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7/31/2019 Isentropic process
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Lec 18: Isentropic processes,TdS relations, entropy changes
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For next time: Read: 7-2 to 7-9 Group project subject selection due on
November 3, 2003
Outline: Entropy generation and irreversible processes Entropy as a property Entropy changes for different substances
Important points: Entropy is a property of a system it is not
conserved and is generated by irreversibleprocesses Know how to identify an isentropic processes Know how to use the tables to find values for
entropy
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Recall we had entropy
2
1 revintT
q
Rlb
Btuor
Kkg
kJ
m
s2 - s1 =
Units
are
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Lets look at a simpleirreversible cycle on a p-v
diagram with two processes
P
1
2
.
.A
B
Let A be
irreversible and Bbe reversible
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Irreversible cycle
0)T
QAB
By Clausius Inequality
Evaluate cyclic integral
0T
Q
T
Q
T
Q2
1 B
2
1 Acycle
(non-rev) (rev)
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Irreversible cycle
For the reversible process, B, dS=Q/dT,thus:
0dST
Q
T
Q2
1
2
1 Acycle
Rearranging and integrating dS:
2
1 AT
QS
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Second Law of Thermodynamics
Entropy is a non-conserved property!
2
1 A
12T
QSSS
This can be viewed as a mathematicalstatement of the second law (for aclosed system).
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We can write entropy change as anequality by adding a new term:
gen
2
1 A
12 S
T
QSS
entropychange
entropytransfer
due toheattransfer
entropyproduction
orgeneration
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Entropy generation
Sgen 0 is an actual irreversible process.
Sgen = 0 is a reversible process.
Sgen 0 is an impossible process.
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TEAMPLAY
Consider the equation
You have probably heard, Entropy alwaysincreases.
Could it ever decrease? What are theconditions under which this could happen(if it can)?
gen
2
1 A
12 S
T
QSS
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Entropy transfer and production
What if heat were transferred from thesystem?
The entropy can actually decrease if
gen
2
1 A
ST
Q
and heat is being transferred awayfrom the system so that Q is negative.
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Entropy Production
Sgen quantifies irreversibilities. Thelarger the irreversibilities, the greaterthe value of the entropy production,Sgen .
A reversible process will have no entropyproduction.
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Entropy transfer and production
S2 S1
> 0, Q could be + or ; if,
because Sgen is always positive.
< 0, if Q is negative and
= 0 if Q = 0 and Sgen = 0.
= 0 if Q is negative and
gen
2
1 A
ST
Q
gen
2
1 A
ST
Q
gen
2
1 A
ST
Q
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Isentropic processes
Note that a reversible (Sgen = 0),adiabatic (Q = 0) process is alwaysisentropic (S1 = S2)
But, if the process is merely isentropicwith S1 = S2, it may not be a reversibleadiabatic process.
For example, if Q 0 and gen
2
1 A
ST
Q
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Entropy generation
Consider
What if we draw our system boundariesso large that we encompass all heat
transfer interactions? We wouldthereby isolate the system.
gen
2
1 A
12 S
T
QSS
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Entropy changes of isolated systems
And then
gen
2
1 A
12 ST
QSS
0
gen12 SSS
But Sgen0. So, the entropy of anisolated system always increases. (Thisis the source of the statement, The world
is running down.)
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Entropy
)ss(xss fgf
)T(s)p,T(s f
It is tabulated just like u, v, and h.
Also,
And, for compressed or subcooled liquids,
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The entropy of a pure substance is determined from the tables, just as forany other property
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Ts Diagram for Water
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TEAMPLAY
Use the tables in your book
Find the entropy of water at 50 kPa and500 C. Specify the units.
Find the entropy of water at 100 C anda quality of 50%. Specify the units.
Find the entropy of water at 1 MPa and
120 C. Specify the units.
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Ts diagrams
pdVw
Work was the area under the curve.
Recall that the P-v diagram was veryimportant in first law analysis, and that
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For a Ts diagram
revintT
QdS
TdSQ revint
2
1
revint TdSQ
Rearrange:
Integrate:
If the internally reversible process also isisothermal at some temperature To:
STdSTQ o
2
1
orevint
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On a T-S diagram, the area under the process curve represents theheat transfer for internally reversible processes
d
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Entropy change of a thermalreservoir
For a thermal reservoir, heat transfer occursat constant temperaturethe reservoirdoesnt change temperature as heat isremoved or added:
TQ
S
Since T=constant:
T
QS
Applies ONLY tothermalreservoirs!!!!
