# 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,

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