# Continuum mechanics/Balance of energy for thermoelasticity

## Balance of energy for thermoelastic materials

Show that, for thermoelastic materials, the balance of energy

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

can be expressed as

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

Proof:

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

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

Plug into energy equation to get

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

Recall,

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

Hence,

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

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

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

Plugging into the energy equation, we have

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

or,

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

## Rate of internal energy/entropy for thermoelastic materials

For thermoelastic materials, the specific internal energy is given by

${\displaystyle e={\bar {e}}({\boldsymbol {E}},\eta )}$

where ${\displaystyle {\boldsymbol {E}}}$ is the Green strain and ${\displaystyle \eta }$ is the specific entropy. Show that

${\displaystyle {\cfrac {d}{dt}}(e-T~\eta )=-{\dot {T}}~\eta +{\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\dot {\boldsymbol {E}}}\qquad {\text{and}}\qquad {\cfrac {d}{dt}}(e-T~\eta -{\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\boldsymbol {E}})=-{\dot {T}}~\eta -{\cfrac {1}{\rho _{0}}}~{\dot {\boldsymbol {S}}}:{\boldsymbol {E}}}$

where ${\displaystyle \rho _{0}}$ is the initial density, ${\displaystyle T}$ is the absolute temperature, ${\displaystyle {\boldsymbol {S}}}$ 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

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

Now, for thermoelastic materials,

${\displaystyle T={\frac {\partial {\bar {e}}}{\partial \eta }}\qquad {\text{and}}\qquad {\boldsymbol {S}}=\rho _{0}~{\frac {\partial {\bar {e}}}{\partial {\boldsymbol {E}}}}~.}$

Therefore,

${\displaystyle {\dot {e}}={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\dot {\boldsymbol {E}}}+T~{\dot {\eta }}~.\qquad \implies \qquad {\dot {e}}-T~{\dot {\eta }}={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\dot {\boldsymbol {E}}}~.}$

Now,

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

Therefore,

${\displaystyle {\dot {e}}-{\cfrac {d}{dt}}(T~\eta )+{\dot {T}}~\eta ={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\dot {\boldsymbol {E}}}\qquad \implies \qquad {{\cfrac {d}{dt}}(e-T~\eta )=-{\dot {T}}~\eta +{\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\dot {\boldsymbol {E}}}~.}}$

Also,

${\displaystyle {\cfrac {d}{dt}}\left({\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\boldsymbol {E}}\right)={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\dot {\boldsymbol {E}}}+{\cfrac {1}{\rho _{0}}}~{\dot {\boldsymbol {S}}}:{\boldsymbol {E}}~.}$

Hence,

${\displaystyle {\dot {e}}-{\cfrac {d}{dt}}(T~\eta )+{\dot {T}}~\eta ={\cfrac {d}{dt}}\left({\cfrac {1}{\rho _{0}}}{\boldsymbol {S}}:{\boldsymbol {E}}\right)-{\cfrac {1}{\rho _{0}}}~{\dot {\boldsymbol {S}}}:{\boldsymbol {E}}\qquad \implies \qquad {{\cfrac {d}{dt}}\left(e-T~\eta -{\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\boldsymbol {E}}\right)=-{\dot {T}}~\eta -{\cfrac {1}{\rho _{0}}}~{\dot {\boldsymbol {S}}}:{\boldsymbol {E}}~.}}$

