Conservation of the Linear Momentum

CH.5. BALANCE PRINCIPLES

Multimedia Course on Continuum Mechanics Overview

 Balance Principles Lecture 1  Convective Flux or Flux by Mass Transport Lecture 2 Lecture 3  Local and Material Derivative of a Volume Integral Lecture 4  Conservation of Mass  Spatial Form Lecture 5  Material Form  Reynolds Transport Theorem Lecture 6  Reynolds Lemma  General Balance Equation Lecture 7  Linear Momentum Balance  Global Form Lecture 8  Local Form

2 Overview (cont’d)

 Angular Momentum Balance  Global Spatial Lecture 9  Local Form  Mechanical Energy Balance  External Mechanical Power Lecture 10  Mechanical Energy Balance  External Thermal Power  Energy Balance  Thermodynamic Concepts Lecture 11  First Law of Thermodynamics  Internal Energy Balance in Local and Global Forms Lecture 12  Second Law of Thermodynamics Lecture 13  Reversible and Irreversible Processes Lecture 14  Clausius-Planck Inequality Lecture 15

3 Overview (cont’d)

 Governing Equations  Governing Equations Lecture 16  Constitutive Equations  The Uncoupled Thermo-mechanical Problem

4 5.1. Balance Principles

Ch.5. Balance Principles

5 Balance Principles

The following principles govern the way stress and deformation vary in the neighborhood of a point with time. REMARK  The conservation/balance principles: These principles are always  Conservation of mass valid, regardless of the type of material and the range of  Linear momentum balance principle displacements or deformations.  Angular momentum balance principle  Energy balance principle or first thermodynamic balance principle

 The restriction principle:  Second thermodynamic law

 The mathematical expressions of these principles will be given in,  Global (or integral) form  Local (or strong) form

6 5.2. Convective Flux

Ch.5. Balance Principles

7 Convection

 The term convection is associated to mass transport, i.e., particle movement.  Properties associated to mass will be transported with the mass when there is mass transport (particles motion) convective transport

 Convective flux of an arbitrary property A through a control surface S :

amountof crossing S Φ= A S unitof time

8 Convective Flux or Flux by Mass Transport

 Consider:  An arbitrary property of a continuum medium (of any tensor order) A  The description of the amount of the property per unit of mass, Ψ () x , t (specific content of the property ) . A  The volume of particles dV crossing a differential surface dS during the interval [] t , t + dt is dV=⋅=⋅ dS dhvn dt dS dm=ρρ dV =vn ⋅ dSdt  Then,  The amount of the property crossing the differential surface per unit of time is: Ψ dm d Φ= =ρ Ψ⋅vn dS S dt 9 Convective Flux or Flux by Mass Transport

 Consider:  An arbitrary property of a A continuum medium (of any tensor order) inflow vn⋅≤0 outflow  The specific content of A (the amount vn⋅≥0 of per unit of mass) Ψ x , t . A ()

 Then,  The convective flux of through a spatial surface, S , with unit normal n is: A v is velocity Φ()t =ρ Ψ⋅vndS Where: S ∫s ρ is density

 If the surface is a closed surface, SV = ∂ , the net convective flux is:

Φ∂ ()t =ρ Ψ⋅vndS = outflow - inflow V ∫∂V

10 Convective Flux

REMARK 1 The convective flux through a material surface is always null. REMARK 2 Non-convective flux (conduction, radiation). Some properties can be transported without being associated to a certain mass of particles. Examples of non-convective transport are: heat transfer by conduction, electric current flow, etc. Non-convective transport of a certain property is characterized by the non- convective flux vector (or tensor) qx () , t : non - convectiveflux =qn⋅=dS ; convectiveflux ρψ vn⋅dS ∫∫ss non-convective flux convective vector flux vector

11 Example

Compute the magnitude and the convective flux Φ S which correspond to the following properties: a) volume b) mass c) linear momentum d) kinetic energy

12 Φ =ρ Ψ⋅ S ()t ∫ vndS Example - Solution s

a) If the arbitrary property is the volume of the particles: ≡ V A The magnitude “property content per unit of mass” is volume per unit of mass, i.e., the inverse of density:

V 1 Ψ= = M ρ

The convective flux of the volume of the particles V through the surface S is:

1 Φ=ρ vn ⋅dS = vn ⋅ dS VOLUME FLUX S ∫∫ssρ

13 Φ =ρ Ψ⋅ S ()t ∫ vndS Example - Solution s

b) If the arbitrary property is the mass of the particles: ≡ M A The magnitude “property per unit of mass” is mass per unit of mass, i.e., the unit value:

M Ψ= =1 M

The convective flux of the mass of the particles M through the surface S is:

Φ=ρρ1 vn ⋅dS = vn ⋅ dS MASS FLUX S ∫∫ss

14 Φ =ρ Ψ⋅ S ()t ∫ vndS Example - Solution s

c) If the arbitrary property is the linear momentum of the particles: ≡ M v A The magnitude “property per unit of mass” is mass times velocity per unit of mass, i.e., velocity:

M v Ψ = = v M

The convective flux of the linear momentum of the particles M v through the surface S is:

Φ =ρ vvn() ⋅ dS MOMENTUM FLUX S ∫s

15 Φ =ρ Ψ⋅ S ()t ∫ vndS Example - Solution s

d) If the arbitrary property is the kinetic energy of the particles: 1 ≡ M v2 A 2 The magnitude “property per unit of mass” is kinetic energy per unit of mass, i.e.: 1 M v2 1 Ψ=2 = v2 M 2 1 The convective flux of the kinetic energy of the particles M v 2 through the surface S is: 2 1 Φ=ρ v2 () vn ⋅ dS KINETIC ENERGY FLUX S ∫s 2

16 5.3. Local and Material Derivative of a Volume Integral Ch.5. Balance Principles

17 Derivative of a Volume Integral

 Consider:  An arbitrary property of a continuum medium (of any tensor order) A  The description of the amount of the property per unit of volume (density of the property ), µ () x , t A REMARK and Ψ are related  The total amount of the property through . in an arbitrary volume, V , is: Q()t Q()() t= ∫ µ x, t dV V Q()tt+∆  The time derivative of this volume integral is: Qt()()+∆ t − Qt Qt′() = lim ∆→t 0 ∆t

18 Local Derivative of a Volume Integral

 Consider: Q()t  The volume integral Q()() t= ∫ µ x, t dV V Q()tt+∆ Control Volume, V  The local derivative of Qt () is: µµ()()xx,,t+∆ t dV − t dV local not ∂ ∫∫ REMARK = µ ()x,t dV = lim VV ∫ ∆→t 0 derivative ∂∆t V t The volume is fixed in space (control volume).  It can be computed as: µµ()()xx,t+∆ t dV − , t dV ∂ Qt()()+∆ t − Qt ∫∫ µ ()x,t dV = lim = lim VV= ∫ ∆→tt00∆→ ∂∆ttV ∆ t [µµ()()xx ,t+∆ t − ,] t dV ∫ µµ()()xx,,tt+∆ − t∂ µ() x , t = lim V = lim dV = dV ∆→tt00∆∫∫∆→ ∆∂ tVV tt  ∂µx,t  ∂t 19 Material Derivative of a Volume Integral

 Consider:  The volume integral Q()() t= ∫ µ x, t dV V

Q()tt+∆  The material derivative of Qt () is: Q()t material not d derivative = ∫ µ ()x,t dV = dt ≡ VVt µµ()()xx,,t+∆ t dV − t dV REMARK ∫∫Vt()+∆ t Vt() = lim The volume is mobile in space ∆→t 0 ∆ t and can move, rotate and  It can be proven that: deform (material volume). dd∂∂µµ   µ()x,t dV =µ dV + ∇⋅() µ vv dV = +∇⋅() µµ  dV = + ∇⋅ v  dV dt ∫∂∂t ∫∫ ∫t  ∫dt  VVt ≡ V V    V V material local convective derivative of derivative of derivative of the integral the integral the integral

20 5.4. Conservation of Mass

Ch.5. Balance Principles

21 Principle of Mass Conservation

 It is postulated that during a motion there are neither mass sources nor mass sinks, so the mass of a continuum body is a conserved quantity (for any part of the body).

 The total mass () t of the system satisfies:M ()()t= tt +∆ >0 MM

 Where: ()()t= ρ x, t dV∀∆ V ⊂ V ∆ tt M ∫ Vt ()()tt+∆ =ρ x, ttdVV +∆ ∀∆ ⊂ V ∆ tt+∆ tt +∆ M ∫ Vtt+∆ 22 Conservation of Mass in Spatial Form

 Conservation of mass requires that the material time derivative of the mass () t be zero for any region of a material volume, M ()()tt+∆ − t d ′()t = lim MM= ρdV=0 ∀∆ V ⊂ V , ∀ t ∫∆⊂≡ M ∆→t 0 ∆t dt Vtt VV  The global or integral spatial form of mass conservation principle: ddµ µµ()xv,t dV =( + ∇⋅ ) dV dt ∫∫dt VVt ≡ V ddρ ρρ(,)xvt dV = + ∇ ⋅dV =0 ∀∆ V ⊂ V , ∀ t ∫∫∆ ⊂≡ ∆⊂  dt VVVtt VVdt  By a localization process we obtain the local or differential spatial form of mass conservation principle: for∆→ V dV(,)x t (localization process) CONTINUITY EQUATION dtρρ(,)xx∂ (,)t +(ρρ∇∇ ⋅vx )( ,t ) = + ⋅(vx )( ,t ) = 0 ∀∈ x Vt , ∀ dt ∂t 23 Conservation of Mass in Material Form

d F 1 d F   =Fv∇ ⋅ ()∇ ⋅=v Consider the relations:  dt F dt  dV= F dV0  The global or integral material form of mass conservation principle can be rewritten as: ddρρ 1dtF ∂ ρ(X ,) t∂ FX( ,) ()(+ρρ ∇⋅vdV = + ) dV = (FX ( ,t ) +ρ) dV0 ∫∫V dt dtF dt ∫∂∂t t ∆VV0  ∂ρ F dV0 (X ,)t ∂ [ρ |FX |]( ,t ) ∂ t ∂t  ∂ → ρ FX() ,t dV0=0 ∀∆ V00 ⊂ V , ∀ t ∫ ∂t ∆VV00⊂  The local material form of mass conservation principle reads : ∂ ρ ρ = ρρ= 0  FX()(),0t tt=0 XFtt=0 () X F() X ρ = ∀∈X Vt, ∀ ∂t  t F 0 =1 t 24 5.5. Reynolds Transport Theorem

Ch.5. Balance Principles

25 dρ +ρ∇ ⋅=v 0 Reynolds Lemma dt

 Consider:  An arbitrary property of a continuum medium (of any tensor order) A  The spatial description of the amount of the property per unit of mass, ψ () x , t (specific contents of ) A  The amount of the property in the continuum body at time t A for an arbitrary material volume is: Q() t= ∫ ρψ dV VVt =

 Using the material time derivative leads to, d d ddψρ  Q′() t =ρψdV =()() ρψ + ρψ ∇∇ ⋅vvdV =ρ + ψ( +⋅ ρ ) dV dt ∫∫dt  ∫dt dt  VVt ≡ V V ddψρ =ρψ +  Thus, dt dt =0 (continuity equation) ddψ REYNOLDS ρψdV= ρ dV dt ∫∫dt LEMMA VVt ≡ V 26 d ∂ µ()xv,t dV= µµ dV + ∇⋅() dV dt ∫∂t ∫∫ VVt ≡ V V Reynolds Transport Theorem

 The amount of the property in the continuum body at time t for A an arbitrary fixed control volume is: Q() t= ∫ ρψ dV V  Using the material time derivative leads to, d ∂ ()ρψ ρψ dV = dV +⋅∇ ()ρ ψ; v dV dt ∫∫∂t ∫ VVt ≡ V V dψ dψ = ∫ ρ dV = ∫ nv⋅()ρψ dV ρ V dt ∂V dt  And, introducing the Reynolds Lemma ∂V and Divergence Theorem: dV dψ ∂ ()ρψ ∫∫ρ dV = dV +⋅ ∫ρψ vn dS VVdt ∂t ∂ V eˆ ρψ REMARK 3 ∫ dV eˆ 2 V

The Divergence Theorem: eˆ1 ∫ ∇⋅vdV = ∫∫ nv ⋅ dS = vn ⋅ dS V ∂∂ VV 27 dψ ∂ (ρψ ) ρ dV = dV +⋅ρψ vn dS ∫∫dt ∂t ∫ Reynolds Transport TheoremVV ∂ V

 The eq. can be rewritten as: ∂ dψ REYNOLDS TRANSPORT ∫∫∫ρψ dV= ρ dV −⋅ ρψ vn dS ∂t VVdt ∂ V THEOREM

Change due to the net outward convective flux of through the Change (per unit of time) of A the total amount of . within boundary ∂ V . A dψ the control volume V at time t. ρ dt

Rate of change of the amount of property ∂V integrated over all particles that are fillingA dV the control volume V at time t.

eˆ ρψ 3 ∫ dV eˆ 2 V

eˆ1

28 Reynolds Transport Theorem

∂ dψ REYNOLDS TRANSPORT ∫∫∫ρψ dV= ρ dV −⋅ ρψ vn dS ∂t VVdt ∂ V THEOREM (integral form)

∂ dψ dψ ρψ dV= ρ dV −⋅ ρψ vn dS ρ ∂t ∫∫∫dt dt VV∂ V ∂V ∂ = (ρψ ) dV =∫∇ ⋅(ρψ v) dV ∫ V V ∂t dV ∂ dψ (ρψ ) dV = [ ρ −∇ ⋅ ( ρψ v)] dV∀∆ V ⊂ V ∀ t ∫∫∂t dt ∆⊂VV ∆⊂VV eˆ ρψ 3 ∫ dV eˆ 2 V ∂ dψ eˆ (ρψ )= ρ −∇ ⋅ ( ρψ vx) ∀∈Vt ∀ 1 ∂t dt REYNOLDS TRANSPORT THEOREM (local form)

29 5.6. General Balance Equation

Ch.5. Balance Principles

30 General Balance Equation

 Consider:  An arbitrary property of a continuum medium (of anyA tensor order)  The amount of the property per unit of mass, ψ ()x,t  The rate of change per unit of time of the amount of in the control volume V is due to: A kt(,)x a) Generation of the property per unit mas and time time due to a source:  A   b) The convective (net incoming) flux across the surface of the volume. source term c) The non-convective (net incoming) flux across the surface of the volume: jx(,)t  A   non-convective  So, the global form of the general balance equation is: flux vector ∂ ρψ dV= ρ k dV − ρψ vn ⋅ dS −⋅ j n dS ∂t ∫∫∫AA ∫ VV∂∂ V V acb

31 ∂ ρψ dV= ρ k dV − ρψ vn ⋅−⋅ dS j n dS ∂t ∫∫AA ∫ ∫ General BalanceVV Equation ∂∂ V V

 The global form is rewritten using the Divergence Theorem and the definition of local derivative: ∂ ∫∫ρψ dV + ρψ vn ⋅= dS ∂t VV∂ ∂ =()()ρψ +⋅∇∇ ρψ vjdV =( ρ k −⋅ ) dV ∫∫AA VV∂t dψ = ρ (Reynolds Theorem) dt dψ ρρ dV=( k −∇ ⋅j ) dV ∀∆ V ⊂ V ∀ t ∫∫AA ∆⊂VV dt ∆⊂VV  The local spatial form of the general balance equation is: REMARK dψ dψ ρ =ρk −⋅∇ j For only convective transport () j0 = then ρ = ρ k dt AA A dt A and the variation of the contents of in a given particle is only due to the internal generation ρ k . A 32 dψ ρ =ρk −⋅∇ j dt AA ∂ dψ Example (ρψ )= ρ −∇ ⋅ ( ρ ψ vx) ∀∈V ∂t dt

If the property is associated to mass ≡ , then: A AM  The amount of the property per unit of mass is ψ = 1 .  The mass generation source term is k = 0 . M  The mass conservation principle states that mass cannot be generated.  The non-convective flux vector is j = 0 . M  Mass cannot be transported in a non-convective form. dψ ρ =ρ k −⋅∇ j =0 dt AA =0 =0 Then, the local spatial form of the general balance equation is: dψ ∂ ∂ρ ρ =(ρψ ) +⋅∇ ( ρψ v) = 0 +⋅∇ (ρv )0 = dt ∂   ∂t t =11= ∂ρρd Two equivalent forms of +∇∇ ⋅()ρρv = + ⋅ vx =0 ∀∈Vt ∀ ∂t dt the continuity equation.

