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)
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.
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?
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
SECOND LAW EFFICIENCY
 rev, A
 TL
 1 
 TH

300 K
  1 
 50%
600
K
A

T 

H
300 K
 rev, B  1  L   1 
 70%
T
1000 K
B
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
th
 II 
 rev
Then, in the example:
 II , A 
0.30
 0.60
0.50
 II , B 
0.30
 0.43
0.70
In other words:
Wu
 II 
WRe v
Work producing devices
WRe v
 II 
Wu
Work consuming devices
 II 
COP
COPRe v
Refrigerators and Heat Pumps
In general
 II 
Exergy Re cov ered
Exergy Destroyed
 1
Exergy Supplied
Exergy Supplied
Closed System
Total useful work delivered in a reversible process to the dead state:
Wuseful  U  U 0   P0 V  V0   T0 S  S 0 
The total exergy for a closed process would be given by:
V2
X  U  U 0   P0 V  V0   T0 S  S0   m
 mgz
2
(kJ)
The total exergy per unit mass:
V2
  u  u 0   P0 v  v0   T0 s  s0  
 gz
2
(kJ/kg)
EXERGY CHANGE FOR A CLOSED SYSTEM
V22  V12
X  m( 2  1 )  U 2  U 1   P0 V2  V1   T0 S 2  S1   m
 mg ( z 2  z1 )
2
X  m(2  1 )  E2  E1   P0 V2  V1   T0 S2  S1 
Per unit mass:
V22  V12
  (2  1 )  u2  u1   P0 v2  v1   T0 s2  s1  
 g ( z2  z1 )
2
  (2  1 )  e2  e1   P0 v2  v1   T0 s2  s1 
Exergy of a flow stream
x flow  Pv  P0v  ( P  P0 )v
x flowing    xno _ flowing  x flow
V2
 u  u0   P0 v  v0   T0 s  s0  
 gz  ( P  P0 )v
2
V2
  h  h0   T0 s  s0  
 gz
2
Exergy change of a flow stream
V22  V12
   2  1 )  h2  h1   T0 s2  s1  
 g ( z2  z1 )
2
Decrease or Exergy Principle
(Exergy Destruction)
W
Q
Isolated
system
System
(closed or
open)
 T0 S gen  X  0
Exergy decreases
X destroyed  T0 S gen
Exergy destruction
Exergy Balance
 Total   Total   Total   Change in the 

 
 
 

 exergy    exergy    exergy    total exergy 
 entering   leaving   destroyed   of the system 

 
 
 

Exergy transfer by heat, work and mass
By Heat:
By Work:
By mass:
 T 
X heat  1  0 Q
T 

X work
W  Wsur


 W

X mass  m
If boundary work
No boundary work
X heat  X work  X destroyed  X
 T0
 1  T
k

Closed system (no mass flowing)

Qk  W  P0 (V2  V1 )  T0 S gen  X



 T0  
dV 
dX



1

Q

W

P

T
S

gen
  T  k  0 dt  0
dt
k 

Sol.
V2
X  U  U 0   P0 V  V0   T0 S  S0   m
 mgz
2
X  mu  u0   P0 mv  v0   T0 ms  s0 
From table A6 & A4, for water:
u=2594.7 kJ/kg
@ 180°C
800 kPa
v=0.2472 m3/kg
s=6.7155 kJ/kg.K
u0 = 104.83 kJ/kh
@ 25°C
100 kPa
v0 = 0.001003 m3/kg
s0 = 0.3672 kJ/kg.K
From table A13 (R-134a superheated vapor):
u=386.99 kJ/kg
@ 180°C
800 kPa
v=0.044554 m3/kg
s=1.3327 kJ/kg.K
u0 = 252.615 kJ/kh
@ 25°C
100 kPa
v0 = 0.23803 m3/kg
s0 = 1.10605 kJ/kg.K
kJ
m3
X w  1kg761.92  104.83  100kPa.1kg0.001127  0.001003
kg
kg
kJ
 298K .1kg2.1392  0.3672
kg.K
X w  622.7kJ
kJ
m3
X R  1kg386.99  252.615  100kPa.1kg0.044554  0.23803
kg
kg
kJ
 298K .1kg1.3327  1.10605
kg.K
X R  47.5kJ
Sol.
X in  X out  X destroyed  X
X destroyed  0
For a reversible process, therefore Wrev=X2-X1
2  1  u2  u1   P0 v2  v1   T0 s2  s1 
u2  u1   CvT2  T1 
Cv = 0.164 Btu/lbm.R
and
but




s2  s1   Cv  ln  T2   R  ln  v2 
 T1 
R = 0.0621 Btu/lbm.R
 v1 
Btu
985  535R 
2  1  0.164
lbm.R

