Exergy Analysis
ME 210 Advanced
Thermodynamics
Definitions




Exergy (also called Availability or Work Potential): the
maximum useful work that can be obtained from a
system at a given state in a given environment; in other
words, the most work you can get out of a system
Surroundings: outside the system boundaries
Environment: the area of the surroundings not affected
by the process at any point (For example, if you have a
hot turbine, the air next to the turbine is warm. The
environment is the area of the surroundings far enough
away that the temperature isn’t affected.)
Dead State: when a system is in thermodynamic
equilibrium with the environment, denoted by a subscript
zero; at this point no more work can be done
Example

A coal-fired furnace is used in a power plant. It delivers
5000 kW at 1000 K. The environment is at 300 K. What
is the exergy of the added heat? You can use two steps
to solve this problem.



Determine the maximum percentage of the heat that can be
converted to work.
Using your answer from the first part, determine the maximum
work possible.
This is the maximum work output possible between the
given state and the dead state, i.e., the heat’s exergy. In
this case, 30% of the 5000 kW is unavailable energy—it
can’t be converted to work.
Why Study Exergy?

In the last several decades, exergy analysis has
begun to be used for system optimization.
 By
analyzing the exergy destroyed by each
component in a process, we can see where we
should be focusing our efforts to improve system
efficiency.
 It can also be used to compare components or
systems to help make informed design decisions.
Reversible Work

Wrev (reversible work): the maximum amount of
work it’s possible to produce (or minimum
necessary to input) in a process between given
initial and final states. Note that this is different
from an isentropic process where we were given
an inlet state and solved for the exit state using
s2=s1. Since the exit and inlet states are both
fixed, the process is not necessarily isentropic.
 What
two conditions will cause a process to be
isentropic?
Irreversibilities
Irreversibility, I: exergy destroyed; wasted
work potential. It represents energy that
could have been converted into work but
was instead wasted
 What are some sources of I?
 To have high system efficiency, we want I
to be as small as possible.

I, cont.




I=Wrev, out –Wu, out
(work output device, like a
turbine) OR
I=Wu, in –Wrev, in (work input device, like a
pump)
Wu: useful work; the amount of work done that
can actually be used for something desirable
Wu=W-Wsurr where W=actual work done
Surroundings Work, Wsurr
Here some work is
used to push the
atmospheric air (the
surroundings) out of
the way; that work
can’t be used for
other purposes.
Wsurr  P0 dV  P0 V2  V1  positive

Surroundings Work, Wsurr, cont.

Here Patm helps push the
piston in; this is gained
work. In a process where
the piston goes in and
out continually, the
surrounding work values
cancel out.
Wsurr  P0 V2  V1  negative

What is Wsurr for a control
volume?
Second Law Efficiency, hII
Thermal efficiency tells us what we get out
compared to what we put in.
 The second law efficiency tells us how
much we get out compared to the
maximum possible we could get out, given
the inlet and exit conditions.

Second Law Efficiency, cont.

hth,max=1-TL/TH=1-300/800=0.635
Say hth=0.45
 hII=0.45/0.625=0.72
 We want a high hth and hII
 Another way to look at this:
for a work output device
hII=Wu/Wrev

Second Law Efficiency, cont.

A general definition:
exergy recovered (what' s available after the process)
h II 
exergy supplied (what' s available at the beginning)
exergy destroyed(I)
 1
exergy supplied
Three Efficiency Definitions

The second two are defined for work
OUTPUT devices
Thermal h th
Wnet

Qin
Wactual
Isentropich s 
Wisentropic
2
nd
Wu
Law h II 
Wrev
Example

A freezer is maintained at 20°F by
removing heat from it at a rate of 75
Btu/min. The power input to the freezer is
0.70 hp, and the surrounding air is 75°F.
Determine a) the reversible power, b) the
irreversibility, an c) the second-law
efficiency of this freezer.
Ref: Cengel & Boles, Thermodynamics, An Engineering Approach, 4th edition, Mc-Graw Hill, 2002.):
Exergy


We can calculate the exergy, X (work potential) at a
given state. The work potential is a function of the total
energy of the system.
X  X KE  X PE  X internal energy  X flow work





(remember that in a control mass, there will be no flow work)
XKE (exergy due to kinetic energy): V2/2 (on a per unit
mass basis
XPE: gZ
Xinternal energy: u-uo+Po(v-vo)-To(s-s0)
To see a derivation of this last equation, see the
appendices on the web site. The “o” stands for the dead
state (atmospheric conditions). If a piston is at
atmospheric pressure and temperature (the dead state),
it can’t do any work.
Exergy of a Closed System

Exergy of a closed system, per unit mass j, can be
found be adding all the terms
2