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The Tds Equations
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Derivation ofTds equations:
dQ dW = dU
For a simple closedsystem:
dW = PdV
The work is given by:
dQ = dU + PdV
Substituting gives:
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More derivation.
For a reversible process:
TdS = dQ
Make the substitution for Q in the energyequation:
PdV+dU=TdS
Or on a per unit mass basis:
Pdv+du=Tds
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Entropy is a property. The Tds expression
that we just derived expresses entropy interms of other properties. The propertiesare independent of path.We can use theTds equation we just derived to calculatethe entropy change between any twostates:
Tds = du +Pdv
Tds = dh - vdP
Starting with enthalpy, it is possible todevelop a second Tds equation:
Tds Equations
L t l k t th t h
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Lets look at the entropy changefor an incompressible
substance:
dT
T
)T(cds
We start with the first Tds equation:
Tds = cv(T)dT + Pdv
For incompressible substances, v const, sodv = 0.
We also know that cv(T) = c(T), so we canwrite:
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Entropy change of anincompressible substance
dTT
)T(css
2
1
T
T
12
1
212
T
Tlncss
Integrating
If the specific heat does not vary with
temperature:
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TEAMPLAY
Work Problem 7-48
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Entropy change for an ideal gas
dTcdh p And
dpp
RTdTcTds p
Tds = dh - vdp
Start with 2nd Tds equation
Remember dh and v for an ideal gas?
v=RT/p
Substituting:
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Change in entropy for an ideal gas
p
dpR
T
dTcds p
Dividing through by T,
Dont forget, cp=cp(T)..a function oftemperature! Integrating yields
1
2
T
T
p12pplnR
TdT)T(css
2
1
Entropy change of an ideal gas
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Entropy change of an ideal gasfor constant specific heats:
approximation
Now, if the temperature range is solimited that c
p constant (and c
v
constant),
1
2pp
T
Tlnc
T
dTc
1
2
1
2p12
p
plnR
T
Tlncss
Entropy change of an ideal gas
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Entropy change of an ideal gasfor constant specific heats:
approximation
Similarly it can be shown from
Tds = du + pdv
that
1
2
1
2v12
vvlnR
TTlncss
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TEAMPLAY
Work Problem 7-62
Entropy change of an ideal gas
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Entropy change of an ideal gasfor variable specific heats: exact
analysis
1
2
T
T
p12 p
p
lnRT
dT
)T(css
2
1
2
1
T
T
p
T
dTc
Integrating..
To evaluate entropy change, well
have to evaluate the integral:
Entropy change of an ideal gas
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122
1
T
0T
p
T
0T
p
T
T
pT
dTcT
dTcT
dTc
)T(s)T(s 1o
2
o
And so(T), the reference entropy, istabulated in the ideal gas tables for areference temperature of T = 0 and p = 1
atm.
Entropy change of an ideal gasfor variable specific heats: exact
analysis
Evaluation of the integral
Entropy change of an ideal gas for
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Entropy change of an ideal gas forvariable specific heats: exact
analysis
Only is tabulated. Theis not.
So,
dTcp dTcv
1
21
o
2
o
12p
plnR)T(s)T(sss
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Entropy change of an ideal gas
Note that the entropy change of an idealgas, unlike h and u, is a function of twovariables.
Only the reference entropy, so, is afunction of T alone.
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Sample Problem
A rigid tank contains 1 lb of carbon monoxideat 1 atm and 90F. Heat is added until thepressure reaches 1.5 atm. Compute:
(a) The heat transfer in Btu.
(b) The change in entropy in Btu/R.
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Draw diagram:
State 1:
P = 1atm
T = 90oF
CO:
m= 1 lbmState 2:
P = 1.5 atm
Rigid Tank => volume isconstant
Heat Transfer
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Assumptions
Work is zero - rigid tank
kinetic energy changes zero
potential energy changes zero
CO is ideal gas
CO in tank is system
Constant specific heats
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Apply assumptions to conservationof energy equation
PE+KE+UWQ
12v TTmc=Q
For constant specific heats, weget:
0 0 0
Need T2> How do we get it?
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Apply ideal gas EOS:
2
1
22
11
mRT
mRT
VP
VP Cancel common
terms...
Solve forT2:
R825R460901.01.5TPPT 112
2
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Solve for heat transfer
R550825Rlb
Btu18.0)lb1(Q
m
m
Btu5.49Q
Now, lets get entropy change...
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For constant specific heats:
1
2
1
2v12
vvRln
TTlncmSS
Since v2 = v1
0
1
2v12
TTlncSS
R550
R825ln
Rlb
Btu18.0)lb1(SS
m
m12
Btu/R073.0SS 12