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Joule−Thomson(Kelvin) ：ガスの液化に使う
T1, V1, P1
T2, V2, P2 P1 > P2
断熱壁
before
after
細孔定圧
定圧
Joule−Thomson(Kelvin) ：ガスの液化に使う
T1, V1, P1
T2, V2, P2 P1 > P2
断熱壁
before
after
細孔定圧
定圧
Joule−Thomson(Kelvin) ：ガスの液化に使う
T1, V1, P1
T2, V2, P2 P1 > P2
断熱壁
before
after
細孔定圧
定圧
Joule−Thomson(Kelvin) ：ガスの液化に使う
T1, V1, P1
T2, V2, P2 P1 > P2
断熱壁
before
after
細孔定圧
定圧
等エンタルピー過程
left: 0 U1
right: U2 0total : U2 - U1
after before
dQ = 0
internal energy
workleft: +P1V1
right: -P2V2
total: +P1V1 -P2V2
U2 - U1 = +P1V1 -P2V2U1 + P1V1 = U2 + P2V2
H1 = H2
気体は仕事をされた気体は仕事をした
：断熱過程
等エンタルピー過程
left: 0 U1
right: U2 0total : U2 - U1
after before
dQ = 0
internal energy
workleft: +P1V1
right: -P2V2
total: +P1V1 -P2V2
U2 - U1 = +P1V1 -P2V2U1 + P1V1 = U2 + P2V2
H1 = H2
気体は仕事をされた気体は仕事をした
：断熱過程
等エンタルピー過程
left: 0 U1
right: U2 0total : U2 - U1
after before
dQ = 0
internal energy
workleft: +P1V1
right: -P2V2
total: +P1V1 -P2V2
U2 - U1 = +P1V1 -P2V2U1 + P1V1 = U2 + P2V2
H1 = H2
気体は仕事をされた気体は仕事をした
：断熱過程
何の変数で全微分形を書いてよい！が，UP THAS VGで書く変数を使うのがBEST！
H = H(S, P )H = H(T, P ) →教科書 µJT
dH = TdS + V dP
↑SがS(T,P)であればOK
理想気体のSは，以下のように書かれる
S(T, V ) = CV lnT
T ∗ + nR lnV
V ∗
∗ : a reference statePV = nRT, P ∗V ∗ = nRT ∗
S(T, P ) = CV lnT
T ∗ + nR ln�
nRT
P
P ∗
nRT ∗
�
= (CV + nR) lnT
T ∗ + nR lnP ∗
P
= CP lnT
T ∗ − nR lnP
P ∗
P⼀一定でTを変化するのとP⼀一定でSを変化は同じ
教科書H(P, T ) :
dH =�
∂H
∂P
�
T
dP +�
∂H
∂T
�
P
dT = 0�
∂T
∂P
�
H
= −�
∂H
∂P
�
T
/
�∂H
∂T
�
P� �� �=CP
dH = TdS + V dP�
∂H
∂P
�
T
= T
�∂S
∂P
�
T
+ V
dG = V dP − SdT →�
∂S
∂P
�
T
= −�
∂V
∂T
�
P�∂H
∂P
�
T
= −T
�∂V
∂T
�
P
+ V = −TnR
P+ V = 0
µJT ≡
教科書H(P, T ) :
dH =�
∂H
∂P
�
T
dP +�
∂H
∂T
�
P
dT = 0�
∂T
∂P
�
H
= −�
∂H
∂P
�
T
/
�∂H
∂T
�
P� �� �=CP
dH = TdS + V dP�
∂H
∂P
�
T
= T
�∂S
∂P
�
T
+ V
dG = V dP − SdT →�
∂S
∂P
�
T
= −�
∂V
∂T
�
P�∂H
∂P
�
T
= −T
�∂V
∂T
�
P
+ V = −TnR
P+ V = 0
µJT ≡
教科書H(P, T ) :
dH =�
∂H
∂P
�
T
dP +�
∂H
∂T
�
P
dT = 0�
∂T
∂P
�
H
= −�
∂H
∂P
�
T
/
�∂H
∂T
�
P� �� �=CP
dH = TdS + V dP�
∂H
∂P
�
T
= T
�∂S
∂P
�
T
+ V
dG = V dP − SdT →�
∂S
∂P
�
T
= −�
∂V
∂T
�
P�∂H
∂P
�
T
= −T
�∂V
∂T
�
P
+ V = −TnR
P+ V = 0
µJT ≡
教科書H(P, T ) :
dH =�
∂H
∂P
�
T
dP +�
∂H
∂T
�
P
dT = 0�
∂T
∂P
�
H
= −�
∂H
∂P
�
T
/
�∂H
∂T
�
P� �� �=CP
dH = TdS + V dP�
∂H
∂P
�
T
= T
�∂S
∂P
�
T
+ V
dG = V dP − SdT →�
∂S
∂P
�
T
= −�
∂V
∂T
�
P�∂H
∂P
�
T
= −T
�∂V
∂T
�
P
+ V = −TnR
P+ V = 0
µJT ≡
dH = TdS + V dP
S = S(T, P )
dS =�
∂S
∂T
�
P
dT +�
∂S
∂P
�
T
dP
dH = T
�∂S
∂T
�
P
dT +�T
�∂S
∂P
�
T
+ V
�dP
TdS = dQ, T
�∂S
∂T
�
P
= CP dT
dG = −SdT + V dP →�
∂S
∂P
�
T
= −�
∂V
∂T
�
P
dH = CP dT +�−T
�∂V
∂T
�
P
+ V
�dP
µJT =�
∂T
∂P
�
H
=�−T
�∂V
∂T
�
P
+ V
�/CP