33 5.7. Linear Momentum Balance

Ch.5. Balance Principles

34 Linear Momentum in Classical Mechanics

 Applying Newton’s 2nd Law to the discrete system formed by n particles, the resulting force acting on the system is: nn ndv Rt= fa = mm =i = () ∑∑iiii ∑  ii=11 = i= 1dt Resulting force mass conservation on the system dm principle: i = 0 nn dt d dmi dt() =∑∑miivv −=i P dt ii=11= dt dt = ()t linear momentum P  For a system in equilibrium, R = 0, ∀ t : dt() P = 0 ()t= cnt CONSERVATION OF THE dt P LINEAR MOMENTUM

35 n Linear Momentum ()tm= v P ∑ ii in Continuum Mechanics i=1

 The linear momentum of a material volume V t of a continuum medium with mass is: M ()()()()t= vx, td= ρ x ,, t vx tdV P ∫∫M M V

d= ρ dV M

36 Linear Momentum Balance Principle

 The time-variation of the linear momentum of a material volume is equal to the resultant force acting on the material volume.

dt() d P = ρ vR dV= () t dt dt ∫ Vt

 Where: body forces R()t=∫∫ρ bt dV + dS VV∂ surface forces

 If the body is in equilibrium, the linear momentum is conserved: dt() R ()t = 0 P = 0 ()t= cnt dt P

37 Global Form of the Linear Momentum Balance Principle

 The global form of the linear momentum balance principle: d dt() = ρρ+ = =P ∀∆ ⊂ ∀ R()t ∫∫ bt dV dS ∫ v dV V V, t ∆⊂ ∂∆⊂ dt ∆ ⊂ ≡ dt VV VV Vtt  VV     t P ( )

 Introducing tn = ⋅ σ and using the Divergence Theorem, ∫∫tn dS=⋅=⋅σ dS ∫ ∇σ dV ∂∂VV V  So, the global form is rewritten: ∫∫ρ bt dV+= dS ∆⊂VV ∂∆⊂VV d = ()ρρ bv +∇σ⋅dV = dV∀∆ V ⊂ V, ∀ t ∫∫dt 38 ∆⊂VV ∆VVVtt ⊂≡ Local Form of the Linear Momentum Balance Principle

 Applying Reynolds Lemma to the global form of the principle: ddv ()∇⋅ σ+ρ bv dV = ρρ dV = dV∀∆ V ⊂ V, ∀ t ∫ dt ∫∫dt ∆⊂VV ∆VVVtt ⊂≡ ∆⊂VV

 Localizing, the linear momentum balance principle reads:

∆→V dV(,)x t dtvx(,) ∇σ⋅ (,)xt +ρρ bx(,) t= = ρax (,)t ∀∈x Vt , ∀ dt LOCAL FORM OF THE LINEAR MOMENTUM BALANCE (CAUCHY’S EQUATION OF MOTION)

39 5.8. Angular Momentum Balance

Ch.5. Balance Principles

40 Angular Momentum in Classical Mechanics

 Applying Newton’s 2nd Law to the discrete system formed by n particles, the resulting torque acting on the system is:

nn dvi MO()tm=∑∑ rfii ×= r i × i = ii=11= dt =0 nn n d dri dd =∑∑rvi ×−mm ii ×=ii v ∑ rvi ×= m ii L dt ii=11= dt dt i= 1dt = v = ()t i angularL momentum dt() M ()t = L O dt

 For a system in equilibrium, M O = 0, ∀ t : dt() CONSERVATION OF THE L = 0 ∀t ()t= cnt dt L ANGULAR MOMENTUM

41 Angular Momentum in Continuum Mechanics

 The angular momentum of a material volume V t of a continuum medium with mass is: M ()t=×=×r vx(),td xρ ()() x,,t vx tdV L ∫∫ M ≡x V M

d= ρ dV M

42 Angular Momentum Balance Principle

 The time-variation of the angular momentum of a material volume with respect to a fixed point is equal to the resultant moment with respect to this fixed point.

dt() d = ×=ρ L rv dV MO ()t dt dt ∫ ≡ VVt ≡ x

 Where: torque due to body forces =×ρ +× MO ()t ∫∫ r b dV rt dS VV∂ torque due to surface forces

43 Global Form of the Angular Momentum Balance Principle

 The global form of the angular momentum balance principle: d (rb×ρρ ) dV +× ( rt ) dS = ( rv × ) dV ∫∫ dt ∫ V ∂≡V VVt

 Introducing tn = ⋅ σ and using the Divergence Theorem, ∫∫ rt× dS = rn ×⋅σσ dS = ∫ r ×TT ⋅ n dS = ∫() r × σ ⋅ n dS = ∂∂VV ∂ V ∂ V =×⋅∫ ()r σ∇T dV V  It can be proven that, REMARK T  is the Levi-Civita ()r×σ∇ ⋅ = rm ×∇⋅σ + ijk  permutation symbol. = = σ memmiˆ i; i ijk jk

44 Global Form of the Angular Momentum Balance Principle

 Applying Reynolds Lemma to the right-hand term of the global form equation: Reynold's Lemma dd↓ d r×ρρ v dV =()rv ×dV = ρ()rv ×=dV dt ∫∫dt ∫dt VVtt≡ VV≡ V ddrv=0 d v =ρρ ×+×vrdV = r × dV ∫∫dt dt dt VV= v

 Then, the global form of the balance principle is rewritten: dv ×ρσ +∇⋅σ + ˆ = × ρ ∫∫rb() ijk jk e i dV r dV VVdt

45 Local Form of the Angular Momentum Balance Principle

 Rearranging the equation: =0 (Cauchy’s Eq.) dv r× ∇ ⋅σ− + ρρ b+ m dV = 0 mx(,)t dV= 0 ∀∆ V ⊂ V, ∀ t ∫∫ ∆⊂VV ∆⊂VVdt

 Localizing

mx(,)t= 0 mi = ijkσ jk =0 ;i , jk , ∈{} 1, 2, 3 ; ∀∈x Vt , ∀ t i =10⇒⇒σ += σ σσ = 123 23 132 32 23 32 σ σσ  =11 =− 11 12 13   i =20⇒⇒σ += σ σσ = σ ≡ σ12 σσ 22 23  231 31 213 13 31 13  = =−   11 σ13 σσ23 33 i =30⇒⇒σ += σ σσ = 312 12 321 21 12 21  = =−  11 T σσ(,)xxxt= (,) t ∀∈ Vtt , ∀ SYMMETRY OF THE CAUCHY’S STRESS TENSOR 46 5.9. Mechanical Energy Balance

Ch.5. Balance Principles

47 Power

 Power, Wt () , is the work performed in the system per unit of time.

 In some cases, the power is an exact time-differential of a function (then termed) energy : E dt() Wt()= E dt  It will be assumed that the continuous medium absorbs power from the exterior through:  Mechanical Power: the work performed by the mechanical actions (body and surface forces) acting on the medium.  Thermal Power: the heat entering the medium.

48 External Mechanical Power

 The external mechanical power is the work done by the body forces and surface forces per unit of time.  In spatial form it is defined as: P() t=ρ bv ⋅ dV +⋅ tv dS e ∫∫VV∂

dr ρρb⋅=⋅ dV bv dV dt =v dr t⋅=⋅dS tv dS dt =v Mechanical Energy Balance

 Using tn = ⋅ σ and the Divergence Theorem, the traction contribution reads, Divergence Theorem ↓ t⋅vndS = ⋅()σ⋅ vdS = ∇⋅()σ ⋅vdV = ()∇⋅σσ ⋅vv +: ∇ dV ∫∫∂∂VV ∫ V ∫ V n⋅σ = l spatial velocity  Taking into account the identity: ld = +  w =0 skew gradient tensor σσ:l= :d + σ :w symmetric

 So, tv⋅dS =()∇ ⋅⋅σσvdV+ : d dV ∫∫∂VV ∫ V

50 dv ∇⋅= σ+ρρ b Mechanical Energy Balance dt

 Substituting and collecting terms, the external mechanical power in spatial form is, ∫ tv⋅ dS ∂V

Pe () t=ρ bv ⋅+ dV ()∇⋅σσ ⋅vddV +: dV = ∫V ∫∫VV dv =∇⋅+⋅+()σσρρbvdvddV:: dV = ⋅+dV σdV ∫V ∫∫VVV ∫        dt    =ρ dv dd112 dt =ρ (vv ⋅= )ρ ( v) dt 22dt v= v

Reynold's Lemma dd11↓ =ρρ22 +=σσ + Pe () t ∫( v)dV ∫:d dV ∫∫( v)dV:d dV Vdt 22 Vdt VV

51 Mechanical Energy Balance. Theorem of the expended power. Stress power

d 1 P( t) =ρρb ⋅ v dV +⋅ t v dS =v2dV +σ :d dV e ∫∫VV∂ dt ∫∫2 VVt ≡ V

external mechanical power P kinetic energy σ entering the medium K stress power d Pt( ) =( t) + P Theorem of the expended e dt K σ mechanical power REMARK The stress power is the mechanical power entering the system which is not spent in changing the kinetic energy. It can be interpreted as the work by unit of time done by the stress in the deformation process of the medium. A rigid solid will produce zero stress power ( d0 = ) .

52 External Thermal Power

 The external thermal power is incoming heat in the continuum medium per unit of time.  The incoming heat can be due to:  Non-convective heat transfer across the volume’s surface. incoming heat − qx(,)t ⋅= n dS ∫  unit of time ∂V heat conduction flux vector

 Internal heat sources heat generated by internal sources ρ r(,)x t dV = ∫  unit of time V specific internal heat production

53 External Thermal Power

 The external thermal power is incoming heat in the continuum medium per unit of time.  In spatial form it is defined as: =ρρ − ⋅ = −⋅∇ Qe () t∫∫∫ r dVqn dS ( rq ) dV V∂ VV    =∫ nq ⋅ dS ∂V =∫(∇ ⋅q) dV V where: qx () , t is the non-convective heat flux vector per unit of spatial surface rt () x , is the internal heat source rate per unit of mass.

54 Total Power

 The total power entering the continuous medium is:

d 1 PQ+= ρρv2dV+σ :d dV + r dV −q ⋅ n dS eedt ∫2 ∫ ∫∫ VVt ≡∂V V V

55 5.10. Energy Balance

Ch.5. Balance Principles

56 Thermodynamic Concepts

 A thermodynamic system is a macroscopic region of the continuous medium, always formed by the same collection of continuous matter (material volume). It can be: ISOLATED SYSTEM OPEN SYSTEM Thermodynamic space

MATTER

HEAT

 A thermodynamic system is characterized and defined by a set of thermodynamic variables µµ 1, 2, .... µ n which define the thermodynamic space.

 The set of thermodynamic variables necessary to uniquely define a system is called the thermodynamic state of a system.

57 Thermodynamic Concepts

 A thermodynamic process is the energetic development of a thermodynamic system which undergoes successive thermodynamic states, changing from an initial state to a final state → Trajectory in the thermodynamic space.  If the final state coincides with the initial state, it is a closed cycle process.

 A state function is a scalar, vector or tensor entity defined univocally as a function of the thermodynamic variables for a given system.  It is a property whose value does not depend on the path taken to reach that specific value.

58 State Function

Is a function φµ () 1 ,..., µ n uniquely valued in terms of the “thermodynamic state”

or, equivalently, in terms of the thermodynamic variables { µµ 12 ,,,  µ n }

 Consider a function φµµ () 12 , , that is not a state function, implicitly defined in the thermodynamic space by the differential form:

δφ= f1()() µµ 12, df µ 1+ 2 µµ 12 , d µ 2  The thermodynamic processes Γ 1 and Γ 2 yield: φ =+= φ δφf (, µ µ ) δµ BA∫∫ΓΓ212 2  11  δφ≠≠ δφ φ' φ ∫∫ΓΓ B B 12 φBA'=+= φ δφf21(, µ µ2 ) δµ2  ∫∫ΓΓ22  For to be a state function, the differential form must be an exact differential: , i.e., must be integrable:  The necessary and sufficient condition for this is the equality of cross-derivatives: ∂f ()µµ,..., ∂f ()µµ,..., in1 =jn1 ∀∈ij,{} 1,... n δφ= d φ ∂∂µµ 59 ji First Law of Thermodynamics

POSTULATES: 1. There exists a state function () t named total energy of the system, such that its E material time derivative is equal to the total power entering the system:

dd1 = + =ρρ2 +σ + −⋅ ()()()t: Pee t Q t ∫vdV ∫:d dV ∫∫ r dVq n dS dt E dt ≡∂2 VVt V VV Qt() Pte() e

2. There exists a function () t named the internal energy of the system, such that: U  It is an extensive property, so it can be defined in terms of a specific internal energy (or internal energy per unit of mass) ut () x , : ()t:= ρ u dV U ∫ REMARK V d and d are exact differentials,  The variation of the total energy of the system is: E K therefore, so is d  = dd − . ddd UEK ()ttt=() + () Then, the internal energy is a dtE dt KU dt state function. 60 Global Form of the Internal Energy Balance

 Introducing the expression for the total power into the first

postulate: = K dd1 ()t =ρρv2dV +σ :d dV + r dV −⋅q n dS dtE dt ∫2 ∫ ∫∫ VVt ≡∂V V V

 Comparing this to the expression in the second postulate: ddd ()ttt=() + () dtE dt KU dt

 The internal energy of the system must be: dd GLOBAL FORM =ρρ =σ + −⋅ ()t∫ u dV ∫:d dV ∫∫ r dVq n dS OF THE INTERNAL dtU dt ≡∂ VVt V V V ENERGY BALANCE Qt Ptσ () , e () stress power external thermal power 61 Local Spatial Form of the Internal Energy Balance

 Applying Reynolds Lemma to the global form of the balance equation, and using the Divergence Theorem: d d du ()t =∫ρρu dV = ∫dV = ∫σ :d dV + ∫∫ρ r dV −⋅q n dS dtU dt ∆ ⊂ ≡ ∆ ⊂ ≡ dt ∆⊂ ∆⊂ ∂∆⊂ Vtt  VV     Vtt VV VV VV VV     ()t ∫ ∇⋅q dV U ∂∆VV ⊂ du ⇒ ∫ρρdV = ∫σ :d dV+ ∫∫ r dV− ∇ ⋅q dV∀∆ V ⊂ V ∀ t ∆⊂VV dt ∆⊂VV ∆⊂VV ∆⊂VV  Then, the local spatial form of the energy balance principle is obtained through localization ∆→ V dV (,) x t as: du LOCAL FORM OF THE ρρ=σ :d +() r −∇⋅ q ∀ x ∈ Vt, ∀ ENERGY BALANCE dt (Energy equation)

62 Second Law of Thermodynamics

 The total energy is balanced in all thermodynamics processes following: ddd Pt()()+==+ Qt E KU eedt dt dt  In an isolated system (no work can enter or exit the system) d dd Pt ()()+== Qt E 0 UK+=0 eedt dt dt

 However, it is not established if the energy exchange can happen in both senses or not: dd dd UK><00 UK<00> dt dt dt dt

 There is no restriction indicating if an imagined arbitrary process is physically possible or not.

63 Second Law of Thermodynamics

 The concept of energy in the first law does not account for the observation that natural processes have a preferred direction of progress. For example:

 If a brake is applied on a spinning wheel, the speed is reduced due to the conversion of kinetic energy into heat (internal energy). This process never occurs the other way round. dd UK><00 dt dt  Spontaneously, heat always flows to regions of lower temperature, never to regions of higher temperature.

64 MMC - ETSECCPB - UPC 24/04/2017 Reversible and Irreversible Processes

 A reversible process can be “reversed” by means of infinitesimal changes in some property of the system.  It is possible to return from the final state to the initial state along the same path.  A process that is not reversible is termed irreversible. REVERSIBLE PROCESS IRREVERSIBLE PROCESS

 The second law of thermodynamics allows discriminating: IMPOSSIBLE thermodynamic processes REVERSIBLE POSSIBLE IRREVERSIBLE

65 Second Law of Thermodynamics

POSTULATES: 1. There exists a state function θ () x , t denoted absolute temperature, which is always positive.

