Continuum mechanics/Specific heats of thermoelastic materials

For thermoelastic materials, show that the specific heats are related by the relation

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} ~.

Proof:

Recall that

C_v := \frac{\partial \hat{e}(\boldsymbol{E},T)}{\partial T} = T~\frac{\partial \hat{\eta}}{\partial T}

and

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} ~.

Therefore,

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} ~.

Also recall that

\eta = \hat{\eta}(\boldsymbol{E}, T) = \tilde{\eta}(\boldsymbol{S}, T) ~.

Therefore, keeping $\boldsymbol{S}$ constant while differentiating, we have

\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} ~.

Noting that $\boldsymbol{E} = \tilde{\boldsymbol{E}}(\boldsymbol{S},T)$ , and plugging back into the equation for the difference between the two specific heats, we have

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} ~.

Recalling that

\frac{\partial \hat{\eta}}{\partial \boldsymbol{E}} = - \cfrac{1}{\rho_0}~\frac{\partial \hat{\boldsymbol{S}}}{\partial T}

we get

{ 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} ~. }

For thermoelastic materials, show that the specific heats can also be related by the equations

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) ~.

We can also write the above as

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

where $\boldsymbol{\alpha}_E := \frac{\partial \boldsymbol{E}}{\partial T}$ is the thermal expansion tensor and $\boldsymbol{\mathsf{C}} := \frac{\partial \boldsymbol{S}}{\partial \boldsymbol{E}}$ is the stiffness tensor.

Proof:

Recall that

\boldsymbol{S} = \rho_0~\frac{\partial \psi}{\partial \boldsymbol{E}} = \rho_0~\boldsymbol{f}(\boldsymbol{E}(\boldsymbol{S},T),T)~.

Recall the chain rule which states that if

g(u,t) = f(x(u,t), y(u,t))

then, if we keep $u$ fixed, the partial derivative of $g$ with respect to $t$ is given by

\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} ~.

In our case,

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}.

Hence, we have

\boldsymbol{S} = g(\boldsymbol{S}, T) = f(\boldsymbol{E}(\boldsymbol{S},T), T) = \rho_0~\boldsymbol{f}(\boldsymbol{E}(\boldsymbol{S},T),T)~.

Taking the derivative with respect to $T$ keeping $\boldsymbol{S}$ constant, we have

\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]

or,

\mathbf{0} = \frac{\partial \boldsymbol{f}}{\partial \boldsymbol{E}}:\frac{\partial \boldsymbol{E}}{\partial T} + \frac{\partial \boldsymbol{f}}{\partial T}~.

Now,

\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}} ~.

Therefore,

\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) ~.

Again recall that,

\frac{\partial \psi}{\partial \boldsymbol{E}} = \cfrac{1}{\rho_0}~\boldsymbol{S} ~.

Plugging into the above, we get

\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} ~.

Therefore, we get the following relation for $\partial \boldsymbol{S}/\partial T$ :

\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} ~.

Recall that

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} ~.

Plugging in the expressions for $\partial \boldsymbol{S}/\partial T$ we get:

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} ~.

Therefore,

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} ~.

Using the identity $(\boldsymbol{\mathsf{A}}:\boldsymbol{B}):\boldsymbol{C} = \boldsymbol{C}:(\boldsymbol{\mathsf{A}}:\boldsymbol{B})$ , we have

{ 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) ~. }

Consider an isotropic thermoelastic material that has a constant coefficient of thermal expansion and which follows the Saint-Venant–Kirchhoff model, i.e,

\boldsymbol{\alpha}_E = \alpha~\boldsymbol{\mathit{1}} \qquad\text{and}\qquad \boldsymbol{\mathsf{C}} = \lambda~\boldsymbol{\mathit{1}}\otimes\boldsymbol{\mathit{1}} + 2\mu~\boldsymbol{\mathsf{I}}

where $\alpha$ is the coefficient of thermal expansion and $3~\lambda = 3~K - 2~\mu$ where $K, \mu$ are the bulk and shear moduli, respectively.

Show that the specific heats related by the equation

C_p - C_v = \cfrac{1}{\rho_0}\left[\alpha~\text{tr}{\boldsymbol{S}} + 9~\alpha^2~K~T\right]~.

Proof:

Recall that,

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 ~.

Plugging the expressions of $\boldsymbol{\alpha}_E$ and $\boldsymbol{\mathsf{C}}$ into the above equation, we have

\begin{align} 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{align}

Therefore,

{ C_p - C_v = \cfrac{1}{\rho_0}\left[\alpha~\text{tr}{\boldsymbol{S}} + 9~\alpha^2~K~T\right]~. }

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