CH.5. BALANCE PRINCIPLES. Multimedia Course on Continuum Mechanics

Transcription

1 CH.5. BALANCE PRINCIPLES Multimedia Course on Continuum Mechanics

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

3 Overview (cont d) Angular Momentum Balance Global Spatial Local Form Mechanical Energy Balance External Mechanical Power Mechanical Energy Balance External Thermal Power Energy Balance Thermodynamic Concepts First Law of Thermodynamics Internal Energy Balance in Local and Global Forms Second Law of Thermodynamics Reversible and Irreversible Processes Clausius-Planck Inequality Lecture 9 Lecture 10 Lecture 11 Lecture 12 Lecture 13 Lecture 14 Lecture 15 3

4 Overview (cont d) Governing Equations Governing Equations Constitutive Equations The Uncoupled Thermo-mechanical Problem Lecture 16 4

5 5.1. Balance Principles Ch.5. Balance Principles 5

6 Balance Principles The following principles govern the way stress and deformation vary in the neighborhood of a point with time. The conservation/balance principles: Conservation of mass Linear momentum balance principle 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 REMARK These principles are always valid, regardless of the type of material and the range of displacements or deformations. 6

7 5.2. Convective Flux Ch.5. Balance Principles 7

8 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 surface S : through a control Φ = S amount ofa crossing unitof time S 8

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

10 Convective Flux or Flux by Mass Transport Consider: An arbitrary property A of a continuum medium (of any tensor order) The specific content of A (the amount of A per unit of mass) Ψ x,t. ( ) inflow vn 0 outflow vn 0 Then, The convective flux of A normal n is: S ( t) Φ = ρ Ψvn s ds through a spatial surface, S, with unit Where: v is velocity ρ is density If the surface is a closed surface, ( t) Φ = ρ Ψvn ds S = = outflow - inflow, the net convective flux is: 10

11 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 nonconvective flux vector (or tensor) : s ( ) q x,t non - convective flux = q n ds ; convective flux = ρψ v n ds non-convective flux vector s convective flux vector 11

12 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

13 Example - Solution S ( t) Φ = ρ Ψvn s ds a) If the arbitrary property is the volume of the particles: A The magnitude property content per unit of mass is volume per unit of mass, i.e., the inverse of density: Ψ= = M 1 ρ The convective flux of the volume of the particles through the surface S is: 1 Φ S = ρ vn ds = s ρ vn s ds OLUME FLUX 13

14 Example - Solution S ( t) Φ = ρ Ψvn s ds b) If the arbitrary property is the mass of the particles: A M 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 = ρ 1 vn ds = ρvn S s s ds MASS FLUX 14

15 Example - Solution S ( t) Φ = ρ Ψvn s ds c) If the arbitrary property is the linear momentum of the particles: A M v The magnitude property per unit of mass is mass times velocity per unit of mass, i.e., velocity: Ψ M v = = M v The convective flux of the linear momentum of the particles surface S is: Φ S s ( ) = ρ vvn ds MOMENTUM FLUX M v through the 15

16 Example - Solution S ( t) Φ = ρ Ψvn s ds d) If the arbitrary property is the kinetic energy of the particles: A M v The magnitude property per unit of mass is kinetic energy per unit of mass, i.e.: 1 2 M v 2 1 Ψ= = v M 2 2 The convective flux of the kinetic energy of the particles surface S is: 1 2 Φ S = ρ v( vn ) ds KINETIC ENERGY FLUX s M v 2 through the 16

17 5.3. Local and Material Derivative of a olume Integral Ch.5. Balance Principles 17

18 Derivative of a olume Integral Consider: An arbitrary property of a continuum medium (of any tensor order) The description of the amount of the property per unit of volume (density of the property A ), µ x,t The total amount of the property in an arbitrary volume,, is: ( ) = µ ( x, ) The time derivative of this volume integral is: A Q t t d ( ) Q t = ( + ) ( ) Q t t Q t lim t 0 t ( ) REMARK and Ψ are related through. Q( t) ( + t) Q t 18

19 Local Derivative of a olume Integral local derivative 19 Consider: The volume integral The local derivative of Q( t) is: µ (, + ) µ (, ) not x x = µ ( x, t) d = lim t t 0 It can be computed as: Q( t+ t) Q( t) µ ( x t) d ( x + ) ( x ) ( x, + ) ( x, ) µ t t d µ t d, = lim = lim = t t 0 t t 0 t ( ) = µ ( x, ) Q t t d t t d t d t [ µ, t t µ, t ] d µ µ µ = lim = lim d = t 0 t t 0 t t µ, t x t Q( t) ( t) Q t+ Control olume, REMARK The volume is fixed in space (control volume). ( x, t+ t) ( x, t) ( x, t) d

20 Material Derivative of a olume Integral Consider: The volume integral ( ) = µ ( x, ) Q t t d The material derivative of not material d derivative = ( x, t) d µ = t µ ( x, t + t) d µ ( x, t) d ( t+ t) () t = lim t 0 t It can be proven that: Q( t) is: Q( t ) Q( t+ t) REMARK The volume is mobile in space and can move, rotate and deform (material volume). d µ dµ (, t) d d ( ) d ( ) d d µ x = µ + µ = + µ = + µ t v v t v t material local convective derivative of derivative of derivative of the integral the integral the integral 20

21 5.4. Conservation of Mass Ch.5. Balance Principles 21

22 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 M t the system satisfies: M ( ) of ( t) = M ( t+ t) > 0 22 Where: ( ) ρ ( x, ) t t ( ) ρ ( x, ) M t = t d M t+ t = t+ t d t+ t t t+ t t+ t

23 Conservation of Mass in Spatial Form 23 Conservation of mass requires that the material time derivative of the mass M t be zero for any region of a material volume, ( ) ( + ) M ( ) M t t t d M ( t) = lim = ρd = 0, t t 0 t t t The global or integral spatial form of mass conservation principle: d dµ (, t) d ( ) d µ x = + µ v t d dρ ρ(,) t d = + ρ d = 0, t x t t v By a localization process we obtain the local or differential spatial form of mass conservation principle: for d (,) x t (localization process) CONTINUITY EQUATION dρ(,) x t ρ(,) x t + ( ρ v)( x, t) = + ( ρv)( x, t) = 0 x, t t

24 Conservation of Mass in Material Form 0 0 Consider the relations: d F 1 = F v ( v) = F d = F d0 d F The global or integral material form of mass conservation principle can be rewritten as: dρ dρ 1 d F ρ( X,) t F ( X,) t ( + ρ v) d = ( ) d ( (, t) ) d + ρ = F X + ρ F t t 0 ρ( X,) t F d0 [ ρ ](, t) F X t t ρ (, t) d0 = 0 0 0, t t F X 24 t The local material form of mass conservation principle reads : ρ t ρ ρ ρ0 ( ) = ( ) = ( ) ( ) F X, 0 t= 0 X F t= 0 t X F X t = 1 ρ, t = X 0 t F t 0

25 5.5. Reynolds Transport Theorem Ch.5. Balance Principles 25

26 Reynolds Lemma dρ + ρ v = 0 26 Consider: An arbitrary property of a continuum medium (of any tensor order) 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 A in the continuum body at time for an arbitrary material volume is: Q( t) = ρψ d Using the material time derivative leads to, Thus, d ρψ d = t A d d dψ dρ Q ( t) = ρψ d ( ρψ ) ( ρψ ) d ρ ψ ( ρ ) d = + v = + + v t dψ dρ = ρ + ψ dψ ρ d REYNOLDS LEMMA t = =0 (continuity equation) t

27 Reynolds Transport Theorem d t µ ( x, t) d = µ d + ( µ v) d t A The amount of the property in the continuum body at time t for an arbitrary fixed control volume is: Q( t) = ρψ d Using the material time derivative leads to, d ( ρψ) ρψ d = d + ( ρ ; ψ ) d t v And, introducing the Reynolds Lemma and Divergence Theorem: dψ ( ρψ ) t ρ d = d + ρψvn ds t REMARK The Divergence Theorem: vd = nv ds = vn ds 27 = dψ ρ d = n ( ρψ v ) ê 3 ê 1 d ê 2 dψ ρ d ρψ d

28 ( ρψ ) dψ ρ d = d + ρψvn ds t Reynolds Transport Theorem The eq. can be rewritten as: t dψ ρψd = ρ d ρψ ds vn Change (per unit of time) of the total amount of A. within the control volume at time t. REYNOLDS TRANSPORT THEOREM Change due to the net outward convective flux of A through the boundary. dψ ρ Rate of change of the amount of property integrated over all particles that are filling the control volume at time t. A d ê 3 ê 2 ρψ d ê 1 28

29 Reynolds Transport Theorem t dψ ρψd = ρ d ρψ ds vn REYNOLDS TRANSPORT THEOREM (integral form) t dψ ρψd = ρ d ρψ ds vn = ( ρψ) d t dψ ( ρψ) d = [ ρ ( ρψ )] d t t v = ( ρψv) d dψ ( ρψ) = ρ ( ρψv) x t t ê 3 ê 1 ê 2 dψ ρ d ρψ d REYNOLDS TRANSPORT THEOREM (local form) 29

30 5.6. General Balance Equation Ch.5. Balance Principles 30

31 General Balance Equation Consider: A An arbitrary property of a continuum medium (of any 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 A in the control volume is due to: a) Generation of the property per unit mas and time time due to a source: b) The convective (net incoming) flux across the surface of the volume. c) The non-convective (net incoming) flux across the surface of the volume: So, the global form of the general balance equation is: 31 t ρψd = ρk d ρψvn ds ja nds A a b c ka (,) x t source term ja (,) x t non-convective flux vector

32 t ρψd = ρk d ρψvn ds j nds A General Balance Equation A t The global form is rewritten using the Divergence Theorem and the definition of local derivative: ρψd + ρψvn ds = = ( ρψ) + ( ρψv) d = ( ρk ) d t A ja dψ = ρ (Reynolds Theorem) dψ ρ d = ( ρk ) d t A ja The local spatial form of the general balance equation is: REMARK dψ For only convective transport ( ja = 0) then ρ = ρk A and the variation of the contents of in a given particle is only due to the internal generation k. ρ A dψ ρ = ρk j A A 32

33 Example dψ ρ = ρk j A dψ ( ρψ ) = ρ ( ρ ψ v) x t A A If the property is associated to mass A M, then: The amount of the property per unit of mass is ψ =1. The mass generation source term is k = 0. The mass conservation principle states that mass cannot be generated. The non-convective flux vector is = 0. Mass cannot be transported in a non-convective form. dψ ρ = ρ k 0 A j A = = 0 = 0 Then, the local spatial form of the general balance equation is: dψ ρ = ( ρψ ) + ( ρψ v) = 0 ρ + ( ρv) = 0 t t = 1 = 1 j M ρ d ρ + ( ρv) = + ρ v = 0 x t t M Two equivalent forms of the continuity equation. 33

34 5.7. Linear Momentum Balance Ch.5. Balance Principles 34

35 Linear Momentum in Classical Mechanics Applying Newton s 2 nd Law to the discrete system formed by n particles, the resulting force acting on the system is: n n n dvi R( t) = i = mi i = mi = f a Resulting force on the system i= 1 i= 1 i= 1 d dm dp = = n n i mivi vi i= 1 i= 1 ( t) mass conservation principle: dm i = 0 = P t linear momentum For a system in equilibrium, R = 0, t : dp ( t) = 0 ( ) P ( t) = cnt CONSERATION OF THE LINEAR MOMENTUM 35

36 Linear Momentum in Continuum Mechanics P ( t) n = miv i i= 1 The linear momentum of a material volume medium with mass M is: t of a continuum P ( ) = v( x, ) M = ρ ( x, ) v( x, ) t t d t t d M dm = ρ d 36

37 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. ( ) dp t d = ρ d = v R Where: ( ) R t = ρ b d + t ds t body forces ( t) surface forces If the body is in equilibrium, the linear momentum is conserved: dp ( t) R ( t ) = 0 = 0 P ( t) = cnt 37

38 Global Form of the Linear Momentum Balance Principle The global form of the linear momentum balance principle: d dp R( t) = ρ b d + t ds = d ρ v = t t, t P ( t) ( t) Introducing t = n σ and using the Divergence Theorem, 38 t ds = n σ ds = σ d So, the global form is rewritten: ρ b d + t ds = d = ( ρ b + σ) d = ρ d, t v t t

39 Local Form of the Linear Momentum Balance Principle Applying Reynolds Lemma to the global form of the principle: d dv σ+ ρ b d = ρ v d = ρ d t ( ), t t Localizing, the linear momentum balance principle reads: d (,) x t dvx (,) t σ (,) xt + ρ bx (,) t = ρ = ρax (,) t x, t LOCAL FORM OF THE LINEAR MOMENTUM BALANCE (CAUCHY S EQUATION OF MOTION) 39

40 5.8. Angular Momentum Balance Ch.5. Balance Principles 40

41 Angular Momentum in Classical Mechanics Applying Newton s 2 nd Law to the discrete system formed by n particles, the resulting torque acting on the system is: dv n n i O( t) = i i = i mi = i= 1 i= 1 M r f r d dr d dl = = = n n n i ri mivi mivi ri mivi i= 1 i= 1 i= 1 = L ( t) angular momentum ( ) = v i dl t MO ( t) = For a system in equilibrium, MO = 0, t : dl ( t) = 0 t L ( t) = cnt =0 CONSERATION OF THE ANGULAR MOMENTUM 41

42 Angular Momentum in Continuum Mechanics The angular momentum of a material volume medium with mass M is: L ( ) = r v( x, ) M = x ρ ( x, ) v( x, ) t of a continuum t t d t t d M x dm = ρ d 42

43 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. ( ) dl t d = = Where: O ( ) x t r ρ v d MO ( t) torque due to body forces M t = r ρ b d + r t ds torque due to surface forces 43

44 Global Form of the Angular Momentum Balance Principle The global form of the angular momentum balance principle: d ( r ρ b) d + ( r t) ds = ( ρ ) d r v Introducing It can be proven that, T ( r σ ) = r σ + m m= meˆ i i ; mi = ijkσ jk t t = n σ and using the Divergence Theorem, T σ σ ( σ ) r t r n r n r n T ds = ds = ds = ds = T ( r σ ) = d REMARK ijk is the Levi-Civita permutation symbol. 44

45 Global Form of the Angular Momentum Balance Principle Applying Reynolds Lemma to the right-hand term of the global form equation: d d d r v d ( r v) d ( r v) d ρ = ρ = ρ = t dr dv dv = ρ v+ r d = r ρ = v Then, the global form of the balance principle is rewritten: =0 t Reynold's Lemma dv r ( ρ b+ σ ) + eˆ ijkσ jk i d = r ρ d d 45

46 Local Form of the Angular Momentum Balance Principle Rearranging the equation: =0 (Cauchy s Eq.) dv r σ + ρ b ρ + m d = 0 (,) t d = mx 0, t 46 Localizing { } mx (, t) = 0 m= σ = 0 ; i, jk, 1, 2, 3 ; x, t i = 1 σ + σ = 0 σ = σ = 1 = 1 i = 2 σ + σ = 0 σ = σ = 1 = 1 i = 3 σ + σ = 0 σ = σ = 1 = 1 i ijk jk t σ σ σ σ σ σ σ σ σ σ T σ(,) x t = σ (,) x t x, t SYMMETRY OF THE CAUCHY S STRESS TENSOR t

47 5.9. Mechanical Energy Balance Ch.5. Balance Principles 47

48 Power Power, time. Wt (), is the work performed in the system per unit of In some cases, the power is an exact time-differential of a function (then termed) energy E : d () t Wt () = E 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

49 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: e ( ) P t = ρ b v d + t v ds dr ρb d = ρb v d = v dr t ds = t v ds = v

50 Mechanical Energy Balance Using t = n σ and the Divergence Theorem, the traction contribution reads, t vds = n ( σ v) ds = ( σ v) d = ( ) + : d σ v σ v n σ Taking into account the identity: l= d+ w σ:l = σ:d + σ:w Divergence Theorem =0 skew symmetric = l spatial velocity gradient tensor So, ( σ) σ : t vds = vd + dd 50

51 Mechanical Energy Balance σ+ ρ b = ρ dv Substituting and collecting terms, the external mechanical power in spatial form is, tv ds P t = ρ b vd + σ vd + σ : dd = e ( ) ( ) dv = ( σ + ρb) vd + : d ρ d : d σ d = + v σ d = ρ dv d 1 d 1 = ρ = ρ Reynold's Lemma ( v v) ( v ) 2 2v= v d 1 2 d 1 2 Pe ( t) = ρ ( v) d + d ρ( v) d d 2 σ:d = + 2 σ:d 2 51

52 Mechanical Energy Balance. Theorem of the expended power. Stress power d 1 2 Pe ( t) = ρb vd + ds ρ v d d t v = + 2 σ :d external mechanical power entering the medium d Pe ( t) = K ( t) + P σ t kinetic energy P σ stress 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 ( d= 0). K Theorem of the expended mechanical power 52

53 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. qx (,) t nds = heat conduction flux vector incoming heat unit of time Internal heat sources ρ r( x,) t d = specific internal heat production heat generated by internal sources unit of time 53

54 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: Qe ( t) = ρ r d qn ds = ( ρ r q) d where: q( x,t) r( x, t) = nq ds = ( q) d is the non-convective heat flux vector per unit of spatial surface is the internal heat source rate per unit of mass. 54

55 Total Power The total power entering the continuous medium is: d 1 v 2 Pe + Qe = ρ d + σ :dd + ρ r d q nds 2 t 55

56 5.10. Energy Balance Ch.5. Balance Principles 56

57 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 MATTER Thermodynamic space HEAT A thermodynamic system is characterized and defined by a set of thermodynamic variables µ 1, µ 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

58 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

59 State Function Is a function φ µ µ uniquely valued in terms of the thermodynamic state or, equivalently, in terms of the thermodynamic variables { µ 1, µ 2,, µ n} Consider a function φ ( µ 1, µ 2), that is not a state function, implicitly defined in the thermodynamic space by the differential form: ( ) ( ) δφ = f1 µ 1, µ 2 dµ 1+ f2 µ 1, µ 2 dµ 2 The thermodynamic processes Γ and Γ yield: 1 2 φb = φa + δφ = f2( µ 1, µ 2) δµ 2 Γ 1 Γ1 ' δφ δφ φ φ Γ 1 Γ B B 2 φb' = φa + δφ = f2( µ 1, µ 2) δµ 2 Γ 2 Γ2 For to be a state function, the differential form must be an exact differential:, i.e., must be integrable: 59 The necessary and sufficient condition for this is the equality of cross-derivatives: f ( ) ( 1,..., i µ 1,..., µ f n j µ µ n) = i, j { 1,... n} δφ = dφ µ µ j ( ) 1,..., n i

60 First Law of Thermodynamics POSTULATES: ( ) 1. There exists a state function E t named total energy of the system, such that its material time derivative is equal to the total power entering the system: d E d 1 t = Pe t + Qe t = d d r d ds + 2 σ :d + ρ q n t P () t Qe () t 2 ( ) : ( ) ( ) ρv ( ) e 2. There exists a function U t named the internal energy of the system, such that: 60 It is an extensive property, so it can be defined in terms of a specific internal energy (or internal energy per unit of mass) u( x, t) : U ( t) : = ρ u d REMARK de and dkare exact differentials, The variation of the total energy of the system is: d E ( t) = d K ( t) + d therefore, so is du = de dk. U ( t) Then, the internal energy is a state function.

