1.1 Exergy : Work Potential of Energy

1

CHAPTER 1

Exergy

1.1 Exergy : work potential of energy

Exergy = maximum useful work that could be obtained from the system at a given state in a specified environment.

Reversible work = maximum useful work that can be obtained as a system undergoes a process between two specified states

Exergy destruction = wasted work potential during a process due to irrevesibilities

Dead state = a system in dead state when it is in thermodynamic equilibrium with its environment. In this state, a system is at the temperature and pressure of the environment. Properties at dead state are denoted by subscript zero (eg P0, T0, h0, u0 and s0). At dead state a system has zero exergy. For example, the atmosphere contains a tremendous amount of energy but no exergy because it is in the dead state.

Surroundings = everything outside the system boundaries

Immediate surroundings = portion of the surroundings that is affected by the process

Environment = region beyond the immediate surroundings whose properties are not affected by the process at any point. Any irresversiblities during a process occur within the system and its immediate surroundings, and the environment is free of any irreversibilities.

Hot potato 70 oC

Immediate surroundings

Environment, 30 oC

30 oC

A system delivers maximum possible work as it undergoes a reversible process from the specified initial state to the state of its environment (dead state). This useful work potential of the system at the specified state is called exergy.

Exergy is not the amount of work a work-producing device actually delivers, but represents the upper limit on the amount of work a device can deliver without violation of thermodynamic laws.

Exergy is a property of system-environment combination, and not of the system alone.

Exergy in an extensive property, with units of energy (J or KJ)

Previously in the 1940s, exergy was known as availability (made popular by MIT), but exergy is used globally since its introduction in Europe in the 1950s.

1.1.1 Exergy associated with kinetic energy

Kinetic energy is a form of mechanical energy. Thus it can be converted to work entirely. The work potential (ie exergy) of the kinetic energy of a sytem is equal to the kinetic energy itself, independent of the temperature and pressure of the environment.

Exergy of kinetic energy : (kJ/kg) ……………………. (6.1)

where V is velocity of system relative to the environment.

1.1.2Exergy associated with potential energy

Potential energy is also a form of mechanical energy, and thus it can be converted to work entirely. The work potential (ie exergy) of the potential energy of a sytem is equal to the potential energy itself, independent of the temperature and pressure of the environment.

Exergy of potential energy : (kJ/kg) ………… ……….. (6.2)

where g = gravitational accelaration (m2s-1)

z = elevation of system relative to a reference level in the environment (m)

Example 1–1

A wind turbine with a rotor diameter of 12 m is installed at a location when the wind is blowing steadily at an average velocity of 10 m/s. Determine the maximum power of the turbine.

The air flow has kinetic energy, and it reaches the dead state when brougt to a complete stop. Exergy of blowing air is,

= = 50 J/kg

Mass flow rate of air through rotor is,

= = 1335 kg/s

Thus,

Maximum power = (ke) = (1335 kg/s)(0.050 kJ/kg) = 66.8 kW

For real wind turbines, irreversibilties mean that actual power output is a lot less this maximum value.

**** Betz’s law states that power output of a wind machine is maximum when the wind is slowed to one-third of its initial velocity. In practice, actual efficiency ranges between 20 and 40 %, with 35 % being the figure for most wind turbines.

Example 1-2

Determine the work potential (exergy) of a hydroelectric turbine shown in the following figure, and the maximum power output when the water mass flow rate through the turbine is 1000 kg/s.

= (9.81 m/s2) (50 m) = 490.5 J/kg

Maximum power = (pe) = (1000 kg/s)(0.4905 kJ/kg) = 490.5 kW

z = 50 m

m

turbine

1.2 Reversible work and irreversibility

Work done by or against the surroundings during a process is called surroundings work, Wsurr. This work which cannot be recovered and utilized for any useful purpose is,

Wsurr = P0 ( V2 – V1 )

The difference between actual work W, and the surroundings work Wsurr is called the usefulwork Wu

Wu = W – Wsurr = W – P0 ( V2 – V1 )

When a system is expanding and doing work, part of the work (Wsurr) is used to overcome the atmospheric pressure and represents a loss. However, when a system is compressed, the atmospheric pressure helps the compression process, thus Wsurr is a gain.

Wsurr has significance only for systems whose volume change during the process (ie. involve moving boundary work). It has no significance for rigid tanks and steady-flow devices (eg turbines, compressors, nozzles, heat exchangers etc.).

