30
Balance Equations and Entropy Inequality J. BLUHM Institut f¨ ur Mechanik Fakult¨atf¨ ur Ingenieurwissenschaften, Abteilung Bauwissenschaften Universit¨ at Duisburg-Essen, 45117 Essen Literature: de Boer, Reint: Vektor- und Tensorrechnung f¨ ur Ingenieure, Springer- Verlag, Berlin · Heidelberg · New York, 1982. Hutter, Kolumban & J ¨ ohnk, Klaus: Continuum methods of physical modeling, Springer-Verlag, Berlin · Heidelberg · New York, 2004. Holzapfel, Gerhard A.: Nonlinear solid mechanics – A continuum ap- proach for engineering. John Wiley & Sons, 2000. M ¨ uller, Ingo: Grundz¨ uge der Thermodynamik, Springer-Verlag, Berlin · Heidelberg · New York, 2001.

Computational Mechanics- Balance Equations

Embed Size (px)

Citation preview

Page 1: Computational Mechanics- Balance Equations

Balance Equations and Entropy Inequality

J. BLUHMInstitut fur Mechanik

Fakultat fur Ingenieurwissenschaften, Abteilung Bauwissenschaften

Universitat Duisburg-Essen, 45117 Essen

Literature:de Boer, Reint: Vektor- und Tensorrechnung fur Ingenieure, Springer-Verlag, Berlin · Heidelberg · New York, 1982.Hutter, Kolumban & Johnk, Klaus: Continuum methods of physicalmodeling, Springer-Verlag, Berlin · Heidelberg · New York, 2004.Holzapfel, Gerhard A.: Nonlinear solid mechanics – A continuum ap-proach for engineering. John Wiley & Sons, 2000.Muller, Ingo: Grundzuge der Thermodynamik, Springer-Verlag, Berlin ·Heidelberg · New York, 2001.

Page 2: Computational Mechanics- Balance Equations
Page 3: Computational Mechanics- Balance Equations

1. Balance equations and entropy inequality

The balance equations as well as the entropy inequality are material indepen-dent realtions:

• Balance of mass – mass conservation, continuity equation;

• Balance of momentum – conservation of momentum,principle of linear momentum,Newton‘s equation of motion;

• Balance of moment of momentum – conservation of angular momentum;

• Balance of energy – first law of thermodynamics;

• Entropy inequality – second law of thermodynamics,Clausius-Duhem inequality.

The structure of the balance laws of mass, momentum, angular momentumand energy for a single body is given by

(B )· = ( E ) ,

where the symbol ( . . . )· denotes the material time derivative of the corre-sponding quantity following the motion of the single body.

1

Page 4: Computational Mechanics- Balance Equations

structure of the (B )· = ( E )balance equationsfor a single body

B E

balance of mass mass 0(M )

balance of momentum external forcesmomentum

( l ) (k )

balance of angular external momentsangular momentum momentum

(h(0) ) (m(0) )

balance of energy internal and increments of thekinetic energy mechanical and

non-mechanicalwork

(E + K ) (W + Q )

Table 1.1: Structure of the balance laws for a single body

2

Page 5: Computational Mechanics- Balance Equations

1.1 Balance of mass

The axiom of the balance of mass (conservation of mass) reads

M = 0 , (1.1)

where

M =

B

ρ dv (1.2)

denotes the mass at time t.

With the material time derivative

M = (

B

ρ dv )· = (

B0

ρ J dV )· =

B0

( ρ J )· dV

=

B0

( ρ J + ρ J ) dV

=

B0

( ρ J + ρ J div x ) dV

=

B0

( ρ + ρ div x ) J dV

=

B

( ρ + ρ div x ) dv

(1.3)

one obtains the following form of the balance of mass:∫

B

( ρ + ρ div x ) dv = 0 . (1.4)

Considering

ρ (x , t ) =∂ρ

∂t+ grad ρ · x (1.5)

the alternative representation form

3

Page 6: Computational Mechanics- Balance Equations

B

(∂ρ

∂t+ grad ρ · x + ρ div x

︸ ︷︷ ︸

= div( ρ x )

) dv = 0

=⇒∫

B

[∂ρ

∂t+ div( ρ x ) ] dv = 0 (1.6)

can be derived. With the divergence theorem∫

B

div( ρ x ) dv =

∂B

ρ x · n da (1.7)

Equation (1.6) can be transferred to the global statement∫

B

∂ρ

∂tdv = −

∂B

ρ x · n da . (1.8)

The relations (1.4) and (1.6) are denoted as the Lagrange and Euler form ofthe balance of mass.