## Energy equation for thermoelastic materials

For thermoelastic materials, show that the balance of energy equation

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

can be expressed as either

${\displaystyle \rho ~C_{v}~{\dot {T}}={\boldsymbol {\nabla }}\cdot ({\boldsymbol {\kappa }}\cdot {\boldsymbol {\nabla T)}}+\rho ~s+{\cfrac {\rho }{\rho _{0}}}~T~{\frac {\partial {\hat {\boldsymbol {S}}}}{\partial T}}:{\dot {\boldsymbol {E}}}}$

or

${\displaystyle \rho ~\left(C_{p}-{\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}\right)~{\dot {T}}={\boldsymbol {\nabla }}\cdot ({\boldsymbol {\kappa }}\cdot {\boldsymbol {\nabla T)}}+\rho ~s-{\cfrac {\rho }{\rho _{0}}}~T~{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}:{\dot {\boldsymbol {S}}}}$

where

${\displaystyle C_{v}={\frac {\partial {\hat {e}}({\boldsymbol {E}},T)}{\partial T}}\qquad {\text{and}}\qquad C_{p}={\frac {\partial {\tilde {e}}({\boldsymbol {S}},T)}{\partial T}}~.}$

For the special case where there are no sources and we can ignore heat conduction (for very fast processes), the energy equation simplifies to

${\displaystyle \rho ~\left(C_{p}-{\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\boldsymbol {\alpha }}\right)~{\dot {T}}=-{\cfrac {\rho }{\rho _{0}}}~T~{\boldsymbol {\alpha }}:{\dot {\boldsymbol {S}}}}$

where ${\displaystyle {\boldsymbol {\alpha }}:={\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}}$ is the thermal expansion tensor which has the form ${\displaystyle {\boldsymbol {\alpha }}=\alpha {\boldsymbol {1}}}$ for isotropic materials and ${\displaystyle \alpha \,}$ 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 ${\displaystyle {\boldsymbol {E}}}$ and ${\displaystyle T}$, then

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

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

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

Since

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

we have, either

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

or

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

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

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

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

${\displaystyle -{\cfrac {\rho }{\rho _{0}}}~T~{\frac {\partial {\hat {\boldsymbol {S}}}}{\partial T}}:{\dot {\boldsymbol {E}}}+\rho ~C_{v}~{\dot {T}}=-{\boldsymbol {\nabla }}\cdot \mathbf {q} +\rho ~s}$

and

${\displaystyle {\cfrac {\rho }{\rho _{0}}}~T~{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}:{\dot {\boldsymbol {S}}}+\rho ~C_{p}~{\dot {T}}-{\cfrac {\rho }{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}~{\dot {T}}=-{\boldsymbol {\nabla }}\cdot \mathbf {q} +\rho ~s~.}$

From Fourier's law of heat conduction

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

Therefore,

${\displaystyle -{\cfrac {\rho }{\rho _{0}}}~T~{\frac {\partial {\hat {\boldsymbol {S}}}}{\partial T}}:{\dot {\boldsymbol {E}}}+\rho ~C_{v}~{\dot {T}}={\boldsymbol {\nabla }}\cdot ({\boldsymbol {\kappa }}\cdot {\boldsymbol {\nabla T)}}+\rho ~s}$

and

${\displaystyle {\cfrac {\rho }{\rho _{0}}}~T~{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}:{\dot {\boldsymbol {S}}}+\rho ~C_{p}~{\dot {T}}-{\cfrac {\rho }{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}~{\dot {T}}={\boldsymbol {\nabla }}\cdot ({\boldsymbol {\kappa }}\cdot {\boldsymbol {\nabla T)}}+\rho ~s~.}$

Rearranging,

${\displaystyle {\rho ~C_{v}~{\dot {T}}={\boldsymbol {\nabla }}\cdot ({\boldsymbol {\kappa }}\cdot {\boldsymbol {\nabla T)}}+\rho ~s+{\cfrac {\rho }{\rho _{0}}}~T~{\frac {\partial {\hat {\boldsymbol {S}}}}{\partial T}}:{\dot {\boldsymbol {E}}}}}$

or,

${\displaystyle {\rho ~\left(C_{p}-{\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}\right)~{\dot {T}}={\boldsymbol {\nabla }}\cdot ({\boldsymbol {\kappa }}\cdot {\boldsymbol {\nabla T)}}+\rho ~s-{\cfrac {\rho }{\rho _{0}}}~T~{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}:{\dot {\boldsymbol {S}}}~.}}$