•   Wikis

# Continuum mechanics/Balance of energy for thermoelasticity: Wikis

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

## 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 η 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 ρ0 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

$C_v = \frac{\partial \hat{e}(\boldsymbol{E},T)}{\partial T} \qquad \text{and} \qquad C_p = \frac{\partial \tilde{e}(\boldsymbol{S},T)}{\partial T} ~.$

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,

${ \rho~\left(C_p - \cfrac{1}{\rho_0}~\boldsymbol{S}:\frac{\partial \tilde{\boldsymbol{E}}}{\partial T}\right) ~\dot{T} = \boldsymbol{\nabla} \cdot (\boldsymbol{\kappa}\cdot\boldsymbol{\nabla T)} + \rho~s -\cfrac{\rho}{\rho_0}~T~\frac{\partial \tilde{\boldsymbol{E}}}{\partial T}:\dot{\boldsymbol{S}} ~. }$