61 Global Form of the Internal Energy Balance Introducing the expression for the total power into the first postulate: = K d d 1 2 E ( t) = ρ v d + d + ρ r d ds 2 σ :d q n t Comparing this to the expression in the second postulate: d E d K d U ( t) = ( t) + ( t) 61 d The internal energy of the system must be: d U ( t) = ρu d = d + ρ r d ds σ :d q n t ( ) Pσ t stress power Q ( t), e external thermal power GLOBAL FORM OF THE INTERNAL ENERGY BALANCE

62 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 U ( t) = u d d d r d ds ρ = ρ = σ :d + ρ q n t t t t q d Then, the local spatial form of the energy balance principle is obtained through localization d (,) x t as: du ρ U () t ( ρ ), = σ :d + r q x t du ρ d = d + ρ r d d t σ :d q LOCAL FORM OF THE ENERGY BALANCE (Energy equation) 62

63 Second Law of Thermodynamics The total energy is balanced in all thermodynamics processes following: de dk du P ( t) + Q ( t) = = + e e In an isolated system (no work can enter or exit the system) de U K e ( ) + ( ) = = 0 P t Q t e d + = 0 However, it is not established if the energy exchange can happen in both senses or not: du dk > 0 < 0 There is no restriction indicating if an imagined arbitrary process is physically possible or not. d du dk < 0 > 0 63

64 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. Spontaneously, heat always flows to regions of lower temperature, never to regions of higher temperature. du dk > 0 < /04/2017 MMC - ETSECCPB - UPC

65 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. REERSIBLE PROCESS IRREERSIBLE PROCESS The second law of thermodynamics allows discriminating: IMPOSSIBLE thermodynamic processes REERSIBLE POSSIBLE IRREERSIBLE 65

66 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: The following inequality holds true: S( t) = ρ s( x, t) d d d r S() t = ρsd ρ d ds θ q n θ = reversible process > irreversible process Global form of the 2 nd Law of Thermodynamics 66

67 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 e d d r S() t = ρsd ρ d ds θ q n θ = reversible process > irreversible process ( ) Q t = ρ r d ds Global form of the 2 nd Law of Thermodynamics qn rate of the total amount of the entity heat, per unit of time, (external thermal power) entering into the system = Γ e ( t) r Γ e ( t) = ρ d ds θ q n θ rate of the total amount of the entity heat per unit of absolute temperature, per unit of time (external heat/unit of temperature power) entering into the system 67

68 Second Law of Thermodynamics Consider the decomposition of entropy into two (extensive) counterparts: Entropy generated inside the continuous medium: Entropy generated by interaction with the outside medium: S S ( ) ( i) ( i = s ) ρ ( x, ) t d ( e) ( e = s ) ρ ( x, ) ( i ) ( ) ( i) ( e) ( e ) ( ) S t = S t + S t ds ds ds = + t d 68

69 Second Law of Thermodynamics If one establishes, ( e) ds r =Γ e = ρ d ds θ q n θ Then the following must hold true: And thus, ( i) ( e) ( i) ( e) ds ds ds r + = ρ d ds θ q n θ ds ds ds ds r q = = ρ d ds 0 t θ n θ REPHRASED SECOND LAW OF THERMODYNAMICS : The internally generated entropy of the system, S = ds ( e) ( i ) () t, never decreases along time 69

70 Local Spatial Form of the Second Law of Thermodynamics The previous eq. can be rewritten as: d ( i) d r q s d s d d ds 0 t ρ = ρ ρ θ n θ t t t t Applying the Reynolds Lemma and the Divergence Theorem: ( i) ρ ds ds r d d d d 0 t ρ ρ q = θ θ ( i) Then, the local spatial form of the second law of thermodynamics is: ρ ds ds r 0, t ρ ρ q = θ θ x = reversible process > irreversible process Local (spatial) form of the 2 nd Law of Thermodynamics (Clausius-Duhem inequality) 70

71 Local Spatial Form of the Second Law of Thermodynamics Considering that, q 1 1 ( ) = q q θ 2 θ θ θ The Clausius-Duhem inequality can be written as ( ) r 1 s + q 0 θ ρθ i = s = s ( i) ds ds r 1 1 = + q q θ 0 2 θ ρθ ρθ ( i) = ( i) = s local CLAUSIUS-PLANCK 1 INEQUALITY 2 s cond q θ 0 ρθ REMARK (Stronger postulate) Internally generated entropy can ( i) be generated locally, s local, or by ( i) thermal conduction, s cond, and both must be non-negative. HEAT FLOW INEQUALITY 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

72 Alternative Forms of the Clausius-Planck Inequality Substituting the internal energy balance equation given by du not ρ = ρu = σ : d+ ρr q into the Clausius-Planck inequality, i ρθ s : = ρθ s ρr+ q 0 local q ρr = σ : d ρu yields, ( σ d ρ u ) ρθ s + : 0 ( u θs ) σ : d 0 ρ + Clausius-Planck Inequality in terms of the specific internal energy 72

73 5.11. Governing Equations Ch.5. Balance Principles 74

74 Governing Equations in Spatial Form ρ+ ρ v = 0 σ + ρb= ρv T σ = σ Conservation of Mass. Continuity Equation. Linear Momentum Balance. First Cauchy s Motion Equation. Angular Momentum Balance. Symmetry of Cauchy Stress Tensor. 1 eqn. 3 eqns. 3 eqns. ρu = σ : d+ ρr q Energy Balance. First Law of Thermodynamics. 1 eqn. 8 PDE + 2 restrictions ( u θ s ) σ :d 0 ρ + 1 q θ 0 2 ρθ Second Law of Thermodynamics. Clausius-Planck Inequality. Heat flow inequality 2 restrictions 75

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 q specific internal energy 1 variable heat flux per unit of surface vector field 3 variables 19 scalar unknowns θ s absolute temperature specific entropy 1 variable 1 variable 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. 76

76 Constitutive Equations in Spatial Form (, θ, ) σ = σ v ζ Thermo-Mechanical Constitutive Equations. 6 eqns. s = s( v, θ, ζ Entropy ) Constitutive Equation. 1 eqn. ( θ) q= q v, = K θ Thermal Constitutive Equation. Fourier s Law of Conduction. 3 eqns. i (,,, ) u = f ρ v θ ζ ( ρθ,, ζ ) = 0 { 1, 2,..., } F i p Heat Kinetic set of new thermodynamic variables: ζ = ζ1, ζ 2,..., ζ p. { } State Equations. REMARK 1 The strain tensor is not considered an unknown as they can be obtained through the motion equations, i.e., ε = ε v. ( ) (1+p) eqns. (19+p) PDE + (19+p) unknowns REMARK 2 These equations are specific to each material. 77

77 The Coupled Thermo-Mechanical Problem ρ+ ρ v = 0 Conservation of Mass. Continuity Mass Equation. 1 eqn. 16 scalar unknowns Linear Momentum Balance. First Cauchy s Motion Equation. 3 eqns. 10 equations σ=σ( ε( v), θ ) Mechanical constitutive equations. 6 eqns. Energy Balance. First Law of Thermodynamics. 1 eqn. Second Law of Thermodynamics. Clausius-Planck Inequality. 2 restrictions. 78 MMC - ETSECCPB - UPC

78 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

79 The Uncoupled Thermo-Mechanical Problem ρ+ ρ v = 0 Conservation of Mass. Continuity Mass Equation. 1 eqn. 10 scalar unknowns Linear Momentum Balance. First Cauchy s Motion Equation. 3 eqns. Mechanical problem σ=σ( ε ( v), θ ) Mechanical constitutive equations. 6 eqns. Energy Balance. First Law of Thermodynamics. Second Law of Thermodynamics. Clausius-Planck Inequality. 1 eqn. 2 restrictions. Thermal problem 80

80 The Uncoupled Thermo-Mechanical Problem Then, the variables involved in the mechanical problem are: ρ density 1 variable Mechanical variables v σ velocity vector field Cauchy s stress tensor field 3 variables 6 variables Thermal variables u q θ s specific internal energy 1 variable heat flux per unit of surface vector field 3 variables absolute temperature 1 variable specific entropy 1 variable 81

81 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 5.2 Mass Transport or Convective Flux Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar In continuum mechanics, the term convection is associated with mass transport in the medium, which derives from the motion of its particles. 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

82 194 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. convective flux of A through S not amount of A crossing S = Φ S = unit of time Continuum Mechanics for Engineers Theory and Problems 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: /rg

83 Mass Transport or Convective Flux 195 Figure 5.3: Cylinder occupied by the particles that have crossed ds in the time interval [t, t + ]. 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, these particles will have followed a pathline dx = v, such that at the instant of time t + 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 + ], these will occupy a cylinder generated by translating the base ds along the directrix dx = v, and whose volume is given by d = ds dh = v n ds. (5.1) Since the volume (d ) of the particles crossing ds in the time interval [t, t + ] is known, the mass crossing ds in this same time interval can be obtained by multiplying (5.1) by the density, dm = ρ d = ρv n ds. (5.2) Finally, the amount of A crossing ds in the time interval [t, t + ] is calculated by multiplying (5.2) by the function Ψ (amount of A per unit of mass), Ψ dm = ρψv n ds. (5.3) Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar Dividing (5.3)by yields the amount of the property that crosses the differential control surface ds per unit of time, d Φ S = Ψ dm = ρψv n ds. (5.4) 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 of A through S } Φ S = S ρψv n ds (5.5) X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi: /rg

84 196 CHAPTER 5. BALANCE PRINCIPLES Example 5.1 Compute the magnitude Ψ and the convective flux Φ S 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 and Ψ = 1 ρ lead to Φ S = Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar S v nds = volume flow rate. 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 Φ S = ρ v nds. 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 Φ S = ρ v(v n) ds. (Note that in this case Ψ and the convective flux Φ S are vectors). d) If the property A is the kinetic energy then Ψ is the kinetic energy per unit of mass. Therefore, A 1 2 m v 2 and Ψ = v 2 lead to Φ S = 2 ρ v 2 (v n) ds. S S S Remark 5.1. In a closed control surface 1, S =, the expression of the convective flux corresponds to the net outflow, defined as the outflow minus the inflow (see Figure 5.4), that is, net convective flux of A not = Φ = ρψv n ds. X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi: /rg

85 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 properties can be transported within a continuous medium in a manner not necessarily associated with mass transport. This form of non-convective transport receives several names (conduction, diffusion, etc.) depending on the physical problem being studied. A typical example is heat flux by conduction. 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 Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar 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: /rg

86 198 CHAPTER 5. BALANCE PRINCIPLES 5.3 Local and Material Derivatives of a olume 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 volume 2, amount of A μ (x,t)= unit of volume. (5.6) Consider an arbitrary volume in space. At time t, the total amount Q(t) of the property contained in this volume is Q(t)= μ (x,t) d. (5.7) To compute the content of property A at a different time t + Δt, the following two situations arise: 1) A control volume 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 t 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 Q(t + Δt) Q(t) (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 Local Derivative Definition 5.2. The local derivative of the volume integral, Q(t)= μ (x,t) d, Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar is the time derivative of Q(t) when the volume is a volume fixed in space (control volume), see Figure 5.5. The notation local derivative not = μ (x,t) d t will be used. 2 μ is related to Ψ =(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: /rg

87 Local and Material Derivatives of a olume Integral 199 Figure 5.5: Local derivative of a volume integral. The amount Q of the generic property A in the control volume at times t and t + Δt is, respectively, Q(t)= μ (x,t) d and Q(t + Δt)= μ (x,t + Δt) d. (5.9) Using (5.9) in addition to the concept of time derivative of Q(t) results in 3 Q (t)= 1 ( ) μ (x,t) d = lim Q(t + Δt) Q(t) = t Δt 0 Δt 1 = lim μ (x,t + Δt) d μ (x,t) d = Δt 0 Δt μ (x,t + Δt) μ (x,t) μ(x,t) = lim d = d, Δt 0 Δt t }{{} }{{} μ(x,t) local derivative t of μ (5.10) which yields the mathematical expression of the local derivative of a volume integral. Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar Local derivative of a volume integral μ(x,t) μ (x,t) d = t t d (5.11) 3 Note that the integration domain does not vary when the volume 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: /rg

88 200 CHAPTER 5. BALANCE PRINCIPLES Material Derivative Definition 5.3. The material derivative of the volume integral, Q(t)= μ (x,t) d, is the time derivative of Q(t) when the volume t is a material volume (mobile in space), see Figure 5.6. The notation material derivative not = d μ (x,t) d t will be used. The content Q of the generic property A in the material volume t at times t and t+δt is, respectively, Q(t)= μ (x,t) d and Q(t + Δt)= μ (x,t + Δt) d. (5.12) t t+δt Then, the material derivative is mathematically expressed as 4 Q (t)= d Q(t + Δt) Q(t) μ (x,t) d = lim = Δt 0 Δt t t 1 = lim μ (x,t + Δt) d μ (x,t) d. Δt 0 Δt t+δt t (5.13) The following step consists in introducing two variable substitutions, each suitable for one of the two integrals in (5.13), which lead to the same integration domain in both expressions. These variable substitutions are given by the equation of motion x = ϕ (X, t), particularized for times t and t + Δt, x t = ϕ (X,t) (dx 1 dx 2 dx 3 ) t = F(X,t) (dx 1 dx 2 dx 3 ), }{{}}{{} d t d 0 Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar x t+δt = ϕ (X,t + Δt) (dx 1 dx 2 dx 3 ) }{{ t+δt = F(X,t + Δt) (dx 1 dx 2 dx 3 ), }}{{} d t+δt d 0 (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: /rg

89 Local and Material Derivatives of a olume Integral 201 Figure 5.6: Material derivative of a volume integral. where the identity d t = F(X,t) d 0 has been taken into account. The variable substitutions in (5.14) are introduced in (5.13), resulting in ( μ (X,t + Δt) d 1 {}}{ μ (x,t) d = lim μ (x(x,t + Δt),t + Δt) F(X,t + Δt) d Δt 0 Δt ) 0 t 0 μ (x(x,t),t) F(X,t) d 0 = }{{} 0 μ (X,t) μ (X,t + Δt) F(X,t + Δt) μ (X,t) F(X,t) = lim d 0 = Δt 0 Δt 0 }{{} ( ) μ (X,t) F(X,t) = d ( ) μ (x,t) F(x,t) t d ( ) = μ F d 0. 0 (5.15) Finally, expanding the last integral in (5.15) 5 and considering the equality d F / = F v yields ( ) d d ( ) dμ μ (x,t) d = μ F d 0 = F + d F μ d 0 = t 0 0 }{{} F v ( ) ( ) dμ dμ = + μ v F d 0 = }{{} + μ v d, 0 d t t (5.16) Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar 5 The change of variable x t = ϕ (X,t) is undone here. X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi: /rg

90 202 CHAPTER 5. BALANCE PRINCIPLES that is 6, d μ (x,t) d t t not = d t μ (x,t) d = ( ) dμ + μ v d. (5.17) Recalling the expression of the material derivative of a property (1.15) results in ( ) d μ μ (x,t) d = + v μ + μ v d = t t }{{} (μv) (5.18) μ = d + (μv) d = μ d + (μv) d, t t 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) d = μ d + (μv) d t t }{{}}{{}}{{} material local convective derivative derivative derivative Remark 5.4. The form of the material derivative, given as a sum of a local derivative and a convective derivative, that appears when differentiating properties of the continuous medium (see Chapter 1, Section 1.4) also appears here when differentiating integrals in the continuous medium. Again, the convective derivative is associated with the existence of a velocity (or motion) in the medium and, thus, with the possibility of mass transport. Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar (5.19) 6 The expression d t μ (x,t) d denotes the time derivative of the integral over the material volume t (material derivative of the volume integral) particularized at time t, when the material volume occupies the spatial volume. X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi: /rg

91 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 t that at times t and t + Δt occupies the volumes in space t and t+δt, respectively (see Figure 5.7). Consider also the spatial description of the density, ρ (x,t). The mass enclosed by the material volume at times t and t + Δt is, respectively, M(t)= t ρ (x,t) d and M(t + Δt)= t+δt ρ (x,t + Δt) d. (5.20) Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar By virtue of the principle of conservation of mass, M(t)=M(t + Δt) must be satisfied 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, M (t)= d ρ d = 0 t. (5.21) t X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi: /rg

92 204 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 ρ d = + ρ v d = 0 Δ t t, t t (Δt ) t (Δt ), (5.22) which must be satisfied for t and, also, for any partial material volume Δ t t that could be considered. In particular, it must be satisfied for each of the elemental material volumes associated with the different particles in the continuous medium that occupy the differential volumes d t. Applying (5.22) on each differential volume d t d (x,t) yields 7 d(x,t) ( ) ( ) dρ dρ (x,t) + ρ v d = + ρ (x,t) v(x,t) d (x,t)=0 x t, t = dρ + ρ v = 0 d x t, t (5.23) Local spatial form of the principle of conservation of mass (mass continuity equation) dρ + ρ v = 0 d x t, t (5.24) 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) results in ρ t Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar + v ρ + ρ v }{{} (ρv) = 0 = ρ t + (ρv)=0, (5.25) 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: /rg

93 Conservation of Mass. Mass continuity Equation 205 ρ t + (ρv)=0 ρ t + (ρv i) = 0 i {1,2,3} x i ρ t + (ρv x) + (ρv y) + (ρv z) x y z = 0 Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar x t, t (5.26) Material Form of the Principle of Conservation of Mass From (5.22) 8, ( ) ( dρ dρ + ρ v d = + ρ 1 ) d F d = F t ( t 1 = F dρ F + ρ d F ) 1 d ( ) d = ρ F F t }{{} t d ( ) ρ F ( ) = ρ (X,t) F(X,t) d 0 Δ 0 0, t, t 0 d = }{{} F d 0 (5.27) where the integration domain is now the volume in the reference configuration, 0. Given that (5.27) must be satisfied for each and every part Δ 0 of 0,a localization process can be applied, which results in 9 ( ) t ρ (X,t) F(X,t) = 0 X 0, t = ρ (X,t) F(X,t) = ρ (X) F(X) t = ρ (X,0) F (X,0) } {{ } not = ρ 0 F 0 = ρ (X,t) F (X,t) } {{ } not = ρ t F t Local material form of the mass conservation principle ρ 0 (X)=ρ t (X) F t (X) X 0, t = ρ 0 F }{{} 0 = ρ t F t. = 1 (5.28) (5.29) 8 Here, the expression deduced in Chapter 2, d F / = 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: /rg

94 206 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 Reynolds Lemma Consider an arbitrary material volume of the continuous medium that at time t occupies the volume in space t. The amount of the generic property A in the material volume t at time t is Q(t)= t ρψ d. (5.30) The variation along time of the content of property A in the material volume t 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 Q (t)= d ( ) d (ρψ) ρψ d = + ρψ v d. (5.31) }{{} t μ Considering the expression of the material derivative of a product of functions, grouping terms and introducing the mass continuity equation (5.24) yields ( d ρψd = ρ dψ +Ψ dρ ) + ρψ v d = t ( = ρ dψ ( ) ) dρ +Ψ + ρ v (5.32) d = }{{} =0 (mass continuity eqn.) Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar Reynolds Lemma d ρψ d = ρ dψ t d. (5.33) Reynolds Theorem Consider the arbitrary volume, fixed in space, shown in Figure 5.8. The amount of property A in this control volume is Q(t)= ρψ d. (5.34) X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi: /rg

95 Balance Equation. Reynolds Transport Theorem 207 Figure 5.8: Reynolds Transport Theorem. The variation of the amount of property A in the material volume t, which instantaneously coincides at time t with the control volume ( t ),isgivenby expression (5.19) of the material derivative of a volume integral (with μ = ρψ) and by (5.11), d (ρψ) ρψ d = d + (ρψ v) d. (5.35) t t Introducing the Reynolds Lemma (5.33) and the Divergence Theorem 10 in (5.35) results in Reynolds d Lemma ρψ d = ρ dψ d = (ρψ) d + (ρψ v)d = t t Divergence Theorem (ρψ) = d + ρψ v n ds, t (5.36) which can be rewritten as follows. Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar 10 The Divergence Theorem provides the following relation between a volume integral and a surface integral of a tensor A. A d = n A ds, where n is the outward unit normal vector in the boundary of the volume. X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi: /rg

96 208 CHAPTER 5. BALANCE PRINCIPLES Reynolds Transport Theorem ρψ d = ρ Ψ d t t }{{}}{{} variation per unit of variation due to the time of the content of change in the content of property A in the property A of the particles in the interior of control volume Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar ρψ v n ds }{{} variation due to the net convective flux of A exiting through the boundary (5.37) The local form of the Reynolds Transport Theorem can be obtained by localizing in (5.36), ρ dψ d = ρ dψ (ρψ) d + (ρψ v)d Δ = t = (ρψ) + (ρψ v) x = t Local form of the Reynolds Transport Theorem (ρψ) = ρ dψ (ρψ v) t x 5.6 General Expression of the Balance Equations (5.38) (5.39) Consider a certain property A of a continuous 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 can be transported both by motion of mass (convective transport) and by non-convective transport. To this aim, the following terms are defined: A source term k A (x,t) (of the same tensor order than property A) that characterizes the internal generation of the property, k A (x,t)= internally generated amount of A unit of mass / unit of time. (5.40) A vector j A (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: /rg

97 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 (see Figure 5.9). Then, the variation per unit of time of property A in volume 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, and 3) the (net incoming) non-convective flux of A through. That is, amount of A generated in due to the internal sources ρk A (x,t)d =, unit of time ρψ v n ds = j A n ds = Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar amount of A exiting through per convective flux unit of time amount of A exiting through per non-convective flux unit of time,, (5.41) X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi: /rg

98 210 CHAPTER 5. BALANCE PRINCIPLES and the expression of the balance of the amount of property A in the control volume results in Global form of the general balance equation ρψ d = ρk A d ρψ v n ds j A n ds t }{{}}{{}}{{}}{{} variation of the variation due variation due to variation due to amount of A in to internal the incoming the incoming per unit of time generation convective flux nonconvective flux (5.42) Using the Divergence Theorem and (5.11), the global form of the general balance equation (5.42) can be written as ρψ d = ρk A d (ρψ v)d j A d = t ( ) (ρψ)+ (ρψ v) d = (ρk A j A )d Δ t (5.43) and localizing in (5.43), the local spatial form of the general balance equation Local spatial form of the general balance equation (ρψ)+ (ρψ v) = ρ dψ = ρk A j A t }{{}}{{}}{{}}{{} ρ dψ variation of the variation variation amount of due to due to nonconvective property (per internal unit of volume generation transport and of time) by a source (5.44) Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet 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: /rg

99 General Expression of the Balance Equations 211 Remark 5.5. Expression (5.42) and, especially, expression (5.44), ρ dψ = ρk A j A, exhibit the negative contribution ( j A ) of the non-convective flux to the variation in content of the property per unit of volume and of time, ρ dψ/. Only when all the flux is convective (by mass transport) can this variation originate solely from the internal generation of this property, ρ dψ = ρk A. 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 k M = 0 since, following the principle of conservation of mass, it is not possible to generate mass. The non-convective mass flux vector is j M = 0 because mass cannot be transported in a non-convective manner. Therefore, (5.44) results in the balance of mass generation, Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar ρ dψ = ρ t + (ρv)=0, 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: /rg

100 212 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 m i, an acceleration a i and is subjected to a force f i (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) yields 11, Figure 5.10 dv i f i = m i a i = m i = d (m iv i ) (5.45) The linear momentum of the particle 12 is defined as the product of its mass by its velocity (m i v i ). 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 R(t)= n i=1 f i = n i=1 m i a i = n i=1 dv i m i = d Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar n dp (t) m i v i =. (5.46) i=1 }{{} P = linear momentum Note that, again, to obtain the last expression in (5.46), the principle of conservation of mass (dm i / = 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 denominated the principle of balance of linear momentum. Remark 5.6. If the system is in equilibrium, R = 0. Then, R(t)=0 t = dp (t) = 0 = n i=1 which is known as the conservation of linear momentum. m i v i = P = const., 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: /rg

101 Balance of Linear Momentum 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 t of the continuous medium with mass M as P (t)= M v dm }{{} ρ d = ρ v d. (5.47) t 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) R(t)= = d ρ v d. t 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 d + t ds. (5.48) } {{ } body forces }{{} surface forces Applying the principle of balance of linear momentum on the resultant in (5.48) yields the integral form of the balance of linear momentum. Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar Global form of the principle of balance of linear momentum ρb d + t ds = d ρv d t (5.49) X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi: /rg

102 214 CHAPTER 5. BALANCE PRINCIPLES Figure 5.11: Forces acting on a material volume of the continuous medium Local Form of the Balance of Linear Momentum Using Reynolds Lemma (5.33) on(5.49) and introducing the Divergence Theorem, results in d ρv d = ρb d + n σ ds = ρ dv }{{} d t t t Divergence = (5.50) Theorem n σ ds = σ d = ( σ + ρb)d + ρ dv d Δ (5.51) Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar and, localizing in (5.51), yields the local spatial form of the balance of linear momentum, also known as Cauchy s equation 13. Local spatial form of the principle of balance of linear momentum (Cauchy s equation) σ + ρb = ρ dv = ρa x, t (5.52) 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: /rg

103 Balance of Angular Momentum Balance of Angular Momentum Consider a discrete system composed of n particles such that for an arbitrary particle i, its position vector is r i, its mass is m i, a force f i acts on it, and it has a velocity v i and an acceleration a i (see Figure 5.12). The moment about the origin of the force acting on this particle is M i = r i f i and the moment about the origin of the linear momentum 14 of the particle is L i = r i m i v i. Considering Newton s second law, the moment M i is 15 Figure 5.12 dv i M i = r i f i = r i m i a i = r i m i (5.53) Extending the previous result to the discrete system formed by n particles, the resultant moment about the origin M O of the forces acting on the system of particles is obtained as 16 M O (t)= d n i=1 n i=1 r i f i = r i m i v i = n i=1 n i=1 r i m i a i = dr i m i v i + }{{} v i }{{} = 0 = M O (t)= d n i=1 n dv i r i m i dv i r i m i i=1 n i=1 r i m i v i }{{} Angular momentum L Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar = dl(t) = (5.54) Equation (5.54) expresses that the resultant moment M O of all the forces acting on the discrete system of particles is equal to the variation 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 (v i v i = 0). X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi: /rg

104 216 CHAPTER 5. BALANCE PRINCIPLES Remark 5.7. If the system is in equilibrium, M O = 0. Then, M O (t)=0 t = dl(t) = 0 = n r i m i v i = L = const., i=1 which is known as the conservation of angular momentum 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 = r v dm = r ρ v d (5.55) }{{} M ρ d 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. M O (t)= dl(t) = d t r ρ v d Continuum Mechanics for Engineers Theory and Problems 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: /rg

