Continuum mechanics/Balance of energy for thermoelasticity

Balance of energy for thermoelastic materials Show that, for thermoelastic materials, the balance of energy $\rho \dot{e}-\mathit{\sigma }:\mathrm{\nabla }𝐯+\mathrm{\nabla }\cdot 𝐪-\rho s=0.$ can be expressed as $\rho T\dot{\eta }=-\mathrm{\nabla }\cdot 𝐪+\rho s.$

Proof:

Since $e=e(𝑭,T)$, we have

$\dot{e}=\frac{\partial e}{\partial 𝑭}:\dot{𝑭}+\frac{\partial e}{\partial \eta }\dot{\eta }.$

Plug into energy equation to get

$\rho \frac{\partial e}{\partial 𝑭}:\dot{𝑭}+\rho \frac{\partial e}{\partial \eta }\dot{\eta }-\mathit{\sigma }:\mathrm{\nabla }𝐯+\mathrm{\nabla }\cdot 𝐪-\rho s=0.$

Recall,

$\frac{\partial e}{\partial \eta }=T\text{and}\rho \frac{\partial e}{\partial 𝑭}=\mathit{\sigma }\cdot {𝑭}^{-T}.$

Hence,

$(\mathit{\sigma }\cdot {𝑭}^{-T}):\dot{𝑭}+\rho T\dot{\eta }-\mathit{\sigma }:\mathrm{\nabla }𝐯+\mathrm{\nabla }\cdot 𝐪-\rho s=0.$

Now, $\mathrm{\nabla }𝐯=𝒍=\dot{𝑭}\cdot {𝑭}^{-1}$. Therefore, using the identity $𝑨:(𝑩\cdot 𝑪)=(𝑨\cdot {𝑪}^T):𝑩$, we have

$\mathit{\sigma }:\mathrm{\nabla }𝐯=\mathit{\sigma }:(\dot{𝑭}\cdot {𝑭}^{-1})=(\mathit{\sigma }\cdot {𝑭}^{-T}):\dot{𝑭}.$

Plugging into the energy equation, we have

$\mathit{\sigma }:\mathrm{\nabla }𝐯+\rho T\dot{\eta }-\mathit{\sigma }:\mathrm{\nabla }𝐯+\mathrm{\nabla }\cdot 𝐪-\rho s=0$

or,

$\rho T\dot{\eta }=-\mathrm{\nabla }\cdot 𝐪+\rho s.$

Rate of internal energy/entropy for thermoelastic materials For thermoelastic materials, the specific internal energy is given by $e=\overline{e}(𝑬,\eta )$ where $𝑬$ is the Green strain and $\eta$ is the specific entropy. Show that $\frac{d}{dt}(e-T\eta )=-\dot{T}\eta +\frac{1}{{\rho }_0}𝑺:\dot{𝑬}\text{and}\frac{d}{dt}(e-T\eta -\frac{1}{{\rho }_0}𝑺:𝑬)=-\dot{T}\eta -\frac{1}{{\rho }_0}\dot{𝑺}:𝑬$ where ${\rho }_0$ is the initial density, $T$ is the absolute temperature, $𝑺$ is the 2nd Piola-Kirchhoff stress, and a dot over a quantity indicates the material time derivative.

Taking the material time derivative of the specific internal energy, we get

$\dot{e}=\frac{\partial \overline{e}}{\partial 𝑬}:\dot{𝑬}+\frac{\partial \overline{e}}{\partial \eta }\dot{\eta }.$

Now, for thermoelastic materials,

$T=\frac{\partial \overline{e}}{\partial \eta }\text{and}𝑺={\rho }_0\frac{\partial \overline{e}}{\partial 𝑬}.$

Therefore,

$\dot{e}=\frac{1}{{\rho }_0}𝑺:\dot{𝑬}+T\dot{\eta }.⟹\dot{e}-T\dot{\eta }=\frac{1}{{\rho }_0}𝑺:\dot{𝑬}.$

Now,

$\frac{d}{dt}(T\eta )=\dot{T}\eta +T\dot{\eta }.$

Therefore,

$\dot{e}-\frac{d}{dt}(T\eta )+\dot{T}\eta =\frac{1}{{\rho }_0}𝑺:\dot{𝑬}⟹\frac{d}{dt}(e-T\eta )=-\dot{T}\eta +\frac{1}{{\rho }_0}𝑺:\dot{𝑬}.$

Also,

$\frac{d}{dt}(\frac{1}{{\rho }_0}𝑺:𝑬)=\frac{1}{{\rho }_0}𝑺:\dot{𝑬}+\frac{1}{{\rho }_0}\dot{𝑺}:𝑬.$

Hence,

$\dot{e}-\frac{d}{dt}(T\eta )+\dot{T}\eta =\frac{d}{dt}(\frac{1}{{\rho }_0}𝑺:𝑬)-\frac{1}{{\rho }_0}\dot{𝑺}:𝑬⟹\frac{d}{dt}(e-T\eta -\frac{1}{{\rho }_0}𝑺:𝑬)=-\dot{T}\eta -\frac{1}{{\rho }_0}\dot{𝑺}:𝑬.$

