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Module 8Module 8
Non equilibriumNon equilibrium Th d iThermodynamics
Lecture 8 1Lecture 8.1Basic PostulatesBasic Postulates
NON-EQUILIRIBIUM THERMODYNAMICS
Steady State processes. (Stationary)Concept of Local thermodynamic eqlbmConcept of Local thermodynamic eqlbm
Extensive tproperty
Heat conducting bar
⎞⎛ Zdefine properties
Specific =z ⎟
⎠⎞
⎜⎝⎛
→ mZ
m 0lim
property
NON-EQLBM THERMODYNAMICS
Postulate I
Although system as a whole is not ineqlbm arbitrary small elements of iteqlbm., arbitrary small elements of itare in local thermodynamic eqlbm &have state fns which depend on statehave state fns. which depend on stateparameters through the same
l ti hi i th f lbrelationships as in the case of eqlbmstates in classical eqlbmth d ithermodynamics.
NON-EQLBM QTHERMODYNAMICS
Postulate II
JFS ∑&κκ JFS ∑=
Entropy gen rate
affinities fluxesgen rate
NON-EQLBM QTHERMODYNAMICS
Purely “resistive” systems
Flux is dependent only on affinityFlux is dependent only on affinityat any i t t
at that i t tinstant instant
System has no “memory”-
NON-EQLBM THERMODYNAMICS
Coupled Phenomenon
( ).;,,, 210 propextensiveFFFJJ −−−= κκ
Since Jk is 0 when affinities are zero,Since Jk is 0 when affinities are zero,
∑∑∑ 1−−−−++= ∑∑∑ jiijk
jij
jjkk FFLFLJ
!21
NON-EQLBM THERMODYNAMICS
where2
; ⎟⎟⎠
⎞⎜⎜⎝
⎛
∂∂∂
=⎟⎟⎠
⎞⎜⎜⎝
⎛
∂∂
= ijj FFJL
FJL κ
κκκ
00⎟⎠
⎜⎝ ∂∂⎟
⎠⎜⎝ ∂ jij FFF
kinetic Coeffkinetic Coeff( )−−−−= ,, 10 FFLL jkjk
Relationship betweenPostulate III
Relationship betweenaffinity & flux from‘other’ sciencesother sciences
NON-EQLBM THERMODYNAMICS
Heat Flux : ( )y
TCyTkJ Q ∂
∂−=
∂∂
−= ρρα
yy ∂∂
( )uu ∂∂ ρMomentum : ( )yu
yuJ M ∂
∂−=
∂∂
−=ρνμ
Mass : ycDJ m ∂
∂−=
Electricity :
y
EJ ∂−= λElectricity :
yJ e ∂
= λ
NON-EQLBM QTHERMODYNAMICS
Postulate IV
Onsager theorem {in the absence of ti fi ld }magnetic fields}
kjjk LL = kjjk
NON-EQLBM QTHERMODYNAMICS
Entropy production in systemsinvolving heat Flowinvolving heat Flow
dx
T1 T2 A1 2x
A
NON-EQLBM QTHERMODYNAMICS
⎟⎠⎞
⎜⎝⎛=
∂∂
−=AQ
xTkJQ
TkJQ ∂
⎠⎝∂ Ax
xT
Tk
TJ Q
s ∂∂
−==
Entropy gen. per unit volume
JJ xss dxx ,−+
dx
NON-EQLBM QTHERMODYNAMICS
JQ ⎬⎫
⎨⎧
−11
dxTT xdxx
Q⎭⎬
⎩⎨
= +
dxdTJd Q⎟
⎞⎜⎛ 1
dxdT
TJ
TdxdJ Q
Q ⋅−=⎟⎠⎞
⎜⎝⎛⋅= 2
1
dTJS Q ⋅=&
dxTSQ ⋅−= 2
NON-EQLBM QTHERMODYNAMICS
Entropy generation due to currentflow : dflow :
Idx
IJ =dEIJ λ−==
AJe = dxA
Je λ==
Heat transfer in element length dx
ddEIQ ⎟
⎠⎞
⎜⎝⎛ −=δelement length dx ⎠⎝
NON-EQLBM QTHERMODYNAMICS
Resulting entropy production perit lunit volume
dEJQS e−==δ&( ) dxTdxAT
Se ==.