Reversible workWrev is the maximum amount of useful work that can be produced (or 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) for a totally reversible process between the initial and final states. When the final state is the dead state, Wrev equals exergy.

Difference between Wrev and Wu is called irreversibility, I

I = Wrev, out – Wu, out or I = Wu, in – Wrev, in

work-producing device work-consuming device

I is equivalent to the exergy destroyed. For totally reversible process, actual and reversible works are identical, thus I = 0 (no entropy generation). I is always positive for actual (irreversible) processes since Wrev Wu for work-producing devices and Wrev Wu for work-consuming devices.

I is wasted work potential or lost opportunity to do work. It respresents energy that could have beeen converted to work but was not. The smaller the irreversibilty, the greater the work that is produced (or the smaller the work that is consumed). Minimizing I improves the performance of a system.

Example 1–3

A heat engine receives heat from a source at 1200 K at a rate of 500 kJ/s and rejects waste heat to a medium at 300 K. The power output is 180 kW. Determine the reversible power and the irreversibility rate for this process.

Wrev = = kW = 375 kW

rev, out – u, out = 375 – 180 = 195 kW

ie 195 kW of power potential is wasted due to irreversibilities.

1.3 Second law efficiency,

For work-producing devices, =

which is applicable to processes (in turbines, piston-cylinder devices etc) as well as to cycles.

For work-consuming non-cyclic (eg compressors) and cyclic (eg refrigerators and heat pumps) devices,

=

and for cyclic devices such as refrigerators and heat pumps, it can also be expressed in terms of Coefficient of Performance COP,

=

Second law efficiency cannot exceed 100 %. Reversible Wrev work is determined by using the same initial and final states as in the actual process.

The definitions above do not apply to devices that do not produce or consume work. For these devices (eg nozzle), we can use the general definition of second-law efficiency of a system during a process

=

where,

exergy supplied = exergy recovered + exergy destroyed

In the worst case (complete destruction of exergy), is zero. In the best case (no exergy destruction), is equal to 1. In general, 0 1.

In a reversible operation, exergy supplied is fully recoverable (irreversibility I = 0). When none of the supplied exergy is recovered, is zero ( I = exergy supplied).

• Second-law efficiency of all reversible devices is 100 %.

Source

1000 K

=70 % 100 %

=70%

Sink

300 K

Second-law efficiency of naturally occuring processes is zero if none of the work potential (exergy) is recovered.

Hot water Atmosphere 25 oC

90 oC

Heat

Exergy supllied and recovered for selected cases.

Exergy supplied / Exergy recovered
Heat engine / (Exergy of heat supplied) –
(Exergy of heat rejected) / Net work output
Heat pump or
refrigerator / Work input / Exergy of heat transferred to the high-temperature medium for heat pump
Exergy of heat transferred from the low-temperature medium for a refrigerator
Heat exchanger (two unmixed fluid streams) / Decrease in exergy of higher-temperature fluid / Increase in exergy of lower-temperature fluid

Example 1–4

An electric resistance heater is used to supply heat to a house maintained at 21 oC, when the outdoor temperature is 10 oC. Determine the second-law efficiency of the heater.

The first-law COP of resistance heater, COPactual = = 1

A reversible heat pump would have a COP of,

(ie if we were to run a reversible heat pump between 10 and 21 C)

COPHP,rev = = 26.7

For the electric heater,

= = 0.037 @ 3.7 %

1.4 Exergy change of a system

Exergy is the maximum amount of useful work that can be obtained as the system is brought to equilibrium with the environment.

1.4.1 Exergy of a flow stream : Flow exergy

The exergy of a flowing fluid is simply the sum of nonflowing fluid exergy and the so-called “flow exergy” (P – P0)V. On a unit mass basis, flow exergy is

= ………. (kJ/kg)

The exergy change of a fluid stream is,

Note : The exergy change of a closed system or a fluid stream represents:

(1)maximum amount of useful work that can be done if it is positive (+ve), or

(2)minimum amount of useful work that needs to be supplied if it is negative (-ve)

as the system changes from state 1 to state 2 in a specified environment. It also represents the reversible work Wrev

Exergy of a closed system cannot be –ve but the exergy of a flow stream can be –ve at pressures below the environment pressure( ie when PP0)

Example 1–5

Refrigerant-134a is compressed from 0.14 MPa and –10 oC to 0.8 MPa and 50 oC steadily by a compressor. The environment is at 20 oC and 95 kPa. Determine (a) the exergy change of the refrigerant and, (b) the minimum work input per unit mass of refrigerant.