Energy equation for thermoelastic materials For thermoelastic materials, show that the balance of energy equation $\rho T\dot{\eta }=-\mathrm{\nabla }\cdot 𝐪+\rho s$ can be expressed as either $\rho C_v\dot{T}=\mathrm{\nabla }\cdot (\mathit{\kappa }\cdot \mathrm{\nabla }𝑻\mathit{)}+\rho s+\frac{\rho }{{\rho }_0}T\frac{\partial \widehat{𝑺}}{\partial T}:\dot{𝑬}$ or $\rho (C_p-\frac{1}{{\rho }_0}𝑺:\frac{\partial \widetilde{𝑬}}{\partial T})\dot{T}=\mathrm{\nabla }\cdot (\mathit{\kappa }\cdot \mathrm{\nabla }𝑻\mathit{)}+\rho s-\frac{\rho }{{\rho }_0}T\frac{\partial \widetilde{𝑬}}{\partial T}:\dot{𝑺}$ where $C_v=\frac{\partial \hat{e}(𝑬,T)}{\partial T}\text{and}{C}_p=\frac{\partial \tilde{e}(𝑺,T)}{\partial T}.$ For the special case where there are no sources and we can ignore heat conduction (for very fast processes), the energy equation simplifies to $\rho (C_p-\frac{1}{{\rho }_0}𝑺:\mathit{\alpha })\dot{T}=-\frac{\rho }{{\rho }_0}T\mathit{\alpha }:\dot{𝑺}$ where $\mathit{\alpha }:=\frac{\partial \widetilde{𝑬}}{\partial T}$ is the thermal expansion tensor which has the form $\mathit{\alpha }=\alpha 1$ for isotropic materials and $\alpha$ is the coefficient of thermal expansion. The above equation can be used to calculate the change of temperature in thermoelasticity.

Proof:

If the independent variables are $𝑬$ and $T$, then

$\eta =\hat{\eta }(𝑬,T)⟹\dot{\eta }=\frac{\partial \hat{\eta }}{\partial 𝑬}:\dot{𝑬}+\frac{\partial \hat{\eta }}{\partial T}\dot{T}.$

On the other hand, if we consider $𝑺$ and $T$ to be the independent variables

$\eta =\tilde{\eta }(𝑺,T)⟹\dot{\eta }=\frac{\partial \tilde{\eta }}{\partial 𝑺}:\dot{𝑺}+\frac{\partial \tilde{\eta }}{\partial T}\dot{T}.$

Since

$\frac{\partial \hat{\eta }}{\partial 𝑬}=-\frac{1}{{\rho }_0}\frac{\partial \widehat{𝑺}}{\partial T};\frac{\partial \hat{\eta }}{\partial T}=\frac{C_v}{T};\frac{\partial \tilde{\eta }}{\partial 𝑺}=\frac{1}{{\rho }_0}\frac{\partial \widetilde{𝑬}}{\partial T};\text{and}\frac{\partial \tilde{\eta }}{\partial T}=\frac{1}{T}(C_p-\frac{1}{{\rho }_0}𝑺:\frac{\partial \widetilde{𝑬}}{\partial T})$

we have, either

$\dot{\eta }=-\frac{1}{{\rho }_0}\frac{\partial \widehat{𝑺}}{\partial T}:\dot{𝑬}+\frac{C_v}{T}\dot{T}$

or

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

The equation for balance of energy in terms of the specific entropy is

$\rho T\dot{\eta }=-\mathrm{\nabla }\cdot 𝐪+\rho s.$

Using the two forms of $\dot{\eta }$, we get two forms of the energy equation:

$-\frac{\rho }{{\rho }_0}T\frac{\partial \widehat{𝑺}}{\partial T}:\dot{𝑬}+\rho C_v\dot{T}=-\mathrm{\nabla }\cdot 𝐪+\rho s$

and

$\frac{\rho }{{\rho }_0}T\frac{\partial \widetilde{𝑬}}{\partial T}:\dot{𝑺}+\rho C_p\dot{T}-\frac{\rho }{{\rho }_0}𝑺:\frac{\partial \widetilde{𝑬}}{\partial T}\dot{T}=-\mathrm{\nabla }\cdot 𝐪+\rho s.$

From Fourier's law of heat conduction

$𝐪=-\mathit{\kappa }\cdot \mathrm{\nabla }T.$

Therefore,

$-\frac{\rho }{{\rho }_0}T\frac{\partial \widehat{𝑺}}{\partial T}:\dot{𝑬}+\rho C_v\dot{T}=\mathrm{\nabla }\cdot (\mathit{\kappa }\cdot \mathrm{\nabla }𝑻\mathit{)}+\rho s$

and

$\frac{\rho }{{\rho }_0}T\frac{\partial \widetilde{𝑬}}{\partial T}:\dot{𝑺}+\rho C_p\dot{T}-\frac{\rho }{{\rho }_0}𝑺:\frac{\partial \widetilde{𝑬}}{\partial T}\dot{T}=\mathrm{\nabla }\cdot (\mathit{\kappa }\cdot \mathrm{\nabla }𝑻\mathit{)}+\rho s.$

Rearranging,

$\rho C_v\dot{T}=\mathrm{\nabla }\cdot (\mathit{\kappa }\cdot \mathrm{\nabla }𝑻\mathit{)}+\rho s+\frac{\rho }{{\rho }_0}T\frac{\partial \widehat{𝑺}}{\partial T}:\dot{𝑬}$

or,

$\rho (C_p-\frac{1}{{\rho }_0}𝑺:\frac{\partial \widetilde{𝑬}}{\partial T})\dot{T}=\mathrm{\nabla }\cdot (\mathit{\kappa }\cdot \mathrm{\nabla }𝑻\mathit{)}+\rho s-\frac{\rho }{{\rho }_0}T\frac{\partial \widetilde{𝑬}}{\partial T}:\dot{𝑺}.$