V
   u  uo   Po  v  vo   To  s  so  
 gZ
2
This gives us the maximum work we could possibly get
out of a system.
Usually we will be more interested in the change in
exergy from the beginning to end of a process.
For a closed system, 2  1    ?
For a control volume



Xcv=Xclosed+Xflow work
y=Xcv/m (exergy per unit mass)
Xflow work=Wflow-Wagainst atmosphere=Pv-Pov
y cv

V2
 u  uo   Po v  Po vo  To s  so  
 gz  Pv  Po vo
2
Now combine terms: u+Pv=h; uo+Povo=ho
y cv
V2
 h  ho   To s  so  
 gz
2
Change in exergy

If we only have one fluid stream
V22  V12
y 2 y 1  y  h2  h1   To s2  s1  
 g z 2  z1 
2

If we have multiple streams
y 

2


V
2
m 2  h2  To s2 
 gz2  


2



2


V
1
m 1  h1  To s1 
 gz1 


2


Exergy Balance




We will use these equations in an exergy balance to
solve for such quantities as reversible work or exergy
destroyed.
Xin-Xout-Xdestroyed=Xsys
Xdestroyed is potential work that was destroyed due to
irreversibilities like friction.
Exergy can be transferred (Xin-Xout) by heat, work, and
mass flow
Exergy Transfer by Heat Transfer

As we add heat to a system, we
increase its ability to do work.
Wmax  X heat  QHh max

 To
 QH 1 
 TH



See Appendix B on web for a
discussion of how to deal with cold
sinks.
Exergy Transfer by Work and Mass
Flow



If we do work on a system, we increase its
ability to do work.
Xwork=W-Wsurr for boundary work
Xwork=W
for all other kinds of work
 Remember

Xmass=my
Wsurr  P0 V2  V1 
Xdestroyed
Xdestroyed=I=ToSgen
 See Appendix C on the web for a
derivation.
 Review from ME 297

 Ssys=Sin-Sout+Sgen
Entropy Generated, Sgen

For a steady-state control volume, this leads us to
S gen 

 m s   m s  
e e
out
in
For a control mass, this becomes
S gen  S 2  S1 

i i

Q k
Tk
Qk
Tk
Here Tk is the temperature of the heat source or heat
sink (not the system temperature).
Final Equation for Xsys for control
mass

 To
1 
 T
k


Qk  W  Po V2  V1   To S gen  X 2  X 1


• Terms in [ ] are W-Wsurr=Wu
• If we want to find Wrev, then ToSgen=0 and
Wu=Wrev
• Note that if heat transfer is to/from the
surroundings, the Q term drops out.
Example

A 12-ft3 rigid tank contains R-134a at 30 psia
and 40% quality. Heat is transferred now to the
refrigerant from a source at 120°F until the
pressure rises to 60 psia. Assuming the
surroundings to be at 75°F, determine a) the
amount of heat transfer between the source and
the refrigerant and b) the exergy destroyed
during the process.
Ref: Cengel and Boles
Final Equation for Xsys for control
volume
For multiple fluid streams, unsteady flow:

 To
1 
 T
k


Qk  W  Po V2  V1   To S gen 


my  m y
i
i
e
For one fluid streams, steady flow:

 To
1 
 T
k


Qk  W  To S gen  m y i y e   0


To find Wrev, set Sgen=0. If adiabatic, Q=0.
e
 X 2  X1
Set up the following problems.
1.
2.
3.
Refrigerant at T1 and P1 is throttled to a pressure of P2.
Find the reversible work and exergy destroyed during
this process. The atmosphere has a temperature of To.
Air at T1 and P1 with a velocity of V1 enters a nozzle
and exits at P2 and T2 with a velocity of V2. There is a
heat loss Q from the nozzle to the surroundings at To.
Find the exergy destroyed during this process.
Air enters a compressor at ambient conditions (To and
Po) and leaves at P2 and T2. The compressor is
deliberately cooled, and there is a rate of heat loss of
Q to the surroundings. The power input to the
compressor is PWR. Find the rate of irreversibility, I,
for this process.
Example

See handout
Exergy Analysis for a Cycle, 1 fluid
stream, steady flow
I  To S gen
for a component :
Q

S gen  m se  si  
Tk
S  S
 S
gen
gen, pump
gen,boiler
 S gen,turbine  S gen,cond.

Q pump  
Qcond., in 
Qboiler  
Qturbine  








 m  s 2  s1 
  s3  s 2 
  s 4  s3 
  s1  s 4 



To  
Tcomb.chamber  
To  
Tlake 


I  To m 



Qout

Tk ,out

Qin 
Tk ,in 
Second Law Efficiency for a Cycle
W net ,actual
W net ,actual
h II  

Wnet ,reversible W net ,actual  I