Chapter 6. Balance Laws

Continuum Mechanics

C. Agelet de SaracibarETS Ingenieros de Caminos, Canales y Puertos, Universidad Politécnica de Cataluña (UPC), Barcelona, Spain

International Center for Numerical Methods in Engineering (CIMNE), Barcelona, Spain

Chapter 6Balance Laws

October 8, 2013 Carlos Agelet de Saracibar 2

Chapter 6 · Balance Laws1. Introduction2. Conservation of mass3. Reynolds transport theorem4. Linear momentum balance5. Angular momentum balance6. Mechanical energy balance7. Assignments8. First law of thermodynamics9. Second law of thermodynamics10. Thermodynamic processes11. Governing equations

Balance Laws > Contents

Contents


IntroductionThe fundamental laws of continuum mechanics are given by four

conservation/balance laws plus a restriction law.

The four conservation/balance laws are: Conservation of mass · Mass continuity equation Linear momentum balance · Cauchy first motion equation Angular momentum balance · Symmetry of Cauchy stress First law of thermodynamics · Energy balance equation

The restriction law is given by: Second law of thermodynamics · Clausius-Planck and heat

conduction inequalities

Balance Laws > Introduction

Introduction


IntroductionThe mathematical expressions arising from the fundamental

laws will be given in: Global (or integral) form

o Global (or integral) spatial formo Global (or integral) material form

Local (or strong) formo Local (or strong) spatial formo Local (or strong) material form

Balance Laws > Introduction

Introduction


Conservation of MassWe assume that during a motion there are neither mass sources

(reservoirs that supply mass), nor mass sinks (reservoirs thatabsorb mass), so the mass of a continuum body is a conserved

quantity.

Then, the mass is independent of the motion and, hence, thematerial time derivative of the mass of a continuum body (or a material volume) has to be zero,

Balance Laws > Conservation of Mass

Conservation of Mass

( ) ( )0 0m mΩ = Ω >

( ) 0d

mdt

Ω =


Mass DensityThe mass at the material (or reference) configuration may be characterized by a continuous positive scalar field, denoted as

, which is a material property called material (orreference) mass density, such that,

The mass at the spatial (or current) configuration may be characterized by a continuous positive scalar field, denoted as

, which is called spatial (or current) mass density, such that,

Note that, taking t=0 as reference time,



( ) ( )0 0dm dVρ= >X X

( )0 0 0ρ ρ= >X

( ) ( ), , 0dm t t dvρ= >x x

( ), 0tρ ρ= >x

( ) ( ) ( )0,0 ,0 0.ρ ρ ρ= = >x X X


Conservation of Mass: Global Material FormThe mass of a continuum body (or a material volume) is a conserved quantity,

Using,

The global material form of the conservation of mass may be written as,



( ) ( ) ( ) ( )0

0 0 , 0m dV t dv mρ ρΩ Ω

Ω = = = Ω >∫ ∫X x

( ), 0dv J t dV= >X

( ) ( )( ) ( )0 0

0 , , , 0dV t t J t dVρ ρΩ Ω

= >∫ ∫X X Xϕϕϕϕ


Conservation of Mass: Local Material FormLet us consider the global material form of the conservation of mass given by,

Localizing the integral expression, the local material form of theconservation of mass reads,



( ) ( )( ) ( )0 0

0 , , , 0dV t t J t dVρ ρΩ Ω

= >∫ ∫X X Xϕϕϕϕ

( ) ( )( ) ( )0 , , , 0t t J tρ ρ= >X X Xϕϕϕϕ


Conservation of Mass: Global Material FormThe material time derivative of the mass of a continuum body(or a material volume) has to be zero,

Using,

The global material form of the conservation of mass may be written as,



( ) ( ), 0d d

m t dvdt dt

ρΩ

Ω = =∫ x

( )( ) ( )0

, , , 0d

t t J t dVdt

ρΩ

=∫ X Xϕϕϕϕ

( ), 0dv J t dV= >X


Conservation of Mass: Local Material FormLet us consider the global material form of the conservation of mass given by,

Localizing the integral expression, the local material form of theconservation of mass reads,



( )( ) ( )

( )( ) ( )( )

0

0

, , , 0

, , , 0

dt t J t dV

dt

dt t J t dV

dt

ρ

ρ

Ω

Ω

=

=

∫

∫

X X

X X

ϕϕϕϕ

ϕϕϕϕ

( )( ) ( )( ), , , 0d

t t J tdt

ρ =X Xϕϕϕϕ


Conservation of Mass: Global Spatial FormThe mass of a continuum body (or a material volume) is a conserved quantity,

Using,

The global spatial form of the conservation of mass may be written as,



( )1 , 0dV J t dv−= >x

( )( ) ( ) ( )1 1

0 , , , 0t J t dv t dvρ ρ− −

Ω Ω= >∫ ∫x x xϕϕϕϕ

( ) ( ) ( ) ( )0

0 0 , 0m dV t dv mρ ρΩ Ω

Ω = = = Ω >∫ ∫X x


Conservation of Mass: Local Spatial FormLet us consider the global spatial form of the conservation of mass given by,

Localizing the integral expression, the local spatial form of theconservation of mass reads,



( )( ) ( ) ( )1 1

0 , , , 0t J t dv t dvρ ρ− −

Ω Ω= >∫ ∫x x xϕϕϕϕ

( )( ) ( ) ( )1 1

0 , , , 0t J t tρ ρ− − = >x x xϕϕϕϕ


Conservation of Mass: Global Spatial FormThe material time derivative of the mass of a continuum body(or a material volume) has to be zero,

The global spatial form of the conservation of mass may be written as,



( ), 0d

t dvdt

ρΩ

=∫ x

( ) ( ), 0d d

m t dvdt dt

ρΩ

Ω = =∫ x


Conservation of Mass: Local Spatial FormLet us consider the global spatial form of the conservation of mass given by,

Localizing the integral expression, the local spatial form of theconservation of mass, or mass continuity equation, reads,



( )

( ) ( )

( ) ( )

0 0 0

0

, 0

div div 0

dt dv

dt

d d ddv JdV J dV J J dV

dt dt dt

JdV dv

ρ

ρ ρ ρ ρ ρ

ρ ρ ρ ρ

Ω

Ω Ω Ω Ω

Ω Ω

=

= = = +

= + = + =

∫

∫ ∫ ∫ ∫

∫ ∫

x

v v

( ) ( ) ( ), , div , 0t t tρ ρ+ =x x v x


Conservation of Mass: Local Spatial FormThe local spatial form of the conservation of mass, or masscontinuity equation, may be written as,

Using the following expressions,

The local spatial form of the conservation of mass, or masscontinuity equation, may be alternatively written as,



( )( ) ( )( )

,div , , 0

tt t

t

ρρ

∂+ =

∂

xx v x

( ) ( ) ( )grad , div div gradt

ρρ ρ ρ ρ ρ

∂= + ⋅ = + ⋅

∂v v v v

( ) ( ) ( ), , div , 0t t tρ ρ+ =x x v x




Material Time DerivativeGiving the spatial description of an arbitrary property, the material time derivative of the property can be written as,Global and Local Spatial Forms

Global and Local Material Forms

( )

( )

div 0

div 0, div 0

ddv dv

dt

t

ρ ρ ρ

ρρ ρ ρ

Ω Ω= + =

∂+ = + =

∂

∫ ∫ v

v v

( )

( )

0 0

0

0

0, 0

d dJdV J dV

dt dt

dJ J

dt

ρ ρ

ρ ρ ρ

Ω Ω= =

= = >

∫ ∫


Convective Flux of an Arbitrary PropertyConsider an arbitrary property of a continuum medium and let us denote as the spatial description of the amount of the property per unit of mass, the spatial density field

and the spatial velocity field.The convective flux of the property through a fixed spatial

surface with unit normal , i.e. the amount of the propertycrossing the spatial surface per unit of time due to theconvective flux, is given by,

Balance Laws > Reynolds Transport Theorem

Convective Flux

( ), tv x

( ), tn x

( ), tρ x

A

( ), tψ x

A

( ) ( ) ( ) ( ) ( ), , , ,s

t t t t t dsφ ρ ψ= ⋅∫ x x v x n xA


Convective Flux of an Arbitrary PropertyThe net outcoming convective flux of the property through a fixed closed spatial surface with unit outward normal , i.e. the net amount of the property leaving the spatial volumeper unit of time due to the convective flux, i.e. outflow (+) plus inflow (-), is given by,


Convective Flux

( ), tn x

A

( ) ( ) ( ) ( ) ( ), , , ,v

t t t t t dsφ ρ ψ∂

= ⋅∫ x x v x n xA

inflowoutflow0⋅ ≤v n

0⋅ ≥v n

A


Mass FluxGiven a spatial density field, denoted as and a spatial

velocity field, denoted as , the mass flux through a fixed

spatial surface with unit normal is given by,

Note that the mass flux may be viewed as a particular case of theconvective flux of an arbitrary property , setting .The net outcoming mass flux through a closed surface with unitoutward normal is given by,