NON-EQLBM QTHERMODYNAMICS
Total entropy prod / unit vol. withboth electric & thermal gradientsboth electric & thermal gradients
ddJdxdE
TJe
dxdT
TJ
SSS QeQ ⋅−⋅−=+= 2&&&
dxTdxT
FJFJ +⋅= eeQQ FJFJ .+⋅=
affinity affinityaffinity affinity
NON-EQLBM QTHERMODYNAMICS
dT1dxdT
TFQ 2
1−=
dE1dxdE
TFe
1−=
dxT
Analysis of thermo-electric ycircuits
Addl. Assumption : Thermo electric phenomena can be taken as LINEARphenomena can be taken as LINEARRESISTIVE SYSTEMS
{higher order∑=→J
jjKK FLJ {higher order terms negligible}
Here K = 1,2 corresp to heat flux “Q”, elec flux “e”
Analysis of thermo-electric ycircuits
Ab ti b itt∴ Above equations can be written aseQeQQQQ FLFLJ += eQeQQQQ
eeeQeQe FLFLJ +=
Substituting for affinities, theexpressions derived earlier, we getg
dXdE
TL
dXdT
TL
J QeQQ
Q1
2 ⋅−−=dXTdXT
dELdTLJ eQ 1
dXTL
dXTJ ee
eQe 2 −⋅−=
Analysis of thermo-electric ycircuits
We need to find values of the kineticcoeffs. from exptly obtainable data.p yDefining electrical conductivity as the elec flux per unit pot gradient
λas the elec. flux per unit pot. gradient under isothermal conditions we get f bfrom above
dEdELJ ee λ−=−=dXdXT
Je λ==
TL λ→ TLee λ=→
E d f L tEnd of Lecture
Lecture 8 2Lecture 8.2
Thermoelectric Thermoelectric hphenomena
Analysis of thermo-electric ycircuits
Th b i ti b ittThe basic equations can be written aseQeQQQQ FLFLJ += eQeQQQQ
eeeQeQe FLFLJ +=
Substituting for affinities, theexpressions derived earlier, we getg
dXdE
TL
dXdT
TL
J QeQQ
Q1
2 ⋅−−=dXTdXT
dELdTLJ eQ 1
dXTL
dXTJ ee
eQe 2 −⋅−=
Analysis of thermo-electric ycircuits
We need to find values of the kineticcoeffs. from exptly obtainable data.p yDefining electrical conductivity as the elec flux per unit pot gradient
λas the elec. flux per unit pot. gradient under isothermal conditions we get f bfrom above
dEdELJ ee λ−=−=dXdXT
Je λ==
TL λ→ TLee λ=→
Analysis of thermo-electric ycircuits
Consider the situation, undercoupled flow conditions, whencoupled flow conditions, whenthere is no current in the material,i.e. Je=0. Using the abovei.e. Je 0. Using the aboveexpression for Je we get
dELdTLdXdE
TL
dXdT
TL eeeQ −−= 20
eQ
LTL
dXdTdXdE
−=⎟⎟⎠
⎞⎜⎜⎝
⎛→ Seebeck
ff teeJ LTdXdTe
⎟⎠
⎜⎝ =0 effect
Analysis of thermo-electric ycircuits
oreQLdE
−=⎟⎞
⎜⎛
eeJ LTdTe
=⎟⎠
⎜⎝ =0
⎞⎛ dE α=⎟⎠⎞
⎜⎝⎛−=
=0eJdTdESeebeck coeff.
2TLTL eeeQ λαα ===>
Using Onsager theorem22TLL eQQe λα==
Analysis of thermo-electric ycircuits
Further from the basic eqs for Je & JQ, for Je = 0JQ, for Je 0 we get
⎟⎟⎠
⎞⎜⎜⎝
⎛⋅−−−=dXdT
LTL
TL
dXdT
TL
J eQQeQQQ 2 ⎟
⎠⎜⎝ dXLTTdXT ee
Q 2
dXdT
TLLLLL QeeQQQee ⋅
+−= 2 dXTLee
Analysis of thermo-electric ycircuits
For coupled systems, we definethermal conductivity asthermal conductivity as
⎬⎫
⎨⎧
−= QJk ( ) 0=⎭
⎬⎩⎨=
eJdXdTk
This givesThis givesLLLL
k eQQeQQee +−2TL
kee
eQQeQQee−=
Analysis of thermo-electric ycircuits
Substituting values of coeff. Lee, LQe, L l l t d b tLeQ calculated above, we get
( )22 TTkTLQQ λαα+= ( )QQ
( )TkT 22 αλ+=
Analysis of thermo-electric ycircuits
Using these expressions for variouskinetic coeff in the basic eqs forkinetic coeff in the basic eqs forfluxes we can write these as :
( ) ⎟⎠⎞
⎜⎝⎛−+−=
dXdET
dXdTTkJQ λαλα 2( )
⎠⎝ dXdXQ
dddXdE
dXdTJ e λλα −−=
dXdX
Analysis of thermo-electric ycircuits
W l it th ith flWe can also rewrite these with fluxesexpressed as fns of correspondingaffinities alone :
JTTkJ α+∂= eQ JT
dXkJ α+−=
dEk λαλQe J
TkdXdE
TkkJ 22 αλ
λααλ
λ+
+⋅+
−=
Using these eqs. we can analyze theeffect of coupling on the primaryeffect of coupling on the primaryflows
PETLIER EFFECTPETLIER EFFECTUnder Isothermal Conditions
a bJdX
dEJe λ−=JQ, ab
JedXe
Heat flux
ebQbeaQa JTJJTJ αα == ;
PETLIER EFFECTPETLIER EFFECTHeat interaction with surroundings
( ) ebabQaQbaQ JTJJJ αα −=−=
eab JeffPeltier π=. eabff
Peltier coeff.
