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 $C_p-C_v=\frac{1}{{\rho }_0}(𝑺-T\frac{\partial \widehat{𝑺}}{\partial T}):\frac{\partial \widetilde{𝑬}}{\partial T}.$

Proof:

Recall that

$C_v:=\frac{\partial \hat{e}(𝑬,T)}{\partial T}=T\frac{\partial \hat{\eta }}{\partial T}$

and

$C_p:=\frac{\partial \tilde{e}(𝑺,T)}{\partial T}=T\frac{\partial \tilde{\eta }}{\partial T}+\frac{1}{{\rho }_0}𝑺:\frac{\partial \widetilde{𝑬}}{\partial T}.$

Therefore,

$C_p-C_v=T\frac{\partial \tilde{\eta }}{\partial T}+\frac{1}{{\rho }_0}𝑺:\frac{\partial \widetilde{𝑬}}{\partial T}-T\frac{\partial \hat{\eta }}{\partial T}.$

Also recall that

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

Therefore, keeping $𝑺$ constant while differentiating, we have

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

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

$C_p-C_v=T\frac{\partial \hat{\eta }}{\partial 𝑬}:\frac{\partial \widetilde{𝑬}}{\partial T}+\frac{1}{{\rho }_0}𝑺:\frac{\partial \widetilde{𝑬}}{\partial T}.$

Recalling that

$\frac{\partial \hat{\eta }}{\partial 𝑬}=-\frac{1}{{\rho }_0}\frac{\partial \widehat{𝑺}}{\partial T}$

we get

$C_p-C_v=\frac{1}{{\rho }_0}(𝑺-T\frac{\partial \widehat{𝑺}}{\partial T}):\frac{\partial \widetilde{𝑬}}{\partial T}.$

Relation between specific heats - 2 For thermoelastic materials, show that the specific heats can also be related by the equations $C_p-C_v=\frac{1}{{\rho }_0}𝑺:\frac{\partial 𝑬}{\partial T}+\frac{\partial 𝑬}{\partial T}:(\frac{{\partial }^2\psi }{\partial 𝑬\partial 𝑬}:\frac{\partial 𝑬}{\partial T})=\frac{1}{{\rho }_0}𝑺:\frac{\partial 𝑬}{\partial T}+\frac{T}{{\rho }_0}\frac{\partial 𝑬}{\partial T}:(\frac{\partial 𝑺}{\partial 𝑬}:\frac{\partial 𝑬}{\partial T}).$ We can also write the above as $C_p-C_v=\frac{1}{{\rho }_0}𝑺:{\mathit{\alpha }}_E+\frac{T}{{\rho }_0}{\mathit{\alpha }}_E:𝖢:{\mathit{\alpha }}_E$ where ${\mathit{\alpha }}_E:=\frac{\partial 𝑬}{\partial T}$ is the thermal expansion tensor and $𝖢:=\frac{\partial 𝑺}{\partial 𝑬}$ is the stiffness tensor.

Proof:

Recall that

$𝑺={\rho }_0\frac{\partial \psi }{\partial 𝑬}={\rho }_0𝒇(𝑬(𝑺,T),T).$

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,

$u=𝑺,t=T,g(𝑺,T)=𝑺,x(𝑺,T)=𝑬(𝑺,T),y(𝑺,T)=T,\text{and}f={\rho }_0𝒇.$

Hence, we have

$𝑺=g(𝑺,T)=f(𝑬(𝑺,T),T)={\rho }_0𝒇(𝑬(𝑺,T),T).$

Taking the derivative with respect to $T$ keeping $𝑺$ constant, we have

$\frac{\partial g}{\partial T}=\frac{\partial 𝑺}{\partial T}={\rho }_0[\frac{\partial 𝒇}{\partial 𝑬}:\frac{\partial 𝑬}{\partial T}+\frac{\partial 𝒇}{\partial T}\frac{\partial T}{\partial T}]$

or,

$\mathrm{𝟎}=\frac{\partial 𝒇}{\partial 𝑬}:\frac{\partial 𝑬}{\partial T}+\frac{\partial 𝒇}{\partial T}.$

Now,

$𝒇=\frac{\partial \psi }{\partial 𝑬}⟹\frac{\partial 𝒇}{\partial 𝑬}=\frac{{\partial }^2\psi }{\partial 𝑬\partial 𝑬}\text{and}\frac{\partial 𝒇}{\partial T}=\frac{{\partial }^2\psi }{\partial T\partial 𝑬}.$

Therefore,

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

Again recall that,

$\frac{\partial \psi }{\partial 𝑬}=\frac{1}{{\rho }_0}𝑺.$

Plugging into the above, we get

$\mathrm{𝟎}=\frac{{\partial }^2\psi }{\partial 𝑬\partial 𝑬}:\frac{\partial 𝑬}{\partial T}+\frac{1}{{\rho }_0}\frac{\partial 𝑺}{\partial T}=\frac{1}{{\rho }_0}\frac{\partial 𝑺}{\partial 𝑬}:\frac{\partial 𝑬}{\partial T}+\frac{1}{{\rho }_0}\frac{\partial 𝑺}{\partial T}.$