105 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) M O (t)= r ρ b d + r t ds, (5.56) 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 r ρ v d = r ρ b d + r t ds t Local Spatial Form of the Balance of Angular Momentum Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar (5.57) The procedure followed to obtain the local spatial form of the balance equation is detailed below. Introducing Reynolds Lemma in (5.57), d r ρv d = d t ( dr ) = ρ v d + }{{} v }{{} = 0 t and expanding the last term in (5.57), r t }{{} n σ ρ (r v)d = ρ d (r v)d = ( ρ r dv ) d = r ρ dv d, (5.58) ds = r n σ ds = [r] [n σ] T ds = = (r σ T ) n ds Divergence Theorem = ( r σ T ) d, (5.59) X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi: /rg

106 218 CHAPTER 5. BALANCE PRINCIPLES where the component [( r σ T ) ] is computed as i ) ] ( ) [(r σ T symb = e ijk x j σ rk = ( ) eijk x j σ rk = i }{{} x r x r σkr T = e ijk x j x r }{{} δ jr σ rk + e ijk x j σ rk x r }{{} [r σ] i Introducing now (5.60)in(5.59) produces = e ijk σ jk +[r σ] i i {1,2,3}. }{{} m i r t ds = m d + (r σ) d m i = e ijk σ jk i, j,k {1,2,3} and, finally, replacing (5.58) and (5.61) in(5.57) yields (5.60) (5.61) r ρ dv d = r ρb d + m d + (r σ)d. (5.62) 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 ( r σ + ρb ρ dv ) d + m d = 0 }{{} = 0 (5.63) = m d = 0 Δ. Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar Then, localizing in (5.63) and considering the value of m in (5.61), yields } m = 0 x = e ijk σ jk = 0 i, j,k {1,2,3} (5.64) m i = e ijk σ jk = 0 i {1,2,3} X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi: /rg

107 Power 219 and particularizing (5.64) for the three possible values of index i: i = 1:e 1 jk σ jk = e 123 }{{} =1 i = 2:e 2 jk σ jk = e 231 }{{} =1 i = 3:e 3 jk σ jk = e 312 }{{} =1 = σ = σ T, σ 23 + e 132 σ 32 = σ 23 σ 32 = 0 σ 23 = σ 32 }{{} = 1 σ 13 = σ 31 σ 13 = 0 σ 31 = σ 13 σ 31 + e 213 }{{} = 1 σ 12 + e 321 }{{} = 1 σ 21 = σ 12 σ 21 = 0 σ 12 = σ 21 Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar = (5.65) which results in the local spatial form of the balance of angular momentum translating into the symmetry of the Cauchy stress tensor Power Local spatial form of the principle of balance of angular momentum (5.66) σ = σ T 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 (or continuous medium) the power W (t) entering the system is defined as work performed by the system W (t)=. unit of time 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, W (t)= de (t). (5.67) 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: /rg

108 220 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 Mechanical Power. Balance of Mechanical Energy Definition 5.8. The mechanical power entering the continuous medium, P e, 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 P e is P e = ρ b v d + t v ds = ρ b v d + n (σ v) ds. (5.68) }{{} n σ Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar ρb dr d = ρb vd t dr ds = t vds Figure 5.14: Continuous medium subjected to body and surface forces. X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi: /rg

109 Power 221 Applying the Divergence Theorem in the last term of (5.68) yields n (σ v)ds = (σ v)d (σ v)= v j (σ ij v j )= v j + σ ij x i x i σ ij x i }{{} [ σ] j }{{} σ ji }{{} [l] ji =( σ) v + σ : l (5.69) and, taking into account the identity l = v = d + w (see Chapter 2) 18, Replacing (5.70)in(5.69) yields n (σ v) ds = σ : l = σ : d + σ : w = σ : d. (5.70) }{{}}{{} d + w = 0 ( σ) v d + σ : d d. (5.71) Introducing (5.71) in(5.68), the mechanical power entering the continuous medium results in 19 P e = ρ b vd + t vds = ρ b vd + ( σ) vd + σ : dd = = ( σ + ρ b) vd + σ : dd = ρ dv vd + σ : dd = ρ d ( ) 1 2 v v d + σ : dd = ρ d ( ) 1 2 v2 d + σ : dd. (5.72) And applying Reynolds Lemma (5.33)in(5.72), the mechanical power entering the system results in Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar Balance of mechanical energy P e = ρ b vd + t vds = d }{{} mechanical power entering the medium t 1 2 ρv2 d } {{ } K = kinetic energy + σ : d d } {{ } stress power (5.73) 18 The tensor σ is symmetric and the tensor w is antisymmetric. Consequently, their product is null,σ : w = The expression d ( ) 1 2 v v = 1 dv 2 v v dv = dv v is used here, in addition to the notation v v = v 2 = v 2. X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi: /rg

110 222 CHAPTER 5. BALANCE PRINCIPLES 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, P e = ρ b v d + t v ds is invested in: a) modifying the kinetic energy of the particles in the continuous medium, kinetic energy not 1 = K = 2 ρ v2 d = dk = d 1 2 ρ v2 d. b) creating stress power, stress power def = σ : d d. 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). Therefore, 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 mechanical energy of a rigid body is recovered Thermal Power Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar Definition The thermal power entering the continuous medium, Q e, 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: /rg

111 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 amount of heat entering the medium q n ds = unit of time 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 (nonconvective) 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. (5.74) Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar Figure 5.15: Non-convective heat flux. X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi: /rg

112 224 CHAPTER 5. BALANCE PRINCIPLES Internal heat sources Figure 5.16: 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 ρrd =. (5.75) unit of time Consequently, the total heat entering the continuous medium per unit of time (or thermal power Q e ) can be expressed as the sum of the contributions of the conduction flux (5.74) and the internal sources (5.75), Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar Heat power entering the medium Q e = ρrd q n ds. (5.76) Then, considering (5.73) and (5.76), the total power entering the continuous medium can be written as follows. Total power entering the system P e + Q e = d 1 2 ρv2 d + σ : dd + t ρrd q nds (5.77) X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi: /rg

113 Energy Balance Energy Balance 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 μ i (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 μ i 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 t A to a final state at time t B (it is a path or continuous segment in the thermodynamic space), see Figure 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 State function: any scalar, vector or tensor function φ (μ 1,...,μ n ) of the thermodynamic variables that can be written univocally in terms of these variables. Consider a thermodynamic space with thermodynamic variables μ i (x,t) i {1,2,...,n} and a function φ (μ 1,...,μ n ) of said variables implicitly defined in terms of a differential form 20 δφ = f 1 (μ 1,...,μ n )dμ f n (μ 1,...,μ n )dμ n. (5.78) Continuum Mechanics for Engineers Theory and Problems 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 (μ 1,...,μ n ) 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: /rg

114 226 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(μ1 B,...,μB n ) not = φ B when its value φ(μ1 A,...,μA n ) not = φ A and the corresponding path (thermodynamic process) A B are known by means of φ B = φ A + B A δφ. (5.79) However, (5.79) does not guarantee that the result φ B is independent of the path (thermodynamic process) followed. In mathematical terms, it does not guarantee that the function φ : R n R defined by (5.79) is univocal (see Figure 5.20) and, thus, that there exists a single image φ (μ 1,...,μ n ) corresponding to each point in the thermodynamic space. Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar Figure 5.20: Non-univocal function of the thermodynamic variables μ 1 and μ 2. X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi: /rg

115 Energy Balance 227 Remark For a function φ (μ 1,...,μ n ), 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 = f 1 (μ 1,...,μ n )dμ f n (μ 1,...,μ n )dμ n f i (μ 1,...,μ n ) = f j (μ 1,...,μ n ) δφ = dφ. i, j {1,...,n} μ j μ i If the differential form (5.78) is an exact differential, (5.79) results in φ B = φ A + B A [ dφ = φ A + Δφ Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar ] B A (5.80) and the value φ B 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. Remark If φ is a state function, then δφ is an exact differential and the integral along the complete closed cycle of the differential δφ is null, A [ ] A δφ = dφ = Δφ = 0. A A }{{} = 0 X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi: /rg

116 228 CHAPTER 5. BALANCE PRINCIPLES Example 5.3 Determine whether the function φ (μ 1, μ 2 ) defined in terms of an exact differential δφ = 4μ 2 dμ 1 + μ 1 dμ 2 can be a state function or not. Solution Following (5.78), f 1 4μ 2 f 2 μ 1 = f 1 μ 2 = 4 f 2 μ 1 = 1 = f 1 μ 2 f 2 μ 1 Then, δφ is not an exact differential (see Remark 5.10) and φ is not a state function 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). δφ 1 = P e = P e 0 (5.81) δφ 2 = Q e = Q e 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, t P e + Q e = de E (t)= (P e + Q e ) + const. (5.82) t 0 Continuum Mechanics for Engineers Theory and Problems 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: /rg

117 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 }{{} ariation of total energy de = P e + Q e = P e }{{} Mechanical work + Q e }{{} Thermal work (5.83) 2) There exists another state function U, named internal energy of the system, such that a) It is an extensive property 21. Then, a specific internal energy u(x,t) (or internal energy per unit of mass) can be defined as U = ρud. (5.84) 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. de = dk + du }{{}}{{} (5.85) Exact Exact differential differential Remark Note that, since the total energy E and the internal energy U of the system have been postulated to be state functions, de and du in (5.85) are exact differentials. Consequently, the term dk = de du in said equation is also an exact differential (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. Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar 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). X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi: /rg

