Exergy Analysis ME 210 Advanced Thermodynamics. Definitions Exergy (also called Availability or Work...
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Transcript of Exergy Analysis ME 210 Advanced Thermodynamics. Definitions Exergy (also called Availability or Work...
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.
positive 1200 VVPdVPWsurr
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.
What is Wsurr for a control volume?
negative 120 VVPWsurr
Second Law Efficiency, II
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.
th,max=1-TL/TH=1-300/800=0.635
Say th=0.45
II=0.45/0.625=0.72
We want a high th and II
Another way to look at this: for a work output device II=Wu/Wrev
Second Law Efficiency, cont.
A general definition:
suppliedexergy
(I) destroyedexergy 1
beginning) at the available s(what' suppliedexergy
process) after the available s(what' recoveredexergy
II
Three Efficiency Definitions
The second two are defined for work OUTPUT devices
rev
uII
isentropic
actual
in
net
W
W
W
W
Q
W
Law 2
Isentropic
Thermal
nd
s
th
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.
(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.
KE PE internal energy flow workX X X X X
Exergy of a Closed System
Exergy of a closed system, per unit mass can be found be adding all the terms
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
2o o o o o
Vu u P v v T s s gZ
2 1 ?
For a control volume Xcv=Xclosed+Xflow work
=Xcv/m (exergy per unit mass)
Xflow work=Wflow-Wagainst atmosphere=Pv-Pov
Now combine terms: u+Pv=h; uo+Povo=ho
oooooooocv vPPvgzV
ssTvPvPuu 2
2
gzV
ssThh ooocv 2
2
Change in exergy
If we only have one fluid stream
If we have multiple streams
12
21
22
121212 2zzg
VVssThh o
1
21
1112
22
222 22gz
VsThmgz
VsThm oo
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.
See Appendix B on web for a discussion of how to deal with cold sinks.
H
oHHheat T
TQQXW 1maxmax
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=m
120 VVPWsurr
Xdestroyed
Xdestroyed=I=ToSgen
See Appendix C on the web for a derivation.
Review from ME 297Ssys=Sin-Sout+Sgen
Entropy Generated, Sgen
For a steady-state control volume, this leads us to
For a control mass, this becomes
Here Tk is the temperature of the heat source or heat sink (not the system temperature).
k
k
inii
outeegen T
QsmsmS
k
kgen T
QSSS 12
Final Equation for Xsys for control mass
12121 XXSTVVPWQT
Tgenook
k
o
• 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
12121 XXmmSTVVPWQT
Teeiigenook
k
o
01
eigenok
k
o mSTWQT
T
For multiple fluid streams, unsteady flow:
For one fluid streams, steady flow:
To find Wrev, set Sgen=0. If adiabatic, Q=0.
Set up the following problems.
1. 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.
2. 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.
3. 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
ink
in
outk
outo
lake
incond
o
turbine
chambercomb
boiler
o
pump
condgenturbinegenboilergenpumpgengen
kiegen
geno
T
Q
T
QmTI
T
Qss
T
Qss
T
Qss
T
Qssm
SSSSS
T
QssmS
STI
,,
.,4134
.2312
.,,,,
:component afor
Second Law Efficiency for a Cycle
IW
W
W
W
actualnet
actualnet
reversiblenet
actualnetII
,
,
,
,