Relation between specific heats - 1
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For thermoelastic materials, show that the specific heats are related by the relation
![{\displaystyle C_{p}-C_{v}={\cfrac {1}{\rho _{0}}}\left({\boldsymbol {S}}-T~{\frac {\partial {\hat {\boldsymbol {S}}}}{\partial T}}\right):{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5db092d1b29240a683d54c3a6014ee499a67bc75)
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Proof:
Recall that
![{\displaystyle C_{v}:={\frac {\partial {\hat {e}}({\boldsymbol {E}},T)}{\partial T}}=T~{\frac {\partial {\hat {\eta }}}{\partial T}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3791b7630949ec5f48c99e7da0fad51f6e0ec529)
and
![{\displaystyle C_{p}:={\frac {\partial {\tilde {e}}({\boldsymbol {S}},T)}{\partial T}}=T~{\frac {\partial {\tilde {\eta }}}{\partial T}}+{\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1852ac16f1f9b047bd5613ebf45f9e3587f13697)
Therefore,
![{\displaystyle C_{p}-C_{v}=T~{\frac {\partial {\tilde {\eta }}}{\partial T}}+{\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}-T~{\frac {\partial {\hat {\eta }}}{\partial T}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8ec31f95962af6d437068ad12364ee523acfdd56)
Also recall that
![{\displaystyle \eta ={\hat {\eta }}({\boldsymbol {E}},T)={\tilde {\eta }}({\boldsymbol {S}},T)~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d928b8f683781f93b871300406ca17d70c1ddacf)
Therefore, keeping
constant while differentiating, we have
![{\displaystyle {\frac {\partial {\tilde {\eta }}}{\partial T}}={\frac {\partial {\hat {\eta }}}{\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+{\frac {\partial {\hat {\eta }}}{\partial T}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a69ef40b66635fd1d61bef5c4cfdc5e56ac1919f)
Noting that
, and
plugging back into the equation for the difference between the two
specific heats, we have
![{\displaystyle C_{p}-C_{v}=T~{\frac {\partial {\hat {\eta }}}{\partial {\boldsymbol {E}}}}:{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}+{\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/440bd61526519406088253f1aae5daf58c3649c6)
Recalling that
![{\displaystyle {\frac {\partial {\hat {\eta }}}{\partial {\boldsymbol {E}}}}=-{\cfrac {1}{\rho _{0}}}~{\frac {\partial {\hat {\boldsymbol {S}}}}{\partial T}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/88ca4c9d6cd4f927a96e4076417b06f0e017457a)
we get
![{\displaystyle {C_{p}-C_{v}={\cfrac {1}{\rho _{0}}}\left({\boldsymbol {S}}-T~{\frac {\partial {\hat {\boldsymbol {S}}}}{\partial T}}\right):{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}~.}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eba4adbc28d703b26d2160c54559192e2acdfa0e)
Relation between specific heats - 2
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For thermoelastic materials, show that the specific heats can also be related by the equations
![{\displaystyle C_{p}-C_{v}={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+{\frac {\partial {\boldsymbol {E}}}{\partial T}}:\left({\frac {\partial ^{2}\psi }{\partial {\boldsymbol {E}}\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}\right)={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+{\cfrac {T}{\rho _{0}}}~{\frac {\partial {\boldsymbol {E}}}{\partial T}}:\left({\frac {\partial {\boldsymbol {S}}}{\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}\right)~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/18a1e9670790061f946e970628b4ce597e8dd400)
We can also write the above as
![{\displaystyle C_{p}-C_{v}={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\boldsymbol {\alpha }}_{E}+{\cfrac {T}{\rho _{0}}}~{\boldsymbol {\alpha }}_{E}:{\boldsymbol {\mathsf {C}}}:{\boldsymbol {\alpha }}_{E}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3cad8654d8b63763f8f57906ae9c68f6c134668c)
where is the thermal expansion tensor and is the stiffness tensor.