2. There exists a state function S named entropy, such that:  It is an extensive property, so it can be defined in terms of a specific entropy or entropy per unit of mass s : S( t )= ∫ ρ s(x , t ) dV V  The following inequality holds true: nd dd r q Global form of the 2 S() t=ρρ s dV ≥ dV −⋅n dS Law of dt dt ∫∫θθ ∫ VV∂ V Thermodynamics = reversible process > irreversible process 66 Second Law of Thermodynamics

SECOND LAW OF THERMODYNAMICS IN CONTINUUM MECHANICS The rate of the total entropy of the system is equal o greater than the rate of heat per unit of temperature

nd dd r q Global form of the 2 S() t=ρρ s dV ≥ dV −⋅n dS Law of dt dt ∫∫θθ ∫ VV∂ V Thermodynamics

= reversible process = Γe ()t > irreversible process rate of the total amount of the entity heat, per unit =ρ −⋅ of time, (external thermal power) entering into the Qe () t∫∫ r dVqn dS VV∂ system

r q rate of the total amount of the entity heat per unit Γ=e ()tρ dV −⋅n dS of absolute temperature, ∫∫θθ per unit of time (external VV∂ heat/unit of temperature power) entering into the system

67 Second Law of Thermodynamics

 Consider the decomposition of entropy into two (extensive) counterparts:  Entropy generated inside the continuous medium:

St() = S()ie() t + S() () t dS dS()ie dS () = + dt dt dt

 Entropy generated by interaction with the outside medium:

S ()ii= ∫ ρ s,() ()x t dV V S ()ee= ∫ ρ s,() ()x t dV V

68 Second Law of Thermodynamics

()e  dS r q If one establishes, =Γ=ρ − ⋅ e ∫∫dVn dS dt VVθθ∂

 Then the following must hold true: dS()ie dS() dS r q + =≥ρ dV −⋅n dS dt dt dt ∫∫θθ VV∂ ()e  And thus, =dS dt dS()ie dS dS() dS r q = − = − ρ dV − ⋅n dS ≥0 ∀∆ V ⊂ V ∀ t ∫∫θθ dt dt dt dt ∆⊂VV ∂∆⊂VV

REPHRASED SECOND LAW OF THERMODYNAMICS : ()i The internally generated entropy of the system , St () , never decreases along time

69 Local Spatial Form of the Second Law of Thermodynamics

 The previous eq. can be rewritten as:

dd ()i  rq ρ s dV = ρρ s dV−  dV − ⋅n dS ≥ 0 ∀∆ V ⊂ V ∀ t dt ∫dt ∫ ∫∫θθ ∆VVVtt ⊂tt ≡ ∆VVV ⊂ ≡ ∆⊂VV ∂∆⊂VV  Applying the Reynolds Lemma and the Divergence Theorem:

ds()i ds r q ρdV= ρρ dV−  dV −∇ ⋅dV ≥0 ∀∆ V ⊂ V ∀ t ∫ ∫ ∫∫θθ ∆⊂VV dt ∆⊂VV dt ∆⊂VV ∆⊂VV 

 Then, the local spatial form of the second law of thermodynamics is: nd ds()i ds r q Local (spatial) form of the 2 ρ= ρρ − −∇ ⋅ ≥0, ∀∈x Vt ∀ Law of Thermodynamics dt dt θθ  (Clausius-Duhem inequality)

= reversible process > irreversible process

70 Local Spatial Form of the Second Law of Thermodynamics

q 11  Considering that, ∇∇⋅() = ⋅−qq ⋅ ∇θ REMARK θθ θ2 (Stronger postulate) Internally generated entropy can ()i  The Clausius-Duhem inequality can be written as be generated locally, s , or by ()i local ()i = s = s thermal conduction, s , and ()i cond ds ds r 11 both must be non-negative. = −+∇∇ ⋅qq − ⋅θ ≥0 dt dt θ ρθ ρθ 2 = ()i ()i slocal = scond CLAUSIUS-PLANCK HEAT FLOW r 1 1 s −+∇ ⋅q ≥0 INEQUALITY −q ⋅≥∇θ 0 INEQUALITY θ ρθ ρθ 2

Because density and absolute temperature are always positive, it is deduced that q ⋅≤ ∇ θ 0 , which is the mathematical expression for the fact that heat flows by conduction from the hot parts of the medium to the cold ones. 71 Alternative Forms of the Clausius-Planck Inequality

 Substituting the internal energy balance equation given by du not ρρ=ur =σ :dq + ρ −∇⋅ ∇⋅qd −ρρru =σ : −  dt

into the Clausius-Planck inequality,

i ρθslocal :0= ρθ sr − ρ +∇⋅q ≥

yields, ρθsu+()σ :0d −≥ ρ −ρθ()us −+σ :0d ≥ Clausius-Planck Inequality in terms of the specific internal energy

72 5.11. Governing Equations

Ch.5. Balance Principles

74 Governing Equations in Spatial Form

Conservation of Mass. ρρ+ ∇⋅ = 1 eqn.  v 0 Continuity Equation.

Linear Momentum Balance. ∇⋅σ +ρρbv =  3 eqns. First Cauchy’s Motion Equation.

T Angular Momentum Balance. σσ= 3 eqns. Symmetry of Cauchy Stress Tensor.

Energy Balance. ρρur =σ :dq + −∇⋅ 1 eqn. First Law of Thermodynamics.

−−+ρθ()usσ :d ≥0 Second Law of Thermodynamics. 1 Clausius-Planck Inequality. 2 restrictions −q ⋅≥∇θ 0 ρθ 2 Heat flow inequality 8 PDE + 2 restrictions

75 Governing Equations in Spatial Form

 The fundamental governing equations involve the following variables: ρ density 1 variable

v velocity vector field 3 variables

σ Cauchy’s stress tensor field 9 variables u specific internal energy 1 variable

q 3 variables heat flux per unit of surface vector field θ absolute temperature 1 variable 19 scalar s specific entropy 1 variable unknowns  At least 11 equations more (assuming they do not involve new unknowns), are needed to solve the problem, plus a suitable set of boundary and initial conditions.

Constitutive Equations in Spatial Form

σσ= ()v,,θ ζ Thermo-Mechanical Constitutive Equations. 6 eqns.

Entropy ss= ()v,,θ ζ Constitutive Equation. 1 eqn.

Thermal Constitutive Equation. q= qv(),θθ = −K ∇ Fourier’s Law of Conduction. 3 eqns.

uf= ()ρθ,,,v ζ Heat State Equations. (1+p) eqns. Fi (){}ρθ, ,ζ = 0 ip ∈ 1,2,..., Kinetic (19+p) PDE + set of new thermodynamic (19+p) unknowns variables:ζ = {} ζζ 12 , ,..., ζ p . REMARK 1 REMARK 2 The strain tensor is not considered an unknown as they These equations are can be obtained through the motion equations, i.e., εε = () v . specific to each material.

77 The Coupled Thermo-Mechanical Problem

Conservation of Mass. ρρ+ ∇⋅ = 1 eqn.  v 0 Continuity Mass Equation.

16 scalar Linear Momentum Balance. 10 3 eqns. unknowns First Cauchy’s Motion Equation. equations

σ=σ( ε (v ),θ ) Mechanical constitutive equations. 6 eqns.

Energy Balance. 1 eqn. First Law of Thermodynamics.

Second Law of Thermodynamics. 2 restrictions. Clausius-Planck Inequality.

78 MMC - ETSECCPB - UPC The Uncoupled Thermo-Mechanical Problem

 The mechanical and thermal problem can be uncoupled if

 The temperature distribution θ () x , t is known a priori or does not intervene in the mechanical constitutive equations.

 Then, the mechanical problem can be solved independently.

79 The Uncoupled Thermo-Mechanical Problem

Conservation of Mass. ρρ+ ∇⋅ = 1 eqn.  v 0 Continuity Mass Equation.

10 scalar Linear Momentum Balance. Mechanical 3 eqns. unknowns First Cauchy’s Motion Equation. problem

σ=σ( ε (v ), θ ) Mechanical constitutive equations. 6 eqns.

Energy Balance. 1 eqn. First Law of Thermodynamics. Thermal problem Second Law of Thermodynamics. 2 restrictions. Clausius-Planck Inequality.

80 The Uncoupled Thermo-Mechanical Problem

 Then, the variables involved in the mechanical problem are:

ρ density 1 variable

Mechanical v velocity vector field 3 variables variables σ Cauchy’s stress tensor field 6 variables u specific internal energy 1 variable

q heat flux per unit of surface vector field 3 variables Thermal variables θ absolute temperature 1 variable

s specific entropy 1 variable

Chapter 5 Balance Principles

5.1 Introduction Continuum Mechanics is based on a series of general postulates or principles that are assumed to always be valid, regardless of the type of material and the range of displacements or deformations. Among these are the so-called balance principles: • Conservation of mass • Balance of linear momentum • Balance of angular momentum • Balance of energy (or first law of thermodynamics) A restriction that cannot be rigorously understood as a balance principle must be added to these laws, which is introduced by the • Second law of thermodynamics Theory and Problems 5.2 Mass Transport or Convective Flux In continuumContinuum mechanics, the Mechanics term convection is for associated Engineers with mass transport in the medium, which derives© X. Oliver from the and motion C. Agelet of its particles. de Saracibar The continuous medium is composed of particles, some of whose properties are associated with the amount of mass: specific weight, angular momentum, kinetic energy, etc. Then, when particles move and transport their mass, a transport of the these properties occurs, named convective transport (see Figure 5.1). Consider A, an arbitrary (scalar, vector or tensor) property of the continuous medium, and Ψ (x,t), the description of the amount of said property per unit of mass of the continuous medium. Consider also S, a control surface, i.e., a surface fixed in space (see Figure 5.2). Due to the motion of the particles in the medium, these cross the surface along time and, in consequence, there exists a certain amount of the property A that, associated with the mass transport, crosses the control surface S per unit of time.

193 194 CHAPTER 5. BALANCE PRINCIPLES

Figure 5.1: Convective transport in the continuous medium.

Definition 5.1. The convective flux (or mass transport flux) of a generic property A through a control surface S is the amount of A that, due to mass transport, crosses the surface S per unit of time. A convective flux not= Φ = amount of crossing S of A through S S unit of time

Theory and Problems

Continuum Mechanics for Engineers © X. Oliver and C. Agelet de Saracibar

Figure 5.2: Convective ﬂux through a control surface.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Mass Transport or Convective Flux 195

Figure 5.3: Cylinder occupied by the particles that have crossed dS in the time interval [t, t + dt].

To obtain the mathematical expression of the convective ﬂux of A through the surface S, consider a differential surface element dS and the velocity vector v of the particles that at time t are on dS (see Figure 5.3). In a time differential dt, these particles will have followed a pathline dx = vdt, such that at the instant of time t +dt they will occupy a new position in space. Taking now into account all the particles that have crossed dS in the time interval [t, t + dt], these will occupy a cylinder generated by translating the base dS along the directrix dx = vdt, and whose volume is given by dV = dS dh = v · n dt dS . (5.1) Since the volume (dV) of the particles crossing dS in the time interval [t, t + dt] is known, the mass crossing dS in this same time interval can be obtained by multiplying (5.1) by the density, dm = ρ dV = ρv · n dt dS . (5.2)

Finally, the amount of A crossing dS in the time interval [t, t + dt] is calculated by multiplying (5.2) by theTheory function Ψ and(amount Problems of A per unit of mass), Ψ dm = ρΨv · n dt dS . (5.3) Continuum Mechanics for Engineers Dividing (5.3)bydt yields© X. Oliverthe amount and of C. the Agelet property de that Saracibar crosses the differential control surface dS per unit of time, Ψ dm d Φ = = ρΨv · n dS . (5.4) S dt Integrating (5.4) over the control surface S results in the amount of the property A crossing the whole surface S per unit of time, that is, the convective ﬂux of the property A through S. convective ﬂux Φ = ρΨv · n dS (5.5) of A through S S S

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 196 CHAPTER 5. BALANCE PRINCIPLES

Example 5.1 – Compute the magnitude Ψ and the convective ﬂux ΦS corresponding to the following properties: a) volume, b) mass, c) linear momentum, d) kinetic energy.

Solution a) If the property A is the volume occupied by the particles, then Ψ is the volume per unit of mass, that is, the inverse of the density. Therefore, A ≡ Ψ = 1 Φ = · = . V and ρ lead to S v ndS volume ﬂow rate S b) If the property A is the mass, then Ψ is the mass per unit of mass, that is, the unit. Therefore,

A ≡ M and Ψ = 1 lead to ΦS = ρ v · ndS . S c) If the property A is the linear momentum (= mass × velocity), then Ψ is the linear momentum per unit of mass, that is, the velocity. Therefore,

A ≡ mv and Ψ = v lead to ΦS = ρ v(v · n) dS . S (Note that in this case Ψ and the convective ﬂux ΦS are vectors).

Remark 5.1. In a closed control surface1, S = ∂V, the expression of the convective flux corresponds to the net outflow, defined as the outflow minus the inflow (see Figure 5.4), that is, A not= Φ = ρΨ · . net convective flux of ∂V v n dS ∂V

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Mass Transport or Convective Flux 197

Figure 5.4: Net outﬂow through a closed control surface.

Remark 5.2. The convective flux of any property through a material surface is always null. Indeed, the convective flux of any property is associated, by definition, with the mass transport (of particles) and, on the other hand, a material surface is always formed by the same particles and cannot be crossed by them. Consequently, there is no mass transport through a material surface and, therefore, there is no convective flux through it.

Remark 5.3. Some propertiesTheory can and be transported Problems within a continuous medium in a manner not necessarily associated with mass transport. This form of non-convective transport receives several names (conduction,Continuum diffusion, etc.) Mechanics depending on the for physical Engineers problem being studied. A typical© example X. Oliver is heat and flux C. by conduction.Agelet de Saracibar The non-convective transport of a property is characterized by the non-convective flux vector (or tensor) q(x,t), which allows defining the (non-convective) flux through a surface S with unit normal vector n as non-convective flux = q · n dS . S

1 Unless stated otherwise, when dealing with closed surfaces, the positive direction of the unit normal vector n is taken in the outward direction of the surface.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 198 CHAPTER 5. BALANCE PRINCIPLES

5.3 Local and Material Derivatives of a Volume Integral Consider A, an arbitrary (scalar, vector or tensor) property of the continuous medium, and μ, the description of the amount of said property per unit of vol- 2 ume , amount of A μ (x,t)= . (5.6) unit of volume Consider an arbitrary volume V in space. At time t, the total amount Q(t) of the property contained in this volume is Q(t)= μ (x,t) dV . (5.7) V To compute the content of property A at a different time t + Δt, the following two situations arise: 1) A control volume V is considered and, therefore, it is ﬁxed in space and crossed by the particles along time.

2) A material volume that at time t occupies the spatial volume Vt ≡ V is considered and, thus, the volume occupies different positions in space along time. Different values of the amount Q(t + Δt) are obtained for each case, and computing the difference between the amounts Q(t + Δt) and Q(t) when Δt → 0 yields Q(t + Δt) − Q(t) Q (t)= lim , (5.8) Δt→0 Δt resulting in two different deﬁnitions of the time derivative, which lead to the concepts of local derivative and material derivative of a volume integral. 5.3.1 Local DerivativeTheory and Problems

DeﬁnitionContinuum 5.2. The local Mechanics derivative of the for volume Engineers integral, © X.Q Oliver(t)= μ and(x,t C.) dV Agelet, de Saracibar V is the time derivative of Q(t) when the volume V is a volume ﬁxed in space (control volume), see Figure 5.5. The notation ∂ not= μ ( , ) local derivative ∂ x t dV t V will be used.

2 μ is related toΨ =(amount of A)/(unit of mass) through μ = ρΨ and has the same tensor order as the property A .

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Local and Material Derivatives of a Volume Integral 199

Figure 5.5: Local derivative of a volume integral.

The amount Q of the generic property A in the control volume V at times t and t + Δt is, respectively, Q(t)= μ (x,t) dV and Q(t + Δt)= μ (x,t + Δt) dV . (5.9) V V

Using (5.9) in addition to the concept of time derivative of Q(t) results in3 ∂ 1 Q (t)= μ (x,t) dV = lim Q(t + Δt) − Q(t) = ∂t Δt→0 Δt V ⎛ ⎞ 1 = lim ⎝ μ (x,t + Δt) dV − μ (x,t) dV⎠ = Δt→0 Δt V V μ ( , + Δ ) − μ ( , ) ∂μ( , ) = x t t x t = x t , limTheoryΔ and ProblemsdV ∂ dV Δt→0 t t V V ∂μ(x,t) local Continuum Mechanics for Engineersderivative © X. Oliver∂t and C. Agelet de Saracibarof μ (5.10) which yields the mathematical expression of the local derivative of a volume integral.

Local derivative of a volume integral ∂ ∂μ(x,t) (5.11) μ (x,t) dV = dV ∂t ∂t V V

3 Note that the integration domain does not vary when the volume V is considered as a control volume and, therefore, is ﬁxed in space.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 200 CHAPTER 5. BALANCE PRINCIPLES

5.3.2 Material Derivative

Deﬁnition 5.3. The material derivative of the volume integral, Q(t)= μ (x,t) dV , V is the time derivative of Q(t) when the volume Vt is a material volume (mobile in space), see Figure 5.6. The notation d material derivative not= μ (x,t) dV dt Vt will be used.

The content Q of the generic property A in the material volume Vt at times t and t+Δt is, respectively, Q(t)= μ (x,t) dV and Q(t + Δt)= μ (x,t + Δt) dV . (5.12)

Vt Vt+Δt

Then, the material derivative is mathematically expressed as4 d Q(t + Δt) − Q(t) Q (t)= μ (x,t) dV = lim = Δt→0 Δ dt ≡ t Vt ⎛ Vt V ⎞ (5.13) 1 = lim ⎝ μ (x,t + Δt) dV − μ (x,t) dV⎠ . Δt→0 Δt VTheoryt+Δt and ProblemsVt The following step consists in introducing two variable substitutions, each suitableContinuum for one of the two Mechanicsintegrals in (5.13), for which Engineers lead to the same integration domain in both expressions.© X. Oliver These and variable C. Agelet substitutions de Saracibar are given by the equation of motion x = ϕ (X,t), particularized for times t and t + Δt, ⎧ ⎪ = ϕ ( , ) → ( ) = | ( , )| ( ) , ⎪ xt X t dx1 dx2 dx3 t F X t dX1 dX 2 dX3 ⎨⎪ dVt dV0 ⎪ = ϕ ( , + Δ ) → ( ) = | ( , + Δ )| ( ) , ⎪ xt+Δt X t t dx1 dx2 dx3 t+Δt F X t t dX1 dX2 dX3 ⎩⎪ dVt+Δt dV0 (5.14)

4 Note that the integration domains are now different at times t and t + Δt.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Local and Material Derivatives of a Volume Integral 201

Figure 5.6: Material derivative of a volume integral.

where the identity dVt = |F(X,t)| dV0 has been taken into account. The variable substitutions in (5.14) are introduced in (5.13), resulting in μ¯ ( , + Δ ) X t t d 1 μ (x,t) dV = lim μ (x(X,t + Δt),t + Δt) |F(X,t + Δt)| dV0 dt Δt→0 Δt Vt V0 − μ ( ( , ), ) | ( , )| = x X t t F X t dV0 V0 μ¯ (X,t) μ¯ (X,t + Δt) |F(X,t + Δt)| − μ¯ (X,t) |F(X,t)| = lim dV0 = Δt→0 Δt V0 ∂ d μ¯ (X,t) |F(X,t)| = μ (x,t) |F(x,t)| ∂t dt d = μ |F| dVTheory0 . and Problems dt V0 (5.15) Finally,Continuum expanding the last Mechanics integral in (5.15) for5 and Engineers considering the equality d |F|/dt = |F| ∇ · v yields© X. Oliver and C. Agelet de Saracibar d d dμ d |F| μ (x,t) dV = μ |F| dV0 = |F| + μ dV0 = dt dt dt dt Vt V0 V0 |F| ∇ · v dμ dμ = + μ∇ · v |F| dV = + μ∇ · v dV , dt 0 dt V V 0 dVt t (5.16)

5 The change of variable xt = ϕ (X,t) is undone here.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 202 CHAPTER 5. BALANCE PRINCIPLES that is6, d not d dμ μ (x,t) dV = μ (x,t) dV = + μ∇ · v dV . (5.17) dt dt dt V V ≡V V t Vt ≡V t

Recalling the expression of the material derivative of a property (1.15) results in ∂μ d μ ( , ) = + · ∇μ + μ∇ · = x t dV ∂ v v dV dt t Vt ≡V V ∇ · (μv) (5.18) ∂μ ∂ = dV + ∇ · (μv) dV = μ dV + ∇ · (μv) dV , ∂t ∂t V V V V where the expression of the local derivative (5.11) has been taken into account. Then, (5.18) produces the expression of the material derivative of a volume integral.

Material derivative of a volume integral d ∂ μ (x,t) dV = μ dV + ∇ · (μv) dV dt ∂t (5.19) ≡ Vt V V V material local convective derivative derivative derivative

Remark 5.4. The formTheory of the material and derivative, Problems given as a sum of a local derivative and a convective derivative, that appears when differentiating properties of the continuous medium (see Chapter 1, Sec- tionContinuum1.4) also appears here Mechanics when differentiating for Engineersintegrals in the continuous medium. Again,© X. Oliverthe convective and C. derivative Ageletis de associated Saracibar with the existence of a velocity (or motion) in the medium and, thus, with the possibility of mass transport.

6 The expression d μ (x,t) dV dt Vt ≡V denotes the time derivative of the integral over the material volume Vt (material derivative of the volume integral) particularized at time t, when the material volume occupies the spatial volume V.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Conservation of Mass. Mass continuity Equation 203

Figure 5.7: Principle of conservation of mass in a continuous medium.

5.4 Conservation of Mass. Mass continuity Equation

Deﬁnition 5.4. Principle of conservation of mass. The mass of a continuous medium (and, therefore, the mass of any material volume belonging to this medium) is always the same.

Consider a material volume Vt that at times t and t + Δt occupies the volumes in space Vt and Vt+Δt, respectively (see Figure 5.7). Consider also the spatial description of the density, ρ (x,t). The mass enclosed by the material volume V at times t and t + Δt is, respectively, M(t)= ρ (x,t) dVTheoryand M( andt + Δt)= Problemsρ (x,t + Δt) dV . (5.20)

Vt Vt+Δt Continuum Mechanics forM Engineers( )=M( + Δ ) By virtue of the principle© of X. conservation Oliver and of C. mass, Agelett de Saracibart t must be satisﬁed.

5.4.1 Spatial Form of the Principle of Conservation of Mass. Mass Continuity Equation The mathematical expression of the principle of conservation of mass of the material volume M(t) is that the material derivative of the integral (5.20)is null, d M (t)= ρ dV = 0 ∀t . (5.21) dt Vt

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 204 CHAPTER 5. BALANCE PRINCIPLES

By means of the expression of the material derivative of a volume integral (5.17), the integral (or global) spatial form of the principle of conservation of mass results in

Global spatial form of the principle of conservation of mass d dρ , (5.22) ρ dV = + ρ ∇ · v dV = 0 ∀ΔV ⊂ V , ∀t dt dt t t Vt Vt (ΔVt ) (ΔVt ) which must be satisﬁed for Vt and, also, for any partial material volume ΔVt ⊂ Vt that could be considered. In particular, it must be satisﬁed for each of the ele- mental material volumes associated with the different particles in the continuous medium that occupy the differential volumes dVt. Applying (5.22) on each dif- 7 ferential volume dVt ≡ dV (x,t) yields dρ dρ (x,t) + ρ∇ · v dV = + ρ (x,t)∇ · v(x,t) dV (x,t)=0 dt dt dV(x,t) ∀x ∈ Vt, ∀t dρ =⇒ + ρ∇ · v = 0 dV ∀x ∈ V , ∀t dt t (5.23)

Local spatial form of the principle of conservation of mass (mass continuity equation) (5.24) dρ + ρ∇ · v = 0 dV ∀x ∈ V , ∀t dt t Theory and Problems which constitutes the so-called mass continuity equation. Replacing the expression of the material derivative of the spatial description of a property (1.15)in (5.24) resultsContinuum in Mechanics for Engineers © X. Oliver and C. Agelet de Saracibar ∂ρ ∂ρ + v · ∇ρ + ρ∇ · v = 0 =⇒ + ∇ · (ρv)=0 , (5.25) ∂t ∂t ∇ · (ρv) which yields an alternative expression of the mass continuity equation.

7 This procedure, which allows reducing a global (or integral) expression such as (5.22)to a local (or differential) one such as (5.24), is named in continuum mechanics localization process.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Conservation of Mass. Mass continuity Equation 205

⎫ ⎪ ∂ρ ⎪ + ∇ · (ρv)=0 ⎪ ∂t ⎪ ⎬⎪ ∂ρ ∂ (ρvi) ∀ ∈ , ∀ + = 0 i ∈{1,2,3} ⎪ x Vt t (5.26) ∂t ∂x ⎪ i ⎪ ⎪ ∂ρ ∂ (ρv ) ∂ (ρvy) ∂ (ρv ) ⎪ + x + + z = 0 ⎭ ∂t ∂x ∂y ∂z

5.4.2 Material Form of the Principle of Conservation of Mass From (5.22)8, dρ dρ 1 d |F| + ρ∇ · v dV = + ρ dV = dt dt |F| dt Vt Vt 1 dρ d |F| 1 d = |F| + ρ dV = ρ |F| dV = |F| dt dt |F| dt Vt Vt | | F dV0 d ρ |F| dt ∂ = ρ (X,t)|F(X,t)| dV ∀ΔV ⊂ V , ∀t , ∂t 0 0 0 V0 (5.27) where the integration domain is now the volume in the reference conﬁgura- tion, V0. Given that (5.27) must be satisﬁed for each and every part ΔV0 of V0,a localization process can be applied, which results in9 ∂ Theory and Problems ρ (X,t)|F(X,t)| = 0 ∀X ∈ V , ∀t ∂t 0 Continuum=⇒ ρ (X,t)|F(X, Mechanicst)| = ρ (X)|F(X)| for∀t Engineers © X. Oliver and C. Agelet de Saracibar =⇒ ρ ( , )| |( , ) = ρ ( , )| |( , ) =⇒ ρ | | = ρ | | . X 0 F X 0 X t F X t 0 F 0 t F t = 1 not= ρ | | not= ρ | | 0 F 0 t F t (5.28) Local material form of the mass conservation principle ρ ( )=ρ ( )| | ( ) ∀ ∈ , ∀ (5.29) 0 X t X F t X X V0 t

8 Here, the expression deduced in Chapter 2, d |F|/dt = |F| · ∇ · v , is considered. 9 ( , )= =⇒ | | = The equality F X 0 1 F 0 1 is used here.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 206 CHAPTER 5. BALANCE PRINCIPLES

5.5 Balance Equation. Reynolds Transport Theorem Consider A, an arbitrary (scalar, vector or tensor) property of the continuous medium, and Ψ (x,t), the description of the amount of said property per unit of mass. Then, ρΨ (x,t) is the amount of this property per unit of volume.

5.5.1 Reynolds’ Lemma Consider an arbitrary material volume of the continuous medium that at time t occupies the volume in space Vt ≡ V. The amount of the generic property A in the material volume Vt at time t is Q(t)= ρΨ dV . (5.30)

Vt ≡V The variation along time of the content of property A in the material volume Vt is given by the time derivative of Q(t), which using expression (5.17)ofthe material derivative of a volume integral (with μ = ρΨ ) results in d d (ρΨ) Q (t)= ρΨ dV = + ρΨ ∇ · v dV . (5.31) dt dt Vt ≡V μ V

Considering the expression of the material derivative of a product of functions, grouping terms and introducing the mass continuity equation (5.24) yields d dΨ dρ ρΨdV = ρ +Ψ + ρΨ ∇ · v dV = dt dt dt Vt ≡V V dΨ dρ (5.32) = ρ +Ψ + ρ∇ · v dV =⇒ dt dt V Theory and Problems =0 (mass continuity eqn.)

ContinuumReynolds’ Mechanics Lemma for Engineers ©d X. Oliver and C. AgeletdΨ de. Saracibar ρΨ dV = ρ dV (5.33) dt dt Vt ≡V V

5.5.2 Reynolds’ Theorem Consider the arbitrary volume V, ﬁxed in space, shown in Figure 5.8. The amount of property A in this control volume is Q(t)= ρΨ dV . (5.34) V

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Balance Equation. Reynolds Transport Theorem 207

Figure 5.8: Reynolds Transport Theorem.

The variation of the amount of property A in the material volume Vt, which in- stantaneously coincides at time t with the control volume V (Vt ≡ V),isgivenby expression (5.19) of the material derivative of a volume integral (with μ = ρΨ) and by (5.11), d ∂ (ρΨ) ρΨ dV = dV + ∇ · (ρΨ v) dV . (5.35) dt ∂t Vt ≡V V V

Introducing the Reynolds’ Lemma (5.33) and the Divergence Theorem10 in (5.35) results in

Reynolds’ d Lemma dΨ ∂ (ρΨ) ρΨ dV = ρ dV = dV + ∇ · (ρΨ v)dV = dt dt ∂t Vt ≡V TheoryV andV ProblemsV Divergence Theorem ∂ (ρΨ) = dV + ρΨ v · n dS , Continuum Mechanics∂t for Engineers © X. Oliver andV C. Agelet de∂V Saracibar (5.36) which can be rewritten as follows.

10 The Divergence Theorem provides the following relation between a volume integral and a surface integral of a tensor A. ∇ · A dV = n · A dS ∀V , V ∂V where n is the outward unit normal vector in the boundary of the volume V.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 208 CHAPTER 5. BALANCE PRINCIPLES

Reynolds Transport Theorem ∂ ∂Ψ ρΨ = ρ − ρΨ · ∂ dV ∂ dV v n dS t t (5.37) V V ∂ V variation per unit of variation due to the variation due to the net time of the content of change in the content of convective ﬂux of A property A in the property A of the parti- exiting through the control volume V cles in the interior of V boundary ∂V

The local form of the Reynolds Transport Theorem can be obtained by localizing in (5.36), dΨ ∂ (ρΨ) ρ dV = dV + ∇ · (ρΨ v)dV ∀ΔV ⊂ V =⇒ dt ∂t V V V (5.38) dΨ ∂ (ρΨ) ρ = + ∇ · (ρΨ v) ∀x ∈ V =⇒ dt ∂t

Local form of the Reynolds Transport Theorem (5.39) ∂ (ρΨ) dΨ = ρ − ∇ · (ρΨ v) ∀x ∈ V ∂t dt

5.6 General Expression of the Balance Equations Consider a certain propertyTheoryA of a continuous and Problems medium and the amount of this property per unit of mass, Ψ (x,t). In the most general case, it can be assumed that there exists an internal source that generates property A and that this property canContinuum be transported both Mechanics by motion of mass for (convective Engineers transport) and by non-convective transport.© To X. this Oliver aim, the and following C. Agelet terms de are Saracibar deﬁned: • A source term kA (x,t) (of the same tensor order than property A) that characterizes the internal generation of the property, internally generated amount of A kA (x,t)= . (5.40) unit of mass / unit of time

• A vector jA (x,t) of non-convective ﬂux per unit of surface (a tensor order higher than that of property A) that characterizes the ﬂux of the property due to non-convective mechanisms (see Remark 5.3).

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 General Expression of the Balance Equations 209

Figure 5.9: An arbitrary control volume used in the deﬁnition of the global form of the general balance equation.

Consider an arbitrary control volume V (see Figure 5.9). Then, the variation per unit of time of property A in volume V will be due to 1) the generation of property A per unit of time due to the source term, 2) the (net incoming) convective ﬂux of A through ∂V, and 3) the (net incoming) non-convective ﬂux of A through ∂V.

That is,

amountTheory of A generated and in ProblemsV due to the internal sources ρkA (x,t)dV = , unit of time V Continuumamount of MechanicsA exiting through for∂V per Engineers convective ﬂux ρΨ v · n dS = © X. Oliver and C. Agelet de Saracibar , unit of time ∂V amount of A exiting through ∂V per non-convective ﬂux jA · n dS = , unit of time ∂V (5.41)

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 210 CHAPTER 5. BALANCE PRINCIPLES and the expression of the balance of the amount of property A in the control volume V results in

Global form of the general balance equation ∂ ρΨ dV = ρkA dV − ρΨ v · n dS − jA · n dS ∂t (5.42) V V ∂ V ∂ V variation of the variation due variation due to variation due to amount of A in V to internal the incoming the incoming per unit of time generation convective ﬂux non- convective ﬂux

Using the Divergence Theorem and (5.11), the global form of the general balance equation (5.42) can be written as ∂ ρΨ dV = ρkA dV − ∇ · (ρΨ v)dV − ∇ · jA dV =⇒ ∂t V V V V (5.43) ∂ (ρΨ)+∇ · (ρΨ v) dV = (ρkA − ∇ · jA)dV ∀ΔV ⊂ V ∂t V V and localizing in (5.43), the local spatial form of the general balance equation

Local spatial form of the general balance equation ∂ dΨ (ρΨ)+∇ · (ρΨ v) = ρ = ρkA − ∇ · jA ∂t dt Theory and Problems (5.44) dΨ variation of the variation variation ρ amount of due to due to non- dt property (per internal convective Continuum Mechanicsunit of volume generation for Engineerstransport © X. Oliverand of time) and C.by Agelet a source de Saracibar is obtained, where the local form of the Reynolds Transport Theorem (5.39) has been taken into account.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 General Expression of the Balance Equations 211

Remark 5.5. Expression (5.42) and, especially, expression (5.44), dΨ ρ = ρkA − ∇ · jA , dt

exhibit the negative contribution (−∇ · jA) of the non-convective ﬂux to the variation in content of the property per unit of volume and of time, ρ dΨ/dt. Only when all the ﬂux is convective (by mass transport) can this variation originate solely from the internal generation of this property, dΨ ρ = ρkA . dt

Example 5.2 – Particularize the local spatial form of the general balance equation for the case in which property A is associated with the mass.

Solution If property A is associated with the mass, A ≡ M, then: • The content of A per unit of mass (mass / unit of mass) is Ψ = 1. • The source term that characterizes the internal generation of mass is kM = 0 since, following the principle of conservation of mass, it is not possible to generate mass. • The non-convective mass ﬂux vector is jM = 0 because mass cannot be transported in a non-convectiveTheory and manner. Problems Therefore, (5.44) results in the balance of mass generation, Continuum MechanicsdΨ ∂ρ for Engineers © X.ρ Oliver= and+ ∇ C.· (ρ Ageletv)=0 , de Saracibar dt ∂t which is one of the forms of the mass continuity equation (5.26).

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 212 CHAPTER 5. BALANCE PRINCIPLES

5.7 Balance of Linear Momentum Consider a discrete system composed of n particles such that the particle i has a mass mi, an acceleration ai and is subjected to a force fi (see Figure 5.10). Newton’s second law states that the force acting on a particle is equal to the mass of this particle times its acceleration. Using the deﬁnition of acceleration as the material derivative of the velocity and considering the principle of conservation of mass (the variation of mass of a particle is null) yields11, Figure 5.10

dv d f = m a = m i = (m v ) (5.45) i i i i dt dt i i The linear momentum of the particle12 is deﬁned as the product of its mass by its velocity (mivi). Then, (5.45) expresses that the force acting on the particle is equal to the variation of the linear momentum of the particle. Applying now Newton’s second law to the discrete system formed by n particles results in n n n dv d n dP (t) R(t)=∑ f = ∑ m a = ∑ m i = ∑ m v = . (5.46) i i i i dt dt i i dt i=1 i=1 i=1 i=1 P = linear momentum Note that, again, to obtain the last expression in (5.46), the principle of conservation of mass (dmi/dt = 0) has been used. Equation (5.46) expresses that the resultant R of all the forces acting on the discrete system of particles is equal to the variation per unit of time of the linear momentum P of the system. This postulate is denominatedTheory the principle and of balance Problems of linear momentum.

RemarkContinuum 5.6. If the system Mechanics is in equilibrium, forR = Engineers0. Then, © X. Oliver and C. Agelet de Saracibar dP (t) n R(t)=0 ∀t =⇒ = 0 =⇒ ∑ mi vi = P = const., dt i=1 which is known as the conservation of linear momentum.

11 The Einstein notation introduced in (1.1) is not used here. 12 In mechanics, the names translational momentum, kinetic momentum or simply momentum are also used to refer to the linear momentum.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Balance of Linear Momentum 213

5.7.1 Global Form of the Balance of Linear Momentum These concepts, corresponding to classical mechanics, can now be extended to continuum mechanics by deﬁning the linear momentum in a material volume Vt of the continuous medium with mass M as P ( )= M = ρ . t v d v dV (5.47) M ρ dV Vt

Deﬁnition 5.5. Principle of balance of linear momentum. The resultant R(t) of all the forces acting on a material volume of the continuous medium is equal to the variation per unit of time of its linear momentum, dP (t) d R(t)= = ρ v dV . dt dt Vt

The resultant of all the forces acting on the continuous medium deﬁned above is also known to be (see Figure 5.11) R(t)= ρb dV + t dS . (5.48) V ∂ V body surface forces forces Applying the principle of balance of linear momentum on the resultant in (5.48) yields the integral form ofTheory the balance and of linear Problems momentum.

GlobalContinuum form of the principle Mechanics of balance of for linear Engineers momentum © X. Oliver and C. Agelet de Saracibar d (5.49) ρb dV + t dS = ρv dV dt V ∂V Vt ≡V

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 214 CHAPTER 5. BALANCE PRINCIPLES

Figure 5.11: Forces acting on a material volume of the continuous medium.

5.7.2 Local Form of the Balance of Linear Momentum Using Reynolds’ Lemma (5.33)on(5.49) and introducing the Divergence The- orem, results in ⎫ d dv ⎪ ρv dV = ρb dV + n · σ dS = ρ dV ⎪ dt dt ⎪ ≡ ≡ ⎬ Vt V V ∂V t Vt V =⇒ (5.50) Divergence ⎪ · σ Theorem= ∇ · σ ⎪ n dS dV ⎭⎪ ∂V V Theory and Problems dv =⇒ (∇ · σ + ρb)dV + ρ dV ∀ΔV ⊂ V (5.51) dt ContinuumV MechanicsV for Engineers and, localizing in (5.51©), yields X. Oliver the local and spatial C. Agelet form of de the Saracibar balance of linear momentum, also known as Cauchy’s equation13.

Local spatial form of the principle of balance of linear momentum (Cauchy’s equation) (5.52) dv ∇ · σ + ρb = ρ = ρa ∀x ∈ V, ∀t dt

13 The Cauchy equation (already stated, but not deduced, in Chapter 4 ) is, thus, identiﬁed as the local spatial form of the balance of linear momentum.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Balance of Angular Momentum 215

5.8 Balance of Angular Momentum Consider a discrete system composed of n particles such that for an arbitrary particle i, its position vector is ri, its mass is mi, a force fi acts on it, and it has a velocity vi and an acceleration ai (see Figure 5.12). The moment about the origin of the force acting on this particle is Mi = ri ×fi and the moment about the origin of the linear 14 momentum of the particle is Li = ri × mivi. Considering Newton’s second law, the moment 15 Figure 5.12 Mi is dv M = r × f = r × m a = r × m i (5.53) i i i i i i i i dt Extending the previous result to the discrete system formed by n particles, the resultant moment about the origin MO of the forces acting on the system of particles is obtained as16 ⎫ n n n ⎪ ( )= × = × = × dvi ⎪ MO t ∑ ri fi ∑ ri mi ai ∑ ri mi ⎪ i=1 i=1 i=1 dt ⎪ ⎬⎪ d n n dr n dv ∑ r × m v = ∑ i ×m v + ∑ r × m i =⇒ dt i i i dt i i i i dt ⎪ i=1 i=1 i=1 ⎪ ⎪ vi ⎪ ⎭ (5.54) = 0

d n dL(t) =⇒ M (t)= ∑ r × m v = O dt i i i dt Theory and i=1 Problems Angular momentum L

EquationContinuum (5.54) expresses Mechanics that the resultant moment for EngineersMO of all the forces acting on the discrete system© X. of particles Oliver and is equal C. Ageletto the variation de Saracibar per unit of time of the moment of linear momentum (or angular momentum), L , of the system. This postulate is named principle of balance of angular momentum.

15 In mechanics, the moment of (linear) momentum is also named angular momentum or rotational momentum. 15 The Einstein notation introduced in (1.1) is not used here. 16 The vector or cross product of a vector times itself is null (vi × vi = 0).

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 216 CHAPTER 5. BALANCE PRINCIPLES

Remark 5.7. If the system is in equilibrium, MO = 0. Then, dL(t) n MO (t)=0 ∀t =⇒ = 0 =⇒ ∑ ri ×mi vi = L = const., dt i=1 which is known as the conservation of angular momentum.

5.8.1 Global Form of the Balance of Angular Momentum Result (5.54) can be extended to a continuous and inﬁnite system of particles (the continuous medium, see Figure 5.13). In such case, the angular momentum is deﬁned as L = × M = × ρ r v d r v dV (5.55) M ρ dV V and the continuous version of the postulate of balance of angular momentum is obtained as follows.

Deﬁnition 5.6. Principle of balance of moment of (linear) momentum or angular momentum. The resultant moment, about a certain point O in space, of all the actions on a continuous medium is equal to the variation per unit of time of the moment of linear momentum about said point. dL(t) d MO (t)= = r × ρ v dV dt dt Vt ≡V Theory and Problems

Continuum Mechanics for Engineers © X. Oliver and C. Agelet de Saracibar

Figure 5.13: Moments acting on a material volume of the continuous medium.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Balance of Angular Momentum 217

The resultant moment of the forces acting on the continuous medium (moment of the body forces and moment of the surface forces) is (see Figure 5.13)

MO (t)= r × ρ b dV + r × t dS , (5.56) V ∂V then, the global form of the principle of balance of the angular momentum results in:

Global form of the principle of balance of angular momentum d (5.57) r × ρ v dV = r × ρ b dV + r × t dS dt Vt ≡V V ∂V

5.8.2 Local Spatial Form of the Balance of Angular Momentum The procedure followed to obtain the local spatial form of the balance equation is detailed below. Introducing Reynolds’ Lemma in (5.57), d d d r × ρv dV = ρ (r × v)dV = ρ (r × v)dV = dt dt dt ≡ ≡ Vt V Vt V V dr dv dv = ρ ×v dV + ρ r × dV = r × ρ dV , (5.58) dt dt dt V V V v Theory and Problems = 0 and expandingContinuum the last term Mechanics in (5.57), for Engineers © X. Oliver and C. Agelet de Saracibar × = × · σ = [ ] × [ · σ ]T = r t dS r n dS r n dS ∂V n · σ ∂V ∂V Divergence (5.59) Theorem = (r × σ T ) · n dS = r × σ T · ∇ dV , ∂V V

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 218 CHAPTER 5. BALANCE PRINCIPLES

× σ T · ∇ where the component r i is computed as ∂ ∂ σ T symb r × σ · ∇ = eijk x j σrk = eijk x j σrk = i ∂xr ∂xr σ T kr (5.60) ∂x ∂σ = e j σ + e x rk = e σ +[r × ∇ · σ ] i ∈{1,2,3} . ijk ∂x rk ijk j ∂x ijk jk i r r [ × ∇ · σ ] mi δ jr r i Introducing now (5.60)in(5.59) produces

r × t dS = m dV + (r × ∇ · σ ) dV ∂V V V (5.61)

mi = eijk σ jk i, j,k ∈{1,2,3} and, ﬁnally, replacing (5.58) and (5.61)in(5.57) yields dv r × ρ dV = r × ρb dV + m dV + (r × ∇ · σ )dV . (5.62) dt V V V V Reorganizing the terms in (5.62) and taking into account Cauchy’s equation (5.52) (local spatial form of the balance of linear momentum) results in dv r × ∇ · σ + ρb − ρ dV + m dV = 0 Theorydt and Problems V V = 0 (5.63) Continuum Mechanics=⇒ for= Engineers∀Δ ⊂ . © X. Oliver andm C.dV Agelet0 deV SaracibarV V

Then, localizing in (5.63) and considering the value of m in (5.61), yields ! m = 0 ∀x ∈ V =⇒ eijk σ jk = 0 i, j,k ∈{1,2,3} (5.64) mi = eijk σ jk = 0 i ∈{1,2,3}

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Power 219 and particularizing (5.64) for the three possible values of index i: ⎫ = σ = σ + σ = σ − σ = ⇒ σ = σ ⎪ i 1:e1 jk jk e123 23 e132 32 23 32 0 23 32 ⎪ ⎪ = =− ⎪ 1 1 ⎬⎪ = σ = σ + σ = σ − σ = ⇒ σ = σ i 2:e2 jk jk e231 31 e213 13 31 13 0 31 13 =⇒ ⎪ =1 =−1 ⎪ ⎪ i = 3:e σ = e σ + e σ = σ − σ = 0 ⇒ σ = σ ⎪ 3 jk jk 312 12 321 21 12 21 12 21 ⎭⎪ =1 =−1 =⇒ σ = σ T , (5.65) which results in the local spatial form of the balance of angular momentum translating into the symmetry of the Cauchy stress tensor17.

Local spatial form of the principle of balance of angular momentum (5.66) σ = σ T

5.9 Power

Definition 5.7. In classical mechanics as well as in continuum mechanics, power is defined as a concept, previous to that of energy, that can be quantified as the ability to perform work per unit of time. Then, for a system (orTheory continuous and medium) Problems the power W (t) entering the system is defined as work performed by the system ContinuumW (t)= Mechanics for Engineers. © X. Oliverunit and of C. time Agelet de Saracibar

In some cases, but not in all, the power W (t) is an exact differential of a function E (t) that, in said cases, receives the name of energy,

dE (t) W (t)= . (5.67) dt

17 The symmetry of the Cauchy stress tensor (already stated, but not deduced, in Chapter 4 ) is, thus, identiﬁed as the local spatial form of the balance of angular momentum.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 220 CHAPTER 5. BALANCE PRINCIPLES

Here, it is assumed that there exist two procedures by which the continuous medium absorbs power from the exterior and performs work per unit of time with this power − Mechanical power, by means of the work performed by the mechanical actions (body and surface forces) acting on the medium. − Thermal power, by means of the heat entering the medium.

5.9.1 Mechanical Power. Balance of Mechanical Energy

Deﬁnition 5.8. The mechanical power entering the continuous medium, Pe, is the work per unit of time performed by all the (body and surface) forces acting on the medium.

Consider the continuous medium shown in Figure 5.14 is subjected to the action of body forces, characterized by the vector of body forces b(x,t), and of surface forces, characterized by the traction vector t(x,t). The expression of the mechanical power entering the system Pe is = ρ · + · = ρ · + · (σ · ) . Pe b v dV t v dS b v dV n v dS (5.68) V ∂V n · σ V ∂V

Theory and Problems

dr Continuum Mechanics for Engineersρb · dV = ρb · vdV © X. Oliver and C. Agelet dedt Saracibar

dr t · dS = t · vdS dt

Figure 5.14: Continuous medium subjected to body and surface forces.

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Applying the Divergence Theorem in the last term of (5.68) yields ⎧ ⎪ · (σ · ) = ∇ · (σ · ) ⎪ n v dS v dV ⎨⎪ ∂V V ∂ ∂σ ∂v ⎪ ∇ · (σ · v)= (σ v )= ij v + σ j =(∇ · σ ) · v + σ : l ⎪ ∂ ij j ∂ j ij ∂ ⎩⎪ xi xi xi σ [∇ · σ ] ji [l] j ji (5.69) and, taking into account the identity l = v ⊗ ∇ = d + w (see Chapter 2)18, σ = σ + σ = σ . : l : d : w : d (5.70) d + w = 0 Replacing (5.70)in(5.69) yields n · (σ · v) dS = (∇ · σ ) · v dV + σ : d dV . (5.71) ∂V V V

Introducing (5.71)in(5.68), the mechanical power entering the continuous 19 medium results in

Pe = ρ b · vdV + t · vdS = ρ b · vdV + (∇ · σ ) · vdV + σ : ddV = V ∂V V V V dv = (∇ · σ + ρ b) · vdV + σ : ddV = ρ · vdV + σ : ddV = dt V V V V d 1 d 1 ρ v · v dV + σ : ddV = ρ v2 dV + σ : ddV . dt 2 dt 2 V V V V (5.72) And applying Reynolds’Theory Lemma (5.33 and)in(5.72 Problems), the mechanical power entering the system results in BalanceContinuum of mechanical Mechanics energy for Engineers © X. Oliver and C. Agelet de Saracibar d 1 P = ρ b · vdV + t · vdS = ρv2 dV + σ : d dV e dt 2 ≡ (5.73) V ∂V Vt V V mechanical power entering K = kinetic stress the medium energy power

18 The tensor σ is symmetric and the tensor w is antisymmetric. Consequently, their product is null,σ : w = 0. d 1 1 dv 1 dv dv 19 The expression v · v = · v + v · = · v is used here, in addition to the dt 2 2 dt 2 dt dt notation v · v = |v|2 = v2.

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Equation (5.73) constitutes the continuum mechanics generalization of the balance of mechanical energy in classical mechanics.

Deﬁnition 5.9. The balance of mechanical energy states that the mechanical energy entering the continuous medium,

Pe = ρ b · v dV + t · v dS V ∂V is invested in: a) modifying the kinetic energy of the particles in the continuous medium, 1 dK d 1 kinetic energy not= K = ρ v2dV =⇒ = ρ v2dV . 2 dt dt 2 V V b) creating stress power, def stress power = σ : d dV . V

Remark 5.8. Considering (5.73), the stress power can be deﬁned as the part of the mechanical power entering the system that is not used in modifying the kinetic energy. It can be interpreted as the work per unit of time (power) performed by the stresses during the deformation process of the medium. In a rigid body there is no deformation nor strain rate (d = 0). There- fore, the stresses do not perform mechanical work and the stress power is null. In this case, all the mechanical power entering the system is invested in modifying the kinetic energy of the system and the balance of mechanicalTheory energy and of a rigid Problems body is recovered.

Continuum Mechanics for Engineers 5.9.2 Thermal Power© X. Oliver and C. Agelet de Saracibar

Deﬁnition 5.10. The thermal power entering the continuous medium, Qe, is the amount of heat per unit of time entering the medium.

The heat entering the medium can be produced by two main causes: a) Heat entering the medium due to the (non-convective) heat ﬂux across the boundary corresponding to the material volume. Note that, since the vol-

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Power 223

ume is a material volume, the heat flux due to mass transport (convective) is null and, thus, all the heat flux entering the medium will be non-convective. b) The existence of heat sources inside the continuous medium. • Non-convective heat flux Consider the spatial description of the vector of non-convective heat flux per unit of surface, q(x,t). Then, the net non-convective heat flux across the boundary of the material volume is (see Figure 5.15) amount of heat exiting the medium q · n dS = unit of time ∂V amount of heat entering the medium (5.74) − q · n dS = unit of time ∂V

Remark 5.9. A typical example of non-convective ﬂux is heat transfer by conduction phenomena. Heat conduction is governed by Fourier’s Law, which provides the vector of heat ﬂux by (non- convective) conduction q(x,t) in terms of the temperature θ (x,t), Fourier’s Law of q(x,t)=−K∇θ (x,t) , heat conduction where K is the thermal conductivity, a material property.