( )baab T αα −=Π Kelvin Relation( )baab Kelvin Relation
PETLIER REFRIGERATORPETLIER REFRIGERATORFbC FebCua ::
0 KVeu FC
07.13 μαα =−
KTAmpJ baQ 270~.20?=
( )( ) WAmpKKV 074.202707.13 =⋅=
μ
PNTeBi:32
−conductorsSemiKvba μαα 423=−
THOMSON EFFECTTHOMSON EFFECTTotal energy flux thro′ conductor is
EJJJ ⋅+= JQ, surrEJJJ eQE ⋅+= Q surr
Je JeUsing the basic JQ
eJQdx
eq. for coupled flowsflows
EJJTTkJ eeE ++∂∂
−= αx eeE ∂
( )JETTk ++∂
−= α( ) eJETx
k ++∂
α
THOMSON EFFECTTHOMSON EFFECTThe heat interaction with thesurroundings due to gradient in JEg g Eis
Jd dxdxJdJJJd E
EEsurrQ xdxx⋅=−=
+, dx
Td ⎫⎧ ∂ ( ) dxJETxTk
dxd
e ⋅⎭⎬⎫
⎩⎨⎧ ++
∂∂
−= α
THOMSON EFFECTTHOMSON EFFECTSince Je is constant thro′ theconductor
dTdkTJd surrQ ∂2
dxdT
dxdk
xTk
dxsurrQ ⋅−
∂∂
−= 2,
⎟⎠⎞
⎜⎝⎛ +++
ddE
ddT
ddTJe αα
⎠⎝ dxdxdxe
THOMSON EFFECTTHOMSON EFFECTUsing the basic eq. for coupled flows,viz. dEdT
dxdE
dxdTJe λλα −−=
above eq. becomes (for homogeneousmaterial, ); constdTconstkmaterial, )..; const
dxconstk
α 2esurrQ JdJT
dJλ
, ee
surrQ
dxJT
dx−=
Thomson heat Joulean heat
THOMSON EFFECTTHOMSON EFFECTibl h ti
ddJT eα reversible heating or
cooling experienceddx due to current flowing
thro′ a temp gradientthro a temp gradient
dxdTJJ eTQ ⋅= γ, dx
Thomson coeff
dT αComparing we get
dTdT αγ =
THOMSON EFFECTTHOMSON EFFECTWe can also get a relationshipbetween Peltier, Seebeck & Thomson,coeff. by differentiating the exp. forπ b derived earlier vizπab derived earlier, viz.
( )Tbaab αα −=Π
( ) ⎟⎠⎞
⎜⎝⎛ −+−=
Π∴
dTd
dTdT
dTd ba
baab αααα( )
⎠⎝ dTdTdT ba
( ) ( )γγαα −+−= ( ) ( )baba γγαα +=
E d f L tEnd of Lecture
Analysis of thermo-electric ycircuits
Ab ti b itt∴ Above equations can be written aseQeQQQQ FLFLJ += eQeQQQQ
eeeQeQe FLFLJ +=
Substituting for affinities, theexpressions derived earlier, we getg
dXdE
TL
dXdT
TL
J QeQQ
Q1
2 ⋅−−=dXTdXT
dELdTLJ eQ 1
dXTL
dXTJ ee
eQe 2 −⋅−=
Analysis of thermo-electric ycircuits
We need to find values of the kineticcoeffs. from exptly obtainable data.p yDefining electrical conductivity as the elec flux per unit pot gradient
λas the elec. flux per unit pot. gradient under isothermal conditions we get f bfrom above
dEdELJ ee λ−=−=dXdXT
Je λ==
TL λ→ TLee λ=→