Continuum mechanics/Specific heats of thermoelastic materials

Relation between specific heats - 1 For thermoelastic materials, show that the specific heats are related by the relation $C_{p}-C_{v}={\cfrac {1}{\rho _{0}}}\left({\boldsymbol {S}}-T~{\frac {\partial {\hat {\boldsymbol {S}}}}{\partial T}}\right):{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}~.$

Proof:

Recall that

$C_{v}:={\frac {\partial {\hat {e}}({\boldsymbol {E}},T)}{\partial T}}=T~{\frac {\partial {\hat {\eta }}}{\partial T}}$

and

$C_{p}:={\frac {\partial {\tilde {e}}({\boldsymbol {S}},T)}{\partial T}}=T~{\frac {\partial {\tilde {\eta }}}{\partial T}}+{\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}~.$

Therefore,

$C_{p}-C_{v}=T~{\frac {\partial {\tilde {\eta }}}{\partial T}}+{\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}-T~{\frac {\partial {\hat {\eta }}}{\partial T}}~.$

Also recall that

$\eta ={\hat {\eta }}({\boldsymbol {E}},T)={\tilde {\eta }}({\boldsymbol {S}},T)~.$

Therefore, keeping ${\boldsymbol {S}}$ constant while differentiating, we have

${\frac {\partial {\tilde {\eta }}}{\partial T}}={\frac {\partial {\hat {\eta }}}{\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+{\frac {\partial {\hat {\eta }}}{\partial T}}~.$

Noting that ${\boldsymbol {E}}={\tilde {\boldsymbol {E}}}({\boldsymbol {S}},T)$, and plugging back into the equation for the difference between the two specific heats, we have

$C_{p}-C_{v}=T~{\frac {\partial {\hat {\eta }}}{\partial {\boldsymbol {E}}}}:{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}+{\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}~.$

Recalling that

${\frac {\partial {\hat {\eta }}}{\partial {\boldsymbol {E}}}}=-{\cfrac {1}{\rho _{0}}}~{\frac {\partial {\hat {\boldsymbol {S}}}}{\partial T}}$

we get

${C_{p}-C_{v}={\cfrac {1}{\rho _{0}}}\left({\boldsymbol {S}}-T~{\frac {\partial {\hat {\boldsymbol {S}}}}{\partial T}}\right):{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}~.}$

Relation between specific heats - 2 For thermoelastic materials, show that the specific heats can also be related by the equations $C_{p}-C_{v}={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+{\frac {\partial {\boldsymbol {E}}}{\partial T}}:\left({\frac {\partial ^{2}\psi }{\partial {\boldsymbol {E}}\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}\right)={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+{\cfrac {T}{\rho _{0}}}~{\frac {\partial {\boldsymbol {E}}}{\partial T}}:\left({\frac {\partial {\boldsymbol {S}}}{\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}\right)~.$ We can also write the above as $C_{p}-C_{v}={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\boldsymbol {\alpha }}_{E}+{\cfrac {T}{\rho _{0}}}~{\boldsymbol {\alpha }}_{E}:{\boldsymbol {\mathsf {C}}}:{\boldsymbol {\alpha }}_{E}$ where ${\boldsymbol {\alpha }}_{E}:={\frac {\partial {\boldsymbol {E}}}{\partial T}}$ is the thermal expansion tensor and ${\boldsymbol {\mathsf {C}}}:={\frac {\partial {\boldsymbol {S}}}{\partial {\boldsymbol {E}}}}$ is the stiffness tensor.

Proof:

Recall that

${\boldsymbol {S}}=\rho _{0}~{\frac {\partial \psi }{\partial {\boldsymbol {E}}}}=\rho _{0}~{\boldsymbol {f}}({\boldsymbol {E}}({\boldsymbol {S}},T),T)~.$

Recall the chain rule which states that if

$g(u,t)=f(x(u,t),y(u,t))$

then, if we keep $u$ fixed, the partial derivative of $g$ with respect to $t$ is given by

${\frac {\partial g}{\partial t}}={\frac {\partial f}{\partial x}}~{\frac {\partial x}{\partial t}}+{\frac {\partial f}{\partial y}}~{\frac {\partial y}{\partial t}}~.$

In our case,

$u={\boldsymbol {S}},~~t=T,~~g({\boldsymbol {S}},T)={\boldsymbol {S}},~~x({\boldsymbol {S}},T)={\boldsymbol {E}}({\boldsymbol {S}},T),~~y({\boldsymbol {S}},T)=T,~~{\text{and}}~~f=\rho _{0}~{\boldsymbol {f}}.$