Provided that the integrand is continuous, from (1.4) and (1.6), respectively,directly one concludes to the local statement of the balance of mass:

ρ + ρ div x = 0 (1.9)

and

∂ρ

∂t+ div( ρ x ) = 0 , (1.10)

respectively. With help of

div x =1

JJ (1.11)

Equation (1.9) can be reformulated:

1

ρρ = −

1

JJ . (1.12)

The time integration of the aformentioned relation yields the integral form oflocal statement of the Lagrange form of the balance of mass:

4

Page 7: Computational Mechanics- Balance Equations

t∫

t0

1

ρρ dt = −

t∫

t0

1

JJ dt

ln ρ |tt0 = − ln J |tt0

ln ρ (t) − ln ρ (t0) = − ln J (t) + ln J (t0)

ln ρ − ln ρ0 = − ln J + 0

lnρ

ρ0= ln

1

J

=⇒

J = detF =ρ0

ρ. (1.13)

The quantity ρ0 = const. denotes the density of the undeformed referenceplacement at time t = t0, i.e. F = I and detF = J = 1.

5

Page 8: Computational Mechanics- Balance Equations

Summary: Balance of mass

Axiom (

B

ρ dv )· = 0

globale statements

reference

placement M =

B0

ρ0 dV = konst.

actual

placement

B

( ρ + ρ div x ) dv = 0

B

∂ρ

∂tdv = −

∂B

ρ x · n da

local statements

referenceplacement ρ0 = konst.

actualplacement ρ + ρ div x = 0

∂ρ

∂t+ div( ρ x ) = 0

integrale

form J = detF =ρ0

ρ

6

Page 9: Computational Mechanics- Balance Equations

1.2 Balance of momentum

The axiom of the balance of momentum reads:

l = k . (1.14)

Therein,

l =

B

ρ xdv ,

k =

B

ρb dv +

∂B

t da =

B

ρb dv +

∂B

σ n da

(1.15)

denote the vector of momentum and the vector of external forces. The externalforces k are to be split into two parts, namely into a part which is caused bythe local external body force ρb integrated over the actual volume and a partcaused by the external contact force t = σ n integrated over the surface of theactual placement. The quantity σ is the Cauchy stress tensor.

With the material time derivative of the momentum

l = (

B

ρ xdv )· = (

B0

ρ x J dV )· =

B0

( ρ J x )· dV

=

B0

( ρ J x + ρ J x + ρ J x ) dV

=

B0

[ ρ J x + ρ J ( div x ) x + ρ J x ] dV

=

B0

[ ρ x + ρ ( div x ) x + ρ x ] J dV

=

B

[ ρ x + ρ ( div x ) x + ρ x ] dv

=

B

[ ( ρ + ρ div x )︸ ︷︷ ︸

= 0

x + ρ x ] dv

=

B

ρ x dv

(1.16)

7

Page 10: Computational Mechanics- Balance Equations

and the divergence theorem∫

∂B

σ n da =

B

div σ dv (1.17)

the balance of momentum con be transferred to∫

B

ρ x dv =

B

( div σ + ρb ) dv

=⇒∫

B

[ div σ + ρ (b − x ) ] dv = o . (1.18)

This is the so-called Lagrange form of the balance equation. Using

ρ x = (ρ x )· − ρ x

=∂( ρ x )

∂t+ [ grad( ρ x ) ] x − ρ x

=∂( ρ x )

∂t+ [ grad( ρ x ) ] x − (− ρ div x) x

=∂( ρ x )

