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![Page 1: 1 Property Relationships Chapter 6. 2 Apply the differential form of the first law for a closed stationary system for an internally reversible process.](https://reader036.fdocuments.in/reader036/viewer/2022062315/5697c0131a28abf838cccfa5/html5/thumbnails/1.jpg)
1
Property Relationships
Chapter 6
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dUWQ revrev intint
TdSQ rev int PdVW rev int
dUPdVTdS
Apply the differential form of the first law for a closed
stationary system for an internally reversible process
The T-ds relations :
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PdvduTds This equation is known as:
First Gibbs equation or First Tds relationshipDivide by T, ..
T
Pdv
T
duds
Although we get this form for internally reversible process, we still can compute s for an irreversible process. This is because S is a point function.
Divide by the mass, you get
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Second T-ds (Gibbs) relationshipRecall that… Pvuh
vdPPdvdudh Take the differential for both sides
Rearrange to find du
Substitute in the First Tds relationship PdvduTds
vdPPdvdhdu
vdPdhTds Second Tds relationship, or Gibbs equation
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T
vdP
T
dhds Divide by T, ..
Thus We have two equations for ds
T
vdP
T
dhds
T
Pdv
T
duds
To find s, we have to integrate these equations. Thus we need a relation between du and T (or dh and T).Now we can find entropy change (the LHS of the entropy balance) for liquids and solids
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6
2 -Entropy Change of Liquids and Solids
T
Pdv
T
duds
Solids and liquids do not change specific volume appreciably with pressure. That means that dv=0, so the first equation is the easiest to use.
0
T
duds
Thus For solids and liquids
Recall also that For solids and liquids , CdTdu
T
CdT
T
duds
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1
2lnT
TCs
Integrate to give…
Only true Only true for solids for solids and and liquids!!liquids!!
What if the process is isentropic?
0ln,01
2
T
TCs
The only way this expression can equal 0 is if,
Hence, for solids and liquids, isentropic processes are also isothermal.
21 TT
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Example(6-7): Effect of Density of a Liquid on Entropy
Liquid methane is commonly used in various cryogenic applications. The critical temperature of methane is 191 K (or -82oC), and thus methane must be maintained below 191 K to keep it in liquid phase. The properties of liquid methane at various temperatures and pressures are given next page.
Determine the entropy change of liquid methane as it undergoes a process from 110 K and 1 MPa to 120 K and 5 MPa
(a) using actual data for methane and
(b) approximating liquid methane as an incompressible substance. What is the error involved in the later case?
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kgKkjsss /270.0875.4145.522
kgKkjT
TC
T
TCs avg /303.0
110
120ln4785.3lnln
1
2
1
2
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Example (6-19): Entropy generated when a hot block is dropped in a lake
A 50-kg block of iron casting at 500 K is dropped in a large lake that is at 285 K. The block reaches thermal equilibrium with lake water.
Assuming an average specific heat of 0.45 kJ/kg.K for the iron, determine:
(a) The entropy change of the iron block,
(b) The entropy change of the water lake,
(c) the entropy generated during this process.
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KkjT
TmCSiron /65.12
500
285ln45.050ln
1
2
(a) The entropy change of the iron block,
(b) The entropy change of the water lake,
we need also to find Q coming out of the system.
UWQ )(0 12 TTmCQ 50 0.45(285 500) 4838Q kJ
KkJT
QS
lakelake /97.16
285
4838
T=500K
Tsurr= 285 K
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Thus Sg = Stot = Ssys + Slake
T=500K
Tsurr= 285 K
System boundary
(c) the entropy generated during this process.
Sg = Stot = -12.6 + 16.97
= 4.32
Choose the iron block and the lake as the system and treat it is an isolated system.
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3 -The Entropy Change of Ideal Gases, first relation
The entropy change of an ideal gas can be obtained by substituting du = CvdT and P /T= R/ into Tds relations:
1
2
2
1
12
lnRT
dTTCss v
Tds du pd v
dT dds C R
T
integrating
First relation
du Pdds
T T
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A second relation for the entropy change of an ideal gas for a process can be obtained by substituting dh = CpdT and /T= R/P into Tds relations:
2
1 1
212 P
PlnR
T
dTTCss p
Tds dh vdp p
dT dpds C R
T p
integrating
Second relation
dh vdpds
T T
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1
22
1
12 ln
RT
dTTCss v
2
1 1
212 P
PlnR
T
dTTCss p
The integration of the first term on the RHS can be done via two methods:
1. Assume constant Cp and constant Cv (Approximate Analysis)
2. Evaluate these integrals exactly and tabulate the data (Exact Analysis)
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Method 1: Constant specific heats (Approximate Analysis)First relation
1
2
1
2 lnlnv
vR
T
TCs v
Only true for ideal gases, assuming constant heat capacitiesSecond relation
Only true for ideal gases, assuming constant heat capacities
1
2
1
2 lnlnP
PR
T
TCs p
1
22
1
12 ln
RT
dTTCss v
2
1 1
212 ln
P
PR
T
dTTCss p
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Sometimes it is more convenient to calculate the change in entropy per mole, instead of per unit mass
1
2
1
212 lnln
v
vR
T
TCsss uv
1
2
1
212 lnln
P
PR
T
TCsss up
kJ/kmol. K
kJ/kmol. K
Ru is the universal gas constant
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Method 2: Variable specific heats (Exact Analysis)
We could substitute in the equations for Cv and Cp, and perform the integrations Cp = a + bT + cT2 + dT3
But this is time consuming. Someone already did the integrations and
tabulated them for us (table A-17) They assume absolute 0 as the starting point
1
22
1 P
PlnR
T
dTCs p We use the
second relation
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191
20
1
0
2 P
PlnRsss
The integral is expressed as:
2
1
)(T
T p T
dTTC
0
1
0
2
2
1
ssT
dT)T(C
T
Tp
T
p T
dT)T(Cs
0
0
Where
is tabulated in Table A-17
Therefore
K.kg/kJ:unit
2
0)(
T
p T
dTTC 1
0)(
T
p T
dTTC
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From this equation, It can be seen that the entropy of an ideal gas is not a function only of the temperature ( as was the internal energy) but also of the pressure or the specific volume.