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Proof:
Recall that
![{\displaystyle {\boldsymbol {S}}=\rho _{0}~{\frac {\partial \psi }{\partial {\boldsymbol {E}}}}=\rho _{0}~{\boldsymbol {f}}({\boldsymbol {E}}({\boldsymbol {S}},T),T)~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/323f9b59f2146e055805327f3f2c59ed6ff1f62d)
Recall the chain rule which states that if
![{\displaystyle g(u,t)=f(x(u,t),y(u,t))}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bde57338fb4bb22cf2700dd322d22e8a5161b0ba)
then, if we keep
fixed, the partial derivative of
with respect
to
is given by
![{\displaystyle {\frac {\partial g}{\partial t}}={\frac {\partial f}{\partial x}}~{\frac {\partial x}{\partial t}}+{\frac {\partial f}{\partial y}}~{\frac {\partial y}{\partial t}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c201dbddbeeed6732dbb3bd260751844b243e111)
In our case,
![{\displaystyle u={\boldsymbol {S}},~~t=T,~~g({\boldsymbol {S}},T)={\boldsymbol {S}},~~x({\boldsymbol {S}},T)={\boldsymbol {E}}({\boldsymbol {S}},T),~~y({\boldsymbol {S}},T)=T,~~{\text{and}}~~f=\rho _{0}~{\boldsymbol {f}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/abe960dea04b44201fe9d7b2ee8b506e6555a1cc)
Hence, we have
![{\displaystyle {\boldsymbol {S}}=g({\boldsymbol {S}},T)=f({\boldsymbol {E}}({\boldsymbol {S}},T),T)=\rho _{0}~{\boldsymbol {f}}({\boldsymbol {E}}({\boldsymbol {S}},T),T)~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2a5152f0d1927fc66e75ca49625137025c67b0cd)
Taking the derivative with respect to
keeping
constant, we have
![{\displaystyle {\frac {\partial g}{\partial T}}={\frac {\partial {\boldsymbol {S}}}{\partial T}}=\rho _{0}~\left[{\frac {\partial {\boldsymbol {f}}}{\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+{\frac {\partial {\boldsymbol {f}}}{\partial T}}~{\frac {\partial T}{\partial T}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0321b722f0347cfe76a9c572b9ec092787b675e1)
or,
![{\displaystyle \mathbf {0} ={\frac {\partial {\boldsymbol {f}}}{\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+{\frac {\partial {\boldsymbol {f}}}{\partial T}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7413d569ecf3cd57c71338496a175b62ebc87a15)
Now,
![{\displaystyle {\boldsymbol {f}}={\frac {\partial \psi }{\partial {\boldsymbol {E}}}}\qquad \implies \qquad {\frac {\partial {\boldsymbol {f}}}{\partial {\boldsymbol {E}}}}={\frac {\partial ^{2}\psi }{\partial {\boldsymbol {E}}\partial {\boldsymbol {E}}}}\quad {\text{and}}\quad {\frac {\partial {\boldsymbol {f}}}{\partial T}}={\frac {\partial ^{2}\psi }{\partial T\partial {\boldsymbol {E}}}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/46188fd4aafa1c96e5ab8a77c3f57c54d951158c)
Therefore,
![{\displaystyle \mathbf {0} ={\frac {\partial ^{2}\psi }{\partial {\boldsymbol {E}}\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+{\frac {\partial ^{2}\psi }{\partial T\partial {\boldsymbol {E}}}}={\frac {\partial }{\partial {\boldsymbol {E}}}}\left({\frac {\partial \psi }{\partial {\boldsymbol {E}}}}\right):{\frac {\partial {\boldsymbol {E}}}{\partial T}}+{\frac {\partial }{\partial T}}\left({\frac {\partial \psi }{\partial {\boldsymbol {E}}}}\right)~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1c25662fff5617acd5335acdd75d2cc1ea0d0c97)
Again recall that,
![{\displaystyle {\frac {\partial \psi }{\partial {\boldsymbol {E}}}}={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/74c73af6e6c83d963acfa9c380bdfd2ea1e9b289)
Plugging into the above, we get
![{\displaystyle \mathbf {0} ={\frac {\partial ^{2}\psi }{\partial {\boldsymbol {E}}\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+{\cfrac {1}{\rho _{0}}}~{\frac {\partial {\boldsymbol {S}}}{\partial T}}={\cfrac {1}{\rho _{0}}}~{\frac {\partial {\boldsymbol {S}}}{\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+{\cfrac {1}{\rho _{0}}}~{\frac {\partial {\boldsymbol {S}}}{\partial T}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bbf66715e08e2ab436f7a6b504cf5907ffe13951)
Therefore, we get the following relation for
:
![{\displaystyle {\frac {\partial {\boldsymbol {S}}}{\partial T}}=-\rho _{0}~{\frac {\partial ^{2}\psi }{\partial {\boldsymbol {E}}\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}=-{\frac {\partial {\boldsymbol {S}}}{\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/61cb884c7c41d5f09ef2749f99d5ea94f6939aa8)
