•   Wikis

# Continuum mechanics/Specific heats of thermoelastic materials: Wikis

Advertisements

Note: Many of our articles have direct quotes from sources you can cite, within the Wikipedia article! This article doesn't yet, but we're working on it! See more info or our list of citable articles.

# Study guide

Up to date as of January 14, 2010

## Relation between specific heats - 1

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

${ C_p - C_v = \cfrac{1}{\rho_0}\left(\boldsymbol{S} -T~\frac{\partial \hat{\boldsymbol{S}}}{\partial T}\right): \frac{\partial \tilde{\boldsymbol{E}}}{\partial T} ~. }$

## Relation between specific heats - 2

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

$C_p - C_v = \cfrac{1}{\rho_0}~\boldsymbol{S}:\frac{\partial \boldsymbol{E}}{\partial T} + \frac{\partial \boldsymbol{E}}{\partial T}:\left(\frac{\partial^2 \psi}{\partial \boldsymbol{E}\partial\boldsymbol{E}}: \frac{\partial \boldsymbol{E}}{\partial T}\right) = \cfrac{1}{\rho_0}~\boldsymbol{S}:\frac{\partial \boldsymbol{E}}{\partial T} + \cfrac{T}{\rho_0}~ \frac{\partial \boldsymbol{E}}{\partial T}:\left(\frac{\partial \boldsymbol{S}}{\partial \boldsymbol{E}}:\frac{\partial \boldsymbol{E}}{\partial T}\right) ~.$

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

${ C_p - C_v = \cfrac{1}{\rho_0}~\boldsymbol{S}:\frac{\partial \boldsymbol{E}}{\partial T} + T~\frac{\partial \boldsymbol{E}}{\partial T}:\left(\frac{\partial^2 \psi}{\partial \boldsymbol{E}\partial\boldsymbol{E}}:\frac{\partial \boldsymbol{E}}{\partial T} \right) = \cfrac{1}{\rho_0}~\boldsymbol{S}:\frac{\partial \boldsymbol{E}}{\partial T} + \cfrac{T}{\rho_0}~ \frac{\partial \boldsymbol{E}}{\partial T}:\left(\frac{\partial \boldsymbol{S}}{\partial \boldsymbol{E}}:\frac{\partial \boldsymbol{E}}{\partial T}\right) ~. }$

## Specific heats of Saint-Venant–Kirchhoff material

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 α is the coefficient of thermal expansion and $3~\lambda = 3~K - 2~\mu$ where K 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,

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

Advertisements

 Got something to say? Make a comment. Your name Your email address Message