∂t+ [ grad( ρ x ) ] x + ρ x div x

︸ ︷︷ ︸

= div( ρ x ⊗ x )

=∂( ρ x )

∂t+ div( ρ x ⊗ x )

(1.19)

one obtains the alternative form∫

B

{div σ + ρb − [∂( ρ x )

∂t+ div( ρ x ⊗ x ) ] }dv = o

→∫

B

[ div σ + ρb −∂( ρ x )

∂t− div( ρ x ⊗ x )] dv = o

→∫

B

[ div(σ − ρ x ⊗ x ) + ρb −∂( ρ x )

∂t] dv = o

=⇒

8

Page 11: Computational Mechanics- Balance Equations

B

[∂( ρ x )

∂t− div(σ − ρ x ⊗ x ) − ρb ] dv = o . (1.20)

Considering the divergence theorem∫

B

div(σ − ρ x ⊗ x ) dv =

∂B

(σ − ρ x ⊗ x )n da (1.21)

the Euler form of the balance of momentum can be gained:

B

(∂( ρ x )

∂t− ρb ) dv =

∂B

(σ − ρ x ⊗ x )n da . (1.22)

Equation (1.18) and (1.20), respectively, directly yields the local stament

div σ + ρ (b − x ) = o (1.23)

and

∂( ρ x )

∂t− div(σ − ρ x ⊗ x ) − ρb = o , (1.24)

respectively, of the balance of momentum.

With the relation

σ =1

JFSFT =

1

JPFT (1.25)

for the Cauchy stress tensor, where

S = JF−1σ FT−1 = F−1 P , P = FS = Jσ FT−1 (1.26)

are the 2. (symmetric) and 1. (non-symmetric) Piola-Kirchhoff stress tensor,and the transport theorem

da = JFT−1 dA (1.27)

regarding surface elements, the volume integral of the divergence of σ can betransferred to

9

Page 12: Computational Mechanics- Balance Equations

B

div σ dv =

∂B

σ n da =

∂B

σ da =

∂B0

Jσ FT−1 dA

=

∂B0

PdA =

∂B0

Pn0 dA

=

B0

DivPdV .

(1.28)

Considering the transport theorem of volume elements and the integral formof the balance of mass,

dv = JdV , ρ = J−1 ρ0 , (1.29)

the balance equation (1.18) can be reformulated as follows:∫

B0

ρ0 xdV =

B

(Div P + ρ0 b ) dV .

=⇒∫

B0

[ Div P + ρ0 (b − x ) ] dV = o . (1.30)

Thus, with respect to the reference placement the local statement of the bal-ance of momentum (Lagrange form) is given by

DivP + ρ0 (b − x ) = o . (1.31)

10

Page 13: Computational Mechanics- Balance Equations

Summary: Balance of momentum

Axiom (

B

ρ x dv )· =

B

ρb dv +

∂B

t da

global statements

reference

placement

B0

[ DivP + ρ0 (b − x ) ] dV = o

actual

placement

B

[ div σ + ρ (b − x ) ] dv = o

B

(∂( ρ x )

∂t− ρb ) dv =

∂B

(σ − ρ x ⊗ x )n da

local statements

referenceplacement Div P + ρ0 (b − x ) = o

actualplacement div σ + ρ (b − x ) = o

∂( ρ x )

∂t− div(σ − ρ x ⊗ x ) − ρb = o

11

Page 14: Computational Mechanics- Balance Equations

1.3 Balance of moment of momentum

For non-polar bodies the axiom of the balance of moment (angular) momentumreads

h(0) = m(0) , (1.32)

where

h(0) =

B

x × ρ xdv ,

m(0) =

B

x × ρb dv +

∂B

x × t da =

B

x × ρb dv +

∂B

x × σ n da

(1.33)

are the moment of momentum with respect to a fixed reference point (0) ofthe configuration space and the momentum of external forces with respect tothe same reference point.

