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Up to date as of January 14, 2010

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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}} ~. }

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