Hence, we have

${\boldsymbol {S}}=g({\boldsymbol {S}},T)=f({\boldsymbol {E}}({\boldsymbol {S}},T),T)=\rho _{0}~{\boldsymbol {f}}({\boldsymbol {E}}({\boldsymbol {S}},T),T)~.$

Taking the derivative with respect to $T$ keeping ${\boldsymbol {S}}$ constant, we have

${\frac {\partial g}{\partial T}}={\frac {\partial {\boldsymbol {S}}}{\partial T}}=\rho _{0}~\left[{\frac {\partial {\boldsymbol {f}}}{\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+{\frac {\partial {\boldsymbol {f}}}{\partial T}}~{\frac {\partial T}{\partial T}}\right]$

or,

$\mathbf {0} ={\frac {\partial {\boldsymbol {f}}}{\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+{\frac {\partial {\boldsymbol {f}}}{\partial T}}~.$

Now,

${\boldsymbol {f}}={\frac {\partial \psi }{\partial {\boldsymbol {E}}}}\qquad \implies \qquad {\frac {\partial {\boldsymbol {f}}}{\partial {\boldsymbol {E}}}}={\frac {\partial ^{2}\psi }{\partial {\boldsymbol {E}}\partial {\boldsymbol {E}}}}\quad {\text{and}}\quad {\frac {\partial {\boldsymbol {f}}}{\partial T}}={\frac {\partial ^{2}\psi }{\partial T\partial {\boldsymbol {E}}}}~.$

Therefore,

$\mathbf {0} ={\frac {\partial ^{2}\psi }{\partial {\boldsymbol {E}}\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+{\frac {\partial ^{2}\psi }{\partial T\partial {\boldsymbol {E}}}}={\frac {\partial }{\partial {\boldsymbol {E}}}}\left({\frac {\partial \psi }{\partial {\boldsymbol {E}}}}\right):{\frac {\partial {\boldsymbol {E}}}{\partial T}}+{\frac {\partial }{\partial T}}\left({\frac {\partial \psi }{\partial {\boldsymbol {E}}}}\right)~.$

Again recall that,

${\frac {\partial \psi }{\partial {\boldsymbol {E}}}}={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}~.$

Plugging into the above, we get

$\mathbf {0} ={\frac {\partial ^{2}\psi }{\partial {\boldsymbol {E}}\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+{\cfrac {1}{\rho _{0}}}~{\frac {\partial {\boldsymbol {S}}}{\partial T}}={\cfrac {1}{\rho _{0}}}~{\frac {\partial {\boldsymbol {S}}}{\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+{\cfrac {1}{\rho _{0}}}~{\frac {\partial {\boldsymbol {S}}}{\partial T}}~.$

Therefore, we get the following relation for $\partial {\boldsymbol {S}}/\partial T$:

${\frac {\partial {\boldsymbol {S}}}{\partial T}}=-\rho _{0}~{\frac {\partial ^{2}\psi }{\partial {\boldsymbol {E}}\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}=-{\frac {\partial {\boldsymbol {S}}}{\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}~.$

Recall that

$C_{p}-C_{v}={\cfrac {1}{\rho _{0}}}\left({\boldsymbol {S}}-T~{\frac {\partial {\boldsymbol {S}}}{\partial T}}\right):{\frac {\partial {\boldsymbol {E}}}{\partial T}}~.$

Plugging in the expressions for $\partial {\boldsymbol {S}}/\partial T$ we get:

$C_{p}-C_{v}={\cfrac {1}{\rho _{0}}}\left({\boldsymbol {S}}+T~\rho _{0}~{\frac {\partial ^{2}\psi }{\partial {\boldsymbol {E}}\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}\right):{\frac {\partial {\boldsymbol {E}}}{\partial T}}={\cfrac {1}{\rho _{0}}}\left({\boldsymbol {S}}+T~{\frac {\partial {\boldsymbol {S}}}{\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}\right):{\frac {\partial {\boldsymbol {E}}}{\partial T}}~.$

Therefore,

$C_{p}-C_{v}={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+T~\left({\frac {\partial ^{2}\psi }{\partial {\boldsymbol {E}}\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}\right):{\frac {\partial {\boldsymbol {E}}}{\partial T}}={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+{\cfrac {T}{\rho _{0}}}~\left({\frac {\partial {\boldsymbol {S}}}{\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}\right):{\frac {\partial {\boldsymbol {E}}}{\partial T}}~.$

