Continuum mechanics/Entropy inequality

Clausius-Duhem inequality

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

\frac{d}{d t} (\int_{Ω} ρ η dV) \geq \int_{\partial Ω} ρ η (u_{n} - 𝐯 \cdot 𝐧) dA - \int_{\partial Ω} \frac{𝐪 \cdot 𝐧}{T} dA + \int_{Ω} \frac{ρ s}{T} dV .

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

ρ \dot{η} \geq - \nabla \cdot (\frac{𝐪}{T}) + \frac{ρ s}{T} .

Proof:

Assume that $Ω$ is an arbitrary fixed control volume. Then $u_{n} = 0$ and the derivative can be taken inside the integral to give

\int_{Ω} \frac{\partial}{\partial t} (ρ η) dV \geq - \int_{\partial Ω} ρ η (𝐯 \cdot 𝐧) dA - \int_{\partial Ω} \frac{𝐪 \cdot 𝐧}{T} dA + \int_{Ω} \frac{ρ s}{T} dV .

Using the divergence theorem, we get

\int_{Ω} \frac{\partial}{\partial t} (ρ η) dV \geq - \int_{Ω} \nabla \cdot (ρ η 𝐯) dV - \int_{Ω} \nabla \cdot (\frac{𝐪}{T}) dV + \int_{Ω} \frac{ρ s}{T} dV .

Since $Ω$ is arbitrary, we must have

\frac{\partial}{\partial t} (ρ η) \geq - \nabla \cdot (ρ η 𝐯) - \nabla \cdot (\frac{𝐪}{T}) + \frac{ρ s}{T} .

Expanding out

\frac{\partial ρ}{\partial t} η + ρ \frac{\partial η}{\partial t} \geq - \nabla (ρ_{η}) \cdot 𝐯 - ρ η (\nabla \cdot 𝐯) - \nabla \cdot (\frac{𝐪}{T}) + \frac{ρ s}{T}

or,

\frac{\partial ρ}{\partial t} η + ρ \frac{\partial η}{\partial t} \geq - η \nabla ρ \cdot 𝐯 - ρ \nabla η \cdot 𝐯 - ρ η (\nabla \cdot 𝐯) - \nabla \cdot (\frac{𝐪}{T}) + \frac{ρ s}{T}

or,

(\frac{\partial ρ}{\partial t} + \nabla ρ \cdot 𝐯 + ρ \nabla \cdot 𝐯) η + ρ (\frac{\partial η}{\partial t} + \nabla η \cdot 𝐯) \geq - \nabla \cdot (\frac{𝐪}{T}) + \frac{ρ s}{T} .

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

\dot{ρ} = \frac{\partial ρ}{\partial t} + \nabla ρ \cdot 𝐯; \dot{η} = \frac{\partial η}{\partial t} + \nabla η \cdot 𝐯 .

Therefore,

(\dot{ρ} + ρ \nabla \cdot 𝐯) η + ρ \dot{η} \geq - \nabla \cdot (\frac{𝐪}{T}) + \frac{ρ s}{T} .

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

ρ \dot{η} \geq - \nabla \cdot (\frac{𝐪}{T}) + \frac{ρ s}{T} .

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

ρ \dot{η} \geq - \nabla \cdot (\frac{𝐪}{T}) + \frac{ρ s}{T}

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

ρ (\dot{e} - T \dot{η}) - σ : \nabla 𝐯 \leq - \frac{𝐪 \cdot \nabla T}{T} .

Proof:

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

ρ \dot{η} \geq - \nabla \cdot (\frac{𝐪}{T}) + \frac{ρ s}{T} or ρ \dot{η} \geq - \frac{1}{T} \nabla \cdot 𝐪 - 𝐪 \cdot \nabla (\frac{1}{T}) + \frac{ρ s}{T} .

Now, using index notation with respect to a Cartesian basis $𝐞_{j}$ ,

\nabla (\frac{1}{T}) = \frac{\partial}{\partial x_{j}} (T^{- 1}) 𝐞_{j} = - (T^{- 2}) \frac{\partial T}{\partial x_{j}} 𝐞_{j} = - \frac{1}{T^{2}} \nabla T .

Hence,

ρ \dot{η} \geq - \frac{1}{T} \nabla \cdot 𝐪 + \frac{1}{T^{2}} 𝐪 \cdot \nabla T + \frac{ρ s}{T} or ρ \dot{η} \geq - \frac{1}{T} (\nabla \cdot 𝐪 - ρ s) + \frac{1}{T^{2}} 𝐪 \cdot \nabla T .

Recall the balance of energy

ρ \dot{e} - σ : \nabla 𝐯 + \nabla \cdot 𝐪 - ρ s = 0 ⟹ ρ \dot{e} - σ : \nabla 𝐯 = - (\nabla \cdot 𝐪 - ρ s) .

Therefore,

ρ \dot{η} \geq \frac{1}{T} (ρ \dot{e} - σ : \nabla 𝐯) + \frac{1}{T^{2}} 𝐪 \cdot \nabla T ⟹ ρ \dot{η} T \geq ρ \dot{e} - σ : \nabla 𝐯 + \frac{𝐪 \cdot \nabla T}{T} .

Rearranging,

ρ (\dot{e} - T \dot{η}) - σ : \nabla 𝐯 \leq - \frac{𝐪 \cdot \nabla T}{T} .