Clausius–Duhem inequality

This inequality is a statement concerning the irreversibility of natural processes, especially when energy dissipation is involved. It was named after the German physicist Rudolf Clausius and French physicist Pierre Duhem.

Clausius–Duhem inequality in terms of the specific entropy

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

${\displaystyle \frac{d}{dt}\left({\int }_{\mathrm{\Omega }}\rho \eta dV\right)\ge {\int }_{\mathrm{\partial }\mathrm{\Omega }}\rho \eta \left(u_n-\mathbf{v}\cdot \mathbf{n}\right)dA-{\int }_{\mathrm{\partial }\mathrm{\Omega }}\frac{\mathbf{q}\cdot \mathbf{n}}{T}dA+{\int }_{\mathrm{\Omega }}\frac{\rho s}{T}dV.}$

In this equation ${\displaystyle t}$ is the time, ${\displaystyle \mathrm{\Omega }}$ represents a body and the integration is over the volume of the body, ${\displaystyle \mathrm{\partial }\mathrm{\Omega }}$ represents the surface of the body, ${\displaystyle \rho }$ is the mass density of the body, ${\displaystyle \eta }$ is the specific entropy (entropy per unit mass), ${\displaystyle u_n}$ is the normal velocity of ${\displaystyle \mathrm{\partial }\mathrm{\Omega }}$, ${\displaystyle \mathbf{v}}$ is the velocity of particles inside ${\displaystyle \mathrm{\Omega }}$, ${\displaystyle \mathbf{n}}$ is the unit normal to the surface, ${\displaystyle \mathbf{q}}$ is the heat flux vector, ${\displaystyle s}$ is an energy source per unit mass, and ${\displaystyle T}$ is the absolute temperature. All the variables are functions of a material point at ${\displaystyle \mathbf{x}}$ at time ${\displaystyle t}$.

In differential form the Clausius–Duhem inequality can be written as

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

where ${\displaystyle \dot{\eta }}$ is the time derivative of ${\displaystyle \eta }$ and ${\displaystyle \mathbf{\nabla }\cdot (\mathbf{a})}$ is the divergence of the vector ${\displaystyle \mathbf{a}}$.

Proof

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

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

Using the divergence theorem, we get

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

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

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

Expanding out

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

or,

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

or,

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

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

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

Therefore,

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

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

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