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EXERGYEXERGYBasic DefinitionsBasic Definitions
ExergyExergy: : is property used to determine the useful work is property used to determine the useful work potential of a given amount of energy at some specified state.potential of a given amount of energy at some specified state.
It does not represent the amount of work that a work-producing It does not represent the amount of work that a work-producing device will actually deliver upon installation. Rather, it represents device will actually deliver upon installation. Rather, it represents the upper limit on the amount of work a device can deliver the upper limit on the amount of work a device can deliver without violating any thermodynamic laws.without violating any thermodynamic laws.A system delivers the maximum possible work as it undergoes a A system delivers the maximum possible work as it undergoes a reversible process from the specified initial state to the state of reversible process from the specified initial state to the state of its environment (dead state)its environment (dead state)
Dead StateDead State: a system is said to be in dead state when it is in : a system is said to be in dead state when it is in thermodynamic equilibrium with its environment. Also it has no thermodynamic equilibrium with its environment. Also it has no potential or kinetic energy. And it is chemically inert (no reaction potential or kinetic energy. And it is chemically inert (no reaction with the environment)with the environment)
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Exergy of potential energyExergy of potential energy: (work potential) of a : (work potential) of a system is equal to the potential energy itself system is equal to the potential energy itself regardless of the temperature and pressure of the regardless of the temperature and pressure of the environment.environment.
Exergy of kinetic energyExergy of kinetic energy: (work potential) of a : (work potential) of a system is equal to the kinetic energy itself system is equal to the kinetic energy itself regardless of the temperature and pressure of the regardless of the temperature and pressure of the environment. (not necessarily true)environment. (not necessarily true)
REVERSIBLE WORK AND IRREVERSIBILITY:REVERSIBLE WORK AND IRREVERSIBILITY:At difference to exergy, actual processes do not occur At difference to exergy, actual processes do not occur
from an initial point to a final point equal to the from an initial point to a final point equal to the dead state. On the other hand isentropic dead state. On the other hand isentropic efficiencies are limited to adibatic processes.efficiencies are limited to adibatic processes.
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Surrounding Work:Surrounding Work: is the work done by or against the is the work done by or against the surroundings during a process. It has significance only for a surroundings during a process. It has significance only for a process where boundary work occurs (closed system).process where boundary work occurs (closed system).
WWsurrsurr = P = P00(V(V22 V V11))
Then the useful work would be:Then the useful work would be:
WWuu = W W = W Wsurrsurr = W P = W P00(V(V22 V V11))
How is WHow is Wsurrsurr for a rigid tank? for a rigid tank?
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Reversible WorkReversible Work: maximum amount of useful work : maximum amount of useful work that can be produced (or the minimum work that that can be produced (or the minimum work that needs to be supplied) as a system undergoes a needs to be supplied) as a system undergoes a process between the specified initial and final process between the specified initial and final states.states.
Whats is the difference between exergy and Whats is the difference between exergy and reversible work?reversible work?
Any difference between reversible and useful work is Any difference between reversible and useful work is due to irreversibilities.due to irreversibilities.
I = WI = Wrev,outrev,out W Wu,outu,out I = W I = Wu,inu,in W Wrev,inrev,in
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SECOND LAW EFFICIENCYSECOND LAW EFFICIENCY
rev
thII
60.050.030.0
, AII
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:
43.070.030.0
, BII
%5060030011,
KK
TT
AH
LArev
%70100030011,
KK
TT
BH
LBrev
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In other words:
v
uII W
W
Re
Work producing devices
u
vII W
WRe
vII COP
COP
Re
Work consuming devices
Refrigerators and Heat Pumps
In general
SuppliedExergyDestroyedExergy
SuppliedExergyeredExergy
II 1covRe
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00000 SSTVVPUUWuseful
Closed SystemTotal useful work delivered in a reversible process to the dead state:
The total exergy for a closed process would be given by:
mgzVmSSTVVPUUX 2
2
00000
The total exergy per unit mass:
gzVssTvvPuu 2
2
00000
(kJ)
(kJ/kg)
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EXERGY CHANGE FOR A CLOSED SYSTEM
)(2
)( 122
122
1201201212 zzmgVVmSSTVVPUUmX
1201201212 )( SSTVVPEEmX
)(2
)( 122
122
1201201212 zzgVVssTvvPuu
1201201212 )( ssTvvPee
Per unit mass:
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Exergy of a flow stream
flowflowingnoflowing xxx _
vPPvPPvx flow )( 00
vPPgzVssTvvPuu )(2 02
00000
gzVssThh 2
2
000
Exergy change of a flow stream
)(2
) 122
122
1201212 zzgVVssThh
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Isolated system
System (closed or
open)
Q
W
Decrease or Exergy Principle (Exergy Destruction)
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00 XST gen
Exergy destructiongendestroyed STX 0
Exergy decreases
Exergy Balance
systemtheof
exergytotaltheinChange
destroyedexergyTotal
leavingexergyTotal
enteringexergyTotal
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Exergy transfer by heat, work and mass
By Heat: QTT
X heat
01
By Work:
By mass:
W
WWX
sur
work
No boundary work
If boundary work
mX mass
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XXXX destroyedworkheat Closed system (no mass flowing)
XSTVVPWQTT
genkk
01200 )(1
dtdXST
dtdVPWQ
TT
genkk
0001
-
mgzVmSSTVVPUUX 2
2
00000
Sol.
