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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=→