118 230 CHAPTER 5. BALANCE PRINCIPLES From (5.83) and considering (5.77), de = P e + Q e = d 1 K = 2 ρv2 d t 1 2 ρv2 d + σ : d d + ρrd q n ds de = dk + du = d 1 2 ρv2 d }{{} dk t + σ : d d + Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar ρrd q n ds } {{ } du (5.86) Global form of the internal energy balance du = d ρud = σ : d d + ρrd q n ds Remark From (5.87) it follows that any variation per unit of time of the internal energy du/ is produced by a generation of stress power, σ : d d, and a variation per unit of time of the content of heat in the medium, ρrd q n ds. (5.87) Applying Reynolds Lemma (5.33) and the Divergence Theorem on (5.87) yields d t ρud = ρ du d = σ : d d + ρrd q d Δ. (5.88) X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi: /rg

119 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) ρ du = σ : d +(ρr q) x, t (5.89) 5.11 Reversible and Irreversible Processes The first law of thermodynamics leads to a balance equation that must be fulfilled for all the physical processes that take place in reality, P e + Q e = de = du + dk. (5.90) In particular, if an isolated system 22 is considered, the time variation of the total energy of the system will be null (de/ = 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/ must be compensated with a variation of kinetic energy dk/ 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/ = dk/ > 0ordU/ = dk/ < 0). That is, it does not establish any restriction that indicates if an imaginary and arbitrary Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar 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: /rg

120 232 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 consisting 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. Consider now the following two processes: Figure ) 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 energy 23, 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 decrease in temperature). The conclusion to this observation is that the second process considered in the example is not a feasible physical process. More generally, only thermodynamic processes that tend to increase the internal energy 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. Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar 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). X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi: /rg

121 Second Law of Thermodynamics. Entropy 233 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 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 (even if a different path can be followed, see Figure 5.24). In general, within a same thermodynamic process there will exist reversible and irreversible sections. Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar 5.12 Second Law of Thermodynamics. Entropy 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 intensive 24 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 (specific 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: /rg

122 234 CHAPTER 5. BALANCE PRINCIPLES 2) There exists a state function named entropy S with the following characteristics: a) It is an extensive variable. This implies that there exists a specific entropy (entropy per unit of mass) s such that b) The inequality s = entropy unit of mass Integral form of the second law of thermodynamics ds = d ρsd ρ r θ d t = S = ρsd. (5.91) Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar q θ n ds (5.92) 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 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, defined in terms of a non-convective flux vector per unit of surface q(x,t). These terms allow computing the amount of heat per unit of time entering a material volume t, which at a certain instant of time occupies the spatial volume t with outward unit normal vector n, as Q e = ρrd q n ds. (5.93) 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: /rg

123 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 unit of time heat/unit of absolute temperature unit of time 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 = ρ r unit of time θ d r r θ Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar q q θ q n ds. (5.94) θ 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 ds ρ r θ d q θ n ds }{{} amount of the property heat / unit of absolute temperature entering the domain per unit of time (5.95) 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) /. S (e) : entropy generated by the interaction of the continuous medium with its exterior. Its variation rate is ds (e) /. X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi: /rg

124 236 CHAPTER 5. BALANCE PRINCIPLES Then, the following is naturally satisfied. ds = ds(e) + ds(i) (5.96) 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) = ρ r θ d q n ds (5.97) θ 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 ρ r θ d q θ n ds 0. (5.98) Remark According to (5.98), the internally generated entropy S (i) of the system (continuous medium) never decreases (ds (i) / 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) / = 0). In this case, the second law of thermodynamics establishes that ds (i) = ds 0 or, in other words, the total entropy of a perfectly isolated system never decreases. This is the starting point of some alternative formulations of the second law of thermodynamics. Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar Reformulation of the Second Law of Thermodynamics In view of the considerations in Section , 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: /rg

125 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)= ρsd. (5.100) 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 can be defined in terms of their respective specific values s (i) and s (e), S (i) = ρs (i) d and S (e) = ρs (e) d (5.101) S = S (i) + S (e) = ds = ds(i) + ds(e) and introducing Reynolds Lemma (5.33) in(5.102) yields ds (i) ds (e) = d = d t t ρs (i) d = ρs (e) d = Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar ρ ds(i) ρ ds(e) d, d. (5.102) (5.103) 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 ds (e) = ρ r θ d q n ds. (5.104) θ 5) The internally generated entropy never diminishes. Based on the variation of its content during the thermodynamic process, the following situations are defined: ds (i) 0 = 0 reversible process > 0 irreversible process < 0 non-feasible process (5.105) X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi: /rg

126 238 CHAPTER 5. BALANCE PRINCIPLES Local Form of the Second Law of Thermodynamics. Clausius-Planck Equation Using (5.101)to(5.104), expression (5.105) is rewritten as d t ρs (i) d = d ds (i) t = ds ds(e) 0 ( ρsd ρ r θ d Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar ) q θ n ds 0 (5.106) Applying Reynolds Lemma (5.33) (on the first and second integral of the lefthand term in (5.106)) and the Divergence Theorem (on the last term) yields ( ρ ds(i) d = ρ ds ) d ρ r ( q ) θ d d 0 Δ θ (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) ρ ds(i) = ρ ds (ρ r ( q )) θ 0 x, t θ Where, again, in (5.108) the sign = corresponds to reversible processes, > corresponds to irreversible processes, and < indicates that the corresponding process is not feasible. Equation (5.108) can be rewritten as follows. ρ ds(i) }{{} not = ṡ(i) ( q ) = 1 θ θ q 1 θ 2 q θ = ρ ds }{{} not = ṡ ρ r θ + 1 θ q 1 θ 2 q θ 0 (5.108) (5.109) X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi: /rg

127 Second Law of Thermodynamics. Entropy 239. s (i) = ṡ r θ + 1 ρθ q }{{}. s (i) local 1 ρθ 2 q θ 0 (5.110) }{{}. s (i) cond Then, a much stronger (more restrictive) formulation of the second law of thermodynamics can be posed. This formulation postulates that the internally generated entropy, ṡ(i), can be generated locally, ṡ(i) local, or by heat conduction,. s (i) cond, and that both contributions to the generation of entropy must be nonnegative. Local internal generation of entropy (Clausius-Planck inequality). s (i) local = ṡ r θ + 1 ρθ q 0 (5.111) Internal generation of entropy by heat conduction. s (i) cond = 1 ρθ 2 q θ 0 (5.112) Remark Equation (5.112) can be interpreted in the following manner: since the density, ρ, and the absolute temperature, θ, are positive magnitudes, said equation can be written as Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar q θ 0, 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: /rg

128 240 CHAPTER 5. BALANCE PRINCIPLES Figure 5.25: Heat flux is opposed to the thermal gradient. Remark 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 Alternative Forms of the Second Law of Thermodynamics Alternative expressions of the Clausius-Planck equation (5.111) in combination with the local form of the energy balance equation (5.89) are often used in continuum mechanics. 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), Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar ρ du not = ρ u. = σ : d + ρr q = ρr q = ρ u. σ : d, (5.113) and, replacing it in the Clausius-Planck equation (5.111), ρθṡ(i) local = ρθṡ (ρr q)=ρθṡ ρ. u + σ : d 0. (5.114) X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi: /rg

129 Second Law of Thermodynamics. Entropy 241 Clausius-Planck equation in terms of the internal energy ρ (. u θṡ)+σ : d 0 (5.115) 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 Differentiating (5.116) with respect to time results in ψ def = u sθ. (5.116). ψ =. u s. θ ṡθ =. u θṡ =. ψ + s. θ (5.117) and, replacing (5.117) in(5.115), yields the Clausius-Planck equation in terms of the Helmholtz free energy, ρθṡ(i) local = ρ ( u. (. θṡ)+σ : d = ρ ψ + s θ. ) + σ : d 0. (5.118) Clausius-Planck equation in terms of the free energy (. ρ ψ + s θ. ) + σ : d 0 (5.119) For the infinitesimal strain case, d = ε. (see Chapter 2, Remark 2.22), and replacing in (5.119) results in Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar Clausius-Planck equation (infinitesimal strain) (. ρ ψ + s θ. ) + σ : ε. 0. (5.120) X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi: /rg

130 242 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 dρ + ρ v 1 equation (5.121) i = 0 x i 2) Balance of linear momentum. Cauchy s equation. σ + ρb = ρ dv σ ji + ρb i = ρ dv 3 equations (5.122) i i {1,2,3} x j 3) Balance of angular momentum. Symmetry of the stress tensor. } σ = σ T 3 equations (5.123) σ 12 = σ 21 ; σ 13 = σ 31 ; σ 23 = σ 32 4) Energy balance. First law of thermodynamics. ρ du = σ : d +(ρr q) ρ du ( = σ ij d ij + ρr q ) 1 equation (5.124) i x i Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi: /rg

131 Continuum Mechanics Equations. Constitutive Equations 243 5) Second law of thermodynamics. Clausius-Planck and heat flux inequalities. ρ ( u. θṡ)+σ : d 0 ρ ( u. θṡ)+σ ijd ij 0 1 restriction 1 (5.125) ρθ 2 q θ restriction ρθ 2 q θ i 0 x i 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 in 25 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. q = K θ q i = K θ 3 equations (5.126) i {1,2,3} x i Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar 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)= s v (see Chapter 2, Section ). X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi: /rg