00000 ssmTvvmPuumX
From table A6 & A4, for water:
u=2594.7 kJ/kg u0 = 104.83 kJ/kh
@ 180C v=0.2472 m3/kg @ 25C v0 = 0.001003 m3/kg
800 kPa s=6.7155 kJ/kg.K 100 kPa s0 = 0.3672 kJ/kg.K
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From table A13 (R-134a superheated vapor):
u=386.99 kJ/kg u0 = 252.615 kJ/kh
@ 180C v=0.044554 m3/kg @ 25C v0 = 0.23803 m3/kg
800 kPa s=1.3327 kJ/kg.K 100 kPa s0 = 1.10605 kJ/kg.K
Kkg
kJkgK
kgmkgkPa
kgkJkgX w
.3672.01392.21.298
001003.0001127.01.10083.10492.76113
kJX w 7.622
Kkg
kJkgK
kgmkgkPa
kgkJkgX R
.10605.13327.11.298
23803.0044554.01.100615.25299.38613
kJX R 5.47
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XXXX destroyedoutin
0destroyedX
Sol.
For a reversible process, therefore Wrev=X2-X1
1201201212 ssTvvPuu 1212 TTCvuu
1
2
1
212 lnln v
vRTTCvss
but
and
Cv = 0.164 Btu/lbm.R R = 0.0621 Btu/lbm.R
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3
3
12
.4039.5125.17.14
125.1ln
.0621.0
535985ln
.164.0535
535985.
164.0
ftpsiBtu
lbmftpsia
RlbmBtu
RlbmBtuR
RRlbm
Btu
lbmBtu77.6012
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Solution:
120C
100
P (kPa)
v (m3/kg)
20 C
1
180 21
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:
P1=140kPa v1=0.1652 m3/kg P2=180kPa v2=0.17563 m3/kg
u1=246.01 kJ/kg u2=331.96 kJ/kg
T1=20C s1=1.0532 kJ/kgK T2=120C s2=1.3118 kJ/kgK
Also at (1) v1 = v1 and P1=P2
a) 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
b) 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:
12012012 ssmTvvmPuumX )0532.13118.1)298)(4.1(1652.017563.0)100)(4.1(01.24696.3314.1 X
kJX 90.13
kJkJkJWXX udestroyed 73.1217.190.13
084.09.13
17.1
XWu
II
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XXXXX destroyedmassworkheat
Exergy Balance Open System
Notice that now we are including the exergy entering and leaving with mass, then:
(kJ)
In rate form:
Fortunately we usually have to deal with steady flow devices, then our equation becomes:
(kW)
1212001 XXXmmVVPWQTT
destroyedoutin
kk
dtdXXmm
dtdVPWQ
TT
destroyedoutin
kk
001
01 0
destroyedoutin
kk
XmmWQTT
-
01 120
destroyedkk
XmWQTT
01 0
outinrevk
k
mmWQTT
0 destroyedrev XW
01 120
destroyedk
k
xwqTT (kJ/kg)
Per unit mass:
(kW)
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)
for a single stream this last equation becomes:
Then:
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or: 01 0
k
kinoutrev Q
TTmmW
Single stream (one inlet - one outlet):
01 012
k
krev Q
TTmW
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: h1=355.46 Btu/lbm s1=0.54379Btu/lbm.R
State (2) sat. vapor: h2=1198.8 Btu/lbm s2=1.5460 Btu/lbm.R
At environment conditions (P0=14.7 psia, To=80F comp. liq.):
h0=48.07 Btu/lbm s0=0.09328 Btu/lbm.R
gendestroyed sTx 0
Tqsssss
Tq
gengen )( 12
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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.
RlbmBtu
RRlbmBtusgen .
124.0960
./34.843)54379.05460.1(
lbmBtu
RlbmBtuRsTx gendestroyed 96.66.
124.0)540(0
12012 ssThh
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a) 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:
Cp=1.134 kJ/kg.K (from table)
Q = -30kW (lost)
PEKEhhmWQ a 12
12 TTCpmQW a
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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:
kWKkgKkJ
skgkWWa 67.4321023903134.14.330
01 012
k
krev Q
TTmW
][ 1201212 ssThhmmW rev
Kkg
kJKkg
kJKkg
kJPPR
TTCpss
.11.0
1200500ln
.287.0
1023903ln
.134.1ln
12ln
1
212
kJKkg
kJKKKkg
kJskgW rev 574]
.11.0[298]1023903[
.134.14.3
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The exergy destroyed will be given by:
Finally, the second law efficiency would be:
kJWWX arevdestroyed 45.14167.43212.574
754.012.57467.432
rev
aII
W
W
EXERGYSlide 2Slide 3Slide 4Slide 5Slide 6SECOND LAW EFFICIENCYSlide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19Slide 20Slide 21Slide 22Slide 23Slide 24Slide 25Slide 26Slide 27Slide 28Slide 29Slide 30