Continuum mechanics/Balance of energy for thermoelasticity

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

Proof:

Since $e = e(\boldsymbol{F}, T)$, we have

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

Plug into energy equation to get

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

Recall,

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

Hence,

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

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

$\boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} = \boldsymbol{\sigma}:(\dot{\boldsymbol{F}}\cdot\boldsymbol{F}^{-1}) = (\boldsymbol{\sigma}\cdot\boldsymbol{F}^{-T}):\dot{\boldsymbol{F}} ~.$

Plugging into the energy equation, we have

$\boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} + \rho~T~\dot{\eta} - \boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} + \boldsymbol{\nabla} \cdot \mathbf{q} - \rho~s = 0$

or,

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

Rate of internal energy/entropy for thermoelastic materials For thermoelastic materials, the specific internal energy is given by $e = \bar{e}(\boldsymbol{E}, \eta)$ where $\boldsymbol{E}$ is the Green strain and $\eta$ is the specific entropy. Show that $\cfrac{d}{dt}(e - T~\eta) = - \dot{T}~\eta + \cfrac{1}{\rho_0}~\boldsymbol{S}:\dot{\boldsymbol{E}} \qquad\text{and}\qquad \cfrac{d}{dt}(e - T~\eta - \cfrac{1}{\rho_0}~\boldsymbol{S}:\boldsymbol{E}) = - \dot{T}~\eta - \cfrac{1}{\rho_0}~\dot{\boldsymbol{S}}:\boldsymbol{E}$ where $\rho_0$ is the initial density, $T$ is the absolute temperature, $\boldsymbol{S}$ 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

$\dot{e} = \frac{\partial \bar{e}}{\partial \boldsymbol{E}}:\dot{\boldsymbol{E}} + \frac{\partial \bar{e}}{\partial \eta}~\dot{\eta} ~.$

Now, for thermoelastic materials,

$T = \frac{\partial \bar{e}}{\partial \eta} \qquad \text{and} \qquad \boldsymbol{S} = \rho_0~\frac{\partial \bar{e}}{\partial \boldsymbol{E}} ~.$

Therefore,

$\dot{e} = \cfrac{1}{\rho_0}~\boldsymbol{S}:\dot{\boldsymbol{E}} + T~\dot{\eta} ~. \qquad \implies \qquad \dot{e} - T~\dot{\eta} = \cfrac{1}{\rho_0}~\boldsymbol{S}:\dot{\boldsymbol{E}} ~.$

Now,

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

Therefore,

$\dot{e} - \cfrac{d}{dt}(T~\eta) + \dot{T}~\eta = \cfrac{1}{\rho_0}~\boldsymbol{S}:\dot{\boldsymbol{E}} \qquad \implies \qquad { \cfrac{d}{dt}(e - T~\eta) = -\dot{T}~\eta + \cfrac{1}{\rho_0}~\boldsymbol{S}:\dot{\boldsymbol{E}} ~. }$

Also,

$\cfrac{d}{dt}\left(\cfrac{1}{\rho_0}~\boldsymbol{S}:\boldsymbol{E}\right) = \cfrac{1}{\rho_0}~\boldsymbol{S}:\dot{\boldsymbol{E}} + \cfrac{1}{\rho_0}~\dot{\boldsymbol{S}}:\boldsymbol{E} ~.$

Hence,

$\dot{e} - \cfrac{d}{dt}(T~\eta) + \dot{T}~\eta = \cfrac{d}{dt}\left(\cfrac{1}{\rho_0} \boldsymbol{S}:\boldsymbol{E}\right) - \cfrac{1}{\rho_0}~\dot{\boldsymbol{S}}:\boldsymbol{E} \qquad \implies \qquad { \cfrac{d}{dt}\left(e - T~\eta - \cfrac{1}{\rho_0}~\boldsymbol{S}:\boldsymbol{E}\right) = - \dot{T}~\eta - \cfrac{1}{\rho_0}~\dot{\boldsymbol{S}}:\boldsymbol{E} ~. }$

Energy equation for thermoelastic materials For thermoelastic materials, show that the balance of energy equation $\rho~T~\dot{\eta} = - \boldsymbol{\nabla} \cdot \mathbf{q} + \rho~s$ can be expressed as either $\rho~C_v~\dot{T} = \boldsymbol{\nabla} \cdot (\boldsymbol{\kappa}\cdot\boldsymbol{\nabla T)} + \rho~s +\cfrac{\rho}{\rho_0}~T~\frac{\partial \hat{\boldsymbol{S}}}{\partial T}:\dot{\boldsymbol{E}}$ or $\rho~\left(C_p - \cfrac{1}{\rho_0}~\boldsymbol{S}:\frac{\partial \tilde{\boldsymbol{E}}}{\partial T}\right) ~\dot{T} = \boldsymbol{\nabla} \cdot (\boldsymbol{\kappa}\cdot\boldsymbol{\nabla T)} + \rho~s -\cfrac{\rho}{\rho_0}~T~\frac{\partial \tilde{\boldsymbol{E}}}{\partial T}:\dot{\boldsymbol{S}}$ where $C_v = \frac{\partial \hat{e}(\boldsymbol{E},T)}{\partial T} \qquad \text{and} \qquad C_p = \frac{\partial \tilde{e}(\boldsymbol{S},T)}{\partial T} ~.$ For the special case where there are no sources and we can ignore heat conduction (for very fast processes), the energy equation simplifies to $\rho~\left(C_p - \cfrac{1}{\rho_0}~\boldsymbol{S}:\boldsymbol{\alpha}\right) ~\dot{T} = -\cfrac{\rho}{\rho_0}~T~\boldsymbol{\alpha}:\dot{\boldsymbol{S}}$ where $\boldsymbol{\alpha} := \frac{\partial \tilde{\boldsymbol{E}}}{\partial T}$ is the thermal expansion tensor which has the form $\boldsymbol{\alpha} = \alpha\boldsymbol{1}$ 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 $\boldsymbol{E}$ and $T$, then