Convective Flux

( ), tv x

( ), tn x

( ), tρ x

( ) ( ) ( ) ( ), , ,Ms

t t t t dsφ ρ= ⋅∫ x v x n x

A 1ψ =

( ) ( ) ( ) ( ), , ,Mv

t t t t dsφ ρ∂

= ⋅∫ x v x n x

( ), tn x


Volume FluxGiven a spatial velocity field, denoted as , the volume flux through a fixed spatial surface with unit normal is givenby,

Note that the volume flux may be viewed as a particular case of the convective flux of an arbitrary property , setting . The net outcoming volume flux through a closed surface withunit outward normal is given by,


Convective Flux

( ), tv x

( ), tn x

( ) ( ) ( ), ,Vs

t t t dsφ = ⋅∫ v x n x

A 1ψ ρ −=

( ) ( ) ( ), ,Vv

t t t dsφ∂

= ⋅∫ v x n x

( ), tn x


Reynolds LemmaConsider an arbitrary property of a continuum medium and let us denote as the spatial description of the amount of the property per unit of mass.The amount of the property at the current time t can be written as,

The material time derivative of the property can be writtenas,


Reynolds Transport Theorem

A

( ), tψ x

A

( )t dvρψΩ

= ∫A

A

( )d

t dvdt

ρψΩ

= ∫A


Reynolds LemmaThe material time derivative of the property may be writtenas,



A

( )

( )

( )

0 0

0 00 0

dt dv

dt

d dJdV J dV

dt dt

ddV dV

dt

dv

ρψ

ρψ ρψ

ρ ψ ρ ψ

ρψ

Ω

Ω Ω

Ω Ω

Ω

=

= =

= =

=

∫

∫ ∫

∫ ∫

∫

A


Reynolds LemmaThe Reynolds Lemma for an arbitrary property takes theform,

Note that Reynolds lemma may be directly obtained using massconservation, taking into account that,



A

ddv dv

dtρψ ρψ

Ω Ω=∫ ∫

( ) ( ), , 0dm t t dvρ= >x x


Reynolds Transport TheoremThe following key expression holds,



( )

( ) ( )

( ) ( )

( ) ( )

grad

grad div

div

d

dt

t

t

t

ρψ ρψ ρψ

ρψ ρψ ρψ

ρψ ρψ ρψ

ρψ ρψ

= −

∂= + ⋅ −

∂

∂= + ⋅ +

∂

∂= +

∂

v

v v

v


Reynolds Transport TheoremThe Reynolds transport theorem for an arbitrary property

may be written as,



A

( )

( ) ( )

( )

div

dt dv

dt

dv

dv dvt

dv dst

ρψ

ρψ

ρψ ρψ

ρψ ρψ

Ω

Ω

Ω Ω

Ω ∂Ω

=

=

∂= +

∂

∂= + ⋅

∂

∫

∫

∫ ∫

∫ ∫

v

v n

A


Reynolds Transport Theorem: Mass ConservationTaking the mass as a particular case of an arbitrary property, theReynolds transport theorem yields a global spatial form of theconservation of mass and take the form,



( ) ( )

( )

0

div

ddv

dt

dv dvt

dv dst

ρ

ρ ρ

ρ ρ

Ω

Ω Ω

Ω ∂Ω

=

∂= +

∂

∂= + ⋅

∂

∫

∫ ∫

∫ ∫

v

v n




Material Time DerivativeGiving the spatial description of an arbitrary property, the material time derivative of the property can be written as,Global Spatial Form

Local Spatial Form

( ) ( )

( )

div

ddv dv

dt

dv dvt

dv dst

ρψ ρψ

ρψ ρψ

ρψ ρψ

Ω Ω

Ω Ω

Ω ∂Ω

=

∂= +

∂

∂= + ⋅

∂

∫ ∫

∫ ∫

∫ ∫

v

v n

( ) ( )divt

ρψ ρψ ρψ∂

= +∂

v


Assignment 6.1 [Classwork]Consider the spatial description of a velocity field given by,

The reference time is t=0. The mass density at the referenceconfiguration is constant. Obtain the spatial mass density and the mass flux through the open cylindrical surface S of the figure.


Assignment 6.1

, , 0t

x y zv ye v y v−= = =

A

A


Balance Laws > Linear Momentum Balance

Linear Momentum Balance

Linear MomentumThe linear momentum of a material volume, denoted as , is defined as a vector-valued function given by,

( )LtM

( ) ( ) ( )

( ) ( )0

0

, ,

,

L t t t dv

t dV

ρ

ρ

Ω

Ω

=

=

∫

∫

M x v x

X V X




Linear MomentumUsing Reynolds Lemma, i.e. conservation of mass, the material

time derivative of the linear momentum takes the form,

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )0 0

0 0

, , , ,

, ,

L

dt t t dv t t dv

dt

dt dV t dV

dt

ρ ρ

ρ ρ

Ω Ω

Ω Ω

= =

= =

∫ ∫

∫ ∫

M x v x x v x

X V X X V X




Resultant ForceThe resultant force acting on a material volume, denoted as is defined as a vector-valued function given by,

( ) ,tF

( ) ( ) ( ) ( )

( ) ( ) ( )0 0

0

, , ,

, ,

t t t dv t ds

t dV t dS

ρ

ρ

Ω ∂Ω

Ω ∂Ω

= +

= +

∫ ∫

∫ ∫

F x b x t x

X B X T X




Linear Momentum Balance LawThe linear momentum balance law states that the time-

variation of the linear momentum of a material volume is equalto the resultant force acting on that material volume.

If the continuum body is in equilibrium, the resultant force iszero and the linear momentum is a conserved quantity,

( ) ( ) ( )L L

dt t t

dt= =M M F

( ) ( )L L

dt t

dt= =M M 0




Linear Momentum Balance Law: Global Spatial FormThe global spatial form of the linear momentum balance lawcan be written as,

( ) ( ) ( ) ( )

( ) ( ) ( )

, , , ,

, , ,

dt t dv t t dv

dt

t t dv t ds

ρ ρ

ρ

Ω Ω

Ω ∂Ω

=

= +

∫ ∫

∫ ∫

x v x x v x

x b x t x




Linear Momentum Balance Law: Global Spatial FormThe surface forces can be written in spatial form as,

Substituting into the global spatial form yields,

divds ds dv∂Ω ∂Ω Ω

= =∫ ∫ ∫t nσ σσ σσ σσ σ

( ) ( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )( )

, , , ,

, , div ,

, , div ,

dt t dv t t dv

dt

t t dv t dv

t t t dv

ρ ρ

ρ

ρ

Ω Ω

Ω Ω

Ω

=

= +

= +

∫ ∫

∫ ∫

∫

x v x x v x

x b x x

x b x x

σσσσ

σσσσ




Linear Momentum Balance Law: Local Spatial FormLocalizing, the local spatial form of the linear momentumbalance law, known as Cauchy’s first equation of motion, can be written as,

If the resultant force is zero, the local spatial form of the linear momentum balance law can be written as,

( ) ( ) ( ) ( ) ( ), , , , div ,t t t t tρ ρ= +x v x x b x x σσσσ

( ) ( ) ( ), , div ,t t tρ= +0 x b x xσσσσ




Linear Momentum Balance Law: Global Material FormThe global material form of the linear momentum balance lawcan be written as,

( ) ( ) ( ) ( )

( ) ( ) ( )

0 0

0 0

0 0

0

, ,

, ,

dt dV t dV

dt

t dV t dS

ρ ρ

ρ

Ω Ω

Ω ∂Ω

=

= +

∫ ∫

∫ ∫

X V X X V X

X B X T X




Linear Momentum Balance Law: Global Material FormThe surface forces can be written in material form as,

Substituting into the global material form yields,

0 0 0

DIVdS dS dV∂Ω ∂Ω Ω

= =∫ ∫ ∫T PN P

( ) ( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )( )

0 0

0 0

0

0 0

0

0

, ,

, DIV ,

, DIV ,

dt dV t dV

dt

t dV t dV

t t dV

ρ ρ

ρ

ρ

Ω Ω

Ω Ω

Ω

=

= +

= +

∫ ∫

∫ ∫

∫

X V X X V X

X B X P X

X B X P X




Linear Momentum Balance Law: Local Material FormLocalizing, the local material form of the linear momentumbalance law can be written as,

If the resultant force is zero, the local material form of the linear momentum balance law can be written as,

( ) ( ) ( ) ( ) ( )0 0, , DIV ,t t tρ ρ= +X V X X B X P X

( ) ( ) ( )0 , DIV ,t tρ= +0 X B X P X






0 0 0 00 0 0

0 0 DIV

ddV dV dV dS

dtρ ρ ρ

ρ ρ

Ω Ω Ω ∂Ω= = +

= +

∫ ∫ ∫ ∫v v b T

v b P

div

ddv dv dv ds

dtρ ρ ρ

ρ ρ

Ω Ω Ω ∂Ω= = +

= +

∫ ∫ ∫ ∫v v b t

v b

σσσσ


Balance Laws > Angular Momentum Balance

Angular Momentum Balance

Angular MomentumThe angular momentum of a material volume about a fixed

spatial point , denoted as , is defined as a vector-valuedfunction given by,

where the vector position is defined as,

( )AtM

( ) ( ) ( ) ( )

