Continuum mechanics/Entropy inequality

Clausius-Duhem inequality The Clausius-Duhem inequality can be expressed in integral form as $\frac{d}{dt}\left(\underset{\mathrm{\Omega }}{\int }\rho \eta \text{dV}\right)\ge \underset{\partial \mathrm{\Omega }}{\int }\rho \eta (u_n-𝐯\cdot 𝐧)\text{dA}-\underset{\partial \mathrm{\Omega }}{\int }\frac{𝐪\cdot 𝐧}{T}\text{dA}+\underset{\mathrm{\Omega }}{\int }\frac{\rho s}{T}\text{dV}.$ In differential form the Clusius-Duhem inequality can be written as $\rho \dot{\eta }\ge -\mathrm{\nabla }\cdot \left(\frac{𝐪}{T}\right)+\frac{\rho s}{T}.$

Proof:

Assume that $\mathrm{\Omega }$ is an arbitrary fixed control volume. Then $u_n=0$ and the derivative can be taken inside the integral to give

$\underset{\mathrm{\Omega }}{\int }\frac{\partial }{\partial t}(\rho \eta )\text{dV}\ge -\underset{\partial \mathrm{\Omega }}{\int }\rho \eta (𝐯\cdot 𝐧)\text{dA}-\underset{\partial \mathrm{\Omega }}{\int }\frac{𝐪\cdot 𝐧}{T}\text{dA}+\underset{\mathrm{\Omega }}{\int }\frac{\rho s}{T}\text{dV}.$

Using the divergence theorem, we get

$\underset{\mathrm{\Omega }}{\int }\frac{\partial }{\partial t}(\rho \eta )\text{dV}\ge -\underset{\mathrm{\Omega }}{\int }\mathrm{\nabla }\cdot (\rho \eta 𝐯)\text{dV}-\underset{\mathrm{\Omega }}{\int }\mathrm{\nabla }\cdot \left(\frac{𝐪}{T}\right)\text{dV}+\underset{\mathrm{\Omega }}{\int }\frac{\rho s}{T}\text{dV}.$

Since $\mathrm{\Omega }$ is arbitrary, we must have

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

Expanding out

$\frac{\partial \rho }{\partial t}\eta +\rho \frac{\partial \eta }{\partial t}\ge -\mathrm{\nabla }({\rho }_{\eta })\cdot 𝐯-\rho \eta (\mathrm{\nabla }\cdot 𝐯)-\mathrm{\nabla }\cdot \left(\frac{𝐪}{T}\right)+\frac{\rho s}{T}$

or,

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

or,

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

Now, the material time derivatives of $\rho$ and $\eta$ are given by

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

Therefore,

$(\dot{\rho }+\rho \mathrm{\nabla }\cdot 𝐯)\eta +\rho \dot{\eta }\ge -\mathrm{\nabla }\cdot \left(\frac{𝐪}{T}\right)+\frac{\rho s}{T}.$

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

$\rho \dot{\eta }\ge -\mathrm{\nabla }\cdot \left(\frac{𝐪}{T}\right)+\frac{\rho s}{T}.$

Clausius-Duhem inequality in terms of internal energy In terms of the specific entropy, the Clausius-Duhem inequality is written as $\rho \dot{\eta }\ge -\mathrm{\nabla }\cdot \left(\frac{𝐪}{T}\right)+\frac{\rho s}{T}$ Show that the inequality can be expressed in terms of the internal energy as $\rho (\dot{e}-T\dot{\eta })-\mathit{\sigma }:\mathrm{\nabla }𝐯\le -\frac{𝐪\cdot \mathrm{\nabla }T}{T}.$

Proof:

Using the identity $\mathrm{\nabla }\cdot (\phi 𝐯)=\phi \mathrm{\nabla }\cdot 𝐯+𝐯\cdot \mathrm{\nabla }\phi$ in the Clausius-Duhem inequality, we get

$\rho \dot{\eta }\ge -\mathrm{\nabla }\cdot \left(\frac{𝐪}{T}\right)+\frac{\rho s}{T}\text{or}\rho \dot{\eta }\ge -\frac{1}{T}\mathrm{\nabla }\cdot 𝐪-𝐪\cdot \mathrm{\nabla }\left(\frac{1}{T}\right)+\frac{\rho s}{T}.$

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

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

Hence,

$\rho \dot{\eta }\ge -\frac{1}{T}\mathrm{\nabla }\cdot 𝐪+\frac{1}{T^2}𝐪\cdot \mathrm{\nabla }T+\frac{\rho s}{T}\text{or}\rho \dot{\eta }\ge -\frac{1}{T}(\mathrm{\nabla }\cdot 𝐪-\rho s)+\frac{1}{T^2}𝐪\cdot \mathrm{\nabla }T.$

Recall the balance of energy

$\rho \dot{e}-\mathit{\sigma }:\mathrm{\nabla }𝐯+\mathrm{\nabla }\cdot 𝐪-\rho s=0⟹\rho \dot{e}-\mathit{\sigma }:\mathrm{\nabla }𝐯=-(\mathrm{\nabla }\cdot 𝐪-\rho s).$

Therefore,

$\rho \dot{\eta }\ge \frac{1}{T}(\rho \dot{e}-\mathit{\sigma }:\mathrm{\nabla }𝐯)+\frac{1}{T^2}𝐪\cdot \mathrm{\nabla }T⟹\rho \dot{\eta }T\ge \rho \dot{e}-\mathit{\sigma }:\mathrm{\nabla }𝐯+\frac{𝐪\cdot \mathrm{\nabla }T}{T}.$

Rearranging,

$\rho (\dot{e}-T\dot{\eta })-\mathit{\sigma }:\mathrm{\nabla }𝐯\le -\frac{𝐪\cdot \mathrm{\nabla }T}{T}.$

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