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# Continuum mechanics/Entropy inequality: Wikis

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# Study guide

Up to date as of January 14, 2010

## Clausius-Duhem inequality

The Clausius-Duhem inequality can be expressed in integral form as

$\cfrac{d}{dt}\left(\int_{\Omega} \rho~\eta~\text{dV}\right) \ge \int_{\partial \Omega} \rho~\eta~(u_n - \mathbf{v}\cdot\mathbf{n})~\text{dA} - \int_{\partial \Omega} \cfrac{\mathbf{q}\cdot\mathbf{n}}{T}~\text{dA} + \int_{\Omega} \cfrac{\rho~s}{T}~\text{dV} ~.$

In differential form the Clusius-Duhem inequality can be written as

$\rho~\dot{\eta} \ge - \boldsymbol{\nabla} \cdot \left(\cfrac{\mathbf{q}}{T}\right) + \cfrac{\rho~s}{T} ~.$

Proof:

Assume that Ω is an arbitrary fixed control volume. Then un = 0 and the derivative can be taken inside the integral to give

$\int_{\Omega} \frac{\partial }{\partial t}(\rho~\eta)~\text{dV} \ge -\int_{\partial \Omega} \rho~\eta~(\mathbf{v}\cdot\mathbf{n})~\text{dA} - \int_{\partial \Omega} \cfrac{\mathbf{q}\cdot\mathbf{n}}{T}~\text{dA} + \int_{\Omega} \cfrac{\rho~s}{T}~\text{dV} ~.$

Using the divergence theorem, we get

$\int_{\Omega} \frac{\partial }{\partial t}(\rho~\eta)~\text{dV} \ge -\int_{\Omega} \boldsymbol{\nabla} \cdot (\rho~\eta~\mathbf{v})~\text{dV} - \int_{\Omega} \boldsymbol{\nabla} \cdot \left(\cfrac{\mathbf{q}}{T}\right)~\text{dV} + \int_{\Omega} \cfrac{\rho~s}{T}~\text{dV} ~.$

Since Ω is arbitrary, we must have

$\frac{\partial }{\partial t}(\rho~\eta) \ge -\boldsymbol{\nabla} \cdot (\rho~\eta~\mathbf{v}) - \boldsymbol{\nabla} \cdot \left(\cfrac{\mathbf{q}}{T}\right) + \cfrac{\rho~s}{T} ~.$

Expanding out

$\frac{\partial \rho}{\partial t}~\eta + \rho~\frac{\partial \eta}{\partial t} \ge -\boldsymbol{\nabla} (\rho_\eta)\cdot\mathbf{v} - \rho~\eta~(\boldsymbol{\nabla} \cdot \mathbf{v}) - \boldsymbol{\nabla} \cdot \left(\cfrac{\mathbf{q}}{T}\right) + \cfrac{\rho~s}{T}$

or,

$\frac{\partial \rho}{\partial t}~\eta + \rho~\frac{\partial \eta}{\partial t} \ge -\eta~\boldsymbol{\nabla} \rho\cdot\mathbf{v} - \rho~\boldsymbol{\nabla} \eta\cdot\mathbf{v} - \rho~\eta~(\boldsymbol{\nabla} \cdot \mathbf{v}) - \boldsymbol{\nabla} \cdot \left(\cfrac{\mathbf{q}}{T}\right) + \cfrac{\rho~s}{T}$

or,

$\left(\frac{\partial \rho}{\partial t} + \boldsymbol{\nabla} \rho\cdot\mathbf{v} + \rho~\boldsymbol{\nabla} \cdot \mathbf{v}\right) ~\eta + \rho~\left(\frac{\partial \eta}{\partial t} + \boldsymbol{\nabla} \eta\cdot\mathbf{v}\right) \ge -\boldsymbol{\nabla} \cdot \left(\cfrac{\mathbf{q}}{T}\right) + \cfrac{\rho~s}{T} ~.$

Now, the material time derivatives of ρ and η are given by

$\dot{\rho} = \frac{\partial \rho}{\partial t} + \boldsymbol{\nabla} \rho\cdot\mathbf{v} ~;~~ \dot{\eta} = \frac{\partial \eta}{\partial t} + \boldsymbol{\nabla} \eta\cdot\mathbf{v} ~.$