( ) ( ) ( )0

0

, , ,

, ,

A t t t t dv

t t dV

ρ

ρ

Ω

Ω

= ×

= ×

∫

∫

M r x x v x

R X X V X

0x

( ) ( ), ,t t= =r r x R X

( ) ( ) ( )0 0, , ,t t t= = − = − =r r x x x X x R Xϕϕϕϕ




Angular MomentumUsing Reynolds Lemma, i.e. conservation of mass, the material

time derivative of the angular momentum takes the form,

( ) ( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

0

0

0

0

, , ,

, , ,

, ,

, ,

A

dt t t t dv

dt

t t t dv

dt t dV

dt

t t dV

ρ

ρ

ρ

ρ

Ω

Ω

Ω

Ω

= ×

= ×

= ×

= ×

∫

∫

∫

∫

M r x x v x

r x x v x

R X X V X

R X X V X




Resultant MomentThe resultant moment about a fixed spatial point , denoted as

is defined as a vector-valued function given by,( ) ,tM

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )0 0

0

, , , , ,

, , , ,

t t t t dv t t ds

t t dV t t dS

ρ

ρ

Ω ∂Ω

Ω ∂Ω

= × + ×

= × + ×

∫ ∫

∫ ∫

M r x x b x r x t x

R X X B X R X T X

0x




Reference Configuration Current Configuration

0ΩΩ

time 0t = time t

1e

2e

3e

1 1,X x

2 2,X x

dA

N

dA

da

n

da

ϕϕϕϕ

X

x

F

0x

0= −r x x

t

T

T

P’

dv

ρb

0ρ B

PdV

0ρ B

0= −r x x

3 3,X x x




Angular Momentum Balance LawThe angular momentum balance law states that the time-

variation of the angular momentum of a material volume abouta fixed spatial point, is equal to the resultant moment about thisfixed spatial point, acting on that material volume.

If the resultant moment about the fixed spatial point is zero, then the angular momentum about this spatial point is a conserved quantity,

( ) ( ) ( )A A

dt t t

dt= =M M M

( ) ( )A A

dt t

dt= =M M 0




Angular Momentum Balance: Global Spatial FormThe global spatial form of the angular momentum balance lawcan be written as,

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( )

( ) ( )

, , , , , ,

, , ,

, ,

dt t t dV t t t dv

dt

t t t dv

t t ds

ρ ρ

ρ

Ω Ω

Ω

∂Ω

× = ×

= ×

+ ×

∫ ∫

∫

∫

r x x v x r x x v x

r x x b x

r x t x




Angular Momentum Balance: Global Spatial FormThe moment of the surface forces can be written in spatial form

as,

div abc cb ads ds dv dvε σ∂Ω ∂Ω Ω Ω

× = × = × +∫ ∫ ∫ ∫r t r n r eσ σσ σσ σσ σ

( )

( )

,

, ,

abc b c a abc b cd d a

abc b cd ad

abc b cd d a abc b d cd a

abc b a abc cb ac

r t ds r n ds

r dv

r dv r dv

r dv dv

ε ε σ

ε σ

ε σ ε σ

ε ε σ

∂Ω ∂Ω

Ω

Ω Ω

Ω Ω

=

=

= +

= ∇⋅ +

∫ ∫

∫

∫ ∫

∫ ∫

e e

e

e e

e eσσσσ




Angular Momentum Balance: Global Spatial FormSubstituting into the global spatial form yields,

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( )

( ) ( )

( )

, , , , , ,

, , ,

, div ,

,abc cb a

dt t t dV t t t dv

dt

t t t dv

t t dv

t dv

ρ ρ

ρ

ε σ

Ω Ω

Ω

Ω

Ω

× = ×

= ×

+ ×

+

∫ ∫

∫

∫

∫

r x x v x r x x v x

r x x b x

r x x

x e

σσσσ




Angular Momentum Balance: Local Spatial FormLocalizing, the local spatial form of the angular momentumbalance law yields,

Using the Cauchy’s first motion equation yields,

and the local spatial form of the angular momentum balance law can be written as,

( ) ( ), ,Tt t=x xσ σσ σσ σσ σ

( )div abc cb aρ ρ ε σ× = × + +r v r b e σσσσ

abc cb aε σ =e 0




Angular Momentum Balance: Global Material FormThe global material form of the angular momentum balance lawcan be written as,

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( )

( ) ( )

0 0

0

0

0 0

0

, , , ,

, ,

, ,

dt t dV t t dV

dt

t t dV

t t dS

ρ ρ

ρ

Ω Ω

Ω

∂Ω

× = ×

= ×

+ ×

∫ ∫

∫

∫

R X X V X R X X V X

R X X B X

R X T X




Angular Momentum Balance: Global Material FormThe moment of the surface forces can be written in material

form as,( )

0 0 0 0

DIVT

abc acbdS dS dV dVε

∂Ω ∂Ω Ω Ω× = × = × +∫ ∫ ∫ ∫r T r PN r P PF e

( )

( )

( )

0 0

0

0 0

0 0

0 0

,

, ,

DIV

abc b c a abc b cD D a

abc b cD aD

abc b cD D a abc b D cD a

abc b a abc bD cD ac

T

abc acb

r T dS r P N dS

r P dV

r P dV r P dV

r dV F P dV

dV dV

ε ε

ε

ε ε

ε ε

ε

∂Ω ∂Ω

Ω

Ω Ω

Ω Ω

Ω Ω

=

=

= +

= ∇⋅ +

= × +

∫ ∫

∫

∫ ∫

∫ ∫

∫ ∫

e e

e

e e

P e e

r P PF e




Angular Momentum Balance: Global Material FormSubstituting into the global material form yields,

( )

0 0

0 0

0

0 0

0 DIV

T

abc acb

ddV dV

dt

dV dV

dV

ρ ρ

ρ

ε

Ω Ω

Ω Ω

Ω

× = ×

= × + ×

+

∫ ∫

∫ ∫

∫

r v r v

r b r P

PF e




Angular Momentum Balance: Local Material FormLocalizing, the local material form of the angular momentumbalance law can be written as,

Using the Cauchy’s first motion equation yields,

and the local material form of the angular momentum balance law can be written as,

( ) ( ) ( ) ( ), , , ,T Tt t t t=P X F X F X P X

( )0 0 DIVT

abc acbρ ρ ε× = × + × +r v r b r P PF e

( )T

abc acbε =PF e 0






0 0

0 0

0 0

0

T T

ddV dV

dt

dV dS

ρ ρ

ρ

Ω Ω

Ω ∂Ω

× = ×

= × + ×

=

∫ ∫

∫ ∫

r v r v

r b r T

PF FP

T

ddV dv

dt

dv ds

ρ ρ

ρ

Ω Ω

Ω ∂Ω

× = ×

= × + ×

=

∫ ∫

∫ ∫

r v r v

r b r t

σ σσ σσ σσ σ


Balance Laws > Mechanical Energy Balance

Kinetic Energy

Kinetic EnergyThe global spatial form of the kinetic energy of a continuum body, denoted as , takes the form,

The global material form of the kinetic energy of a continuum body, denoted as , takes the form,

( )tK

( ) ( ) ( ) ( ) ( ) ( )21 1

, , , , ,2 2

t t t dv t t t dvρ ρΩ Ω

= = ⋅∫ ∫x v x x v x v xK

( ) ( ) ( ) ( ) ( ) ( )0 0

2

0 0

1 1, ,

2 2, dV t t dVt tρ ρ

Ω Ω= ⋅= ∫ ∫X X V X V XV XK

( )tK



Internal Mechanical Power

Internal Mechanical PowerThe internal mechanical power per unit of spatial volume, is thework done by the stresses per unit of time and unit of spatialvolume, and can be written as,

( ) ( )

( ) ( )

( ) ( ) ( )

( ) ( )

1

1 1 1 1

1 1 1

1

1 1 1 1

1 1 1

: :

: : : :

: : :

T T

T T

ab ab ab ab

T T

aA Ab ab aA Ab ab aA ab bA aA aA

T T

aA AB Bb ab AB Aa ab bB AB AB

J

J J J J

J J J

d J d

J P F d J P F l J P l F J P F

J F S F d J S F d F J S E

σ τ

−

− − − −

− − −

−

− − − −

− − −

=

= = = =

= = =

=

= = = =

= = =

d d

PF d PF l P lF P F

FSF d S F dF S E

σ τσ τσ τσ τ




Internal Mechanical PowerThe internal mechanical power in the continuum body, denotedas , i.e. work done per unit of time by the stresses, can be written as,