Theory and Problems

Continuum Mechanics for Engineers © X. Oliver and C. Agelet de Saracibar

Figure 5.15: Non-convective heat ﬂux.

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Figure 5.16: Internal heat sources.

• Internal heat sources Heat can be generated (or absorbed) in the interior of the continuous medium due to certain phenomena (chemical reactions, etc.). Consider a scalar function r (x,t) that describes in spatial form the heat generated by the internal sources per unit of mass and unit of time (see Figure 5.16). Then, the heat entering the system, per unit of time, due to the existence of internal heat sources is heat generated by the internal sources ρrdV = . (5.75) unit of time V Consequently, the total heat entering the continuous medium per unit of time (or thermal power Qe) can be expressed as the sum of the contributions of the conduction ﬂuxTheory (5.74) and and the internal Problems sources (5.75),

Heat power entering = ρ − · . Continuum MechanicsQe forrdV Engineersq n dS (5.76) the medium© X. Oliver andV C. Agelet∂V de Saracibar

Then, considering (5.73) and (5.76), the total power entering the continuous medium can be written as follows.

Total power entering the system d 1 P + Q = ρv2 dV + σ : ddV + ρrdV − q · ndS (5.77) e e dt 2 Vt ≡V V V ∂V

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Energy Balance 225

5.10 Energy Balance 5.10.1 Thermodynamic Concepts • Thermodynamic system: a certain amount of continuous matter always formed by the same particles (in the case studied here, a material volume). • Thermodynamic variables: a set of macroscopic variables that characterize the system and intervene in all the physical processes to be studied. They are designated by μi (x,t) i ∈ {1,2,...,n}. • State, independent or free variables: a subset of the group of thermodynamic variables in terms of which all the other variables can be expressed. • Thermodynamic state: a thermodynamic state is defined when a certain value is assigned to the state variables and, therefore, to all the thermodynamic variables. In a hyperspace (thermodynamic space) defined by the thermodynamic variables μi i ∈ {1,2,...,n} (see Figure 5.17), a thermodynamic state is represented by a point. • Thermodynamic process: the energetic development of a thermodynamic system that undergoes successive thermodynamic states, changing from an initial state at time tA to a final state at time tB (it is a path or continuous segment in the thermodynamic space), see Figure 5.18. • Closed cycle: A thermodynamic process in which the final thermodynamic state coincides with the initial thermodynamic state (all the thermodynamic variables recover their initial value), see Figure 5.19. • State function: any scalar, vector or tensor function φ (μ1,...,μn) of the thermodynamic variables that can be written univocally in terms of these variables. Consider a thermodynamic space with thermodynamic variables μi (x,t) i ∈ {1,2,...,n} and a function φ (μ1,...,μn) of said variables implicitly defined in terms of a differentialTheory form20 and Problems δφ = f1 (μ1,...,μn)dμ1 + ... + fn (μ1,...,μn)dμn . (5.78) Continuum Mechanics for Engineers © X. Oliver and C. Agelet de Saracibar

Figure 5.17: Thermodynamic process.

20 In continuum mechanics thermodynamics it is common to mathematically describe a function φ (μ1,...,μn) of the thermodynamic variables in terms of a differential form δφ.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 226 CHAPTER 5. BALANCE PRINCIPLES

Figure 5.18: Thermodynamic space. Figure 5.19: Closed cycle.

Consider also a given thermodynamic process A → B in the space of the thermodynamic variables. Equation (5.78) provides the value of the function φ(μB,...,μB) not= φ φ(μA,...,μA) not= φ 1 n B when its value 1 n A and the corresponding path (thermodynamic process) A → B are known by means of

φB = φA + δφ . (5.79) A

However, (5.79) does not guarantee that the result φB is independent of the path (thermodynamic process) followed. In mathematical terms, it does not guarantee that the function φ : Rn → R deﬁned by (5.79) is univocal (see Figure 5.20) and, thus, that there exists a single image φ (μ1,...,μn) corresponding to each point in the thermodynamic space.

Theory and Problems

Continuum Mechanics for Engineers © X. Oliver and C. Agelet de Saracibar

Figure 5.20: Non-univocal function of the thermodynamic variables μ1 and μ2.

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Remark 5.10. For a function φ (μ1,...,μn), implicitly described in terms of a differential form δφ,tobeastate function (that is, for it to be univocal), said differential form must be an exact differential δφ = dφ. In other words, the differential form δφ must be integrable. The necessary and sufﬁcient condition for a differential form such as (5.78) to be an exact differential is the equality of mixed partial derivatives, ⎫ ⎪ ⎬⎪ δφ = f1 (μ1,...,μn)dμ1 + ... + fn (μ1,...,μn)dμn ⇔ δφ = φ . ∂ f (μ ,...,μ ) ∂ f (μ ,...,μ ) ⎪ d i 1 n = j 1 n ∀i, j ∈{1,...,n} ⎭⎪ ∂μj ∂μi

If the differential form (5.78) is an exact differential, (5.79) results in

B B φB = φA + dφ = φA + Δφ (5.80) A A and the value φB is independent of the integration path. Then, function φ is said to be a state function that depends only on the values of the state variables and not on the thermodynamic process. Theory and Problems Remark 5.11. If φ is a state function, then δφ is an exact differential andContinuum the integral along Mechanics the complete closed for cycle Engineers of the differential δφ is null, © X. Oliver and C. Agelet de Saracibar A " A δφ = dφ = Δφ = 0 . A A = 0

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 228 CHAPTER 5. BALANCE PRINCIPLES

Example 5.3 – Determine whether the function φ (μ1, μ2) deﬁned in terms of an exact differential δφ = 4μ2 dμ1 + μ1 dμ2 can be a state function or not.

Solution Following (5.78), ∂ f1 = f ≡ 4μ ∂μ 4 ∂ ∂ 1 2 =⇒ 2 =⇒ f1 = f2 ∂μ ∂μ f2 ≡ μ1 ∂ f 2 1 2 = 1 ∂μ1 Then, δφ is not an exact differential (see Remark 5.10) and φ is not a state function.

5.10.2 First Law of Thermodynamics Experience shows that the mechanical power (5.73) is not an exact differential and, therefore, the mechanical work performed by the system in a closed cycle is not null. The same happens with the thermal power (5.76). "

δφ1 = Pe dt =⇒ Pe dt = 0 " (5.81) δφ2 = Qe dt =⇒ Qe dt = 0

However, there exists experimental evidence that proves that the sum of the mechanical and thermal powers, that is, the total power entering the system (5.77) (see Figure 5.21), is, in effect, an exact differential and, thus, a state function E that corresponds to the concept of energy can be deﬁned in terms of it,

Theory andt Problems

Pe dt + Qe dt = dE ⇒ E (t)= (Pe + Qe) dt + const. (5.82) Continuum Mechanicst0 for Engineers © X. Oliver and C. Agelet de Saracibar

Figure 5.21: Total power entering the system.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Energy Balance 229

The ﬁrst law of thermodynamics postulates the following: 1) There exists a state function E, named total energy of the system, such that its variation per unit of time is equal to the sum of the mechanical and thermal powers entering the system.

dE = P + Q dt e e E = + (5.83) d Pe dt Q e dt Variation of Mechanical Thermal total energy work work

2) There exists another state function U, named internal energy of the system, such that a) It is an extensive property21. Then, a speciﬁc internal energy u(x,t) (or internal energy per unit of mass) can be deﬁned as U = ρudV . (5.84) V

b) The variation of the total energy of the system E is equal to the sum of the variation of the internal energy U and the variation of the kinetic energy K. E = K + U d d d (5.85) Exact Exact differential differential Theory and Problems Remark 5.12. Note that, since the total energy E and the internal energy U of the system have been postulated to be state functions, dEContinuumand dU in (5.85) are Mechanics exact differentials. for Consequently, Engineers the term dK = dE − dU in© said X. equation Oliver is and also C. an exact Agelet differential de Saracibar (because the difference between exact differentials is also an exact differential) and, thus, is a state function. Then, it is conﬁrmed that (5.85) indirectly postulates the character of state function and, therefore, the energetic character of K.

21 A certain property is extensive when the complete content of the property is the sum of the content of the property in each of its parts. An extensive property allows deﬁning the content of this property per unit of mass (speciﬁc value of the property) or per unit of volume (density of the property).

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From (5.83) and considering (5.77), ⎫ E ⎪ d d 1 2 σ ⎪ = Pe + Qe = ρv dV + σ : d dV + ρrdV− q · n dS⎪ dt dt 2 ⎬ ≡ Vt V V V ∂V ⇒ 1 2 ⎪ K = ρv dV ⎪ 2 ⎭ V dE dK dU d 1 = + = ρv2dV + σ : d dV + ρrdV− q · n dS dt dt dt dt 2 V V V ∂V dK dU dt dt (5.86)

Global form of the internal energy balance dU d (5.87) = ρudV= σ : d dV + ρrdV− q · n dS dt dt Vt ≡V V V ∂V

Remark 5.13. From (5.87) it follows that any variation per unit of time of the internal energy dU/dt is produced by − a generation ofTheory stress power, andσ : Problemsd dV, and V − a variation per unit of time of the content of heat in the medium, Continuumρ − · Mechanics for Engineers rdV ©q X.n dS Oliver. and C. Agelet de Saracibar V ∂V

Applying Reynolds’ Lemma (5.33) and the Divergence Theorem on (5.87) yields d du ρudV= ρ dV = σ : d dV + ρrdV− ∇·q dV ∀ΔV ⊂ V . dt dt Vt ≡V V V V V (5.88)

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Reversible and Irreversible Processes 231

Finally, localizing in (5.88) results in the local spatial form of the internal energy balance.

Local spatial form of the internal energy balance (energy equation) (5.89) du ρ = σ : d +(ρr − ∇ · q) ∀x ∈ V, ∀t dt

5.11 Reversible and Irreversible Processes The first law of thermodynamics leads to a balance equation that must be ful- filled for all the physical processes that take place in reality, dE dU dK P + Q = = + . (5.90) e e dt dt dt In particular, if an isolated system22 is considered, the time variation of the total energy of the system will be null (dE/dt = 0 ⇒ the total energy is conserved). Therefore, the energy balance equation (5.90), established by the first law of thermodynamics, imposes that any variation of internal energy dU/dt must be compensated with a variation of kinetic energy dK/dt of equal value but of opposite sign, and vice-versa (see Figure 5.22). What the first law of thermodynamics does not establish is whether this (kinetic and internal) energy exchange in an isolated system can take place equally in both directions or not (dU/dt = −dK/dt > 0ordU/dt = −dK/dt < 0). That is, it does not establish any restriction that indicates if an imaginary and arbitrary Theory and Problems

Figure 5.22: Isolated thermodynamic system.

22 An isolated thermodynamic system is a system that cannot exchange energy with its exterior. In a strict sense, the only perfectly isolated system is the universe, although one can think of quasi-isolated or imperfectly isolated smaller systems.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 232 CHAPTER 5. BALANCE PRINCIPLES process that implies an energy exchange in a certain direction is physically possible or not. It only establishes the fulﬁllment of the energy balance (5.90) in the event that the process takes place. However, experience shows that certain processes that could be imagined theoretically never take place in reality. Suppose, for example, the isolated system in Figure 5.23 consist- ing of − a rigid (non-deformable) wheel that spins with angular velocity ω, and − a brake that can be applied on the wheel at a certain instant of time. Figure 5.23

Consider now the following two processes: 1) At a certain instant of time the brake acts, the rotation speed of the wheel ω decreases and, thus, so does its kinetic energy (dK < 0). On the other hand, due to the friction between the brake and the wheel, heat is generated and there is an increase of the internal energy (dU > 0). Experience shows that this process, in which the internal energy increases at the expense of decreasing the kinetic energy23, can take place in reality and, therefore, is a physically feasible process. 2) Maintaining the brake disabled, at a certain instant of time the wheel spontaneously increases its rotation speed ω and, thus, its kinetic energy increases (dK > 0). According to the first law of thermodynamics, the internal energy of the system will decrease (dU < 0). However, experience shows that this (spontaneous) increase of speed never takes place, and neither does the decrease in the amount of heat of the system (which would be reflected in a decreaseTheory in temperature). and Problems The conclusion to this observation is that the second process considered in the exampleContinuum is not a feasible Mechanics physical process for. More Engineers generally, only thermodynamic processes that© tend X. toOliver increase and the C. internal Agelet energy de Saracibar and decrease the kinetic energy, and not the other way round, are feasible for the system under consideration. It is concluded, then, that the first law of thermodynamics is only applicable when a particular physical process is feasible, and the need to determine when a particular physical process is feasible, or if a physical process is feasible in one direction, in both or in none, is noted. The answer to this problem is provided by the second law of thermodynamics.

23 The wheel, being a non-deformable medium, has null stress power (see Remark 5.8) and all the variation of internal energy of the system derives from a variation of its heat content (see Remark 5.13).

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Figure 5.24: Reversible (left) and irreversible (right) processes.

The previous considerations lead to the classiﬁcation, from a thermodynamic point of view, of the possible physical processes in feasible and non-feasible processes and, in addition, suggest classifying the feasible processes into reversible and irreversible processes.

Definition 5.11. A thermodynamic process A → B is a reversible process when it is possible to return from the final thermodynamic state B to the initial thermodynamic state A along the same path (see Figure 5.24). A thermodynamic process A → B is an irreversible process when it is not possible to return from the final thermodynamic state B to the initial thermodynamic state A, along the same path (evenifa different path can be followed, see Figure 5.24).

In general, within a sameTheory thermodynamic and Problemsprocess there will exist reversible and irreversible sections. Continuum Mechanics for Engineers 5.12 Second Law© of X. Thermodynamics. Oliver and C. Agelet Entropy de Saracibar 5.12.1 Second Law of Thermodynamics. Global form The second law of thermodynamic postulates the following: 1) There exists a state function named absolute temperature θ (x,t) that is intensive24 and strictly positive (θ > 0).

24 A certain property is intensive when the complete content of the property is not the sum of the content of the property in each of its parts. Contrary to what happens with extensive properties, in this case the content of the property cannot be deﬁned per unit of mass (spe- ciﬁc value of the property) or per unit of volume (density of the property). Temperature is a paradigmatic example of intensive property.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 234 CHAPTER 5. BALANCE PRINCIPLES

2) There exists a state function named entropy S with the following character- istics: a) It is an extensive variable. This implies that there exists a speciﬁc entropy (entropy per unit of mass) s such that entropy s = =⇒ S = ρsdV. (5.91) unit of mass V b) The inequality

Integral form of the second law of thermodynamics dS d r q = ρsdV ≥ ρ dV − · n dS (5.92) dt dt θ θ Vt ≡V V ∂V

is satisﬁed, where: − The sign = corresponds to reversible processes. − The sign > corresponds to irreversible processes. − The sign < cannot occur and indicates that the corresponding process is not feasible.

5.12.2 Physical Interpretation of the Second Law of Thermodynamics As discussed Section 5.9.2, the magnitude heat in the system is characterized by a) A source term (or generation of heat per unit of mass and unit of time) r (x,t), defined in the interior of the material volume. b) The non-convective flux (heat flux by conduction) across the boundary of the material surface,Theory defined in terms and of Problems a non-convective flux vector per unit of surface q(x,t). TheseContinuum terms allow computing Mechanics the amount of for heat Engineersper unit of time entering a material volume Vt, which© atX. aOliver certain instant and C. of timeAgelet occupies de Saracibar the spatial volume Vt ≡ V with outward unit normal vector n,as

Qe = ρrdV− q · n dS . (5.93) V ∂V

Consider now a new magnitude deﬁned as heat per unit of absolute temperature in the system. If θ (x,t) is the absolute temperature, the amount of said magnitude will be characterized by a) A source term r/θ corresponding to the generation of heat per unit of absolute temperature, per unit of mass and unit of time.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Second Law of Thermodynamics. Entropy 235

b) A non-convective ﬂux vector q/θ of the heat per unit of absolute temperature.

Magnitude Source term Non-convective ﬂux vector heat r q unit of time heat/unit of absolute temperature r q unit of time θ θ

Similarly to (5.93), the new source term r/θ and non-convective ﬂux vector q/θ allow computing the amount of heat per unit of absolute temperature entering the material volume per unit of time as (heat/unit of temperature) entering V r q = ρ dV − · n dS . (5.94) unit of time θ θ V ∂V

Observing now (5.94), the second term in this expression is identiﬁed as the magnitude deﬁned in (5.92). This circumstance allows interpreting the second law of thermodynamics establishing that the generation of entropy per unit of time in a continuous medium is always larger than or equal to the amount of heat per unit of temperature entering the system per unit of time.