With the material time derivative of h(0),

h(0) = (

B

x × ρ x dv )· = (

B0

x × ρ x J dV )· =

B0

(x × ρ J x )· dV

=

B0

[ x × ρ J x + x × ( ρ J x )· ] dV

=

B0

[ x × ρ J x + x × ( ρ J x + ρ J x + ρ J x ) ] dV

=

B0

{ x × ρ J x + x × [ ρ J x + ρ J ( div x ) x + ρ J x ] }dV

=

B0

{ x × ρ x + x × [ ρ x + ρ ( div x ) x + ρ x ] } J dV

=

B

{ x × ρ x︸ ︷︷ ︸

= o

+ x × [ ( ρ + ρ div x )︸ ︷︷ ︸

= o

x + ρ x ] }dv

=

B

x × ρ xdv ,

(1.34)

12

Page 15: Computational Mechanics- Balance Equations

and the divergence theorem∫

∂B

x × σ n da =

B

(x × div σ + I × σ ) dv (1.35)

the balance of moment of momentum can be transferred to∫

B

x × ρ x dv =

B

x × ρb dv +

B

(x × div σ + I × σ ) dv

→∫

B

x × ρ x dv =

B

[x × ( div σ + ρb ) + I × σ ] dv

=⇒∫

B

{x × [ div σ + ρ (b − x ) ] + I × σ }dv = o . (1.36)

Considering the local statement of the balance of momentum,

div σ + ρ (b − x ) = o , (1.37)

the balance of moment of monmentum simplifies to∫

B

I × σ dv = o . (1.38)

Thus, the local statement of the balance equation reads:

I × σ = o . (1.39)

The aformentioned equation is fulfill if

σ = (σ)T , (1.40)

i.e. the Cauchy stress tensor is symmetric.

13

Page 16: Computational Mechanics- Balance Equations

Summary: Balance of moment of momentum

Axiom (

B

x × ρ x dv )· =

B

x × ρb dv +

∂B

x × t da

global statements

referenceplacement

actual

placement

B

I × σ dv = o

local statements

referenceplacement

actualplacement I × σ = o =⇒ σ = (σ)T

14

Page 17: Computational Mechanics- Balance Equations

1.4 Balance of energy (first law of thermodynamic)

For the description of thermo-mechanical effects, the balance of energy is ofmain interest regarding the coupling of thermal fields (temperature, heat flux,internal heat source) and mechanical fields (motion, velocity, acceleration, de-formations, deformation velocities). The axiom of the balance of energy (firstlaw of thermodynamic) reads:

E + K = W + Q . (1.41)

The scalar quantities

E =

B

ρ εdv ,

K =

B

1

2ρ x · xdv ,

W =

B

x · ρb dv +

∂B

x · t da =

B

x · ρb dv +

∂B

x · σ n da

=

B

x · ρb dv +

∂B

x · σ da ,

Q =

B

ρ r dv −

∂B

q · n da =

B

ρ r dv −

∂B

q · da

(1.42)

denote the internal energy (E), where ε is the specific internal energy, thekinetic energy (K), the increment of the mechanical work of the external forces(W) and the increment of the non-mechanical work Q. The non-mechanicalwork Q consists of two parts caused by the local external heat supply r = r(x, t)per mass element ρ dv and the heat influx vector q = q(x, t).

With the material time derivations

E = (

B

ρ εdv )· =

B0

( ρ ε J )· dV

=

B0

( ρ ε J + ρ ε J + ρ ε J ) dV

(1.43)

15

Page 18: Computational Mechanics- Balance Equations

=

B0

( ρ ε J + ρ ε J + ρ ε J div x ) dV

=

B

( ρ ε + ε ( ρ + ρ div x )︸ ︷︷ ︸

= 0

dv =

B

ρ ε dv ,

K = (

B

1

2ρ x · x dv )· =

B0

1

2( ρ J x · x )· dV

=

B0

1

2( ρ J x · x + ρ J x · x + ρ J x · x + ρ J x · x ) dV

=

B0

1

2( ρ J x · x + ρ J x · x div x + 2 ρ J x · x ) dV

=

B

1

2( ρ x · x + ρ x · x div x + 2 x · x ) dv

=

B

1

2[ x · x ( ρ + ρ div x )

︸ ︷︷ ︸

= 0

+ 2 ρ x · x ] dv =

B

ρ x · x dv

(1.43)

and the relations∫

∂B

x · σ da =

∂B

σT x · da =

B

div(σT x ) dv

=

B

( div σ · x + σ · grad x ) dv

=

B

( div σ · x + σ · L ) dv ,

∂B

q · da =

B

div qdv

(1.44)

the balance of energy (1.41) can be transferred to

16

Page 19: Computational Mechanics- Balance Equations

B

ρ ε dv +

B

ρ x · x dv =

=

B

[ ( div σ + ρb ) · x + σ · L ] dv +

B

( ρ r − divq ) dv .

(1.45)

Thus, the local statement of the balance of energy reads

ρ ε = [ div σ + ρ (b − x ) ] · x + σ · L + ρ r − div q . (1.46)

Considering the local statement of the balance of momentum,

div σ + ρ (b − x ) = o ,

see (1.23), one obtains the following representation form:

ρ ε − σ · L − ρ r + div q = 0 . (1.47)

Figure 1.1: Illustration of heat flux and internal heat source

Taking into account the symmetric characteristic of the stress tensor σ = σT

and the additive decomposition of L = D + W into a symmetric and a non-symmetric part (D = DT, W = −WT), the scalar product σ · L in (1.47)results in

17

Page 20: Computational Mechanics- Balance Equations

σ · L = σ · (D + W ) = σ · D + σ · W︸ ︷︷ ︸

= 0

= σ · D .(1.48)

Thus, one obtains the following statement of the balance of energy:

ρ ε − σ · D − ρ r + div q = 0 . (1.49)

The introduction of the Helmholtz free energy function ψ = ψ(x, t),

ψ = ε − Θ η , (1.50)

and the corresponding material time derivative,

ψ = ε − Θ η − Θ η , (1.51)

yields the alternative form

ρ ( ψ + Θ η + Θ η ) − σ · D − ρ r + div q = 0 (1.52)

of the local balance of energy. The quantities η = η(x, t) and Θ = Θ(x, t) arethe specific entropy and the absolute temperature.

18

Page 21: Computational Mechanics- Balance Equations

Summary: Balance of energy

Axiom (

B

ρ εdv )· + (

B

1

2ρ x · xdv )· =

=

B

x · ρb dv +

∂B

x · t da +

B

ρ r dv −

∂B

q · n da

global statements

actual

placement

B

ρ ε dv +

B

ρ x · x dv =

=

B

[ ( div σ + ρb ) · x + σ · L ] dv +

B

( ρ r − div q ) dv

B

ρ ε dv =

B

σ · Ddv +

B

( ρ r − div q ) dv

local statements

actualplacement ρ ε − σ · D − ρ r + div q = 0

ρ ( ψ + Θ η + Θ η ) − σ · D − ρ r + div q = 0

19

Page 22: Computational Mechanics- Balance Equations

Remark:The kinetic energy of a body can be transformed into potential energy andbackwards. In this case it is a matter of purely mechanical conversion ofenergy.

The kinetic energy can not disappear, but the energy distributes to the atoms,i.e. the kinetic energy of the disordered motion of atoms can be transformedinto potential energy between the atoms. One can not see this energy. There-fore, one says the kinetic energy is transformed into internal energy. Thisenergy can be felt and measured as heat. The temperature is a quantity forthe (medial) kinetic energy of the atoms.

Contributions of the internal energy:

– kinetic energy of molecules (see example),– potential energy of molecules

(liquid or vapor expand⇒ increasing distance of the molecules ⇒ evaporation),

– chemically bond energy of molecules.

20

Page 23: Computational Mechanics- Balance Equations

1.4.1 Balance of energy – alternative representation forms

As aforementioned, the axiom of the balance of energy (first law of thermody-namics) reads:

E + K = W + Q , (1.53)

where

E =

B

ρ εdv ,

K =

B

1

2ρ x · xdv ,

W =

B

x · ρb dv +

∂B

x · t da =

B

x · ρb dv +

∂B

x · σ n da ,

Q =

B

ρ r dv −

∂B

q · n da .

(1.54)

Taking into account the balance equation of mass, the material time derivativesof the internal and kinetic energy E and K are given by

E = (

B

ρ εdv )· =

B

ρ εdv ,

K = (

B

1

2ρ x · x dv )· =

B

ρ x · xdv .

(1.55)

With∫

∂B

x · σ n da =

B

( div σ · x + σ · L ) dv (1.56)

and the local statements of the balance of momentum and balance of momentof momentum (symmetric characteristic of the stress tensor σ = σ

T) the rateof the mechanical work W ca be reformulated:

21

Page 24: Computational Mechanics- Balance Equations

W =

B

x · ρb dv +

B

( div σ · x + σ · D ) dv

=

B

[ ( div σ + ρb )︸ ︷︷ ︸

= ρ x

· x + σ · D ] dv

=

B

( ρ x · x + σ · D ) dv =

B

( ρ x · x ) dv + Win .

(1.57)

The quantity

Win =

B

σ · Ddv (1.58)

denotes the rate of the internal mechanical work. With the additive decom-position of the stress tensor σ and the symmetric part of the spatial velocitygradient D into deviatorical and spherical parts,

σ = σD +

1

3(σ · I ) I = σ

D − p I , p = −1

3(σ · I ) ,

D = DD +1

3(D · I ) I ,

(1.59)

and the relation (. . . )D ·I = 0, the rate of the mechanical work can be expressedas

W =

B

σ · Ddv =

B

[ σD − p I ] · [DD +1

3(D · I ) I ] dv

=

B

[ σD · DD − p (DD · I )︸ ︷︷ ︸

= 0

+1

3(σ

D · I )︸ ︷︷ ︸

= 0

(D · I ) −

−1

3p (D · I ) ( I · I )

︸ ︷︷ ︸

= 3

] dv

=

B

(σD · DD − pD · I ) dv

=

B

(σD · DD − p div x ) dv

=

B

(σD · DD − p J−1 J ) dv .

(1.60)

22

Page 25: Computational Mechanics- Balance Equations

With help of∫

∂B

q · n da =

B

div qdv (1.61)

the rate of the non-mechanical work can be presented as

Q =

B

( ρ r − div q ) dv . (1.62)

With the details mentioned before, one obtains the following global statementof the balance of energy:

E + K =

B

ρ ε dv +

B

ρ x · xdv

=

B

( ρ x · x + σD · DD − p J−1 J ) dv +

B

( ρ r − div q ) dv

= W + Q .

(1.63)

Thus, the material time derivative of the internal energy reads:

E =

B

ρ ε dv = W + Q − K

=

B

ρ εdv

=

B

(σD · DD − p J−1 J ) dv +

B

( ρ r − div q ) dv

= Win + Q .

(1.64)

23

Page 26: Computational Mechanics- Balance Equations

1.4.2 Balance of energy for gas and non-viscous fluids

With respect to the description of the behavior of gas and non-viscous fluidsthe deviatorical part of the stresses can be approximately neglected, i.e.

σD = 0 . (1.65)

In this case, the expression for the rate of the internal work results in

Win = −

B

p J−1 J dv = −

B0

p J dV , (1.66)

where regarding the second term the transport theorem dv = JdV has beenused. With the assumption that the pressure is constant in space one obtains

Win = −p

B0

J dV = −p (

B0

J dV )· = −p (

B

dv )·

= −p V = −pdv

dt.

(1.67)

Insertion this relation into the expression for E yields:

E = Win + Q = −p V + Q . (1.68)

24

Page 27: Computational Mechanics- Balance Equations

1.5 Entropy inequality (second law of thermodynamics)

The entropy is a measure of a part of heat quantity which can not transformedinto mechanical work due to the equal distributions of the molecules of the sys-tem. Thus, the entropy inequality (second law of thermodynamics) is definedas follows

H ≥

B

1

Θρ r dv −

∂B

1

Θq · n da , (1.69)

where

H =

B

ρ η dv (1.70)

denote the entropy and η the specific entropy. Considering the materiel timederivatives

H = (

B

ρ η dv )· =

B0

( ρ η J )· dv

=

B0

( ρ η J + ρ η J + ρ η J ) dv

=

B0

( ρ η J + ρ η J + ρ η J div x ) dv

=

B

[ ρ η + η ( ρ + ρ div x )︸ ︷︷ ︸

= 0

] dv =

B

ρ η dv

(1.71)

and the relation∫

∂B

1

Θq · n da =

B

div(1

Θq ) dv (1.72)

the entropy inequality results in∫

B

ρ η dv ≥

B

[1

Θρ r − div (

1

Θq ) ] dv . (1.73)

With help of (1.73), the local statement

25

Page 28: Computational Mechanics- Balance Equations

ρ η −1

Θρ r + div (

1

Θq ) ≥ 0 . (1.74)

is gained. Using the relation

div (1

Θq ) = −

1

Θ2gradΘ · q +

1

Θdiv q (1.75)

one obtains the following form of the second law of thermodynamics:

ρ η −1

Θ( ρ r − div q ) −

1

Θ2q · gradΘ ≥ 0 . (1.76)

Taking into account the balance of energy, see (1.49),

ρ ε − σ · D − ρ r + div q = 0

and

ρ r − div q = ρ ε − σ · D ,

respectively one obtains the following representation form of the inequality(1.76):

ρ η −1

Θ( ρ ε − σ · D ) −

1

Θ2q · gradΘ ≥ 0 . (1.77)

The multiplication of the aforementioned inequality with Θ (regarding theabsolute temperature it is essential: Θ ≥ 0) yields

Θ ρ η − ρ ε + σ · D −1

Θq · grad Θ ≥ 0

=⇒

− ρ ( ε − Θ η ) + σ · D −1

Θq · grad Θ ≥ 0 . (1.78)

The inequality (1.78) can be reformulated as follows:

− ρ ( ε − Θ η − Θ η + Θ η ) + σ · D −1

Θq · gradΘ ≥ 0

=⇒

− ρ [ ( ε − Θ η )· + Θ η ] + σ · D −1

Θq · grad Θ ≥ 0 . (1.79)

26

Page 29: Computational Mechanics- Balance Equations

The expression ( ε−Θ η )· represents a total differential. Thus, these quantitiesin the brackets can be combined to one quantity. This quantity

ψ = ε − Θ η (1.80)

is called as Helmholtz free energy, compare (1.50). Finally, with (1.80) thelocal statement

− ρ ( ψ + Θ η ) + σ · D −1

Θq · grad Θ ≥ 0 . (1.81)

of the entropy inequality can be derived.

Special cases:For processes with constant temperature in space, the global statement of thesecond law of thermodynamics, see (1.69) and (1.73), respectively the globalstatement of the second law of thermodynamics simplifies to

B

ρ η dv ≥1

Θ

B

( ρ r − div q ) dv (1.82)

and

H ≥1

ΘQ , (1.83)

respectively, where (1.62) has taken into consideration. This relation is thestarting point regarding the formulation of the well-known Gibbs equation inthe framework of the gas dynamics.

Remark:With respect to the derivation of the local statement (1.52) of the balanceof energy, the introduced relationship for the Helmholtz free energy funktionψ = ε−Θ η seemed arbitrary. The expression for ψ recently arises in connectionwith the reformulation with the entropy inequality. The reformulation of theentropy implies the change of variables, i.e. the substitution of the variableset {ε, η} by the set {ψ, η}. The approach is called Legendre transformation.

27

Page 30: Computational Mechanics- Balance Equations

Summary: Entropy inequality

Axiom (

B

ρ η dv )· ≥

B

1

Θρ r dv −

∂B

1

Θq · n da

global statement

actual

placement

B

ρ η dv ≥

B

[1

Θρ r − div (

1

Θq ) ] dv

local statement

actual

placement ρ η −1

Θρ r + div (

1

Θq ) ≥ 0

− ρ ( ε − Θ η ) + σ · D −1

Θq · grad Θ ≥ 0

− ρ ( ψ + Θ η ) + σ · D −1

Θq · grad Θ ≥ 0

28