( )

0

0

0

1

1

1

:

: :

: :

: :

int t dv

J dv dV

J dv dV

J dv dV

Ω

−

Ω Ω

−

Ω Ω

−

Ω Ω

=

= =

= =

= =

∫

∫ ∫

∫ ∫

∫ ∫

d

d d

P F P F

S E S E

P σσσσ

τ ττ ττ ττ τ

( )int tP




Material Time DerivativeGiving the spatial description of an arbitrary property, the material time derivative of the property can be written as,Global Spatial Form

Global Material Forms

( ) :int t dvΩ

= ∫ dσσσσP

( )0

0

0

:

:

:

int t dV

dV

dV

Ω

Ω

Ω

=

=

=

∫

∫

∫

d

P F

S E

ττττP



External Mechanical Power

External Mechanical PowerThe external mechanical power, denoted as , is the workdone per unit of time by the body forces and surface forces, and can be written in global spatial and material forms, respectively, as,

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )0 0

0

, , , , ,

, , , ,

extt t t t dv t t ds

t t dV t t dS

ρ

ρ

Ω ∂Ω

Ω ∂Ω

= ⋅ + ⋅

= ⋅ + ⋅

∫ ∫

∫ ∫

x b x v x t x v x

X B X V X T X V X

P

( )exttP



Mechanical Energy Balance

Mechanical Energy BalanceThe external mechanical power, denoted as , i.e. workdone per unit of time by the body forces and surface forces, can be written in global spatial form as,

The external mechanical power of the surface forces can be written in global spatial form as,

( )extt dv dsρ

Ω ∂Ω= ⋅ + ⋅∫ ∫b v t vP

( )exttP

( )div div : grad

ds ds

dv dv dv

∂Ω ∂Ω

Ω Ω Ω

⋅ = ⋅

= ⋅ = ⋅ +

∫ ∫

∫ ∫ ∫

t v v n

v v v

σσσσ

σ σ σσ σ σσ σ σσ σ σ




Then, using the local spatial form of the linear momentum

balance equation, the external mechanical power can be writtenin global spatial form as,

( )

( )

( ) ( )

2

2

div : grad

div : grad

1: :

2

1:

2

ext

int

t dv ds

dv dv dv

dv dv

d ddv dv dv dv

dt dt

d ddv dv t t

dt dt

ρ

ρ

ρ

ρ ρ

ρ

Ω ∂Ω

Ω Ω Ω

Ω Ω

Ω Ω Ω Ω

Ω Ω

= ⋅ + ⋅

= ⋅ + ⋅ +

= + ⋅ +

= ⋅ + = +

= + = +

∫ ∫

∫ ∫ ∫

∫ ∫

∫ ∫ ∫ ∫

∫ ∫

b v t v

b v v v

b v v

vv l v d

v d

P

K P




σσσσ




Mechanical Energy BalanceThe mechanical energy balance states that the external

mechanical power supplied to the continuum body is spent in changing its kinetic energy and doing an internal mechanical

power.

( ) ( ) ( )ext int

dt t t

dt= +P K P




Mechanical Energy Balance: Global Spatial FormThe global spatial form of the mechanical energy balance can be written as,

( )

21:

2

ext t dv ds

ddv dv

dt

ρ

ρ

Ω ∂Ω

Ω Ω

= ⋅ + ⋅

= +

∫ ∫

∫ ∫

b v t v

v d

P

σσσσ




Mechanical Energy Balance: Global Material FormThe global material form of the mechanical energy balance can be written as,

( )0 0

0 0

0 0

0 0

0

2

0

2

0

2

0

1:

2

1:

2

1:

2

ext t dV dS

ddV dV

dt

ddV dV

dt

ddV dV

dt

ρ

ρ

ρ

ρ

Ω ∂Ω

Ω Ω

Ω Ω

Ω Ω

= ⋅ + ⋅

= +

= +

= +

∫ ∫

∫ ∫

∫ ∫

∫ ∫

b v T v

v d

v P F

v S E

P

ττττ




Quasistatic ProblemIf the material time derivative of the kinetic energy is zero (ornegligible), the problem is called quasistatic and the mechanical

energy balance reads,

( ) ( )ext intt t=P P




Free Vibration ProblemIf the external mechanical power is zero (or negligible), theproblem is called free vibration problem and the mechanical


( ) ( )0 int

dt t

dt= +K P




Conservative Mechanical SystemIf both the external and internal mechanical power derive from a potential, i.e. a potential energy for the external loading and a strain energy, such that,

the problem is said to be conservative and the mechanical


( ) ( ) ( ) ( ),ext ext int int

d dt t t t

dt dt= − Π = ΠP P

( ) ( ) ( )

( ) ( ) ( )( ) 0

ext int

int ext

d d dt t t

dt dt dt

dt t t

dt

− Π = + Π

+ Π + Π =

K

K




Conservative Mechanical SystemThe total potential energy, denoted as , is defined as thesum of the potential energy of the external loading, denoted as

and the strain energy, denoted as , yielding,

In a conservative mechanical system the sum of the kinetic

energy and the total potential energy is a conserved quantity,

( ) ( ) constantt t+ Π =K

( )tΠ

( )ext tΠ ( )int tΠ

( ) ( ) ( )( ) ( ) ( )( ) 0int ext

d dt t t t t

dt dt+ Π + Π = + Π =K K


Balance Laws > Assignments

Assignment 6.2

Assignment 6.2A volume flux Q of an incompressible fluid flows in stationary

conditions through the pipeline of the figure. Velocity and pressure distributions at the sections A and B are uniform. Thepipeline is fixed through a rigid bar OP. The weights of thepipeline, rigid bar and fluid are neglected.

O

pA vA

pB

R

θ

P

SA

MF

eA

vB

SB

eB



Assignment 6.2

1) Obtain the velocities at the sections A and B in terms of Q.2) Obtain the reaction force F and moment M at the point O.3) Obtain the values of the angle θ that maximize and minimize

the reaction at the point O.4) Obtain the external power needed to keep the volume flux Q

if the fluid is an incompressible ideal fluid with a sphericalstress state given byen by .p= − 1σσσσ

O

pA vA

pB

R

θ

P

SA

MF

eA

vB

SB

eB



Assignment 6.2

The local spatial form of the conservation of mass or mass

continuity equation, plus the incompressibility condition yields,

The global spatial form of the conservation of mass for anincompressible medium reads,

Then the velocities at the sections A and B are given by,

div 0, 0 div 0ρ ρ ρ+ = = ⇒ =v v

div

0

A B

V V

S S

A A B B

dV dS

dS dS

v S v S

∂= ⋅

= ⋅ + ⋅

= − + =

∫ ∫

∫ ∫

v v n

v n v n

,A A B B A A B Bv S v S Q v Q S v Q S= = ⇒ = =



Assignment 6.2

The global spatial form of the linear momentum balance for a stationary motion reads,

/ fV

V

ddV

dt

dVt

ρ

ρ

=

∂=

∂

∫

∫

R v

v ( ) ( )

( ) ( )2 2

2 2

A B

V V

S S

A A A B B B A B

A B

dS dS

dS dS

Q Qv S v S

S S

ρ ρ

ρ ρ

ρ ρ ρ ρ

∂ ∂+ ⋅ = ⋅

= ⋅ + ⋅

= − + = − +

∫ ∫

∫ ∫

v v n v v n

v v n v v n

e e e e



Assignment 6.2

The resultant force acting on the volume V of fluid, taking intoaccount that the weight of the fluid is negligible, reads,

/ fV

dVρ= ∫R b

/

/

2 2

A B

A B

V V

wall S S

wall f p p

wall f A A A B B B

A B

A B

dS dS

dS dS dS

p S p S

Q Q

S Sρ ρ

∂ ∂+ =

= + +

= + +

= + −

= − +

∫ ∫

∫ ∫ ∫

t t

t t t

R R R

R e e

e e



Assignment 6.2

The resultant force of the wall of the pipeline acting on the

volume V of fluid, reads,

Using the action-reaction principle, the resultant force of the

fluid acting on the wall of the pipeline, reads,

/ /

2 2

wall f f A A A B B B

A A A B B B

A B

p S p S

Q Qp S p S

S Sρ ρ

= − +

= − + + +

R R e e

e e

/ /

2 2

f wall wall f

A A A B B B

A B

Q Qp S p S

S Sρ ρ

= −

= + − +

R R

e e



Assignment 6.2

The global spatial form of the angular momentum balance aboutthe point O, for a stationary motion, reads,

/

O

fV

V

ddV

dt

dVt

ρ

ρ

= ×

∂= ×

∂

∫

∫

M r v

r v ( ) ( )

( )A

V V

S

dS dS

dS

ρ ρ

ρ

∂ ∂+ × ⋅ = × ⋅

= × ⋅

∫ ∫

∫

r v v n r v v n

r v v n ( )