The function s° represents only the temperature-dependent part of entropy
1
20
1
0
2 P
PlnRsss
Temperature dependencePressure
dependence
Is s = f (T) only? like u for an ideal gas. Let us see
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How about the other relation
1
22
1 v
vlnR
T
dTCs v
We can develop another relation for the entropy changed based on the above relation
but this will require the definition of another function and tabulating it which is not practical.
T
v T
dT)T(C?
0
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6-4 Isentropic Processes The entropy of a fixed mass can be
changed by
1. Heat transfer,
2. Irreversibilities It follows that the entropy of a system
will not change if we have
1. Adiabatic process,
2. Internally reversible process. Therefore, we define the following:
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Isentropic Processes of Ideal Gases
Many real processes can be modeled as isentropic
Isentropic processes are the standard against which we should measure efficiency
We need to develop isentropic relationships for ideal gases, just like we developed for solids and liquids
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1
2
1
2 lnlnv
vR
T
TCs v
For the isentropic case, S=0. Thus
1
2
1
2 lnlnv
vR
T
TCv
Constant specific heats (1st relation)
vC
R
v v
v
v
v
C
R
T
T
2
1
1
2
1
2 lnlnln
Recall
Recall also from ch 2, the following relations..…
11 kC/CC/RCCR vpvvp
1
2
1
1
2
k
v
v
T
T Only applies to ideal gases, with constant specific heats
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0lnln1
2
1
2
P
PR
T
TCs p
pC
R
p P
P
P
P
C
R
T
T
1
2
1
2
1
2 lnlnln
k
k
P
P
T
T1
1
2
1
2
Only applies to ideal gases, with constant specific heats
Constant specific heats (2nd relation)
Recall..… or1/ kCR v
1/ p
kR C
k
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Since…
k
k
P
P
T
T1
1
2
1
2
1
2
1
1
2
k
v
v
T
Tand
k
kk
P
P
v
v1
1
2
1
2
1
Which can be simplified to…
1
2
2
1
P
P
v
vk
Third isentropic relationship
HENCE
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1
2
1
1
2
k
v
v
T
T k
k
P
P
T
T1
1
2
1
2
1
2
2
1
P
P
v
vk
constantTvk 1
Compact form
constantTP k
k
1
constantPvk
Full form of Isentropic relations of Ideal Gases
Valid for only for 1- Ideal gas 2- Isentropic process 3- Constant specific heats
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That works if the specific heat constants can be approximated as constant, but what if
that’s not a good assumption?
1
201
02 ln
P
PRsss
We need to use the exact treatment 0
1
201
02 ln
P
PRss
This equation is a good way to evaluate property changes, but it can be tedious if you know the volume ratio instead of the pressure ratio
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1
201
02 ln
P
PRss
R
ss
P
P 01
02
1
2 exp
02
20
1 1
exp
exp
sRP
P sR
s20 is only a
function of temperature!!!
1
201
02 ln
P
P
R
ss Rename the exponential term as Pr , (relative pressure) which is only a function of temperature, and is tabulated on the ideal gas tables
02
20
11
exp
exp
r
r
sR P
PsR
2 2
1 1
r
r
P P
P P
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1
2
1
2
r
r
P
P
P
P
You can use either of the following 2 equations
1
201
02 ln
P
PRss
This is good if you know the pressure ratio but how about if you know only the volume ratioIn this case, we use the ideal gas law
2
22
1
11
T
vP
T
vP
2
1
1
2
1
2
P
P
T
T
v
v
2
1
1
2
r
r
P
P
T
T
1
1
2
2
T
P
P
T r
r
1
2
r
r
v
v
where
rr P/Tv
Remember, these relationships only hold for ideal gases and isentropic processes
1
2
1
2
r
r
v
v
v
v
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Example (6-10):
Isentropic Compression of Air in a Car Engine
Air is compressed in a car engine from 22oC and 95 kPa in a reversible and adiabatic manner. If the compression ratio V1/V2 of this piston-cylinder device is 8, determine the final temperature of the air. <Answer: 662.7 K>
Sol: