θ = θ (s,υ). (1.4) θ (s,υ) υ υ

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Abigail Higgins
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1 1.1 Thermodynamics of Fluids 1 Specific Heats of Fluids Elastic Solids Using the free energy per unit mass ψ as a thermodynamic potential we have the caloric equation of state ψ = ψ ˆ (θυ) (1.1.1) the thermal equations of state s = ŝ(θυ) = ψ ˆ (θυ) p = ˆp(θυ) = (1.1.2) ψ ˆ (θυ) where θ is the absolute temperature s is the entropy per unit mass υ is the specific volume p is the pressure. Other combinations of thermodynamic state variables may be used. The internal energy per unit mass is u =ψ + sθ. (1.3) Solving the relation s = ŝ(θυ) for θ we have Therefore u can be expressed by θ = θ (sυ). (1.4) u(sυ) u = ˆ ψ ( θ (sυ)υ) + s θ (sυ) = u(sυ) (1.1.5) u(sυ) = ψ ˆ (θυ) = s θ (sυ) θ (sυ) = ψ ˆ (θυ) υ + s θ (sυ) = p s θ (sυ) + s θ (sυ) + ψ ˆ (θυ) + θ (sυ) + θ (sυ) = θ (sυ) θ (sυ) + s θ (sυ) + s θ (sυ) = p(sυ) (1.1.6) (1.7) where we have used (1.1.2). That is corresponding to the caloric equation of state we have thermal equations of state u = u(sυ) (1.1.8) 1 For the derivation of the general equations see the book Continuum Mechanics by Ellis H. Dill CRC Press 2007: Section for fluids Section 2.1 for elastic solids. 1

2 θ = θ (sυ) = u(sυ) p = p(sυ) = u(sυ). (1.1.9) The enthalpy per unit mass is h = u + pυ. (1.10) Solving the equation of state p = p(sυ) forυ : The enthalpy can therefore be expressed as υ = υ(s p). (1.11) h(s p) h(s p) h = u(sυ(s p)) + pυ(s p) = h(s p) (1.1.12) = u(sυ) + u(sυ) (s p) s s (s p) (s p) = θ p + p s s = u(sυ) (s p) (s p) + p s = θ (s p) (s p) + p +υ(s p) (s p) (s p) = p + p +υ(s p) = υ(s p). (1.1.13) (1.14) That is corresponding to the caloric equation of state we have thermal equations of state h = h(s p) (1.15) h(s p) θ = θ (s p) = h(s p) υ = υ(s p) =. (1.16) The free enthalpy per unit mass (Gibbs function) is g = h sθ (1.17) Solving the equation of state θ = θ (s p) for s : s = s(θ p). (1.18) Therefore the free enthalpy can be expressed as 2

3 g(θ p) g = h( s(θ p) p) θ s(θ p) = g(θ p) (1.1.19) = g(θ p) h(s p) s s(θ p) θ s(θ p) s(θ p) = θ s(θ p) θ s(θ p) s(θ p) = s(θ p) = h(s p) + h(s p) s s(θ p) θ s(θ p) = υ +θ s(θ p) θ s(θ p) = υ(s p) (1.1.20) (1.21) That is corresponding to the caloric equation of state we have thermal equations of state g = g(θ p) (1.22) s = s(θ p) = g(θ p) υ = υ(s p) = g(θ p). (1.23) Relations between the various equations of state are expressed by Maxwell s reciprocal relations: = 2 ψ θ = υ = 2 u s = υ = + 2 h s = p (1.1.24) The specific heat at constant volume is The specific heat at constant pressure is = 2 g θ =. p c υ = u υ = u υ υ = θ υ. (1.1.25) 3

4 c p = h p = h s p s p = θ p. (1.1.26) The relation between the specific heats is found by using (1.1.12): c p = u υ + u θ p + p p = c υ + u + u s + p s s υ θ p = c υ + p +θ s + p θ p = c υ +θ υ p where we have used the Maxwell reciprocal relation s θ = υ. (1.27) The (volumetric) coefficient of thermal expansion is defined by α = 1. (1.28) υ p The coefficient of thermal stress at constant volume is Therefore the specific heats are related by β = υ. (1.29) c p c υ = θυαβ. (1.30) 4

5 1.2 Thermoelasticity The free energy per unit mass is a function of absolute temperature the strain tensor E: ψ = ˆ ψ (θe). (2.1) The Kirchhoff stress tensor S the entropy per unit mass are determined by s = ψ ˆ (θe) ψ ˆ (θe) S = ρ 0. (2.2) Let us introduce measures per unit reference volume: Then Ψ = ρ 0 ψ η = ρ 0 s U = ρ 0 u H = ρ 0 h G = ρ 0 g. (2.3) η = ˆΨ(θE) S = + ˆΨ(θE) = ˆη(θE) = Ŝ(θE). (2.4) Solve the thermal equation of state η = ˆη(θE) for θ : The internal energy per unit volume is then given by θ = θ (ηe) (2.5) U (ηe) U = Ψ + ηθ = ˆΨ( θ (ηe)e) + η θ (ηe) = U (ηe) (2.6) U (ηe) = ˆΨ(θE) = η θ (ηe) = ˆΨ( θ (ηe)e) = S η θ (ηe) θ (ηe) + η θ (ηe) + η θ (ηe) + ˆΨ( θ (ηe)e) + η θ (ηe) + η + θ (ηe) = θ (ηe). = S(ηE). θ (ηe) + η θ (ηe) (2.7) (2.8) 5