22

BS

B B z z

B

dS

Qv S R R

S

ρ

ρ ρ

+ × ⋅

= − = −

∫ r v v n

e e



Assignment 6.2

The resultant moment about the point O acting on the volume V of fluid, taking into account that the weight of the fluid is

negligible, reads,

/

O

fV

dVρ= ×∫M r b

/

/

2

A B

A B

V V

wall S S

O O O

wall f p p

O

wall f B B z

z

B

dS dS

dS dS dS

p S R

QR

Sρ

∂ ∂+ × = ×

= × + × + ×

= + +

= +

= −

∫ ∫

∫ ∫ ∫

r t r t

r t r t r t

M M M

M e

e



Assignment 6.2

The resultant moment of the wall of the pipeline acting on the

volume V of fluid about the point O, reads,

Using the action-reaction principle, the resultant moment of the

fluid acting on the wall of the pipeline about the point O, reads,

2

/ /

O O

wall f f B B z B B z

B

Qp S R p S R

Sρ

= − = − +

M M e e

/ /

2

O O

f wall wall f

B B z

B

Qp S R

Sρ

= −

= +

M M

e



Assignment 6.2

Equilibrium of forces and moments at point O

aF

2

/

O

f wall B B

B

QM p S R

Sρ

= +

2

, /a f wall A A

A

QR p S

Sρ

= +

2

, /b f wall B B

B

QR p S

Sρ

= +

bF

M

O



Assignment 6.2

The equilibrium of forces and moments about the point O on the

pipeline, taking into account that the weight of the pipeline and

the rigid bar are negligible, reads,

The reaction force F and moment M, at the point O, read,

/f wall+R W

/

O O

f wall W

+ =

+

F 0

M M + =M 0

( ) ( )2 2

/

2

/

f wall A A A A B B B B

O

f wall B B z

B

Q S p S Q S p S

Qp S R

S

ρ ρ

ρ

= − = − + + +

= − = − +

F R e e

M M e



Assignment 6.2

The reactions force F at the point O, reads,

The norm of the reaction force F will take a maximum for anangle such that,

The norm of the reaction force F will take a minimum for anangle such that,

( ) ( )2 2

0 0

A A A A B B B BQ S p S Q S p Sρ ρ

> >

= − + + +F e e

3

2B A

πθ= − ⇒ =e e

2B A

πθ= ⇒ =e e



Assignment 6.2

The external mechanical power needed to keep the volume flux Q, taking into account that the fluid is incompressible, stationary

and the stress state is spherical, is given by,

2

2

1:

2

1

2

ext intV V

V

d ddV dV

dt dt

dVt

ρ

ρ

= + = +

∂=

∂

∫ ∫

∫

v d

v

P K P σσσσ

21tr

2V VdS p dVρ

∂+ ⋅ + −∫ ∫v v n d

2 2

3

2 2

1 1

2 2

1 1 1

2

A BS S

B A

dS dS

QS S

ρ ρ

ρ

= ⋅ + ⋅

= −

∫ ∫v v n v v n



Assignment 6.3

h

h

p = patm ≈ 0

C h/2

h/2v2v1

A

E

B

D

F

x

z

y

Assignment 6.3 [Classwork]An incompressible fluid flows in stationary conditions throughthe pipeline of the figure. Velocities and pressures distributionsare uniform at the sections AE and CD. Pressure on the walls isassumed to be uniform. There is a basculant barrier BC with a hinge on B. An horizontal force F, acting on the point C, iskeeping the barrier in vertical position. Body forces in the fluid are neglected. The weight of the barrier is also neglected.

h

h

p = patm ≈ 0

C h/2

h/2v2v1

A

E

B

D

F

x

z

y



Assignment 6.3

1) Obtain the velocity v2 at the section CD in terms of thevelocity v1 at the section AE.

2) Obtain the resultant force and moment acting on the fluid at the point B.

3) Obtain the resultant force and moment of the fluid on the

barrier at the point B.4) Obtain the force F and the reaction at the point B.5) Obtain the external mechanical power needed, assuming the

stress tensor in the fluid is spherical,given by .p= − 1σσσσ



Assignment 6.4

Assignment 6.4 [Homework]The figure shows the longitudinal section of a pump with a valveOA of weight W per unit of width (normal to the plane of thefigure). There is a hinge on O. The velocity of the pump is V.

v1

αaP2

b a

a/2 a/2

W

V

P1

v2

P2

B

A

O

B

A

Detail

O

αa

P1

The fluid is incompressibleand the motion stationary.Uniform pressure distribu-tions on the sections OB and AB are p1 and p2=0, respectively. Velocity dis-tributions at the sectionsOB and AB are uniform. Body forces in the fluid are negligible.



Assignment 6.4

Assignment 6.4 [Homework]1) Obtain the uniform velocities v1 and v2 at the sections OB

and AB, respectively, in terms of the velocity v of thepumping tool

2) Obtain the resultant of the forces per unit of width given bythe fluid on the valve OA

3) Obtain the moment per unit of width at the point O of theforces given by the fluid on the valve OA

4) Obtain the weight W of the valve OA per unit of width. Environmental pressure p2 is neglected.


Balance Laws > First Law of Thermodynamics

External Thermal Power

External Thermal Power: Global Spatial FormThe external thermal power is defined as the amount of heatthat enters into a material volume per unit of time. We will assume that heat may enter into a material volume dueto: Internal heat sources characterized by a scalar-valued

function, given in spatial description by , representingthe heat source per unit of mass.

Non-convective heat flux through the boundary, characterizedby a vector-valued function, given in spatial description by

, representing the non-convective outward heat flux

per unit of spatial surface.

( ),r tx

( ), tq x




External Thermal Power: Global Spatial FormThe global spatial form of the external thermal power is definedas,

where is the heat source per unit of mass and isthe non-convective outward heat flux per unit of spatial surface, both of them given in spatial description.

( ) ( ) ( ) ( )

( ) ( ) ( )

, , ,

, , div ,

extQ t t r t dv t ds

t r t dv t dv

ρ

ρ

Ω ∂Ω

Ω Ω

= − ⋅

= −

∫ ∫

∫ ∫

x x q x n

x x q x

( ),r tx ( ), tq x




External Thermal Power: Global Material FormUsing the conservation of mass yields,

Using Nanson’s formula, the heat flux through the spatialboundary of the continuum body may be written as,

where is the nominal heat flux, i.e. heat flux per unit of material surface, given by,

0 0

:Tds J dS dS−

∂Ω ∂Ω ∂Ω⋅ = ⋅ = ⋅∫ ∫ ∫q n q F N Q N

Q

1 1, A Aa aJ Q JF q− −= =Q F q

( ) ( ) ( ) ( )0

0, , ,t r t dv R t dVρ ρΩ Ω

=∫ ∫x x X X




External Thermal Power: Global Material FormThe global material form of the external thermal power takesthe form,

where is the heat source per unit of mass and isthe nominal heat flux or heat flux per unit of material surface, both of them given in material description.

( ) ( ) ( ) ( )

( ) ( ) ( )0 0

0 0

0

0

, ,

, DIV ,

extQ t R t dV t dS

R t dV t dV

ρ

ρ

Ω ∂Ω

Ω Ω

= − ⋅

= −

∫ ∫

∫ ∫

X X Q X N

X X Q X

( ),R tX ( ), tQ X



Total External Power

Total External Power: Global Spatial FormThe global material form of the total external power, i.e. mechanical plus thermal external power, can be written as,

( ) ( )

21:

2

ext extt Q t dv ds

r dv ds

ddv dv

dt

rdv ds

ρ

ρ

ρ

ρ

Ω ∂Ω

Ω ∂Ω

Ω Ω

Ω ∂Ω

+ = ⋅ + ⋅

+ − ⋅

= +

+ − ⋅

∫ ∫

∫ ∫

∫ ∫

∫ ∫

b v t v

q n

v d

q n

P

σσσσ




Total External Power: Global Material Form (I)The global material form of the total thermal power, i.e. mechanical plus thermal external power, can be written as,

( ) ( )0 0

0 0

0 0

0 0

0

0

2

0

0

1:

2

ext extt Q t dV dS

r dV dS

ddV dV

dt

rdV dS

ρ

ρ

ρ

ρ

Ω ∂Ω

Ω ∂Ω

Ω Ω

Ω ∂Ω

+ = ⋅ + ⋅

+ − ⋅

= +

+ − ⋅

∫ ∫

∫ ∫

∫ ∫

∫ ∫

b v T v

Q N

v d

Q N

P

ττττ




Total External Power: Global Material Form (II)The global material form of the total thermal power, i.e. mechanical plus thermal external power, can be written as,

( ) ( )0 0

0 0

0 0

0 0

0

0

2

0

0

1:

2

ext extt Q t dV dS

r dV dS

ddV dV

dt

rdV dS

ρ

ρ

ρ

ρ

Ω ∂Ω

Ω ∂Ω

Ω Ω

Ω ∂Ω

+ = ⋅ + ⋅

+ − ⋅

= +

+ − ⋅

∫ ∫

∫ ∫

∫ ∫

∫ ∫

b v T v

Q N

v P F

Q N

P




Total External Power: Global Material Form (III)The global material form of the total thermal power, i.e. mechanical plus thermal external power, can be written as,