Therefore,

$\left(\dot{\rho} + \rho~\boldsymbol{\nabla} \cdot \mathbf{v}\right)~\eta + \rho~\dot{\eta} \ge -\boldsymbol{\nabla} \cdot \left(\cfrac{\mathbf{q}}{T}\right) + \cfrac{\rho~s}{T} ~.$

From the conservation of mass $\dot{\rho} + \rho~\boldsymbol{\nabla} \cdot \mathbf{v} = 0$. Hence,

${ \rho~\dot{\eta} \ge -\boldsymbol{\nabla} \cdot \left(\cfrac{\mathbf{q}}{T}\right) + \cfrac{\rho~s}{T} ~. }$

### Clausius-Duhem inequality in terms of internal energy

In terms of the specific entropy, the Clausius-Duhem inequality is written as

$\rho~\dot{\eta} \ge - \boldsymbol{\nabla} \cdot \left(\cfrac{\mathbf{q}}{T}\right) + \cfrac{\rho~s}{T}$

Show that the inequality can be expressed in terms of the internal energy as

$\rho~(\dot{e} - T~\dot{\eta}) - \boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} \le - \cfrac{\mathbf{q}\cdot\boldsymbol{\nabla} T}{T} ~.$

Proof:

Using the identity $\boldsymbol{\nabla} \cdot (\varphi~\mathbf{v}) = \varphi~\boldsymbol{\nabla} \cdot \mathbf{v} + \mathbf{v}\cdot\boldsymbol{\nabla} \varphi$ in the Clausius-Duhem inequality, we get

$\rho~\dot{\eta} \ge - \boldsymbol{\nabla} \cdot \left(\cfrac{\mathbf{q}}{T}\right) + \cfrac{\rho~s}{T} \qquad\text{or}\qquad \rho~\dot{\eta} \ge - \cfrac{1}{T}~\boldsymbol{\nabla} \cdot \mathbf{q} - \mathbf{q}\cdot\boldsymbol{\nabla} \left(\cfrac{1}{T}\right) + \cfrac{\rho~s}{T} ~.$

Now, using index notation with respect to a Cartesian basis $\mathbf{e}_j$,

$\boldsymbol{\nabla} \left(\cfrac{1}{T}\right) = \frac{\partial }{\partial x_j}\left(T^{-1}\right)~\mathbf{e}_j = -\left(T^{-2}\right)~\frac{\partial T}{\partial x_j}~\mathbf{e}_j = -\cfrac{1}{T^2}~\boldsymbol{\nabla} T ~.$

Hence,

$\rho~\dot{\eta} \ge - \cfrac{1}{T}~\boldsymbol{\nabla} \cdot \mathbf{q} + \cfrac{1}{T^2}~\mathbf{q}\cdot\boldsymbol{\nabla} T + \cfrac{\rho~s}{T} \qquad\text{or}\qquad \rho~\dot{\eta} \ge -\cfrac{1}{T}\left(\boldsymbol{\nabla} \cdot \mathbf{q} - \rho~s\right) + \cfrac{1}{T^2}~\mathbf{q}\cdot\boldsymbol{\nabla} T ~.$

Recall the balance of energy

$\rho~\dot{e} - \boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} + \boldsymbol{\nabla} \cdot \mathbf{q} - \rho~s = 0 \qquad \implies \qquad \rho~\dot{e} - \boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} = - (\boldsymbol{\nabla} \cdot \mathbf{q} - \rho~s) ~.$

Therefore,

$\rho~\dot{\eta} \ge \cfrac{1}{T}\left(\rho~\dot{e}-\boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v}\right) + \cfrac{1}{T^2}~\mathbf{q}\cdot\boldsymbol{\nabla} T \qquad \implies \qquad \rho~\dot{\eta}~T \ge \rho~\dot{e}-\boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} + \cfrac{\mathbf{q}\cdot\boldsymbol{\nabla} T}{T} ~.$

Rearranging,

${ \rho~(\dot{e} - T~\dot{\eta}) - \boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} \le - \cfrac{\mathbf{q}\cdot\boldsymbol{\nabla} T}{T} ~. }$

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