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

$\frac{\partial 𝑺}{\partial T}=-{\rho }_0\frac{{\partial }^2\psi }{\partial 𝑬\partial 𝑬}:\frac{\partial 𝑬}{\partial T}=-\frac{\partial 𝑺}{\partial 𝑬}:\frac{\partial 𝑬}{\partial T}.$

Recall that

$C_p-C_v=\frac{1}{{\rho }_0}(𝑺-T\frac{\partial 𝑺}{\partial T}):\frac{\partial 𝑬}{\partial T}.$

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

$C_p-C_v=\frac{1}{{\rho }_0}(𝑺+T{\rho }_0\frac{{\partial }^2\psi }{\partial 𝑬\partial 𝑬}:\frac{\partial 𝑬}{\partial T}):\frac{\partial 𝑬}{\partial T}=\frac{1}{{\rho }_0}(𝑺+T\frac{\partial 𝑺}{\partial 𝑬}:\frac{\partial 𝑬}{\partial T}):\frac{\partial 𝑬}{\partial T}.$

Therefore,

$C_p-C_v=\frac{1}{{\rho }_0}𝑺:\frac{\partial 𝑬}{\partial T}+T(\frac{{\partial }^2\psi }{\partial 𝑬\partial 𝑬}:\frac{\partial 𝑬}{\partial T}):\frac{\partial 𝑬}{\partial T}=\frac{1}{{\rho }_0}𝑺:\frac{\partial 𝑬}{\partial T}+\frac{T}{{\rho }_0}(\frac{\partial 𝑺}{\partial 𝑬}:\frac{\partial 𝑬}{\partial T}):\frac{\partial 𝑬}{\partial T}.$

Using the identity $(𝖠:𝑩):𝑪=𝑪:(𝖠:𝑩)$, we have

$C_p-C_v=\frac{1}{{\rho }_0}𝑺:\frac{\partial 𝑬}{\partial T}+T\frac{\partial 𝑬}{\partial T}:(\frac{{\partial }^2\psi }{\partial 𝑬\partial 𝑬}:\frac{\partial 𝑬}{\partial T})=\frac{1}{{\rho }_0}𝑺:\frac{\partial 𝑬}{\partial T}+\frac{T}{{\rho }_0}\frac{\partial 𝑬}{\partial T}:(\frac{\partial 𝑺}{\partial 𝑬}:\frac{\partial 𝑬}{\partial T}).$

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, ${\mathit{\alpha }}_E=\alpha 1\text{and}𝖢=\lambda 1\otimes 1+2\mu 𝖨$ where $\alpha$ is the coefficient of thermal expansion and $3\lambda =3K-2\mu$ where $K,\mu$ are the bulk and shear moduli, respectively. Show that the specific heats related by the equation $C_p-C_v=\frac{1}{{\rho }_0}[\alpha \text{tr}𝑺+9{\alpha }^{2}KT].$

Proof:

Recall that,

$C_p-C_v=\frac{1}{{\rho }_0}𝑺:{\mathit{\alpha }}_E+\frac{T}{{\rho }_0}{\mathit{\alpha }}_E:𝖢:{\mathit{\alpha }}_E.$

Plugging the expressions of ${\mathit{\alpha }}_E$ and $𝖢$ into the above equation, we have

$\begin{array}{cc}{C}_p-C_v& =\frac{1}{{\rho }_0}𝑺:(\alpha 1)+\frac{T}{{\rho }_0}(\alpha 1):(\lambda 1\otimes 1+2\mu 𝖨):(\alpha 1)\\ & =\frac{\alpha }{{\rho }_0}\text{tr}𝑺+\frac{{\alpha }^{2}T}{{\rho }_0}1:(\lambda 1\otimes 1+2\mu 𝖨):1\\ & =\frac{\alpha }{{\rho }_0}\text{tr}𝑺+\frac{{\alpha }^{2}T}{{\rho }_0}1:(\lambda \text{tr}11+2\mu 1)\\ & =\frac{\alpha }{{\rho }_0}\text{tr}𝑺+\frac{{\alpha }^{2}T}{{\rho }_0}(3\lambda \text{tr}1+2\mu \text{tr}1)\\ & =\frac{\alpha }{{\rho }_0}\text{tr}𝑺+\frac{3{\alpha }^{2}T}{{\rho }_0}(3\lambda +2\mu )\\ & =\frac{\alpha \text{tr}𝑺}{{\rho }_0}+\frac{9{\alpha }^{2}KT}{{\rho }_0}.\end{array}$

Therefore,

$C_p-C_v=\frac{1}{{\rho }_0}[\alpha \text{tr}𝑺+9{\alpha }^{2}KT].$