$\eta = \hat{\eta}(\boldsymbol{E}, T) \qquad \implies \qquad \dot{\eta} = \frac{\partial \hat{\eta}}{\partial \boldsymbol{E}}:\dot{\boldsymbol{E}} + \frac{\partial \hat{\eta}}{\partial T}~\dot{T} ~.$

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

$\eta = \tilde{\eta}(\boldsymbol{S}, T) \qquad \implies \qquad \dot{\eta} = \frac{\partial \tilde{\eta}}{\partial \boldsymbol{S}}:\dot{\boldsymbol{S}} + \frac{\partial \tilde{\eta}}{\partial T}~\dot{T} ~.$

Since

$\frac{\partial \hat{\eta}}{\partial \boldsymbol{E}} = -\cfrac{1}{\rho_0}~\frac{\partial \hat{\boldsymbol{S}}}{\partial T} ~;~~ \frac{\partial \hat{\eta}}{\partial T} = \cfrac{C_v}{T} ~;~~ \frac{\partial \tilde{\eta}}{\partial \boldsymbol{S}} = \cfrac{1}{\rho_0}~\frac{\partial \tilde{\boldsymbol{E}}}{\partial T} ~;~~\text{and}~~ \frac{\partial \tilde{\eta}}{\partial T} = \cfrac{1}{T}\left(C_p - \cfrac{1}{\rho_0} \boldsymbol{S}:\frac{\partial \tilde{\boldsymbol{E}}}{\partial T}\right)$

we have, either

$\dot{\eta} = -\cfrac{1}{\rho_0}~\frac{\partial \hat{\boldsymbol{S}}}{\partial T}:\dot{\boldsymbol{E}} + \cfrac{C_v}{T}~\dot{T}$

or

$\dot{\eta} = \cfrac{1}{\rho_0}~\frac{\partial \tilde{\boldsymbol{E}}}{\partial T}:\dot{\boldsymbol{S}} + \cfrac{1}{T}\left(C_p - \cfrac{1}{\rho_0} \boldsymbol{S}:\frac{\partial \tilde{\boldsymbol{E}}}{\partial T}\right)~\dot{T} ~.$

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

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

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

$-\cfrac{\rho}{\rho_0}~T~\frac{\partial \hat{\boldsymbol{S}}}{\partial T}:\dot{\boldsymbol{E}} + \rho~C_v~\dot{T} = - \boldsymbol{\nabla} \cdot \mathbf{q} + \rho~s$

and

$\cfrac{\rho}{\rho_0}~T~\frac{\partial \tilde{\boldsymbol{E}}}{\partial T}:\dot{\boldsymbol{S}} + \rho~C_p~\dot{T} - \cfrac{\rho}{\rho_0}~\boldsymbol{S}:\frac{\partial \tilde{\boldsymbol{E}}}{\partial T}~\dot{T} = - \boldsymbol{\nabla} \cdot \mathbf{q} + \rho~s ~.$

From Fourier's law of heat conduction

$\mathbf{q} = - \boldsymbol{\kappa}\cdot\boldsymbol{\nabla} T ~.$

Therefore,

$-\cfrac{\rho}{\rho_0}~T~\frac{\partial \hat{\boldsymbol{S}}}{\partial T}:\dot{\boldsymbol{E}} + \rho~C_v~\dot{T} = \boldsymbol{\nabla} \cdot (\boldsymbol{\kappa}\cdot\boldsymbol{\nabla T)} + \rho~s$

and

$\cfrac{\rho}{\rho_0}~T~\frac{\partial \tilde{\boldsymbol{E}}}{\partial T}:\dot{\boldsymbol{S}} + \rho~C_p~\dot{T} - \cfrac{\rho}{\rho_0}~\boldsymbol{S}:\frac{\partial \tilde{\boldsymbol{E}}}{\partial T}~\dot{T} = \boldsymbol{\nabla} \cdot (\boldsymbol{\kappa}\cdot\boldsymbol{\nabla T)} + \rho~s ~.$

Rearranging,

${ \rho~C_v~\dot{T} = \boldsymbol{\nabla} \cdot (\boldsymbol{\kappa}\cdot\boldsymbol{\nabla T)} + \rho~s +\cfrac{\rho}{\rho_0}~T~\frac{\partial \hat{\boldsymbol{S}}}{\partial T}:\dot{\boldsymbol{E}} }$

or,

${ \rho~\left(C_p - \cfrac{1}{\rho_0}~\boldsymbol{S}:\frac{\partial \tilde{\boldsymbol{E}}}{\partial T}\right) ~\dot{T} = \boldsymbol{\nabla} \cdot (\boldsymbol{\kappa}\cdot\boldsymbol{\nabla T)} + \rho~s -\cfrac{\rho}{\rho_0}~T~\frac{\partial \tilde{\boldsymbol{E}}}{\partial T}:\dot{\boldsymbol{S}} ~. }$