737027320ln(18.4298)7020(18.4

)T

ln(cT)TT(c2

po21p

−=++

×−−=

=

67.2743 =ψ−ψSimilarly:

kW21967.2710)9.12(5.4)(m)(mX 4321des 31

=×+−×=

ψ−ψ+ψ−ψ=•••

)(

60

Example: A vertical piston–cylinder device initially contains 0.1 m3 of helium at 20°C. The mass of the piston is such that it pmaintains a constant pressure of 300 kPa inside. A valve is now opened, and helium is allowed to escape until the volume inside the cylinder is decreased by one-half. Heat transfer takes placethe cylinder is decreased by one half. Heat transfer takes place between the helium and its surroundings at 20°C and 95 kPa so that the temperature of helium in the cylinder remains constant. Determine (a) the maximum work potential of the helium at theDetermine (a) the maximum work potential of the helium at the initial state and (b) the exergy destroyed during this process. Assume Helium is an ideal gas with constant specific heats.

61

Solution:

The work potential of helium at the initial state is simply the initial exergy of helium and is determined from the closed-system exergy relationhelium, and is determined from the closed system exergy relation,

62

Mass balance:

Solve the problem using EXERGY BALANCE

Energy balance:

since the boundary work and ΔU combine into ΔH for constant pressure expansion and compression processes.

since the initial, final, and the exit states are identical and thus s = s = sare identical and thus se = s2 = s1. Therefore, this discharge process is reversible, and

63

Example: A 0.2-m3 rigid tank initially contains saturated refrigerant-134a vapor at 1 MPa The tank is connected by a valve to a supply134a vapor at 1 MPa. The tank is connected by a valve to a supply line that carries refrigerant-134a at 1.4 MPa and 60°C. The valve is now opened, and the refrigerant is allowed to enter the tank. The valve is closed when one-half of the volume of the tank is filled withvalve is closed when one-half of the volume of the tank is filled with liquid and the rest with vapor at 1.2 MPa. The refrigerant exchanges heat during this process with the surroundings at 25°C. Determine (a) the amount of heat transfer and (b) the exergy destruction(a) the amount of heat transfer and (b) the exergy destruction associated with this process.

64

Solution: The direction of heat transfer is from the tank (will be verified).

The properties of refrigerant are (Tables A-11 through A-13)

Mass balance:

Energy balance:

65

66

Alternative solution: use the extended system for exergy balance

1122desto

ii mmXQ)TT1(m φ−φ=−−−ψ

0.0

1122destii T

918.15)ss(T)hh( oiooii

=

−−−=ψ

)(P)(φ 00082870v

405.86h

324305.0s

C25@o

C25@o

o

o

=

=

92.13)vv(P)uu( o1oo11

=

−+−=φ

)vv(P)uu( o2oo22 −+−=φ

0008287.0v C25@o o =

05.091117

983.5x2 ==

97.119−=

kJ1599mmmX 1122iidest =φ+φ−ψ= 555.123u

001684.0v91.117

MPa2.1@2

MPa2.1@2

=

=

67

Summary• Exergy: Work potential of energy

Exergy (work potential) associated with kinetic and potential energy

• Reversible work and irreversibilityReversible work and irreversibility• Second-law efficiency• Exergy change of a system

Exergy of a fixed mass: Nonflow (or closed system) exergyExergy of a flow stream: Flow (or stream) exergy

• Exergy transfer by heat, work, and massgy y , ,• The decrease of exergy principle and exergy destruction• Exergy balance: Closed systems• Exergy balance: Control volumes

Exergy balance for steady-flow systemsReversible work

68

Second-law efficiency of steady-flow devices