132 244 CHAPTER 5. BALANCE PRINCIPLES 7) Constitutive equations (per se) 26. Thermomechanical constitutive f i (σ, ε (v),θ, μ)=0 i {1,...,6} 6 equations equations Entropy constitutive equation } s = s(ε (v),θ, μ)=0 1 equation (5.127) where μ = { μ 1,...,μ p } 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 eqns. of state i (ρ,θ, μ)=0 i {1,2,...,p} (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 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 general for all the continuous medium, regardless of the material that constitutes the medium, and for any range of displacements and strains. Conversely, the constitutive equations (5.126) to (5.128) are specific to the material or the type of continuous medium being studied (solid, fluid, gas) and differentiate them from one another. Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar 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: /rg

133 Continuum Mechanics Equations. Constitutive Equations 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 circumstances 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 considered 27 Mass continuity dρ equation: + ρ v = 0 (1 eqn) Cauchy s equation: σ + ρb = ρ dv (3 eqn) 10 equations, Mechanical f i (σ, ε (v)) = 0 (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 (x,t) 6 unknowns The problem defined by equations (5.129) and (5.130) constitutes the socalled mechanical problem, which involves the variables (5.130) (named mechanical variables) that, moreover, are the real interest in 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. Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar 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. X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi: /rg

134 246 CHAPTER 5. BALANCE PRINCIPLES Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi: /rg

135 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 Mass continuity equation = ρ t + (ρv)=0 = (ρv)=0. Stationary flux = ρ t = 0 Resulting, thus, what had to be proven, (ρv)=0 = (ρv) d = ρv n ds = 0. Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi: /rg

136 248 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 F ext/ f = d ρv d = (ρv) d + ρv(n v) ds = ρv(n v) ds. t S Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar Note that the velocity vector of the fluid along the surfaces S lat 1 and S lat 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 S 2 and S 4 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 F ext/ f = ρv(n v) ds = ρve( e ve) ds = ρv 2 Se. S X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi: /rg

137 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, F ext/ 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 = ρv 2 S + ps. Problem 5.3 A volume flow rate Q circulates, in steady-state regime, through a pipe from end A (with cross-section S A ) to end B (with cross-section S B < S A ). The pipe is secured at point O by a rigid element P O. Determine: a) The entry and exit velocities v A and v B 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. Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar Hypotheses: 1) The water is a perfect fluid (σ ij = pδ ij ) 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: /rg

138 250 CHAPTER 5. BALANCE PRINCIPLES Solution a) The incompressible character of water implies that the density is constant for a same particle and, therefore, dρ/ = 0. Introducing this into the mass continuity equation (5.24), results in v = 0 v d = 0. [1] 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 = ) to be able to apply the Divergence Theorem, v d = n v ds [2] where n is the outward unit normal vector in the boundary of the volume. 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. The volume the defined by the water contained inside the pipe between the crosssections S A and S B is taken as control volume. Consider, in addition, the unit vectors e A and e B 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 is applied only on cross-sections S A and S B since n v = 0 on the walls of the pipe, that is, n and v are perpendicular to one another. n v ds = n v ds+ n v ds = ( e A ) v A e A ds+ S A S B S A Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar S B e B v B e B ds = 0 = v A S A + v B S B = 0 = v A S A = v B S B = Q It is verified, thus, that the flow rate at the entrance and exit of the pipe are the same, v A = Q S A ; v B = Q S B. [3] X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi: /rg

139 Problems and Exercises 251 b) The balance of linear momentum equation (5.49) must be applied to find the value of the force F, R = ρb d + t ds = d ρv d, [4] 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 d = ρv d + ρv(n v) ds. [5] t 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 S A and S B 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 A S B = ρv A e A ( e A v A e A ) ds+ ρv B e B (e B v B e B ) ds S A S B R = ρv 2 A S A e A + ρv 2 B S B e B. [6] Introducing [3] in [6] allows expressing the resultant force R in terms of Q, ( R = ρq 2 1 e A + 1 ) e B. S A S B Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet 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 (S A, S B and S lat, where this last one corresponds to the lateral surface of the walls), X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi: /rg

140 252 CHAPTER 5. BALANCE PRINCIPLES R = ρb d + = t ds = S A p A e A ds+ t ds = t ds+ t ds+ t ds = S A S B S lat S B p B ( e B ) ds+ R p/ f. Here, R p/ 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. R p/ f = R p A e A ds p B ( e B ) ds S A S B R p/ f = ρv 2 A S A e A + ρv 2 B S B e B p A S A e A + p B S B e B R p/ f = ( ρv 2 A + p A) SA e A ( ρv 2 B + p B) SB e B [7] Introducing [3], R p/ f can be expressed in terms of Q, R p/ f = (ρ Q2 ) + p A S A e A S A (ρ Q2 ) + p B S B e B. S B Now the relation between R p/ 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 R f /p = R p/ f. Since it is the only action on the body, and taking into account that the weight of the pipe is negligible, this force must be compensated by an exterior action F for the body to be in equilibrium. Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar R f /p + F = 0 = F = R f /p = R p/ f Introducing [7], the value of F is finally obtained as F = ( ρv 2 A + p A ) SA e A + ( ρv 2 B + p B ) SB e B. X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi: /rg

141 Problems and Exercises 253 Using [3], the force F is expressed in terms of Q, ) ) F = (ρ Q2 + p A S A e A + (ρ Q2 + p B S B e B S A S B. [8] 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 F A and F B, which multiply the vectors ( e A ) and e B, respectively. The vector ( e A ) is fixed and does not depend on θ but e B does vary with θ. The scalars F A and F B are constant values. Therefore, the maximum and minimum values of F will be obtained when F A and F B either completely add or subtract one another, respectively. That is, when the vectors ( e A ) and e B are parallel to each other. Taking into account [3] and [8], the maximum and minimum values are found to be: Minimum value of F Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar θ = π 2 F min = ρq 2 ( 1 S B 1 S A ) + p B S B p A S A X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi: /rg

142 254 CHAPTER 5. BALANCE PRINCIPLES Maximum value of F θ = 3π 2 F min = ρq 2 ( 1 S B + 1 S A ) + p B S B + p A S A c) The balance of angular momentum equation (5.57) must be applied to find the moment M about point O, M liq = r ρb d + r t ds = d r ρv d, [9] where M liq 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 d = r ρv d + (r ρv)(n v) ds. [10] t 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 M liq = (r ρv)(n v) ds+ (r ρv)(n v) ds, [11] S A S B where the following must be taken into account: 1. The solution to each integral can be determined considering the resultant of the velocities in the middle point of each cross-section since the velocity distributions are uniform and parallel in both cases. 2. For cross-section S A, 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 S A and the velocity vector are null. 3. For cross-section S B, vectors r and v belong to the plane of the paper and, thus, their cross product has the direction of the vector ( e z ). In addition, they are perpendicular to each other, so the module of their cross product is the product of their modules. Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi: /rg

143 Problems and Exercises 255 Applying these considerations to [11] yields M liq = Rρ v B ( e z )(e B v B e B ) ds S B M liq = ρ v 2 B RS B e z = ρ Q2 S B Re z [12] 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. M liq = r ρb d + r t ds = r t ds = = r t ds+ r t ds r t ds = S A S B S lat = 0 + Rp B e z ds+ M p/ f = Rp B S B e z + M p/ f, S B where M p/ f is the moment exerted by the pipe on the fluid. To determine its expression, [12] is used, M p/ f = M liq Rp B S B e z = ρ v 2 B RS B e z Rp B S B e z, M p/ f = RS B ( ρ v 2 B + p B ) ez = R (ρ Q2 ) + p B S B e z. [13] S B Introducing the action and reaction law will allow obtaining the moment exerted by the fluid on the pipe, M p/ f = M f /p. Considering the pipe and the rigid element P O as a single body in equilibrium and neglecting the weight of the pipe, Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar M f /p + M = 0 = M = M f /p = M p/ f. Finally, the value of the moment M is obtained, using [13]. ) ( ) M = RS B ρv 2 B + p B ez = R (ρ Q2 + p B S B e z S B 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: /rg

144 256 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. W = d 1 2 ρv2 d + σ : d d [14] The stress power in an incompressible perfect fluid is null, σ : d d = 0. This is proven as follows. ( 1 ( )) σ : d = p 1 : d = ptr(d)= ptr l + l T = 2 v x v x v x x y z = ptr(l)= ptr v y v y v y = x y z v z v z v z x y z ( vx = p x + v y y + v ) z = p v = 0, z where [1] has been applied in relation to the incompressibility condition, to conclude that the divergence of the velocity is null. Applying the Reynolds Transport Theorem (5.39) on the term of the material derivative of the kinetic energy in [14] results in W = d Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar 1 2 ρv2 d = t 1 2 ρv2 d ρv2 (n v) ds. 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: /rg

145 Problems and Exercises 257 W = S A = S A 1 2 ρv2 (n v) ds+ S B 1 2 ρv2 A ( v A ) ds+ S B 1 2 ρv2 (n v) ds = 1 2 ρv2 B (v B ) ds = 1 2 ρv3 A S A ρv3 B S B Then, by means of [3], the final result is obtained. W = 1 ( 1 2 ρq3 SB 2 1 ) SA 2 Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi: /rg

146 258 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. Determine: Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar a) The exit velocity v 2 in terms of the entrance velocity v 1 (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: /rg