EXERGY

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EXERGY

• Basic Definitions
• Exergy is property used to determine the useful
  work potential of a given amount of energy at
  some specified state.
• It does not represent the amount of work that a
  work-producing device will actually deliver upon
  installation. Rather, it represents the upper
  limit on the amount of work a device can deliver
  without violating any thermodynamic laws.
• A system delivers the maximum possible work as it
  undergoes a reversible process from the specified
  initial state to the state of its environment
  (dead state)
• Dead State a system is said to be in dead state
  when it is in thermodynamic equilibrium with its
  environment. Also it has no potential or kinetic
  energy. And it is chemically inert (no reaction
  with the environment)

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• Exergy of potential energy (work potential) of a
  system is equal to the potential energy itself
  regardless of the temperature and pressure of the
  environment.
• Exergy of kinetic energy (work potential) of a
  system is equal to the kinetic energy itself
  regardless of the temperature and pressure of the
  environment. (not necessarily true)
• REVERSIBLE WORK AND IRREVERSIBILITY
• At difference to exergy, actual processes do not
  occur from an initial point to a final point
  equal to the dead state. On the other hand
  isentropic efficiencies are limited to adibatic
  processes.

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• Surrounding Work is the work done by or against
  the surroundings during a process. It has
  significance only for a process where boundary
  work occurs (closed system).
• Wsurr P0(V2 V1)
• Then the useful work would be
• Wu W Wsurr W P0(V2 V1)
• How is Wsurr for a rigid tank?

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• Reversible Work 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.
• Whats is the difference between exergy and
  reversible work?
• Any difference between reversible and useful work
  is due to irreversibilities.
• I Wrev,out Wu,out I Wu,in Wrev,in

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SECOND LAW EFFICIENCY
In cases like this the first-law efficiency alone
is not a realistic measure of performance of
engineering devices. The second-law efficiency is
defined as the ratio of the thermal efficiency to
the maximum possible thermal efficiency under the
same conditions Then, in the example
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In other words
Work producing devices
Work consuming devices
Refrigerators and Heat Pumps
In general
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Closed System Total useful work delivered in a
reversible process to the dead state
The total exergy for a closed process would be
given by
(kJ)
The total exergy per unit mass
(kJ/kg)
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EXERGY CHANGE FOR A CLOSED SYSTEM
Per unit mass
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Exergy of a flow stream
Exergy change of a flow stream
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Decrease or Exergy Principle (Exergy Destruction)
W
Q
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Exergy decreases
Exergy destruction
Exergy Balance
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Exergy transfer by heat, work and mass
By Heat
If boundary work
By Work
No boundary work
By mass
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Closed system (no mass flowing)
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Sol.
From table A6 A4, for water
u2594.7 kJ/kg u0 104.83
kJ/kh _at_ 180C v0.2472 m3/kg _at_ 25C v0
0.001003 m3/kg 800 kPa s6.7155 kJ/kg.K
100 kPa s0 0.3672 kJ/kg.K
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From table A13 (R-134a superheated vapor)
u386.99 kJ/kg u0 252.615 kJ/kh _at_
180C v0.044554 m3/kg _at_ 25C v0
0.23803 m3/kg 800 kPa s1.3327 kJ/kg.K
100 kPa s0 1.10605 kJ/kg.K
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Sol.
For a reversible process, therefore WrevX2-X1
but
and
Cv 0.164 Btu/lbm.R R 0.0621
Btu/lbm.R
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Solution
First the process is at constant volume until the
pressure is enough to move the piston (1-1),
then the process is at constant pressure (1-2)
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• States 1 and 2 are in the region of superheated
  vapor, the summarized data from table is
• P1140kPa v10.1652 m3/kg P2180kPa
  v20.17563 m3/kg
• u1246.01 kJ/kg u2331.96
  kJ/kg
• T120C s11.0532 kJ/kgK T2120C
  s21.3118 kJ/kgK
• Also at (1) v1 v1 and P1P2
• The work done is the boundary work, from (1) to
  (1) is zero since it is a constant volume
  process from (1) to (2) is a constant pressure
  process. Then the boundary work is given by
• Wb P2.m.(v2-v1) (180)(1.4)(0.17563 0.1652)
  2.63 kJ
• Doing a energy balance we obtain
• Q-W m(u1 u1) m(u2 u1) m(u2 u1)
• Then Q m(u2 u1) W (1.4)(331.96 246.01)
  2.63 kJ
• Q 122.96 kJ

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Since there is no kinetic nor potential energy
involved the exergy change can be expressed
by The useful work at the exit is given by
the boundary work minus the work against the
environment Wu Wb m.P0(v2-v1) 2.63kJ
(1.4)(100)(0.175630.1652) Wu 1.17 kJ From the
total exergy change the only amount of useful
work is 1.17kJ everything else is the exergy
destroyed, therefore d) The second law
efficiency is given by
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Exergy Balance Open System
Notice that now we are including the exergy
entering and leaving with mass, then
(kJ)
In rate form
(kW)
Fortunately we usually have to deal with steady
flow devices, then our equation becomes
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for a single stream this last equation becomes
(kW)
Per unit mass
(kJ/kg)
Previous equations can be used to determine the
reversible work by making the exergy destruction
term equal to zero since (i.e. no
irreversibilities implies no exergy destruction)
Then
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or
Single stream (one inlet - one outlet)
Example
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Solution We need to determine the exergy
destroyed during this process. In this case the
easiest way is by This equation leads us to
find the entropy generated which is given by
doing an entropy balance From table at 200
psia State (1) sat. liquid h1355.46 Btu/lbm
s10.54379Btu/lbm.R State (2) sat. vapor
h21198.8 Btu/lbm s21.5460 Btu/lbm.R At
environment conditions (P014.7 psia, To80F
comp. liq.) h048.07 Btu/lbm s00.09328
Btu/lbm.R
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We need to determine q in the previous equation,
we obtain this by doing an energy balance q w
(h2 h1) since there is no work we have q
1198.8 355.46 843.34 Btu/lbm Now we can
determine the entropy generated 960R is the
absolute gas temperature (500F) And the exergy
destroyed will be The exergy 9or work
potential) of the steam is given by Therefore
the temperature of the gases does not affect the
exergy of the steam. However it does affect sgen
and therefore xdestroyed too.
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• We find the actual work by doing an ener4gy
  balance
• Kinetic and potential energy changes are assumed
  to be zero.
• Consider specific heats for the enthalpy change.
  Solving for work we have
• Cp1.134 kJ/kg.K (from table)
• Q -30kW (lost)

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Substitue in the previous equation to
obtain We have an expression for the
reversible work from the exergy balance for a
single stream The ideal situation for a
turbine occurs when there are no heat losses,
therefore The entropy change is obtained
from Then the rev. work is
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The exergy destroyed will be given by Finally,
the second law efficiency would be