Using the identity $({\boldsymbol {\mathsf {A}}}:{\boldsymbol {B}}):{\boldsymbol {C}}={\boldsymbol {C}}:({\boldsymbol {\mathsf {A}}}:{\boldsymbol {B}})$, we have

${C_{p}-C_{v}={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+T~{\frac {\partial {\boldsymbol {E}}}{\partial T}}:\left({\frac {\partial ^{2}\psi }{\partial {\boldsymbol {E}}\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}\right)={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+{\cfrac {T}{\rho _{0}}}~{\frac {\partial {\boldsymbol {E}}}{\partial T}}:\left({\frac {\partial {\boldsymbol {S}}}{\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}\right)~.}$

Specific heats of Saint-Venant–Kirchhoff material Consider an isotropic thermoelastic material that has a constant coefficient of thermal expansion and which follows the Saint-Venant–Kirchhoff model, i.e, ${\boldsymbol {\alpha }}_{E}=\alpha ~{\boldsymbol {\mathit {1}}}\qquad {\text{and}}\qquad {\boldsymbol {\mathsf {C}}}=\lambda ~{\boldsymbol {\mathit {1}}}\otimes {\boldsymbol {\mathit {1}}}+2\mu ~{\boldsymbol {\mathsf {I}}}$ where $\alpha$ is the coefficient of thermal expansion and $3~\lambda =3~K-2~\mu$ where $K,\mu$ are the bulk and shear moduli, respectively. Show that the specific heats related by the equation $C_{p}-C_{v}={\cfrac {1}{\rho _{0}}}\left[\alpha ~{\text{tr}}{\boldsymbol {S}}+9~\alpha ^{2}~K~T\right]~.$

Proof:

Recall that,

$C_{p}-C_{v}={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\boldsymbol {\alpha }}_{E}+{\cfrac {T}{\rho _{0}}}~{\boldsymbol {\alpha }}_{E}:{\boldsymbol {\mathsf {C}}}:{\boldsymbol {\alpha }}_{E}~.$

Plugging the expressions of ${\boldsymbol {\alpha }}_{E}$ and ${\boldsymbol {\mathsf {C}}}$ into the above equation, we have

${\begin{aligned}C_{p}-C_{v}&={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:(\alpha ~{\boldsymbol {\mathit {1}}})+{\cfrac {T}{\rho _{0}}}~(\alpha ~{\boldsymbol {\mathit {1}}}):(\lambda ~{\boldsymbol {\mathit {1}}}\otimes {\boldsymbol {\mathit {1}}}+2\mu ~{\boldsymbol {\mathsf {I}}}):(\alpha ~{\boldsymbol {\mathit {1}}})\\&={\cfrac {\alpha }{\rho _{0}}}~{\text{tr}}{\boldsymbol {S}}+{\cfrac {\alpha ^{2}~T}{\rho _{0}}}~{\boldsymbol {\mathit {1}}}:(\lambda ~{\boldsymbol {\mathit {1}}}\otimes {\boldsymbol {\mathit {1}}}+2\mu ~{\boldsymbol {\mathsf {I}}}):{\boldsymbol {\mathit {1}}}\\&={\cfrac {\alpha }{\rho _{0}}}~{\text{tr}}{\boldsymbol {S}}+{\cfrac {\alpha ^{2}~T}{\rho _{0}}}~{\boldsymbol {\mathit {1}}}:(\lambda ~{\text{tr}}{\boldsymbol {\mathit {1}}}~{\boldsymbol {\mathit {1}}}+2\mu ~{\boldsymbol {\mathit {1}}})\\&={\cfrac {\alpha }{\rho _{0}}}~{\text{tr}}{\boldsymbol {S}}+{\cfrac {\alpha ^{2}~T}{\rho _{0}}}~(3~\lambda ~{\text{tr}}{\boldsymbol {\mathit {1}}}+2\mu ~{\text{tr}}{\boldsymbol {\mathit {1}}})\\&={\cfrac {\alpha }{\rho _{0}}}~{\text{tr}}{\boldsymbol {S}}+{\cfrac {3~\alpha ^{2}~T}{\rho _{0}}}~(3~\lambda +2\mu )\\&={\cfrac {\alpha ~{\text{tr}}{\boldsymbol {S}}}{\rho _{0}}}+{\cfrac {9~\alpha ^{2}~K~T}{\rho _{0}}}~.\end{aligned}}$

Therefore,

${C_{p}-C_{v}={\cfrac {1}{\rho _{0}}}\left[\alpha ~{\text{tr}}{\boldsymbol {S}}+9~\alpha ^{2}~K~T\right]~.}$

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