Recall that
![{\displaystyle C_{p}-C_{v}={\cfrac {1}{\rho _{0}}}\left({\boldsymbol {S}}-T~{\frac {\partial {\boldsymbol {S}}}{\partial T}}\right):{\frac {\partial {\boldsymbol {E}}}{\partial T}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/af251df4128358fa0887d9dfe83c5d7c8814042d)
Plugging in the expressions for
we get:
![{\displaystyle C_{p}-C_{v}={\cfrac {1}{\rho _{0}}}\left({\boldsymbol {S}}+T~\rho _{0}~{\frac {\partial ^{2}\psi }{\partial {\boldsymbol {E}}\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}\right):{\frac {\partial {\boldsymbol {E}}}{\partial T}}={\cfrac {1}{\rho _{0}}}\left({\boldsymbol {S}}+T~{\frac {\partial {\boldsymbol {S}}}{\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}\right):{\frac {\partial {\boldsymbol {E}}}{\partial T}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/481510dd74ee4fe83e3a9134a7b0941e7bcf5302)
Therefore,
![{\displaystyle C_{p}-C_{v}={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+T~\left({\frac {\partial ^{2}\psi }{\partial {\boldsymbol {E}}\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}\right):{\frac {\partial {\boldsymbol {E}}}{\partial T}}={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+{\cfrac {T}{\rho _{0}}}~\left({\frac {\partial {\boldsymbol {S}}}{\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}\right):{\frac {\partial {\boldsymbol {E}}}{\partial T}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6f365ab2be848883a7226515f767d9e103aa755c)
Using the identity
, we have
![{\displaystyle {C_{p}-C_{v}={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+T~{\frac {\partial {\boldsymbol {E}}}{\partial T}}:\left({\frac {\partial ^{2}\psi }{\partial {\boldsymbol {E}}\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}\right)={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+{\cfrac {T}{\rho _{0}}}~{\frac {\partial {\boldsymbol {E}}}{\partial T}}:\left({\frac {\partial {\boldsymbol {S}}}{\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}\right)~.}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7a9bbc8d03aba1ceabadb5afa42d37d16d92a585)
Specific heats of Saint-Venant–Kirchhoff material
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Proof:
Recall that,
![{\displaystyle C_{p}-C_{v}={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\boldsymbol {\alpha }}_{E}+{\cfrac {T}{\rho _{0}}}~{\boldsymbol {\alpha }}_{E}:{\boldsymbol {\mathsf {C}}}:{\boldsymbol {\alpha }}_{E}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/70d866f8c36766f37dc9b8cbc09a6bf3e4ae3937)
Plugging the expressions of
and
into the above
equation, we have
![{\displaystyle {\begin{aligned}C_{p}-C_{v}&={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:(\alpha ~{\boldsymbol {\mathit {1}}})+{\cfrac {T}{\rho _{0}}}~(\alpha ~{\boldsymbol {\mathit {1}}}):(\lambda ~{\boldsymbol {\mathit {1}}}\otimes {\boldsymbol {\mathit {1}}}+2\mu ~{\boldsymbol {\mathsf {I}}}):(\alpha ~{\boldsymbol {\mathit {1}}})\\&={\cfrac {\alpha }{\rho _{0}}}~{\text{tr}}{\boldsymbol {S}}+{\cfrac {\alpha ^{2}~T}{\rho _{0}}}~{\boldsymbol {\mathit {1}}}:(\lambda ~{\boldsymbol {\mathit {1}}}\otimes {\boldsymbol {\mathit {1}}}+2\mu ~{\boldsymbol {\mathsf {I}}}):{\boldsymbol {\mathit {1}}}\\&={\cfrac {\alpha }{\rho _{0}}}~{\text{tr}}{\boldsymbol {S}}+{\cfrac {\alpha ^{2}~T}{\rho _{0}}}~{\boldsymbol {\mathit {1}}}:(\lambda ~{\text{tr}}{\boldsymbol {\mathit {1}}}~{\boldsymbol {\mathit {1}}}+2\mu ~{\boldsymbol {\mathit {1}}})\\&={\cfrac {\alpha }{\rho _{0}}}~{\text{tr}}{\boldsymbol {S}}+{\cfrac {\alpha ^{2}~T}{\rho _{0}}}~(3~\lambda ~{\text{tr}}{\boldsymbol {\mathit {1}}}+2\mu ~{\text{tr}}{\boldsymbol {\mathit {1}}})\\&={\cfrac {\alpha }{\rho _{0}}}~{\text{tr}}{\boldsymbol {S}}+{\cfrac {3~\alpha ^{2}~T}{\rho _{0}}}~(3~\lambda +2\mu )\\&={\cfrac {\alpha ~{\text{tr}}{\boldsymbol {S}}}{\rho _{0}}}+{\cfrac {9~\alpha ^{2}~K~T}{\rho _{0}}}~.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6ab4607613e3804cf51ed913c0fae9162a236c6b)
Therefore,
![{\displaystyle {C_{p}-C_{v}={\cfrac {1}{\rho _{0}}}\left[\alpha ~{\text{tr}}{\boldsymbol {S}}+9~\alpha ^{2}~K~T\right]~.}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e45735dd6a108445fbdce226d8a2dafabd0d4ac7)