Global form of the second law of thermodynamics Theory and Problems dS r q ≥ ρ dV − · n dS dt θ θ (5.95) Continuum MechanicsV ∂V for Engineers © X.amount Oliver of and the property C. Agelet de Saracibar “heat / unit of absolute temperature” entering the domain V per unit of time

Consider now the decomposition of the total entropy of the system S into two distinct components: • S(i): entropy generated (produced) internally by the continuous medium. Its generation rate is dS(i)/dt. • S(e): entropy generated by the interaction of the continuous medium with its exterior. Its variation rate is dS(e)/dt.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 236 CHAPTER 5. BALANCE PRINCIPLES

Then, the following is naturally satisﬁed.

dS dS(e) dS(i) = + (5.96) dt dt dt

Now, if one establishes that the variation rate of the entropy generated by the interaction with the exterior coincides with the magnitude heat per unit of absolute temperature in (5.93), dS(e) r q = ρ dV − · n dS (5.97) dt θ θ V ∂V and, taking into account (5.95)to(5.97), the variation per unit of time of the internally generated entropy results in ⎛ ⎞ dS(i) dS dS(e) dS r q = − = − ⎝ ρ dV − · n dS⎠ ≥ 0 . (5.98) dt dt dt dt θ θ V ∂V

Remark 5.14. According to (5.98), the internally generated entropy S(i) of the system (continuous medium) never decreases (dS(i)/dt ≥ 0). In a perfectly isolated system (strictly speaking, only the universe is a perfectly isolated system) there is no interaction with the exterior and the variation of entropy due to interaction with the exterior is null, (dS(e)/dt = 0). In this case, the second law of thermodynamics establishesTheory that and Problems dS(i) dS = ≥ 0 Continuum Mechanicsdt dt for Engineers or, in other words,©the X. total Oliver entropy and of C. a Agelet perfectly de isolated Saracibar system never decreases. This is the starting point of some alternative for- mulations of the second law of thermodynamics.

5.12.3 Reformulation of the Second Law of Thermodynamics In view of the considerations in Section 5.12.2, the second law of thermodynamics can be reformulated as follows: 1) There exists a state function named absolute temperature such that it is always strictly positive, θ (x,t) > 0 . (5.99)

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Second Law of Thermodynamics. Entropy 237

2) There exists a state function named entropy that is an extensive variable and, thus, can be defined in terms of a specific entropy (or entropy per unit of mass) s(x,t) as S(t)= ρsdV. (5.100) V 3) Entropy can be generated internally, S(i), or produced by interaction with the exterior, S(e). Both components of the entropy are extensive variables and their content in a material volume V can be defined in terms of their respective specific values s(i) and s(e), ( ) ( ) ( ) ( ) S i = ρs i dV and S e = ρs e dV (5.101) V V

(i) (e) ( ) ( ) dS dS dS S = S i + S e =⇒ = + (5.102) dt dt dt and introducing Reynolds’ Lemma (5.33)in(5.102) yields (i) (i) dS d ( ) ds = ρs i dV = ρ dV , dt dt dt V ≡V V t (5.103) (e) (e) dS d ( ) ds = ρs e dV = ρ dV . dt dt dt Vt ≡V V 4) The variation of external entropy (generated by the interaction with the exterior) is associated with the variation of the magnitude heat per unit of absolute temperature, and is deﬁned as Theory and Problems dS(e) r q = ρ dV − · n dS . (5.104) dt θ θ Continuum MechanicsV ∂ forV Engineers © X. Oliver and C. Agelet de Saracibar 5) The internally generated entropy never diminishes. Based on the variation of its content during the thermodynamic process, the following situations are deﬁned: ⎧ ⎨ = 0 reversible process dS(i) ≥ 0 → > 0 irreversible process (5.105) dt ⎩ < 0 non-feasible process

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 238 CHAPTER 5. BALANCE PRINCIPLES

5.12.4 Local Form of the Second Law of Thermodynamics. Clausius-Planck Equation Using (5.101)to(5.104), expression (5.105) is rewritten as

dS(i) dS dS(e) = − ≥ 0 dt dt dt (5.106) d ( ) d r q ρs i dV = ρsdV− ρ dV − · n dS ≥ 0 dt dt θ θ Vt ≡V Vt ≡V V ∂V Applying Reynolds’ Lemma (5.33) (on the ﬁrst and second integral of the left- hand term in (5.106)) and the Divergence Theorem (on the last term) yields ds(i) ds r q ρ dV = ρ dV − ρ dV − ∇ · dV ≥ 0 ∀ΔV ⊂ V dt dt θ θ V V V V (5.107) and localizing in (5.107), the local form of the second law of thermodynamics or Clausius-Duhem equation is obtained.

Local form of the second law of thermodynamics (Clausius-Duhem inequality) (5.108) ds(i) ds r q ρ = ρ − ρ − ∇ · ≥ 0 ∀x ∈ V, ∀t dt dt θ θ

Where, again, in (5.108) the sign = corresponds to reversibleTheoryprocesses, and Problems > corresponds to irreversible processes, and

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Second Law of Thermodynamics. Entropy 239

.( ) . r 1 1 s i = s − + ∇ · q − q · ∇θ ≥ 0 (5.110) θ ρθ ρθ2 .(i) .(i) slocal scond Then, a much stronger (more restrictive) formulation of the second law of thermodynamics can be posed. This formulation postulates that the internally .(i) .(i) generated entropy, s , can be generated locally, slocal, or by heat conduction, .(i) scond, and that both contributions to the generation of entropy must be non- negative.

Local internal generation of entropy (Clausius-Planck inequality) (5.111) .(i) = .− r + 1 ∇ · ≥ slocal s θ ρθ q 0

Internal generation of entropy by heat conduction

.(i) 1 (5.112) s = − q · ∇θ ≥ 0 cond ρθ2

Remark 5.15. Equation (5.112) can be interpreted in the following manner: since the density,Theoryρ, and and the absolute Problems temperature, θ, are positive magnitudes, said equation can be written as Continuum Mechanicsq · ∇θ ≤ 0 , for Engineers © X. Oliver and C. Agelet de Saracibar which establishes that the non-convective heat flux, q, and the temperature gradient, ∇θ, are vectors that have opposite directions (their dot product is negative). In other words, (5.112) is the mathematical expression of the experimentally verified fact that heat flows by conduction from the hottest to the coldest parts in the medium (see Figure 5.24), characterizing as non-feasible those processes in which the contrary occurs.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 240 CHAPTER 5. BALANCE PRINCIPLES

Figure 5.25: Heat ﬂux is opposed to the thermal gradient.

Remark 5.16. In the context of Fourier’s Law of heat conduction, q = −K ∇θ (see Remark 5.9), expression (5.112) can be written as ! q · ∇θ ≤ 0 =⇒−K |∇θ|2 ≤ 0 =⇒ K ≥ 0 q = −K∇θ

which reveals that negative values of the thermal conductivity K lack physical meaning.

5.12.5 Alternative FormsTheory of the Second and Law Problems of Thermodynamics Alternative expressions of the Clausius-Planck equation (5.111) in combination with theContinuum local form of the energy Mechanics balance equation for (5.89 Engineers) are often used in continuum mechanics. © X. Oliver and C. Agelet de Saracibar • Clausius-Planck equation in terms of the speciﬁc internal energy A common form of expressing the Clausius-Planck equation is doing so in terms of the speciﬁc internal energy u(x,t) in (5.84). This expression is obtained using the local spatial form of the energy balance equation (5.89),

du not . . ρ = ρu = σ : d + ρr − ∇ · q =⇒ ρr − ∇ · q = ρu − σ : d , (5.113) dt and, replacing it in the Clausius-Planck equation (5.111),

ρθ .(i) = ρθ .− (ρ − ∇ · )=ρθ .− ρ . + σ ≥ . slocal s r q s u : d 0 (5.114)

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Second Law of Thermodynamics. Entropy 241

Clausius-Planck equation in terms of the internal energy . . (5.115) −ρ (u − θs)+σ : d ≥ 0

• Clausius-Planck equation in terms of the Helmholtz free energy Another possibility is to express the Clausius-Planck equation in terms of the (speciﬁc) Helmholtz free energy ψ (x,t), which is deﬁned in terms of the internal energy, the entropy and the temperature as

def ψ = u − sθ . (5.116)

Differentiating (5.116) with respect to time results in ...... ψ = u − sθ − sθ =⇒ u − θs = ψ + sθ (5.117) and, replacing (5.117)in(5.115), yields the Clausius-Planck equation in terms of the Helmholtz free energy, . ρθ .(i) = −ρ ( . − θ .)+σ = −ρ ψ. + θ + σ ≥ . slocal u s : d s : d 0 (5.118)

Clausius-Planck equation in terms of the free energy . . (5.119) −ρ ψ + sθ + σ : d ≥ 0

. For the inﬁnitesimal strainTheory case, d = andε (see ChapterProblems2, Remark 2.22), and replacing in (5.119) results in

Clausius-Planck equation (inﬁnitesimal strain) Continuum Mechanics for Engineers © X.. Oliver. and. C. Agelet de Saracibar. (5.120) −ρ ψ + sθ + σ : ε ≥ 0

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 242 CHAPTER 5. BALANCE PRINCIPLES

5.13 Continuum Mechanics Equations. Constitutive Equations At this point it is convenient to summarize the set of (local) differential equations provided by the balance principles. 1) Conservation of mass. Mass continuity equation. ⎫ ⎪ dρ ⎪ + ρ∇ · v = 0 ⎬ dt → ⎪ 1 equation (5.121) dρ ∂vi ⎪ + ρ = 0 ⎭ dt ∂xi

2) Balance of linear momentum. Cauchy’s equation. ⎫ ⎪ dv ⎪ ∇ · σ + ρb = ρ ⎬ dt → 3 equations (5.122) ∂σ ⎪ ji dvi ⎪ + ρbi = ρ i ∈{1,2,3} ⎭ ∂x j dt

3) Balance of angular momentum. Symmetry of the stress tensor. ! σ = σ T → 3 equations (5.123) σ12 = σ21 ; σ13 = σ31 ; σ23 = σ32 Theory and Problems 4) Energy balance. First law of thermodynamics. ⎫ Continuum Mechanics⎪ for Engineers du © X. Oliver and C.⎪ Agelet de Saracibar ρ = σ : d +(ρr − ∇ · q) ⎬ dt → 1 equation (5.124) ∂ ⎪ du qi ⎪ ρ = σij dij+ ρr − ⎭ dt ∂xi

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Continuum Mechanics Equations. Constitutive Equations 243

5) Second law of thermodynamics. Clausius-Planck and heat ﬂux inequalities. ⎫ . . ⎬ −ρ (u − θs)+σ : d ≥ 0 → . . ⎭ 1 restriction −ρ (u − θs)+σijdij ≥ 0 ⎫ ⎪ (5.125) 1 ⎪ − q · ∇θ ≥ 0 ⎬ ρθ2 → 1 restriction ∂θ ⎪ − 1 ≥ ⎭⎪ 2 qi 0 ρθ ∂xi These add up to a total of 8 partial differential equations (PDEs) and two restrictions. Counting the number of unknowns that intervene in these equations results in25 ⎫ ⎪ ρ → 1 unknown ⎪ ⎪ → ⎪ v 3 unknowns ⎪ ⎪ σ → 9 unknowns ⎬⎪ → u 1 unknown ⎪ 19 unknowns ⎪ q → 3 unknowns ⎪ ⎪ θ → ⎪ 1 unknown ⎪ ⎭⎪ s → 1 unknown Therefore, it is obvious that additional equations are needed to solve the problem. These equations, which receive the generic name of constitutive equations and are speciﬁc to the material that constitutes the continuous medium, are 6) Fourier’s law of heat conduction. Theory and⎫ Problems ⎪ q = −K ∇θ ⎬ ∂θ → 3 equations (5.126) Continuum Mechanics⎭⎪ for Engineers qi = −K© X. Oliveri ∈{1, and2,3} C. Agelet de Saracibar ∂xi

25 The six components of the strain rate tensor d in (5.124) and (5.125) are not considered unknowns because they are assumed to be implicitly calculable in terms of the velocity v by means of the relation d(v)=∇sv (see Chapter 2, Section 2.13.2).

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 244 CHAPTER 5. BALANCE PRINCIPLES

7) Constitutive equations (per se)26. ⎫ Thermo- ⎬⎪ mechanical f (σ ,ε (v),θ, μ)=0 i ∈{1,...,6}→6 equations constitutive ⎭⎪ i equations ! Entropy constitutive s = s(ε (v),θ, μ)=0 → 1 equation equation (5.127) # $ where μ = μ1,...,μp are a set of new thermodynamic variables (p new unknowns) introduced by the thermo-mechanical constitutive equations. 8) Thermodynamic equations of state. ⎫ Caloric ⎪ u = g(ρ,ε (v),θ, μ) ⎬ eqn. of state → ( + ) ⎪ 1 p eqns. Kinetic ⎭⎪ F (ρ,θ, μ)=0 i ∈{1,2,...,p} eqns. of state i (5.128) There is now a set of (1 + p) equations and (1 + p) unknowns that, with the adequate boundary conditions, constitute a mathematically well-deﬁned problem.

Remark 5.17. The mass continuity equation, Cauchy’s equation, the symmetry of the stress tensor, the energy balance and the inequalities of the second law of thermodynamics (equations (5.121)to(5.125)) are valid and generalTheory for all the continuous and Problems medium, regardless of the material that constitutes the medium, and for any range of displacements and strains. Conversely, the constitutive equations (5.126)to (5.128Continuum) are speciﬁc to the Mechanics material or the type for of Engineers continuous medium being studied (solid,© X.ﬂuid, Oliver gas) and and differentiate C. Agelet them de Saracibar from one another.

26 The strains ε often intervene in the thermo-mechanical constitutive equations. However, these are not considered as additional unknowns because they are assumed to be implicitly calculable in terms of the equation of motion which, in turn, can be calculated by integration of the velocity ﬁeld, ε = ε (v) (see Chapters 1 and 2).

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Continuum Mechanics Equations. Constitutive Equations 245

5.13.1 Uncoupled Thermo-Mechanical Problem To solve the general problem in continuum mechanics, a system of partial differential equations must be solved, which involve the (1 + p) equations and the (1 + p) unknowns discussed in the previous section. However, under certain cir- cumstances or hypotheses, the general problem can be decomposed into two smaller problems (each of them involving a smaller number of equations and unknowns), named mechanical problem and thermal problem, and that can be solved independently (uncoupled) from one another. For example, consider the temperature distribution θ (x,t) is known a priori, or that it does not intervene in a relevant manner in the thermo-mechanical constitutive equations (5.127), and that, in addition, said constitutive equations do not involve new thermodynamic variables (μ = {/0}). In this case, the following set of equations are considered27 ⎫ ⎪ ρ ⎪ Mass continuity d + ρ∇ · = ⎪ v 0 (1 eqn) ⎪ equation: dt ⎪ ⎬⎪ ∇ · σ + ρ = ρ dv → 10 equations , Cauchy’s equation: b (3 eqn) ⎪ dt ⎪ ⎪ ⎪ Mechanical f (σ ,ε (v)) = 0 ⎪ i (6 eqn) ⎭ constitutive equations: i ∈{1,...,6} (5.129) which involve the following unknowns. ⎫ ρ (x,t) → 1 unknown ⎬⎪ v(x,t) → 3 unknowns 19 unknowns (5.130) ⎭⎪ σ (xTheory,t) → 6 unknowns and Problems

The problem deﬁned by equations (5.129) and (5.130) constitutes the so- called mechanicalContinuum problem, Mechanics which involves the for variables Engineers (5.130) (named mechanical variables) that,© moreover, X. Oliver are and the C. real Agelet interest de in Saracibar many engineering problems. The mechanical problem constitutes, in this case, a system of reduced differential equations, with respect to the general problem, and can be solved independently of the rest of equations of said problem.

27 For simplicity, it is assumed that the symmetry of the stress tensor (5.123) is already imposed. Then this equation is eliminated from the set of equations and the number of unknowns of σ is reduced from 9 to 6 components.

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Theory and Problems

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Problems and Exercises 247

PROBLEMS

Problem 5.1 – Justify whether the following statements are true or false. a) The mass flux across a closed material surface is null only when the motion is stationary. b) The mass flux across a closed control surface is null when this flux is stationary.

Solution a) The statement is false because a material surface is always constituted by the same particles and, therefore, cannot be crossed by any particle throughout its motion. For this reason, the mass flux across a material surface is always null, independently of the motion being stationary or not. b) The statement is true because the application of the mass continuity equation on a stationary flux implies ⎫ ⎪ ∂ρ ⎪ =⇒ + ∇ · (ρ )= ⎬⎪ Mass continuity equation ∂ v 0 t =⇒ ∇ · (ρv)=0 . ⎪ ∂ρ ⎪ Stationary flux =⇒ = 0 ⎭ Theory∂t and Problems Resulting, thus, what had to be proven, Continuum∇ · (ρ )= = Mechanics⇒ ∇ · (ρ ) for= Engineersρ · = . v 0© X. Oliver andv C.dV Ageletv den SaracibardS 0 V ∂V

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 248 CHAPTER 5. BALANCE PRINCIPLES

Problem 5.2 – A water jet with cross-section S, pressure p and velocity v, impacts perpendicularly on a disc as indicated in the ﬁgure below. Determine the force F in steady-state regime that must be exerted on the disc for it to remain in a ﬁxed position (consider the atmospheric pressure is negligible).