Btu  985 
Btu
1.5  
535R 0.164
ln 

0
.
0621
ln




lbm.R  535 
lbm.R  12  


ft 3 
Btu


 14.7 psia 1.5  12
3 
lbm  5.4039 psi. ft 
2  1  60.77
Btu
lbm
Solution:
P
(kPa)
1’
180
100
2
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)
1
v (m3/kg)
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
u1=246.01 kJ/kg
T1=20°C
s1=1.0532 kJ/kgK
v2=0.17563 m3/kg
u2=331.96 kJ/kg
T2=120°C
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
Since there is no kinetic nor potential energy involved the exergy
change can be expressed by:
X  mu2  u1   mP0 v2  v1   mT0 s2  s1 
X  1.4331.96  246.01  (1.4)(100)0.17563  0.1652  (1.4)(298)1.3118 1.0532)
X  13.90kJ
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)(0.17563–0.1652)
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:
X destroyed  X  Wu  13.90kJ  1.17kJ  12.73kJ
d) The second law efficiency is given by:
 II 
Wu 1.17

 0.084
X 13.9
Exergy Balance Open System
X heat X work  X mass  X destroyed  X
Notice that now we are including the exergy entering and
leaving with mass, then:

T0 
Qk  W  P0 V2  V1    m   m  X destroyed  X 2  X 1
in
out
k 
 1  T

(kJ)
In rate form:




 T0  
dV 
dX



1

Q

W

P

m


m


X

destroyed
  T  k  0 dt  

dt
in
out
k 

(kW)
Fortunately we usually have to deal with steady flow devices,
then our equation becomes:




 T0  
m   m  X destroyed  0
 1  T  Q k  W  
in
out
k 

for a single stream this last equation becomes:



 T0  
 1  T  Q k  W  m 2  1   X destroyed  0
k 

(kW)
Per unit mass:
 T0 
 1  T qk  w   2  1   xdestroyed  0
k 

(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)
Wrev

X destroyed  0
Then:



 T0  
m   m  0
 1  T  Q k W rev 
in
out
k 

or:

 T0  
  m   m   1   Q k  0
out
in
 Tk 

W rev

Single stream (one inlet - one outlet):

W rev
Example:
 T0  
 m 2  1    1   Q k  0
 Tk 

Solution:
We need to determine the exergy destroyed during this process. In
this case the easiest way is by:
xdestroyed  T0 s gen
This equation leads us to find the entropy generated which is given
by doing an entropy balance:
q
 s gen  s
T

s gen
q
 ( s2  s1 ) 
T
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=80°F comp. liq.):
h0=48.07 Btu/lbm
s0=0.09328 Btu/lbm.R
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:
s gen  (1.5460  0.54379) 
843.34 Btu / lbm.R
Btu
 0.124
960 R
lbm.R
960R is the absolute gas temperature (500°F)
And the exergy destroyed will be:
Btu 
Btu

xdestroyed  T0 s gen  (540 R) 0.124
  66.96
lbm.R 
lbm

The exergy 9or work potential) of the steam is given by:
  h2  h1   T0 s2  s1 
Therefore the temperature of the gases does not affect the exergy of the
steam. However it does affect sgen and therefore xdestroyed too.
a) We find the actual work by doing an ener4gy balance:



Q W a  mh2  h1   KE  PE
Kinetic and potential energy changes are assumed to be zero.
Consider specific heats for the enthalpy change. Solving for work we
have:



W a  Q m CpT2  T1 
Cp=1.134 kJ/kg.K (from table)
Q = -30kW (lost)
Substitue in the previous equation to obtain:
Wa  30kW  3.4
kg
kJ
903  1023K  432.67kW
1.134
s
kgK
We have an expression for the reversible work from the exergy
balance for a single stream:

W rev
 T0  
 m 2  1    1   Q k  0
 Tk 

The ideal situation for a turbine occurs when there are no heat losses,
therefore:



W rev  m 2  1   mh2  h1  T0 [ s2  s1 ]
The entropy change is obtained from:
s2  s1   Cp ln T 2  R ln P2
T1
P1
 1.134
kJ
903
kJ
500
kJ
ln
 0.287
ln
 0.11
kg.K 1023
kg.K 1200
kg.K
Then the rev. work is:

W rev  3.4
kg 
kJ
kJ 
1.134
[903  1023]K  298K [0.11
]   574kJ
s 
kg.K
kg.K 
The exergy destroyed will be given by:



X destroyed  W rev  W a  574.12  432.67  141.45kJ
Finally, the second law efficiency would be:

 II 
Wa

W rev

432.67
 0.754
574.12