( ) ( )0 0

0 0

0 0

0 0

0

0

2

0

0

1:

2

ext extt Q t dV dS

r dV dS

ddV dV

dt

rdV dS

ρ

ρ

ρ

ρ

Ω ∂Ω

Ω ∂Ω

Ω Ω

Ω ∂Ω

+ = ⋅ + ⋅

+ − ⋅

= +

+ − ⋅

∫ ∫

∫ ∫

∫ ∫

∫ ∫

b v T v

Q N

v S E

Q N

P



First Law of Thermodynamics

First Law of ThermodynamicsFirst Postulate. There exist a state function called total energy, denoted as , such that its material time derivative is equalto the total external power supplied to the system, i.e. theexternal mechanical plus thermal power,

( )tE

( ) ( ) ( ): ext ext

dt P t Q t

dt= +E




First Law of ThermodynamicsSecond Postulate. There exist a state function called internal

energy, denoted as , which is an extensive property, i.e. there exist a specific internal energy or internal energy per unit

of mass, denoted as , such that,

( )tU

( ) ( ), ,e e t E t= =x X

( ) ( ) ( ) ( ) ( )0

0: , , ,t t e t dv E t dVρ ρΩ Ω

= =∫ ∫x x X XU




First Law of ThermodynamicsTher first law of thermodynamics states that the material time derivative of the total energy is equal to sum of the material time derivative of the kinetic energy and the material time derivative of the internal energy,

( ) ( ) ( )d d d

t t tdt dt dt

= +E UK




Using the first postulate and the mechanical energy balance, thefirst law of thermodynamics reads,

yielding the following internal energy balance law,

( ) ( ) ( )

( ) ( )

( ) ( ) ( )

ext ext

int ext

d d dt t t

dt dt dt

P t Q t

dt t Q t

dt

= +

= +

= + +

E UK

K P

( ) ( ) ( )int ext

dt t Q t

dt= +U P



Energy Balance

Energy Balance Law: Global Spatial FormThe internal energy balance law in global spatial form can be written as,

( ) ( ) ( )

:

int ext

dt t Q t

dt

dedv edv

dt

dv r dv ds

ρ ρ

ρ

Ω Ω

Ω Ω ∂Ω

= +

=

= + − ⋅

∫ ∫

∫ ∫ ∫d q n

σσσσ

U P



Energy Balance

Energy Balance Law: Local Spatial FormUsing the divergence theorem, the internal energy balance lawin global spatial form can be written as,

Localizing, the local spatial form of the energy balance lawreads,

:

: div

dedv edv

dt

dv r dv ds

dv r dv dv

ρ ρ

ρ

ρ

Ω Ω

Ω Ω ∂Ω

Ω Ω Ω

=

= + − ⋅

= + −

∫ ∫

∫ ∫ ∫

∫ ∫ ∫

d q n

d q

σσσσ

σσσσ

: dive rρ ρ= + −d q σσσσ



Energy Balance

Energy Balance Law: Global Material FormThe internal energy balance law in global material form can be written as,

( ) ( ) ( )

0 0

0 0 0

0 0 0

0 0 0

0 0

0

0

0

:

:

:

int ext

dt t Q t

dt

dedV edV

dt

dV r dV dS

dV r dV dS

dV r dV dS

ρ ρ

ρ

ρ

ρ

Ω Ω

Ω Ω ∂Ω

Ω Ω ∂Ω

Ω Ω ∂Ω

= +

=

= + − ⋅

= + − ⋅

= + − ⋅

∫ ∫

∫ ∫ ∫

∫ ∫ ∫

∫ ∫ ∫

d Q N

P F Q N

S E Q N

ττττ

U P



Energy Balance

Energy Balance Law: Global Material FormUsing the divergence theorem, the internal energy balance lawin global material form can be written as,

0 0

0 0 0

0 0 0

0 0 0

0 0

0

0

0

: DIV

: DIV

: DIV

dedV edV

dt

dV r dV dV

dV r dV dV

dV r dV dV

ρ ρ

ρ

ρ

ρ

Ω Ω

Ω Ω Ω

Ω Ω Ω

Ω Ω Ω

=

= + −

= + −

= + −

∫ ∫

∫ ∫ ∫

∫ ∫ ∫

∫ ∫ ∫

d Q

P F Q

S E Q

ττττ



Energy Balance

Energy Balance Law: Local Material FormLocalizing, the internal energy balance law in local material form

can be written as,

Note that the local material form of the energy balance equationcould have been also obtained from the local spatial form using,

0 0

0

0

: DIV

: DIV

: DIV

e r

r

r

ρ ρ

ρ

ρ

= + −

= + −

= + −

d Q

P F Q

S E Q

ττττ

0 , DIV div , : : : :J J Jρ ρ= = = = =Q q d P F S E d τ στ στ στ σ


Energy Balance


Material Time DerivativeGiving the spatial description of an arbitrary property, the material time derivative of the property can be written as,Local Spatial Form

Local Material Forms

0 0

0

0

: DIV

: DIV

: DIV

e r

r

r

ρ ρ

ρ

ρ

= + −

= + −

= + −

d Q

P F Q

S E Q

ττττ



Balance Laws > Second Law of Thermodynamics

Second Law of Thermodynamics

Second Law of ThermodynamicsFirst Postulate. There exist a state function called absolute

temperature, denoted as , which is always a positive scalar-valued function.

Second Postulate. There exist a state function called entropy, denoted as , which is an extensive property, i.e. there exista specific entropy or entropy per unit of mass, denoted as

, such that,

( ) ( ), ,t tθ θ= = Θx X

( )tH

( ) ( ), ,t tη η= = Ξx X

( ) ( ) ( ) ( ) ( )0

0, , ,t t t dv t dVρ η ρΩ Ω

= = Ξ∫ ∫x x X XH

( ) ( ), , 0t tθ θ= = Θ >x X




Second Law of ThermodynamicsThe second law of thermodynamics states that any admissible

thermodynamic process has to satisfy the following inequality, written in global spatial form as,

or in global material form as,

( )1 1d

t rdv dsdt

ρθ θΩ ∂Ω

≥ − ⋅∫ ∫ q nH

( )0 0

0

1 1dt rdV dS

dtρ

θ θΩ ∂Ω≥ − ⋅∫ ∫ Q NH




Second Law of ThermodynamicsAdmissible thermodynamic processes can be classified as reversible and irreversible processes.A thermodynamic process is said to be reversible if the followingcondition holds,

A thermodynamic process is said to be irreversible if thefollowing condition holds,

( )0 0

0

1 1 1 1dt rdv ds rdV dS

dtρ ρ

θ θ θ θΩ ∂Ω Ω ∂Ω> − ⋅ = − ⋅∫ ∫ ∫ ∫q n Q NH

( )0 0

0

1 1 1 1dt rdv ds rdV dS

dtρ ρ

θ θ θ θΩ ∂Ω Ω ∂Ω= − ⋅ = − ⋅∫ ∫ ∫ ∫q n Q NH




Second Law of Thermodynamics: Global Spatial FormThe global spatial form of the second law of thermodynamicscan be written as,

( )

1 1

d dt dv dv

dt dt

rdv ds

ρη ρη

ρθ θ

Ω Ω

Ω ∂Ω

= =

≥ − ⋅

∫ ∫

∫ ∫ q n

H




Second Law of Thermodynamics: Global Spatial FormUsing the divergence theorem, the global spatial form of thesecond law of thermodynamics can be written as,

2

1 1

1 1div

1 1 1div grad

ddv dv

dt

rdv ds

rdv dv

rdv dv dv

ρη ρη

ρθ θ

ρθ θ

ρ θθ θ θ

Ω Ω

Ω ∂Ω

Ω Ω

Ω Ω Ω

=

≥ − ⋅

= −

= − + ⋅

∫ ∫

∫ ∫

∫ ∫

∫ ∫ ∫

q n

q

q q




Second Law of Thermodynamics: Local Spatial FormLocalizing, the local spatial form of the second law of thermodynamics can be written as,

Multiplying by the absolute temperature yields,

2

1 1 1div gradrρη ρ θ

θ θ θ≥ − + ⋅q q

1div gradrρθη ρ θ

θ≥ − + ⋅q q




Clausius-Duhem Inequality: Local Spatial FormThe local spatial form of the Clausius-Duhem inequality statesthat the dissipation rate per unit of spatial volume is a non-negative scalar-valued quantity and it can be written as,

A stronger assumption is usually introduced, yielding,

1: div grad 0rρθη ρ θ

θ= − + − ⋅ ≥q qD

00

1: div grad 0

1: div 0, : grad 0

int

cond

int cond

int cond

r

r

ρθη ρ θθ

ρθη ρ θθ

≥≥

= + = − + − ⋅ ≥

= − + ≥ = − ⋅ ≥

q q

q q

DD

D D D

D D




Heat Conduction Inequality: Local Spatial FormThe local spatial form of the heat conduction inequality statesthat the projection of the heat flux per unit of spatial surface onthe direction of the spatial gradient of the temperature is a non-

positive scalar-valued quantity, i.e. heat flux takes place from thehot to the cold and not the other way around,

1: grad 0 grad 0

condθ θ

θ= − ⋅ ≥ ⇒ ⋅ ≤q qD




Heat Conduction Inequality: Local Spatial FormUsing Fourier law for heat conduction for an isotropic continuum medium, the second law of thermodynamics yields the followingrestriction on the admissible values of the spatial thermal

conductivity parameter,

1: grad 0 grad grad 0 0

condk kθ θ θ

θ= − ⋅ ≥ ⇒ ⋅ ≥ ⇒ ≥qD




Clausius-Planck Inequality: Local Spatial Form (I)The local spatial form of the Clausius-Planck inequality statesthat the internal dissipation rate per unit of spatial volume is a non-negative scalar-valued quantity and it can be written as,

The local spatial form of the Clausius-Planck inequality for a reversible process takes the form,

The local spatial form Clausius-Planck inequality for anirreversible process takes the form,

: div 0int

rρθη ρ= − + ≥qD

: div 0int

rρθη ρ= − + =qD

: div 0int rρθη ρ= − + >qD




Clausius-Planck Inequality: Local Spatial Form (II)Using the local spatial form of the internal energy balance

equation and the Clausius-Planck inequality given by,

the local spatial form of the Clausius-Planck inequality can be written in terms of the internal energy per unit of mass as,

: div

: div 0int

e r

r

ρ ρ

ρθη ρ

= + −

= − + ≥

d q

q

D

σσσσ

( ): : 0int

eρ θη= − − ≥d D σσσσ

div :r eρ ρ− = −q d σσσσ




Clausius-Planck Inequality: Local Spatial Form (III)Introducing the free energy per unit of mass, denoted as , defined as,

the local spatial form of the Clausius-Planck inequality can be written in terms of the free energy per unit of mass as,

( ): : 0int ρ ψ ηθ= − + ≥d D σσσσ

: eψ θη= −

ψ




Second Law of Thermodynamics: Global Material FormThe global material form of the second law of thermodynamicsstates that,

( )0

0 0

0 0

0

1 1

d dt dV dV

dt dt

r dV dS

ρ η ρ η

ρθ θ

Ω Ω

Ω ∂Ω

= =

≥ − ⋅

∫ ∫

∫ ∫ Q N

H




Second Law of Thermodynamics: Global Material FormUsing the divergence theorem, the global material form of thesecond law of thermodynamics can be written as,

0

0 0

0 0

0 0 0

0 0

0

0

0 2

1 1

1 1DIV

1 1 1DIV GRAD

ddV dV

dt

rdV dS

rdV dV

rdV dV dV

ρ η ρ η

ρθ θ

ρθ θ

ρ θθ θ θ

Ω Ω

Ω ∂Ω

Ω Ω

Ω Ω Ω

=

≥ − ⋅

= −

= − + ⋅

∫ ∫

∫ ∫

∫ ∫

∫ ∫ ∫

Q N

Q

Q Q




Second Law of Thermodynamics: Local Material FormLocalizing, the local material form of the second law of thermodynamics can be written as,

Multiplying by the absolute temperature yields,

0 0 2

1 1 1DIV GRADrρ η ρ θ

θ θ θ≥ − + ⋅Q Q

0 0

1DIV GRADrρ θη ρ θ

θ≥ − + ⋅Q Q




Clausius-Duhem Inequality: Local Material FormThe local material form of the Clausius-Duhem inequality statesthat the dissipation rate per unit of material volume is a non-negative scalar-valued quantity and it can be written as,

A stronger assumption is usually considered, yielding,

0 0 0

1: DIV GRAD 0rρ θη ρ θ

θ= − + − ⋅ ≥Q QD

0

0

0 0 0 0 0

00

0 0 0 0

1: DIV GRAD 0

1: DIV 0, : GRAD 0

int cond

int

cond

int cond

r

r

ρ θη ρ θθ

ρ θη ρ θθ

≥≥

= + = − + − ⋅ ≥

= − + ≥ = − ⋅ ≥

Q Q

Q Q

DD

D D D

D D




Clausius-Planck Inequality: Local Material Form (I)The local material form of the Clausius-Planck inequality statesthat the internal dissipation rate per unit of material volume is a non-negative scalar-valued quantity and it can be written as,

The local material form Clausius-Planck inequality for a reversible process takes the form,

The local material form Clausius-Planck inequality for anirreversible process takes the form,

0 0 0: DIV 0int

rρ θη ρ= − + ≥QD

0 0 0: DIV 0int

rρ θη ρ= − + =QD

0 0 0: DIV 0int

rρ θη ρ= − + >QD




Clausius-Planck Inequality: Local Material Form (II)Using the local material form of the internal energy balance

equation and the Clausius-Planck inequality yields,

0 0

0

0

0 0 0

: DIV

: DIV

: DIV

: DIV 0int

e r

r

r

r

ρ ρ

ρ

ρ

ρ θη ρ

= + −

= + −

= + − = − + ≥

d Q

P F Q

S E Q

Q

D

ττττ

0 0

0

0

DIV :

:

:

r e

e

e

ρ ρ

ρ

ρ

− = −

= −

= −

Q d

P F

S E

ττττ




Clausius-Planck Inequality: Local Material Form (II)The local material form of the Clausius-Planck inequality can be written in terms of the internal energy per unit of mass as,

( )

( )

( )

0 0

0

0

: :

:

: 0

inte

e

e

ρ θη

ρ θη

ρ θη

= − −

= − −

= − − ≥

d

P F

S E

D ττττ




Clausius-Planck Inequality: Local Material Form (III)Introducing the free energy per unit of mass, denoted as , defined as,

the local material form of the Clausius-Planck inequality can be written in terms of the free energy per unit of mass as,

: eψ θη= −

ψ

( )

( )

( )

0 0

0

0

: :

:

: 0

intρ ψ ηθ

ρ ψ ηθ

ρ ψ ηθ

= − +

= − +

= − + ≥

d

P F

S E

D ττττ


Clausius-Planck Inequality


Material Time DerivativeGiving the spatial description of an arbitrary property, the material time derivative of the property can be written as,Local Spatial Forms

Local Material Forms

( ) ( )( ) ( )( ) ( )

0 0 0

0 0

0 0

: : :

: :

: : 0

inte

e

e

ρ θη ρ ψ ηθ

ρ θη ρ ψ ηθ

ρ θη ρ ψ ηθ

= − − = − +

= − − = − +

= − − = − + ≥

d d

P F P F

S E S E

D τ ττ ττ ττ τ

( ) ( ): : : 0int

eρ θη ρ ψ ηθ= − − = − + ≥d d D σ σσ σσ σσ σ

: div 0int rρθη ρ= − + ≥qD

0 0 0: DIV 0int

rρ θη ρ= − + ≥QD


Balance Laws > Thermodynamic Processes

Adiabatic Process

Adiabatic Process: Local Spatial FormA thermodynamic process is said to be adiabatic if the net heat

transfer to or from the continuum body is zero.The internal dissipation rate per unit of spatial volume for anadiabatic process can be written as,

The stress power per unit of spatial volume for an adiabaticprocess is equal to the material time derivative of the internal

energy per unit of spatial volume,

0, div 0 :intr rρθη ρ= = ⇒ = −q D div+ q 0ρθη= ≥

( ): : 0 :int e eρθη ρ θη ρ= = − − ≥ ⇒ =d d D σ σσ σσ σσ σ



Adiabatic Process

Adiabatic Process: Local Material Form (I)A thermodynamic process is said to be adiabatic if the net heat

transfer to or from the continuum body is zero.The internal dissipation rate per unit of material volume for anadiabatic process can be written as,

The stress power per unit of material volume for an adiabaticprocess is equal to the material time derivative of the internal

energy per unit of material volume,

0 0 00, DIV 0 :int

r rρ θη ρ= = ⇒ = −Q D DIV+ Q 0 0ρ θη= ≥

( )0 0 0 0: : 0 :int

e eρ θη ρ θη ρ= = − − ≥ ⇒ =d d D τ ττ ττ ττ τ



Adiabatic Process

Adiabatic Process: Local Material Form (II)A thermodynamic process is said to be adiabatic if the net heat




0 0 00, DIV 0 :int


( )0 0 0 0: : 0 :int

e eρ θη ρ θη ρ= = − − ≥ ⇒ =P F P F D



Adiabatic Process

Adiabatic Process: Local Material Form (III)A thermodynamic process is said to be adiabatic if the net heat




0 0 00, DIV 0 :int


( )0 0 0 0: : 0 :int

e eρ θη ρ θη ρ= = − − ≥ ⇒ =S E S E D



Isentropic Process

Isentropic Process: Local Spatial FormA thermodynamic process is said to be isentropic if it takes place at constant entropy. The internal dissipation rate per unit of spatial volume for anisentropic process can be written as,

The stress power per unit of spatial volume for an isentropicprocess can be written as,

0 :intη ρθη= ⇒ = D div div 0r rρ ρ− + = − + ≥q q

: div :int r eρ ρ θη= − + = − −q d D σσσσ ( ) 0

: dive rρ ρ

≥ ⇒

= − +d qσσσσ



Isentropic Process

Isentropic Process: Local Material Form (I)A thermodynamic process is said to be isentropic if it takes place at constant entropy. The internal dissipation rate per unit of material volume for anisentropic process can be written as,

The stress power per unit of material volume for an isentropicprocess can be written as,

0 00 :int

η ρ θη= ⇒ = D 0 0DIV DIV 0r rρ ρ− + = − + ≥Q Q

0 0 0: DIV :int

r eρ ρ θη= − + = − −Q d D ττττ ( )0 0

0

: DIVe rρ ρ

≥ ⇒

= − +d Qττττ



Isentropic Process

Isentropic Process: Local Material Form (II)A thermodynamic process is said to be isentropic if it takes place at constant entropy. The internal dissipation rate per unit of material volume for anisentropic process can be written as,


0 00 :int


0 0 0: DIV :int

r eρ ρ θη= − + = − −Q P F D ( )0 0

0

: DIVe rρ ρ

≥ ⇒

= − +P F Q



Isentropic Process

Isentropic Process: Local Material Form (III)A thermodynamic process is said to be isentropic if it takes place at constant entropy. The internal dissipation rate per unit of material volume for anisentropic process can be written as,


0 00 :int


0 0 0: DIV :int

r eρ ρ θη= − + = − −Q S E D ( )0 0

0

: DIVe rρ ρ

≥ ⇒

= − +S E Q



Isentropic and Adiabatic Process

Isentropic + Adiabatic Process: Local Spatial FormA thermodynamic process is said to be isentropic and adiabaticif it takes place at constant entropy and the net heat flux to orfrom the continuum body is zero. The internal dissipation rate per unit of spatial volume for anisentropic and adiabatic process is zero and, then, the process isreversible.

The stress power per unit of spatial volume for an isentropic and adiabatic process is equal to the material time derivative of the

internal energy per unit of spatial volume,

0, 0, div 0 :intrη ρθη= = = ⇒ =q D rρ− div+ q 0=

: :int

eρ θη= − −d D σσσσ ( ) 0 : eρ= ⇒ =d σσσσ



Isentropic/Adiabatic Reversible Process

Isentropic/Adiabatic Reversible ProcessIf a thermodynamic process is adiabatic and reversible, then theprocess is also isentropic,

If a thermodynamic process is isentropic and reversible, then theprocess is also adiabatic,

For reversible thermodynamic processes, any adiabatic processis also isentropic, and any isentropic process is also adiabatic .

0, 0 :int int

η ρθη= = ⇒ = D D div 0rρ− + =q

0, 0, div 0 :int int

r rρθη ρ= = = ⇒ = −q D D div+ q 0=



Isothermal Process

Isothermal Process: Local Spatial FormA thermodynamic process is said to be isothermal if it takesplace at constant temperature. The internal dissipation rate per unit of spatial volume for anisothermal process is given by,

0 : :int

θ ρ ψ ηθ= ⇒ = − +d D σσσσ ( ) : 0ρψ= − ≥d σσσσ


Balance Laws > Governing Equations

Governing Equations: Coupled TM Problem

Governing Equations: Spatial Form Conservation of mass. Mass continuity

❶ ❶❸ Balance of linear momentum. Cauchy first motion

❸ ❾ Balance of angular momentum. Symmetry of Cauchy stress

❸ Balance of energy

❶ ❶❸ Clausius-Planck and heat conduction inequalities

❶❶

div ρ ρ+ =b vσσσσ

T=σ σσ σσ σσ σ


div 0ρ ρ+ =v

: div 0, : grad 0int conrρθη ρ θ= − + ≥ = − ⋅ ≥q qD D




Constitutive Equations: Spatial Form Thermo-mechanical constitutive equations

❻ ❶

❶ Thermal constitutive equation. Fourier law

❸ State equations

❶

❶

( )

( )

,

, ,

θ π

η η θ π

=

=

v

v

σ σ ,σ σ ,σ σ ,σ σ ,

( ) ( ), , gradθ θ θ= = −q q v k v

( )

( )

,

,

e e θ

π π ρ θ

=

=

v




Governing Equations: Material Form (I) Conservation of mass. Mass continuity

❶ ❶ Balance of linear momentum. Cauchy first motion

❸ ❾❸ Balance of angular momentum. Symmetry restriction 1st P-K



❶❶

0 0DIV ρ ρ+ =P b v

T T=PF FP

0 0: DIVe rρ ρ= + −P F Q

0Jρ ρ=

0 0 0 0: DIV 0, : GRAD 0int cond

rρ θ η ρ θ= − + ≥ = − ⋅ ≥Q QD D




Constitutive Equations: Material Form (I) Thermo-mechanical constitutive equations

❻ ❶

❶ Thermal constitutive equation. Material Fourier law

❸ State equations

❶

❶

( )

( )

,

, ,

T θ π

η η θ π

=

=

PF v

v

τ ,τ ,τ ,τ ,

( ) ( ), , GRADθ θ θ= = −Q Q v K v

( )

( )

,

,

e e θ

π π ρ θ

=

=

v




Governing Equations: Material Form (II) Conservation of mass. Mass continuity


❸ ❾❸ Balance of angular momentum. Symmetry of Kirchhoff stress



❶❶

0 0: DIVe rρ ρ= + −d Q ττττ

0Jρ ρ=


rρ θ η ρ θ= − + ≥ = − ⋅ ≥Q QD D

( ) 0 0DIV T ρ ρ− + =F b vττττ

T=τ ττ ττ ττ τ




Constitutive Equations: Material Form (II) Thermo-mechanical constitutive equations

❻ ❶


❸ State equations

❶

❶

( )

( )

,

, ,

θ π

η η θ π

=

=

v

v

τ τ ,τ τ ,τ τ ,τ τ ,

( ) ( ), , GRADθ θ θ= = −Q Q v K v

( )

( )

,

,

e e θ

π π ρ θ

=

=

v




Governing Equations: Material Form (III) Conservation of mass. Mass continuity


❸ ❾❸ Balance of angular momentum. Symmetry of 2nd P-K stress



❶❶

0 0: DIVe rρ ρ= + −S E Q

0Jρ ρ=


rρ θ η ρ θ= − + ≥ = − ⋅ ≥Q QD D

( ) 0 0DIV ρ ρ+ =FS b v

T=S S




Constitutive Equations: Material Form (III) Thermo-mechanical constitutive equations

❻ ❶


❸ State equations

❶

❶

( )

( )

,

, ,

θ π

η η θ π

=

=

v

v

τ τ ,τ τ ,τ τ ,τ τ ,

( ) ( ), , GRADθ θ θ= = −Q Q v K v

( )

( )

,

,

e e θ

π π ρ θ

=

=

v



Governing Equations: Mechanical Problem

Mechanical Problem: Spatial Form Conservation of mass. Mass continuity


❸ ❾ Balance of angular momentum. Symmetry of Cauchy stress

❸ Mechanical constitutive equation

❻


T=σ σσ σσ σσ σ

div 0ρ ρ+ =v

( )= vσ σσ σσ σσ σ




Mechanical Problem: Spatial Form Conservation of mass. Mass continuity


❸ ❻ Mechanical constitutive equation

❻


div 0ρ ρ+ =v

( )= vσ σσ σσ σσ σ




Mechanical Problem: Material Form (I) Conservation of mass. Mass continuity




❻

0 0DIV ρ ρ+ =P b v

T T=PF FP

0Jρ ρ=

( )T =PF vττττ




Mechanical Problem: Material Form (II) Conservation of mass. Mass continuity




❻


T=τ ττ ττ ττ τ

0Jρ ρ=

( )= vτ ττ ττ ττ τ




Mechanical Problem: Material Form (II) Conservation of mass. Mass continuity


❸ ❻❸ Mechanical constitutive equation

❻


0Jρ ρ=

( )= vτ ττ ττ ττ τ




Mechanical Problem: Material Form (III) Conservation of mass. Mass continuity


❸ ❾❸ Balance of angular momentum. Symmetry of 2nd P-K stress


❻

( ) 0 0DIV ρ ρ+ =FS b v

T=S S

0Jρ ρ=

( )=S S v




Mechanical Problem: Material Form (III) Conservation of mass. Mass continuity


❸ ❻❸ Mechanical constitutive equation

❻

0Jρ ρ=

( )=S S v

( ) 0 0DIV ρ ρ+ =FS b v

Chapter 6. Balance Laws

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Transcript of Chapter 6. Balance Laws