Assume: steady operating conditions and negligible kinetic and potential energies.

Obtain properties at inlet (1) and exit (2) states,

h1 = 246.36, s1 = 0.9724 h2 = 286.69, s2 = 0.9802

=

= + 0 + 0

= (286.69 – 246.36) – (293)(0.9802 – 0.9724) = 38.0 kJ/kg

Therefore the exergy of R134a increases during compression by 38 kJ/kg. This is reversible work, which is the minimum work input for work-consuming devices. Thus

wmin = = 38 kJ/kg

1.5 Exergy transfer by heat, work, and mass

1. Exergy, like energy, can be transferred to or from a system in three forms: heat, work and mass flow.
2. Exergy transfer occurs at the system boundary when exergy crosses it.
3. Exergy transfer into a system is exergy gain, and exergy transfer from the system is exergy lost.
4. Exergy transfer for fixed mass (closed system) involves only heat transfer and work transfer.

1.5.1 Exergy transfer by heat

Heat transfer Q at a location at thermodynamic temperature T is accompanied by exergy transfer Xheat in the amount of

……. (kJ)

When TT0 – heat transfer to a system increases exergy of system, and heat transfer from a system decreases exergy of system.

When TT0 – heat transfer to a system decreases exergy of system, and heat transfer from a system increases exergy of system.

Q is not heat transfer between system and the environment due difference beween T and T0 !

Heat transfer Q at a location at temperature T is always accompanied by:

(1) entropy transfer in the amount of

(2) exergy transfer in the amount of

1.5.2 Exergy transfer by work, W

Xwork = W – Wsurr …….. for boundary work

= W ……. for other forms of work

where Wsurr = P0 ( V2 – V1 )

P0 = atmospheric pressure

V1, V2 = initial and final volumes of system, respectively

Notes:

1. Exergy transfer with shaft work = shaft work, Wshaft
2. Exergy transfer with electrical work = electrical work, Welectrical
3. Piston-cylinder device; expansion process: W =

Xwork = W – Wsurr

W actual is positive, Wsurr is positive, thus Xwork is positive (out of system)

4. Piston-cylinder device; compression process: W =

Xwork = W – Wsurr

W actual is negative, Wsurr is negative, thus Xwork is negative (into system)

Magnitude of Xwork < magnitude of W

1.5.3 Exergy transfer by mass, m

Mass flow transports exergy, entropy and energy into or out of a system. Exergy transfer by mass is

Exergy of a system increases by when m kg of mass enters, and decreases by the same amount when m kg of mass leaves the system.

NOTES:

1. For adiabatic systems, Xheat = 0
2. When no mass crosses the boundary (closed systems), Xmass = 0
3. The total exergy transfer is zero for isolated systems since no heat, work or mass crosses the boundary.

1.6 The increase of exergy principle and exergy destruction

One of the statements of the scond law – the increase of entropy principle states that entropy generation Sgen must be positive (actual processes) or zero (reversible processes), but Sgen cannot be negative.

An alternative statement of the second law is called – the decrease of exergy principle, which is a counterpart of the increase of entropy principle,

It states that: “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 decreases and exergy is destroyed during an actual process. For an isolated system,

decrease in exergy = exegy destroyed

Irreversibilities such as friction, mixing, chemical reactions, heat transfer through a finite temperature difference, unrestrained expansion, nonquasiequilibrium compression or expansion always generate entropy, and entropy generation leads to exergy destruction. The exergy destroyed is,

It is a positive quantity for any actual process and equal to zero for a reversible process. Exergy destroyed is lost work potential and is also called the irreversibilty or lost work.

The above two equations are applicable to any kind of system undergoing any kind ofprocess, since any system and its surroundings can be enclosed by a sufficiently large arbitrary boundary across which no heat, work and mass transfer occur, thus the system and its surroundings constitute an isolated system.

The decrease of exergy principle is summarized as follows,

Xdestroyed > 0 ………… irreversible process

Xdestroyed = 0 ………… reversible process

Xdestroyed < 0 ……….… impossible process

which can be used to determine whether a process is reversible, irreversible or impossible.

1.7 Exergy balance : control volumes

Explicitly, the exergy balance for a control volume can be written as

Xheat – Xwork + Xmass,in – Xmass,out - Xdestroyed = ( X2 – X1 )CV

taking Qin as positive, and Wout as positive direction of heat and work transfer, respectively.