V =nRT
P, −T
nR
P+ V = −V + V = 0
別解
If the gas temperature is then μJT is since is thus must be so the gas
below the inversion temperature positive always negative negative cools
above the inversion temperature negative always negative positive warms
∂P ∂T
For an ideal gas, μJT is always equal to zero: ideal gases neither warm nor cool upon being expanded at constant enthalpy.
Joule-Thomson coefficients for various gases at atmospheric pressure.
理想気体ではJT過程で
温度は変化しないが実在気体では下がるガスの液化に使われている
P1 > P2
https://en.wikipedia.org/wiki/Joule%E2%80%93Thomson_effect
If the gas temperature is then μJT is since is thus must be so the gas
below the inversion temperature positive always negative negative cools
above the inversion temperature negative always negative positive warms
∂P ∂T
For an ideal gas, μJT is always equal to zero: ideal gases neither warm nor cool upon being expanded at constant enthalpy.
Joule-Thomson coefficients for various gases at atmospheric pressure.
理想気体ではJT過程で
温度は変化しないが実在気体では下がるガスの液化に使われている
P1 > P2
https://en.wikipedia.org/wiki/Joule%E2%80%93Thomson_effect
If the gas temperature is then μJT is since is thus must be so the gas
below the inversion temperature positive always negative negative cools
above the inversion temperature negative always negative positive warms
∂P ∂T
For an ideal gas, μJT is always equal to zero: ideal gases neither warm nor cool upon being expanded at constant enthalpy.
Joule-Thomson coefficients for various gases at atmospheric pressure.
理想気体ではJT過程で
温度は変化しないが実在気体では下がるガスの液化に使われている
P1 > P2
https://en.wikipedia.org/wiki/Joule%E2%80%93Thomson_effect
If the gas temperature is then μJT is since is thus must be so the gas
below the inversion temperature positive always negative negative cools
above the inversion temperature negative always negative positive warms
∂P ∂T
For an ideal gas, μJT is always equal to zero: ideal gases neither warm nor cool upon being expanded at constant enthalpy.
Joule-Thomson coefficients for various gases at atmospheric pressure.
理想気体ではJT過程で
温度は変化しないが実在気体では下がるガスの液化に使われている
P1 > P2
https://en.wikipedia.org/wiki/Joule%E2%80%93Thomson_effect
potential energy↑
kinetic energy↓temperature ↓
LNG : Liquid Natural gas （CH4, C2H6,C3H8, C4H10)エタンのJT係数 > 0
Experimental results for heat capacity and Joule-Thomson coefficient of ethane at zero pressureKonrad Bier, Joachim Kunze, Gerd Maurer, Holger Sand J. Chem. Eng. Data, 1976, 21 (1), pp 5–7 DOI: 10.1021/je60068a033
天然ガス液化の原理は、気体が膨張して密度が減少するときに、希薄な密度を維持するために内部エネルギーの一部が使用されて温度が低下するという、ジュール・トムソン効果を利用するものである
http://oilgas-info.jogmec.go.jp/pdf/0/598/200503_001a.pdf
