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LightofAether
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Homework Statement
Determine the thermodynamic restrictions for a rigid heat conductor defined by the constitutive equations:
[tex]\DeclareMathOperator{\grad}{grad}\psi = \hat{\psi}\left(\theta,\grad \theta, \grad \grad \theta\right) \\
\eta = \hat{\eta}\left(\theta,\grad \theta, \grad \grad \theta\right)\\
\textbf{q} = \hat{\textbf{q}}\left(\theta,\grad \theta, \grad \grad \theta\right)[/tex]
Homework Equations
[/B]
[tex]\rho \left( \dot{\psi}+\dot{\theta} \eta \right)-\textbf{T}:\textbf{D}+\frac{\textbf{q}}{\theta}\cdot \grad\theta\leq0[/tex]
The Attempt at a Solution
I have already found the thermodynamic restrictions for [itex]\psi[/itex] because it's straightforward (take the material derivative and apply the chain rule), but I don't know where to start for [itex]\eta[/itex] or [itex]\textbf{q}[/itex]. We're using the Coleman-Noll approach and I understand the procedure once I have [itex]\psi = \hat{\psi}\left(\theta,\grad \theta, \grad \grad \theta\right)=something[/itex], but I'm struggling with finding a good starting place for [itex]\eta[/itex]. From what my professor has said, it seems like they can be arbitrary as long as they contain [itex]\theta,\grad \theta, \grad \grad \theta[/itex]. That doesn't seem like a very good way to go about this, though. A result of plugging in [itex]\dot{\psi}[/itex] into the relevant equation above (with [itex]\textbf{T}:\textbf{D}=0[/itex] because it's rigid) is [itex]\hat{\eta}=-\frac{\partial \hat{\psi(\theta)}}{\partial \theta}[/itex]. Can I just plug that into the relevant equation above while keeping [itex]\dot{\psi}[/itex] as [itex]\dot{\psi}[/itex] to find the thermodynamic restrictions for [itex]\eta[/itex]? The equation for [itex]\textbf{q}[/itex] is probably [itex]\textbf{q}=-\textbf{K}\textbf{g}[/itex].
What do you think? Am I on the right track?
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