Balance of energy for thermoelastic materials
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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~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8701bfecbbbe89347c5e7b4aee1df73e0de9785a)
can be expressed as
![{\displaystyle \rho ~T~{\dot {\eta }}=-{\boldsymbol {\nabla }}\cdot \mathbf {q} +\rho ~s~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9a855f11467c25fdce54f037bc5f9b23e01fcb86)
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Proof:
Since
, we have
![{\displaystyle {\dot {e}}={\frac {\partial e}{\partial {\boldsymbol {F}}}}:{\dot {\boldsymbol {F}}}+{\frac {\partial e}{\partial \eta }}~{\dot {\eta }}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7d988ff46782cbdec0139db23d6adb0532568419)
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~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aefa9a5b02a934adccad45998d09ce2b5d2ac595)
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}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/88c38adccce273014d3ce65007bc3635efd8b241)
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~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a1f996e64732714afa1e79c3c5398033bb7797c3)
Now,
. Therefore, using the
identity
, 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}}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9ac128150043c087f6193591f54318325dfb180b)
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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0d34060d6c8e83be8162a5bd5d94a82a855f5055)
or,
![{\displaystyle {\rho ~T~{\dot {\eta }}=-{\boldsymbol {\nabla }}\cdot \mathbf {q} +\rho ~s~.}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ff772922a4b71c572778da067d730662a041e715)
Rate of internal energy/entropy for thermoelastic materials
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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 }}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/940ccca86f3ccd7f712cc55ebf7858cfb6128be4)
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}}}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a3a31415f6fe9fd7137362a38fac853e204f687f)
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}}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f94d372b61783630e9350748a4f3a747fe973a18)
Now,
![{\displaystyle {\cfrac {d}{dt}}(T~\eta )={\dot {T}}~\eta +T~{\dot {\eta }}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b3f4b81a1cbbb772e09a0dab6bfab5957bc1eae4)
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}}}~.}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/360f1c9a16bff1554b732e808edc0a2853a2687c)
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}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/83218dbf833524ff261930a139b62011589ada60)
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}}~.}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ba9ef2f292b9ddd7028f14a23d0e880e6ed0c193)
Energy equation for thermoelastic materials
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For thermoelastic materials, show that the balance of energy
equation
![{\displaystyle \rho ~T~{\dot {\eta }}=-{\boldsymbol {\nabla }}\cdot \mathbf {q} +\rho ~s}](https://wikimedia.org/api/rest_v1/media/math/render/svg/42c7608771be3053887bd02796f3e550ff61170c)
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}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/86c4304e748434d801d09c19cc2742a5d551df58)
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}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e218f8e45493545426b6fbc9101ef49e17e15a78)
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}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f234f6374a613a2c19ba2ee76c510ace7731722e)
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}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/288e22055f5e1b0fc34366c303778be82127cae9)
where is the thermal expansion tensor which has the form for isotropic materials and is the coefficient of thermal expansion. The above equation can be used to calculate the change of temperature in thermoelasticity.
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Proof:
If the independent variables are
and
, 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}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1be416a5a8b9fa5219285a3683cd8af5b1bedfda)
On the other hand, if we consider
and
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}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/abd2f540b22ca043c5c01892379d169f0e5929db)
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)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9f5f7418d221fe431f41318b3d7a10d34e7a2faf)
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/637a1a2b186a4f2f63ee7b74e92b2f7ab4872f55)
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}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c90ead39941542e7bd5ba555e2b5cc131e894a08)
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~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9a855f11467c25fdce54f037bc5f9b23e01fcb86)
Using the two forms of
, 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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/77d79a406e1007561321fc1bf0492810e0ba847b)
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~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ef1d90a82965f8c187437f9a9376a85c6684632b)
From Fourier's law of heat conduction
![{\displaystyle \mathbf {q} =-{\boldsymbol {\kappa }}\cdot {\boldsymbol {\nabla }}T~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aea104b856222829c1b86e5ec1065bddd78ac169)
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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e4cfe79df1d7a655ee29904aa6a0e27d87a12472)
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~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/15c210563bf5331dad043b15d4514672d36d7a1d)
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}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8e9d9da412f8066dacbb87f6853326b82e24e7b2)
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}}}~.}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9edd6f9fbf68482deebc131985c41c5cdcbf7566)