Expanding,

= ( X2 – X1 )CV

or in rate form,

=

1.7.1 Exergy balance for steady-flow systems

Most control volumes in practice such as turbines, compressors, nozzles,diffusers, heat exchangers, pipes, and ducts operate steadily. They experience no changes in mass, energy, entropy and exergy contents as well their volumes. Therefore dVcv/dt =0 and dXcv/dt =0. In rate form the general exergy balance for a steady-flow process reduces to,

= 0

For single-stream (one-inlet and one-outlet) steady-flow device, the relation above reduces to

= 0 ……………****

where subscripts 1 and 2 are inlet and exit states, and change of flow exergy is,

=

Dividing eqn **** by mass flow rate gives exergy balance on a unit-mass basis

= o ………… kJ/kg

Notes: For adiabatic single-stream device with no work interactions, , which indidates that the specific exergy of the fluid must decrease as it flows through a work-free adiabatic device or remain the same in the limiting case of a reversible process.

1.7.2 Reversible work, Wrev

The preceding exergy balance relations can be used to determine the reversible work by setting the exergy destroyed equal to zero. The work W in that case becomes the reversible work.

…….... General case W = Wrev when Xdestroyed = 0

For a single-stream steady-flow device, the reversible power is

…………….. Single stream: ………...(kW)

which reduces for an adibatic device to

..…. Adiabatic, single stream :

Reversible work is maximum work output for work-producing devices (eg turbines) and it represents the minimum work input for work-consuming devices (eg compressors).

1.7.3 Second-law efficiency of steady-flow devices

The second-law efficiency of various steady-flow devices can be determined from its general definition,

Neglecting changes in kinetic and potential energies, for an adiabatic turbine,

where sgen = s2 – s1

For an adiabatic compressor with negligible kinetic and potential energies,

where sgen = s2 – s1

For an adiabatic heat exchanger with two unmixed fluid streams, the exergy supplied is the decrease in exergy of the hot stream, and the exergy recovered is the increase in the exergy of the cold stream, provided that the cold stream is at a higher temperature than the surroundings. For the heat exchanger,

where,

T0

Hot stream 1 2

4 3 Cold stream , T3T0

For an adiabatic mixing chamber where hot stream 1 is mixed with cold stream 2, forming a mixture 3, the exergy supplied is the sum of exergies of the hot and cold streams, and the exergy recovered is the exergy of the mixture. The second-law efficiency of the mixingchamber becomes

where and

Example 1–6

Steam enters a turbine steadily at 3 MPa and 450 oC at a rate of 8 kg/s and exits at 0.2 MPa and 150 oC. The steam is losing heat to the surrounding air at 100 kPa and 25 oC at a rate of 300 kW, and kinetic and potential energies are negligible. Determine (a) the actual power output, (b) the maximum possible power output, (c) the second-law efficiency, (d) the exergy destroyed, and (e) the exergy of the steam at the inlet conditions.

3 MPa 1 300 kW

450 C

steam turbine

W

T0 = 25 C

P0 = 100 kPa

0.2 MPa

2 150 C

Taking the turbine as the system, which is a control volume.

Inlet, h1 = 3344.9, s1 = 7.0856

Exit, h2 = 2769.1 , s2 = 7.2810

Dead state (environment), h0hf,25C = 104.83 , s0sf,25C = 0.3672 (Table A-4)

(a) Actual power is determined from steady-flow energy equation,

ie = – 300 + 8 (3344.9 – 2769.1) = 4306 kW

(b) The maximum power output (reversible power) is determined from the rate form exergy balance applied on the extended system (system + immediate surroundings), whose boudary is at the environment temperature T0, and setting exergy destruction to zero.

. 0 0

= 0

Rate of net exergy transfer

by heat, work and mass

ie 0

0 0

=

= 8((3344.9 – 2769.1) –298(7.0856 – 7.2810)) = 4665 kW

(c) The second-law efficiency, = = 0.923 @ 92.3 %

(d) = 4665 – 4306 = 359 kW =

0 0

(e)

= (3344.9 – 104.83) – 298(7.0856 – 0.3672) = 1238 kJ/kg

Example 1–7

Water at 200 kPa and 10 C enters a mixing chamber at a rate of 150 kg/min, where it is mixed adiabatically with steam entering at 200 kPa and 150 C. The mixture leaves the chamber at 200 kPa and 70 C, and heat is being lost to the surrounding air at T0 = 20 C at a rate of 190 kJ/min. Neglecting changes in kinetic and potential energies, determine the reversible power and the rate of exergy destruction for this process.

190 kJ/min