6 That is θ = U (ηe) S = U (ηe) = θ (ηe) = S(ηE). Solve the equation of state S = S(ηE) for E to obtain E = E(ηS) (2.9) The enthalpy per unit volume is then given by That is H = U S :E = U (ηe(ηs)) S :E(ηS) = H(ηS) (2.10) H(ηS) H(ηS) = U (ηe) = θ + S : (ηs) = U (ηe) = S : (ηs) + U (ηe) : (ηs) θ = H(ηS) (ηs) S : (ηs) S : (ηs) E = H(ηS) S : (ηs) = θ (ηs) = E(ηS). S : (ηs) = θ (ηs) E(ηS) E(ηS) = E(ηS). (2.11) (2.12) (2.13) Invert the equation θ = θ (ηs) to obtain η = η(θs) (2.14) The free enthalpy per unit volume is then given by G = H ηθ = H( η(θs)s) θ η(θs) = G(θS) (2.15) G(θS) = H( η(θs)s) (θs) θ η(θs) η(θs) = θ η(θs) θ η(θs) η(θs) = η(θs) (2.16) 6

7 That is G(θS) = H(ηS) + H(ηS) (θs) θ η(θs) = E +θ η(θs) θ η(θs) = E(θS). η = G(θS) E = G(θS) = η(θs) = E(θS). (2.17) (2.18) The Maxwell reciprocal relations for elasticity are = ˆΨ(θE) = ˆΨ(θE) = θ E = U (ηe) = U (ηe) = η E = H(ηS) = H(ηS) = η S = G(θS) = G(θS) =. θ S (2.19) The specific heat at constant strain is The specific heat at constant stress is C E = U = U (ηe) E ˆη(θE) = θ E. (2.20) C S = H = H(ηS) (θs) = θ. (2.21) S S Alternative constitutive formulas for the internal energy are U = U (ηe) = U ( ˆη(θE)E) = U ˆ (θe) = U ˆ (θ E(θS)) = U (θs) = U (θ (ηs)s) = U (ηs). (2.22) From (2.10) 7

8 Using a Maxwell reciprocal relation C S = U (θs) S : E(θS) = U ˆ (θe) + U ˆ (θe) : E(θS) S : E(θS) = C E + U (ηe) + U (ηe) ˆη(θE) S : E(θS) = C E + S +θ ˆη(θE) S : E(θS) = C E +θ ˆη(θE) : E(θS). C S C E = θ Ŝ(θE) : E(θS) (2.23) (2.24) 1.3 Small Deformations 2 Let θ 0 denote the reference temperature T = θ θ 0 denote the temperature change. We consider the case when T << θ 0 E = ε << 1. The constitutive relation for the free energy is Ψ = ˆΨ(T E). (3.1) Exp the constitutive function ˆΨ as a power series in the strain retain quadratic terms: Ψ = ˆΨ 0 (θ) ˆβ(θ) :E + 1 E :Ĉ(θ) :E (3.2) 2 where From (2.4) From (2.20) ˆΨ 0 (T ) = ˆΨ(T 0) ˆβ(T ) = ˆΨ(T E) E=0 Ĉ(T ) = 2 ˆΨ(T E) 2. E=0 S = Ŝ(T E) = ˆβ(T ) + Ĉ(T ) : E η = ˆη(T E) = ˆΨ 0 (T ) T (3.3) + ˆβ(T ) T :E 1 2 E : Ĉ(T ) T :E. (3.4) 2 Sections of my Continuum Mechanics. 8

9 For isotropic materials ( ) 2 ˆΨ0 (T ) C E = T +θ 0 T 2 ˆβ(T ) = ˆβ(T )1 + 2 ˆβ(T ) T 2 :E 1 2 E : 2 Ĉ(T ) T 2 :E Ĉ(T ) = 2 ˆµ(T )I + ˆλ(T )11. (3.5) Consider the special case when the dependence of the moduli µ λ on temperature is neglected ˆβ(T ) is a linear function: Then ˆβ(T ) = 3καT (3.6) κ = λ µ. (3.7) S = 3καT1 + λ ( tre)1 + 2µE (3.8) η = ˆΨ 0 (T ) T + 3κα ( tre). (3.9) The inverse of (3.8) is λ E = αt1 2µ + 3λ tr S 2µ 1 + S 2µ (3.10) so that α is the coefficient of thermal expansion. From (3.5) C E = θ 0 2 ˆΨ0 (T ) T 2 (3.11) for T << θ 0. From (2.24) C S = C E + 9κα 2 θ 0. (3.12) where θ is the absolute temperature κ is the bulk modulus α is the coefficient of lineal expansion. 9

In this section, thermoelasticity is considered. By definition, the constitutive relations for Gradθ. This general case

Section.. Thermoelasticity In this section, thermoelasticity is considered. By definition, the constitutive relations for F, θ, Gradθ. This general case such a material depend only on the set of field