Lecture Notes on Thermodynamics 2008
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Transcript of Lecture Notes on Thermodynamics 2008
Lecture Notes on Thermodynamics Lecture Notes on Thermodynamics 20082008Chapter 10 Thermodynamics Chapter 10 Thermodynamics
RelationsRelationsProf. Man Y. Kim, Autumn 2008, Prof. Man Y. Kim, Autumn 2008, ⓒⓒ[email protected], Aerospace Engineering, Chonbuk National University, Korea [email protected], Aerospace Engineering, Chonbuk National University, Korea
2 Propulsion and Combustion Lab.
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Two Important Partial Derivative RelationsConsider a variable z which is a continuous function of x
and y : ,z f x y and
y x
z zdz dx dy
x y
If we take y and z as independent variables :
,x f y z and yz
x xdx dy dz
y z
(*)
(**)
Substitute eq.(**) into (*) :
y y yz x
z x z x zdz dy dz
x y y z x
1
y y yz x
z x z x zdy dz
x y y z x
Since there are only 2 independent variables,
Reciprocity relation
Cyclic relation
1
1
y y y y
x zz z zx
xx
1yxy z x z
yz x zzy
x zxy yx
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Maxwell Relations
2 Gibbs equations in Chapter 6 :
du Tds Pdv
dh Tds vdP
Maxwell Relations : Four equations relating the properties P, v, T, and s for a simple compressible system of fixed chemical composition
Helmholtz free energy :
a u TsGibb’s free energy :
g h Ts da du Tds sdT
da sdT Pdv dg dh Tds sdT
dg sdT vdP
Since u, h, a, and g are total derivative ;
s v
s P
T v
T P
du Tds Pdv
dh Tds vdP
da sdT Pdv
dg sdT vd
T Pv s
T vP s
s Pv T
s vP
PT
yx
M Ndz Mdx Ndy
y x
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Clapeyron Equation
Let’s consider the 3rd Maxwell relation ;
Clapeyron Equation : P, v, T 를 통해 증발엔탈피 ( ) 와 같은 상변화와 관계있는 엔탈피 변화를 구하는 관계식fgh
T v
s Pv T
상변화가 일어나는 동안 압력은 온도에만 의존하고 비체적에는 무관한 포화압력을 유지함 . 즉 , sat satP f T
v sat
P dPT dT
등온 액체 - 증기 상변화과정에 대해서 세번째 Maxwell 관계식을 적분하면 ;
fgg f g f
sat sat fg
sdP dPs s v v
dT dT v이 과정 동안 압력도 일정하게 유지되므로 ,
g g
fg fgff
dh Tds vdP dh Tds h Ts
따라서 ,
fg
sat fg
hdPdT Tv
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Clapeyron-Clausius Equation
저압상태일 때 ;
Clapeyron-Clausius Equation : Clapeyron 방정식에 약간의 근사를 사용하여 액체 - 증기와 고체 - 증기의 상변화에 적용함 .
증기를 이상기체로 가정하면 ;
g ff g gv v v v
g
RTv
P따라서 ,
2 2
fgf fg
sat s
g
atsat fg
hdPdT T
Ph hdP dP dTTv dT RT P R
작은온도구간에 대하여 는 어떤 평균값으로 일정하므로 ,
2
1 1 2
1 1ln fg
sat sat
hPP R T T
fgh
윗 식에서 를 ( 승화엔탈피 ) 로 대치함으로서 고체 -증기 영역에서도 사용함 .
fgh igh
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Relations between du, dh, ds, Cv and Cp (1/6)• Change of Internal Energy
du Tds Pdv
If,
, vv TT
u uu u T v dT dv
udu C dT
vT vdv
If, ,
v T
s ss s T v ds dT dv
T v
v TT v
s sT dT dv Pd
s sdu T dTv
T vT P dv
T v
since
Therefore,
v
v
T vT
s
CsT T
uT T
v v TP
PP
Finally,
vv
Pdu C dT T P dv
Tand
2 2
1 12 1
T v
vT v v
Pu u C dT T P dv
T
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Relations between du, dh, ds, Cv and Cp (2/6)• Change of Internal Energy - Example
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Relations between du, dh, ds, Cv and Cp (3/6)• Change of enthalpy
dh Tds vdP
If,
, pP TT
h hh h T P dT dP
hdh C dT
PT PdP
If, ,
P T
s ss s T P ds dT dP
T P
P P TT
s sT dT dP vd
s sdh T dT v
PP
TT PT dP
since
Therefore,
T
p
T P
P
CsT T
hv T v
PvT
TsP
Finally,
pP
vdh C dT v T dP
Tand
2 2
1 12 1
T P
pT P P
vh h C dT v T dP
T
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Relations between du, dh, ds, Cv and Cp (4/6)• Change of Entropy
If,
, v
vv T
s ss s T v dT dv
T vC P
ds dT dvT T
Therefore, 2 2
1 12 1
T vv
T v v
C Ps s dT dv
T T
If, , p
P T P
Cs s vs s T P ds dT dP dT dP
T P T TTherefore,
2 2
1 12 1
T Pp
T P P
C vs s dT dP
T T
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Relations between du, dh, ds, Cv and Cp (5/6)• Specific Heat (1/2)
2
21T
v
v
v v
v v
C PT
v TC CP P
ds dTv
dT TT
vT T
2
22p p
P PT
p
P
C Cv v C vT
P Tds dT dP
T T P T T T
Substitute (3’) into (3) ;
1 32 p v
P v
v PT C C dT T dP T dv
T TTake
, 3
v T
P PP P T v dP dT dv
T v
p vP v T v
P v P T v
P v
v P P PC C dT T dT dv T dv
T T v T
v P v P PT dT T dv
T T T v T
v PT dT
T T
p vP v
v PC C T
T T
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Relations between du, dh, ds, Cv and Cp (6/6)
• Specific Heat (2/2)
2 2
p vP v P T
v P v P TC C T
T T T v
We know the cyclic relations as :
1v P T v P T
P T v P v PT v P T T v
where,
1
1P
T
vT
vP
: volume expansivity
: isothermal compressibility
Comments :
1 0
2 0
3 ,
p v
p v
p v
C C
C C asT
for incompressibleliquidandsolid v constant C C
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Joule-Thomson Coefficient (1/2)
• Joule-Thomson Coefficient : 교축 (h=constant) 과정 중의 유체의 온도 변화
0
0
0JT JT
h
temperatureincreasesT
temperatureremainsconstantP
temperaturedecreases
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Joule-Thomson Coefficient (2/2)
10 p JT
P h Pp
v T vdh C dT v T dP v T
T P C T
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No Homework !