Solution Taking into account the Reynolds Transport Theorem (5.39) and that the problem is in steady-state regime, the forces acting on the ﬂuid are d ∂ ∑ = ρ = (ρ ) + ρ ( · ) = ρ ( · ) . Fext/ f v dV ∂ v dV v n v dS v n v dS dt V V t ∂V S

Theory and Problems

Note that the velocity vector of the fluid along the surfaces Slat−1 and Slat−3 is perpendicular to the outward unit normal vector of the volume that encloses the fluid, therefore, v · n = 0. The same happens in the walls of the disc. The vectors v and n in sections S2 and S4 are not perpendicular but, because there exists symmetry and v is perpendicular to F, they do not contribute components to the horizontal forces. Therefore, the only forces acting on the fluid are = ρ ( · ) = ρ (− · ) = −ρ 2 . ∑Fext/ f v n v dS ve e ve dS v Se ∂V S

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Problems and Exercises 249

On the other hand, the external force, the pressure of the water jet and the atmospheric pressure (which is negligible) also act on the ﬂuid, = − + + = − + . ∑Fext/ f Fe atmospheric pressure forces pSe Fe pSe Equating both expressions and isolating the value of the module of the force F ﬁnally results in F = ρv2S + pS .

Problem 5.3 – A volume ﬂow rate Q circulates, in steady-state regime, through a pipe from end A (with cross-section SA) to end B (with cross-section SB < SA). The pipe is secured at point O by a rigid element P − O. Determine:

a) The entry and exit velocities vA and vB in terms of the ﬂow rate. b) The values of the angle θ that maximize and minimize the reaction force F at O, and the corresponding values of said reaction force. c) The values of the angle θ that maximize and minimize the reaction moment M about O, and the corresponding values of said reaction moment. d) The power W of the pump needed to provide the ﬂow rate Q.

Theory and Problems

Hypotheses:

1) The water is a perfect ﬂuid (σij = −pδij) and incompressible. 2) The weight of the pipe and the water are negligible.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 250 CHAPTER 5. BALANCE PRINCIPLES

Solution a) The incompressible character of water implies that the density is constant for a same particle and, therefore, dρ/dt = 0. Introducing this into the mass continuity equation (5.24), results in ∇ · v = 0 ⇐⇒ ∇ · v dV = 0 ∀V . [1] V The adequate integration volume must now be defined. To this aim, a control volume such that its boundary is a closed surface must be found (S = ∂V) to be able to apply the Divergence Theorem, ∇ · v dV = n · v dS ∀V [2] V ∂V where n is the outward unit normal vector in the boundary of the volume V. Then, by means of [1] and [2], the conclusion is reached that the net outflow across the contour of the control volume is null, n · v dS = 0 ∀V . ∂V The volume the defined by the water contained inside the pipe between the cross- sections SA and SB is taken as control volume. Consider, in addition, the unit vectors eA and eB perpendicular to said cross-sections, respectively, and in the direction of the flow of water. Then, the following expression is deduced. Note that the extended integral on the boundary ∂V is applied only on cross-sections SA and SB since n · v = 0 on the walls of the pipe, that is, n and v are perpendicular to one another. Theory and Problems · = · + · = (− ) · + · = n vContinuumdS n v dS Mechanicsn v dS eA forvAeA EngineersdS eB vBeB dS 0 © X. Oliver and C. Agelet de Saracibar ∂V SA SB SA SB

=⇒−vASA + vBSB = 0 =⇒ vASA = vBSB = Q

It is veriﬁed, thus, that the ﬂow rate at the entrance and exit of the pipe are the same,

Q Q vA = ;vB = . [3] SA SB

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Problems and Exercises 251 b) The balance of linear momentum equation (5.49) must be applied to find the value of the force F, d R = ρb dV + t dS = ρv dV , [4] dt V ∂V V where R is the total resultant of the forces acting on the fluid. On the other hand, expanding the right-hand term in [4] by means of the Reynolds Transport Theorem (5.39), yields d ∂ ρv dV = ρv dV + ρv(n · v) dS . [5] dt ∂t V V ∂V The problem is being solved for a steady-state regime, i.e., the local derivative of any property is null. In addition, the flow is known to exist solely through sections SA and SB since n and v are perpendicular to one another on the walls of the pipe. Therefore, according to [4] and [5], R = ρv(n · v) dS+ ρv(n · v) dS = S S A B

= ρvAeA (−eA · vAeA) dS+ ρvBeB (eB · vBeB) dS

SA SB

= −ρ 2 + ρ 2 R vA SA eA vB SB eB. [6]

Introducing [3] in [6] allowsTheory expressing and the resultant Problems force R in terms of Q, = −ρ 2 − 1 + 1 . ContinuumR MechanicsQ eA foreB Engineers © X. Oliver andSA C.S AgeletB de Saracibar Now the different forces that compose R must be analyzed. According to the statement of the problem, body forces can be neglected (b = 0). Therefore, only surface forces must be taken into account, that is, the forces applied on the boundary of the control volume (SA, SB and Slat, where this last one corresponds to the lateral surface of the walls),

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R = ρb dV + t dS = t dS = t dS+ t dS+ t dS = V ∂V ∂V S S S A B lat = + (− ) + . pAeA dS pB eB dS Rp/ f

SA SB

Here, Rp/ f represents the forces exerted on the ﬂuid by the walls of the pipe, which initially are unknown but can be obtained using [6] as follows. = − − (− ) Rp/ f R pAeA dS pB eB dS

SA SB = −ρ 2 + ρ 2 − + Rp/ f vA SA eA vB SB eB pA SA eA pB SB eB = − ρ 2 + − ρ 2 + Rp/ f vA pA SA eA vB pB SB eB [7]

Introducing [3], Rp/ f can be expressed in terms of Q, 2 2 = − ρ Q + − ρ Q + . Rp/ f pA SA eA pB SB eB SA SB

Now the relation between Rp/ f and the unknown being sought, F, must be found. To this aim, the action and reaction law is considered, and the pipe and the rigid element P − O are regarded as a single body. Under these conditions, the force exerted by the ﬂuid on the pipe is = − . TheoryR f /p andRp/ Problemsf Since it is the only action on the body, and taking into account that the weight of the pipeContinuum is negligible, this Mechanics force must be compensated for Engineers by an exterior action F for the body to be in equilibrium.© X. Oliver and C. Agelet de Saracibar + = =⇒ = − = R f /p F 0 F R f /p Rp/ f Introducing [7], the value of F is ﬁnally obtained as = − ρ 2 + + ρ 2 + . F vA pA SA eA vB pB SB eB

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Problems and Exercises 253

Using [3], the force F is expressed in terms of Q, Q2 Q2 F = − ρ + pA SA eA + ρ + pB SB eB . [8] SA SB

There are two possible ways of obtaining the maximum and minimum of |F| in terms of θ: 1) Determine the expression of |F| and search for its extremes by imposing that its derivative is zero (this option not recommended). 2) Direct method, in which the two vectors acting in the value of F are analyzed (this option developed below).

According to [7], the value of F depends on the positive scalar values FA and FB, which multiply the vectors (−eA) and eB, respectively.

The vector (−eA) is ﬁxed and does not depend on θ but eB does vary with θ. The scalars FA and FB are constantTheory values. and Therefore, Problems the maximum and minimum values of F will be obtained when FA and FB either completely add or subtract one another, respectively. That is, when the vectors (−eA) and eB are parallel to each other.Continuum Taking into account Mechanics [3] and [8], the for maximum Engineers and minimum values are found to be: © X. Oliver and C. Agelet de Saracibar − Minimum value of F π θ = 2 | | = ρ 2 1 − 1 + − F min Q pB SB pA SA SB SA

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 254 CHAPTER 5. BALANCE PRINCIPLES

− Maximum value of F π θ = 3 2 | | = ρ 2 1 + 1 + + F min Q pB SB pA SA SB SA c) The balance of angular momentum equation (5.57) must be applied to ﬁnd the moment M about point O, d M = r × ρb dV + r × t dS = r × ρv dV , [9] liq dt V ∂V V where Mliq is the resultant moment of the moments acting on the ﬂuid. On the other hand, expanding the right-hand term in [9] by means of the Reynolds Transport Theorem (5.39), yields d ∂ r × ρv dV = r × ρv dV + (r × ρv)(n · v) dS . [10] dt ∂t V V ∂V As in b), because the problem is in steady-state regime, the local derivative is null. Again, n and v are perpendicular to one another on the walls of the pipe and, thus, considering [9] and [10], results in the expression

Mliq = (r × ρv)(n · v) dS+ (r × ρv)(n · v) dS , [11]

SA SB where the following mustTheory be taken into and account: Problems 1. The solution to each integral can be determined considering the resultant ofContinuum the velocities in the middleMechanics point of each for cross-section Engineers since the velocity distributions are uniform© X. Oliver and parallel and in C. both Agelet cases. de Saracibar 2. For cross-section SA, the resultant of the velocity vector applied on the center of the cross-section acts on point O and, therefore, does not generate any moment because the cross product of the position vector at the center of SA and the velocity vector are null. 3. For cross-section SB, vectors r and v belong to the plane of the paper and, thus, their cross product has the direction of the vector (−ez). In addition, they are perpendicular to each other, so the module of their cross product is the product of their modules.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Problems and Exercises 255

Applying these considerations to [11] yields

Mliq = Rρ vB (−ez)(eB · vB eB) dS

SB 2 = −ρ 2 = −ρ Q Mliq vB RSB ez Rez [12] SB

The following step consists in studying the contributions of the body forces, which in this case are null (b = 0), and of the surface forces.

Mliq = r × ρb dV + r × t dS = r × t dS = V ∂V ∂V = r × t dS+ r × t dS− r × t dS =

SA SB Slat = + + = + , 0 RpB ez dS Mp/ f RpB SB ez Mp/ f

SB where Mp/ f is the moment exerted by the pipe on the ﬂuid. To determine its expression, [12] is used, = − = −ρ 2 − , Mp/ f Mliq RpB SB ez vB RSB ez RpB SB ez 2 = − ρ 2 + = − ρ Q + Mp/ f RSB vB pB ez R pB SB ez. [13] SB

Introducing the action andTheory reaction law andwill allow Problems obtaining the moment exerted by the ﬂuid on the pipe, = − . Mp/ f M f /p Continuum Mechanics− for Engineers Considering the pipe and© the X. rigid Oliver element andP C.O Ageletas a single de Saracibarbody in equilibrium and neglecting the weight of the pipe, + = =⇒ = − = . M f /p M 0 M M f /p Mp/ f Finally, the value of the moment M is obtained, using [13]. 2 = − ρ 2 + = − ρ Q + M RSB vB pB ez R pBSB ez SB

Note that this result does not depend on the angle θ and, therefore, its module will have a constant value.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 256 CHAPTER 5. BALANCE PRINCIPLES d) To determine the value of the power W needed to provide a volume ﬂow rate Q the balance of mechanical energy equation (5.73) is used. d 1 W = ρv2 dV + σ : d dV [14] dt 2 V V The stress power in an incompressible perfect ﬂuid is null, σ : d dV = 0 . V This is proven as follows.

1 σ : d = −p 1 : d = −pTr(d)=−pTr l + lT = 2 ⎡ ⎤ ∂v ∂v ∂v ⎢ x x x ⎥ ⎢ ∂ ∂ ∂ ⎥ ⎢ x y z ⎥ ⎢ ∂ ∂ ∂ ⎥ = − ( )=− ⎢ vy vy vy ⎥ = pTr l pTr⎢ ⎥ ⎢ ∂x ∂y ∂z ⎥ ⎣ ⎦ ∂vz ∂vz ∂vz ∂x ∂y ∂z ∂v ∂v ∂v = −p x + y + z = −p ∇ · v = 0 , ∂x ∂y ∂z where [1] has been applied in relation to the incompressibility condition, to con- clude that the divergenceTheory of the velocity and is null. Problems Applying the Reynolds Transport Theorem (5.39) on the term of the material derivative of the kinetic energy in [14] results in Continuumd 1 Mechanics∂ 1 for Engineers1 W = ρ©v2 X.dV Oliver= andρv2 C.dV Agelet+ ρ dev2 ( Saracibarn · v) dS . dt 2 ∂t 2 2 V V ∂V And, again, considering the problem is in steady-state regime and that n and v are perpendicular to one another on the walls of the pipe, the expression of the incoming power W is determined.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Problems and Exercises 257

1 1 W = ρv2 (n · v) dS+ ρv2 (n · v) dS = 2 2 SA SB 1 1 1 1 = ρv2 (−v ) dS+ ρv2 (v ) dS = ρv3 S + ρv3 S 2 A A 2 B B 2 A A 2 B B SA SB Then, by means of [3], the ﬁnal result is obtained. = 1ρ 3 1 − 1 W Q 2 2 2 SB SA

Theory and Problems

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 258 CHAPTER 5. BALANCE PRINCIPLES

EXERCISES

5.1 – Justify why the following statements are true. a) In an incompressible flow, the volume flow rate across a control surface is null. b) In a steady-state flow, the mass flux across a closed control surface is null. c) In an incompressible fluid in steady-state regime, the density is uniform only when the density at the initial time is uniform.

5.2 – The figure below shows the longitudinal cross-section of a square pipe. Water flows through this pipe, entering through section AE and exiting through section CD. The exit section includes a floodgate BC that can rotate around hinge B and is maintained in vertical position by the action of force F.

Theory and Problems

a) The exit velocity v2 in terms of the entrance velocity v1 (justify the expression used). b) The resultant force and moment at point B of the actions exerted on the fluid by the interior of the pipe. c) The resultant force and moment at point B of the actions exerted by the fluid on floodgate BC. d) The value of the force F and the reactions the pipe exerts on floodgate BC.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Problems and Exercises 259

e) The power of the pump needed to maintain the ﬂow. Additional hypotheses: 1) Steady-state regime 2) Incompressible ﬂuid 3) The pressures acting on the lateral walls of the pipe are assumed constant and equal to the entrance pressure p. 4) The exit pressure is equal to the atmospheric pressure, which is negligible.

5) Perfect fluid: σij = −pδij 6) The weights of the fluid and the floodgate are negligible.

5.3 – The figure below shows the longitudinal cross-section of a pump used to inject an incompressible fluid, fitted with a retention valve OA whose weight, per unit of width (normal to the plane of the figure), is W. Consider a steady- state motion, driven by the velocity of the piston V and the internal uniform pressure P1. The external uniform pressure is P2.

Theory and Problems

Determine:

a) The uniform velocities v1 and v2 in terms of V (justify the expression used).

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 260 CHAPTER 5. BALANCE PRINCIPLES

b) The resultant force, per unit of width, exerted by the fluid on the valve OA. c) The resultant moment about O, per unit of width, exerted by the fluid on the valve OA. d) The value of W needed for the valve OA to maintain its position (as shown in the figure) during the injection process. Additional hypotheses: 1) The body forces of the fluid are negligible.

2) Perfect ﬂuid: σij = −pδij Perform the analysis by linear meter.

5.4 – A perfect and incompressible fluid flows through the pipe junction shown in the figure below. The junction is held in place by a rigid element O − D.

Theory and Problems

Determine:

a) The entrance velocities (vA and vB) and the exit velocity (vC) in terms of the volume flow rate Q (justify the expression used). b) The resultant force and moment at O of the actions exerted on the fluid by the interior of the pipes in the junction. c) The reaction force and moment at D of the rigid element. d) The power W of the pump needed to provide the volume flow rates indicated in the figure.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Problems and Exercises 261

Additional hypotheses: 1) The weights of the ﬂuid and the pipes are negligible.

5.5 – The front and top cross-sections of an irrigation sprinkler are shown in the figure below. A volume flow rate Q of water enters through section C at a pressure P and exits through sections A and B at an atmospheric pressure Patm. The flow is assumed to be in steady-state regime.

Determine: a) The entrance and exit velocities (justify the expression used). b) The resultant force and moment at point O of the actions exerted on the fluid by the interior walls of the sprinkler. c) The reaction that must be exerted on point O to avoid the sprinkler from moving in the verticalTheory direction. and Problems d) The angular acceleration of the sprinkler’s rotation α. To this aim, as- Continuumsume that I0 and I1 are, Mechanics respectively, the for central Engineers moments of inertia about point O of the empty© X. sprinkler Oliver and the C. sprinkler Agelet defull Saracibar of water. e) The power needed to provide a volume flow rate 2Q, considering that W ∗ is the power of the pump needed to provide a volume flow rate Q. Additional hypotheses: 1) Incompressible fluid

2) Perfect ﬂuid: σij = −pδij 3) The weights of the sprinkler and the water inside it are negligible.

4) SA = SB = S and SC = S∗ 5) m = Iα

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961