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The Second Law of ThermodynamicsEntropy and Work
Chapter 7c
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Work Done during a ProcessIn Chapter 4 we found the work done
in a closed system due to moving boundaries and expressed it in terms of the fluid properties
In the processes describing the behavior of many engineering devices there are no moving boundaries
2
1PdvWb
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For example, a steam turbine does not have any moving boundaries
Steam Turbine Work
Steam turbines do produce work
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Work Done During a ProcessIt would be useful to be able to
express the work done during a steady flow process, in terms of system properties
Recall that steady flow systems work best when they have no irreversibilities
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dpedkedhwq outrevinrev ,, Energy Balance for a steady flow device
Tdsqrev
vdPdhTds
dpedkevdPw outrev ,
,
e
rev out iw vdP ke pe
Often the change in kinetic energy and potential energy is 0
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“All” we have to do now is integrate!!
,
e
rev out iw vdP
In order to integrate, we need to know the relationship between v and P
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For solids and liquids
v is constant
,rev out e iw v P P ke pe
,
e
rev out iw vdP ke pe
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Steady flow of a liquid through a pipe or a nozzle
,rev out e iw v P P ke pe There is no work!!
2 2
02
e ie i e i
V Vv P P g z z
Bernoulli’s equation
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Steady Flow of a Liquid through a pump or a turbine
,rev out e iw v P P
Note that the work term is smallest when v is small, so for a pump (which uses work) you want v to be small. For a turbine (which produces work) you want v to be big.
Or..
,rev in e iw v P P
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Compressor Work
,
e
rev in iw vdP
We integrated this equation for v = constant, which is good for liquids – but what about gases?Consider an ideal gas, at constant TP
RTv
, ln erev in
i
Pw RTP
Remember, this is only true for the isothermal case, for an ideal gas
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Compressor Work
,
e
rev in iw vdP
Another special case is isentropic
We derived the isentropic relationships earlier in this chapterCPvk
kk PCv 11 Rearrange to find v, plug in and integrate
1 11
1 1
, 11
k kk e i
rev ink
P Pw C
Now its “just” algebra, to rearrange into a more useful form
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1 1 1 1
, 11
k k k ke e i i
rev ink
C P P C P Pw
, 1 11 1 1
e i e ie e i irev in
k k
R T T kR T Tv P v Pwk
vPC
CPvkk
k
11
1 11
1 1
, 11
k kk e i
rev ink
P Pw C
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, 1i e i
rev ini i
kRT T Tw T Tk
k
k
PP
TT
1
1
2
1
2
1
, 11
kk
i erev in
i
kRT Pw
k P
Remember, this equation only applies to the isentropic case, for an ideal gas, assuming constant specific heats
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Compressor Work
,
e
rev in iw vdP
Another special case is polytropic
Back in Chapter 4 we said that in a polytropic process Pvn is a constantCPvn
This is exactly the same as the isentropic case, but with n instead of k!!
, 1 11 1 1
e i e ie e i irev in
n n
R T T nR T Tv P v Pwn
1
, 11
nn
i erev in
i
nRT Pw
n P
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The area to the left of each line represents the work, vdP
Note, it takes less work for an isothermal process
You should compress isothermallycompress isothermally, and you should use an isentropicisentropic process in a turbineturbine!!
Pv Diagram for Isentropic, Polytropic and Isothermal compression, for the same final and initial pressures
Inlet
Exit
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How do you keep a compression process isothermal?
The gas will heat up as it is compressed, so it needs to be cooled
Intercooling is difficultInstead, multistage compression is
more common, with cooling between steps
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Two stage Compressor
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How do you decide how to break up the compression load?You save the most work by
intercooling, when each compressor carries the same load
1
, 11
kk
i erev in
i
kRT Pwk P
Since you cool back to T1 between stages, the only things that change in this equation are the P’s
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1
, 11
kk
i erev in
i
kRT Pw
k P
For the work done by each stage to be equal, the pressure ratio must be equal
1
11
kk
i xstage
i
kRT Pw
k P
1
11
kk
i e
x
kRT Pk P
e x
x i
P PP P
Or…x e iP P P
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Isentropic Efficiency of Steady Flow Devices
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Efficiency We’d like a measure of efficiency to
compare real devices to the best we can do
There are always irreversibilities that downgrade performance
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Most steady flow devices are intended to operate under adiabatic conditions
If a device is reversible and adiabatic, it is isentropic
Real devices are never really isentropic
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Isentropic EfficiencyLets compare how well real devices
work to how well comparable isentropic devices workSame inlet conditionsSame outlet conditionsTurbine, Compressor and Nozzle
actual result isentropic resultII
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Turbines
work turbineisentropic work turbineactual
Turbine
s
aTurbine w
w
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esi
eaiTurbine hh
hh
ei hhw
pekehmWQ
Remember, the work done in a turbine can be found from the energy balance
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Isentropic Efficiencies of Compressors and PumpsRatio of the work required to raise the
pressure of a gas to a specified value, in a isentropic manner, to the actual work
a
scompressor w
w
Note that this equation is arranged so that it is always less than one!!
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eai
esiCompressor hh
hh
ei hhw
pekehmWQ
Remember, the work done by a compressor can be found from the energy balance
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eai
esiCompressor hh
hh
Applies to both gases and liquids
)( ies PPvw Isentropic work for a liquid
eai
ieCompressor hh
PPv
)( Only applies
to a liquid
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Sometimes compressors are cooled intentionally – Why?
Cooling reduces the specific volume, resulting in less work required for compression
For compressors that are intentionally cooled, the isothermal model is more realistic
a
tCompressor w
w
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Isentropic Efficiency of NozzlesThe objective of a nozzle is to increase
the kinetic energy of the gasUsually, the inlet velocity is low
enough that we can consider it to have zero kinetic energy
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22eaV 22
esV
keh
2
2
22
22
es
ea
ies
iea
s
aNozzle
V
V
VV
VVkeke
esi
eaiNozzle hh
hh
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Entropy Balance
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Entropy Balance
There is no such thing as the conservation of entropy
Entropy of the universe always increases
Real processes always generate entropy
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system theofentropy total
in the Change
GeneratedEntropyTotal
OutEntropyTotal
InEntropyTotal
systemgenoutin SSSS
Often called the entropy balance
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systemgenoutin SSSS
12 SS
Remember, entropy is a state function. It doesn’t change unless the state of the system changes!!!
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How Does Entropy Enter and Leave a System?Heat Transfer
S=Q/T (T=constant) If T is not constant S=Qk/Tk
There is no entropy transfer with work!!
Mass FlowSmass=ms
Entropy Generation
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Closed Systems
12 SSSTQ
genk
kThere is no mass transfer in a closed system
In an isolated system there is no heat transfer, so the equation becomes…
12 SSSSgen
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The Universe is an Isolated System
gssurroundinsystemgen SSSS
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Control Volumes
cvgeneeiik
k SSsmsmTQ
cvgeneeiik
k SSsmsmTQ
Simplify this equation, depending on the process
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Surface area for heat transfer is 30 m^2Thermal Conductivity is 0.69 W/(m C)
Inside
T= 27 C
Outside
T=0 C
Inside Surface T = 20 C
Outside Surface T = 5 C
Consider Example 7-17Entropy Generation in a Wall
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Determine the rate of heat transfer through the wall
W1035
xTkAQ
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Determine the rate of entropy generation in the wall
systemgenoutin SSSS 0
The state of the system does not change with time
0
genout
out
in
in STQ
TQ
0K 278 W1035
K 293 W1035
genS
W/K191.0genS
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Determine the rate of entropy generation for the process
systemgenoutin SSSS 0
Consider an extended system – wall plus the surrounding air
0
genout
out
in
in STQ
TQ
0K 273 W1035
K 300 W1035
genS
W/K341.0genS
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Where is entropy generated?
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Reducing the Cost of Compressed AirSkimRepair Air LeaksInstall High Efficiency MotorsUse a small motor at high capacity,
instead of a large motor at low capacity
Use outside air for compressor intakeReduce the air pressure setting
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SummarySteady Flow work for a reversible process
,
e
rev out iw vdP ke pe
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SummaryFor incompressible substances
The work term is smallest when v is small, so for a pump (which uses work) you want v to be small. For a turbine (which produces work) you want v to be big.
,rev out e iw v P P ke pe
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Summary
•We looked at three special cases of the work equation• Isothermal• Isentropic• Polytropic
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SummaryIsothermal Compression
, ln erev in
i
Pw RTP
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SummaryIsentropic Compression
1
, 11
kk
i erev in
i
kRT Pwk P
, 11 1
e ie e i irev in
k
kR T Tv P v Pwk
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SummaryPolytropic Compression
1
, 11
nn
i erev in
i
nRT Pwn P
, 11 1
e ie e i irev in
n
nR T Tv P v Pwn
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Summary
•The work input to a compressor can be reduced by using multistage compression with intercooling. For maximum savings from the work input, the pressure ratio across each stage of the compressor must be the same.
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Summary
•Most steady-flow devices operate under adiabatic conditions, and the ideal process for these devices is the isentropic process.
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In the relations above, h2a and h2s are the enthalpy values at the exit state for actual and isentropic processes, respectively.
SummaryIsentropic or Adiabatic Efficiency
Actual turbine work wa h1 - h2aIsentropic turbine work ws h1 - h2s
= = =~
Isentropic compressor work ws h2s - h1Actual compressor work wa h2a - h1
= = =~
Actual KE at nozzle exit V2a h1 - h2aIsentropic KE at nozzle exit h1 - h2s2V2s
= = =~2
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SummaryEntropy Balance
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SummaryEntropy Balance – Rate form
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SummaryEntropy Balance – Steady-flow