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Conduction preserves energy while entropy increases

Physical scope

Exact balance-law example with Fourier conduction on a periodic cell; no empirical measurement or claim about a named real fluid.

Example

Assume the Axiom of Choice: every set-indexed family of nonempty sets has a choice function. Let $L_{1}, L_{2}, L_{3} > 0$ be side lengths in metres and let $Q : = \prod_{j = 1}^{3} [0, L_{j}]$ be a periodic cell, with opposite faces identified as in the periodic boundary-data model. Let $ρ > 0$ be constant mass density in kilograms per cubic metre, $c > 0$ constant specific heat in joules per kilogram kelvin, $κ > 0$ constant thermal conductivity in watts per metre kelvin, $T_{*} > ∣ a ∣ > 0$ a reference temperature and amplitude in kelvins, and $m \in Z^{3} ∖ {0}$ . Put $k_{j} : = 2 π m_{j} / L_{j}$ in inverse metres, define $k : = (k_{1}, k_{2}, k_{3}) \in R^{3}$ , and put $λ : = κ ∣ k ∣^{2} / (ρ c)$ in inverse seconds. Choose $0 < δ < λ^{- 1} \log (T_{*} / ∣ a ∣)$ , put $I : = (- δ, \infty)$ , and, for $(t, x) \in I \times R^{3}$ , define

$T (t, x) : = T_{*} + a e^{- λ t} \cos (k \cdot x), u : = 0, σ : = 0, b : = 0, r : = 0, q : = - κ \nabla T,$

$e : = c T, s : = c \log (T / T_{*}) .$

Then $T > 0$ , these fields satisfy the smooth balance equations on the periodic cell, and total internal energy $\int_{Q} ρ e d x$ is constant. The physical entropy density satisfies

$\partial_{t} (ρ s) + div (q / T) = κ \frac{∣ \nabla T ∣^{2}}{T^{2}} \geq 0,$

and the total entropy $\int_{Q} ρ s d x$ is strictly increasing for every finite $t \geq 0$ . This is an exact conditional calculation for the stated balance-law and caloric model; it is not an experimental observation or a constitutive claim about a particular substance.

Example references

Powers, Lecture Notes on Intermediate Fluid Mechanics, §§4.4–4.5

Mei, MIT OpenCourseWare, Heat and Energy Conservation, §4.1

Facts & Assumptions

Given: The constants, periodic cell and smooth fields in the Example; the adopted periodic data; and the energy and entropy balance model; the Axiom of Choice.

[A1]

The model parameters satisfy $ρ, c, κ > 0$ , $T_{*} > ∣ a ∣ > 0$ , and $m \neq 0$ ; the caloric laws are $e = c T$ and $s = c \log (T / T_{*})$ .

[A4]

The Axiom of Choice: the adopted Axiom of Choice implies countable choice by applying its choice function to the set of values of a sequence of nonempty sets.

[A2]

Specified wall, thermal, periodic and initial data: opposite faces of the cell are identified by translations and the fields and required traces are periodic, so paired fluxes cancel.

[A3]

Single-component nonpolar continuum and material balances: the adopted material entropy inequality has outward entropy flux $q / T$ and supply $ρ r / T$ .

[F1]

Kinetic/internal/total energy and boundary work: in the smooth positive-density balance branch, $ρ D_{t} e = σ : D - div q + ρ r$ and total energy is conserved on compatible periodic domains when $r = 0$ and there is no nonconservative body work.

[F2]

Divergence on a bounded C1 Euclidean domain: under countable choice, a $C^{1}$ vector field on the closure of a bounded $C^{1}$ region has volume divergence integral equal to its outward surface flux.

Proof

1.1givenA1algebra

Since $∣ a e^{- λ t} ∣ < ∣ a ∣ e^{λ δ} < T_{*}$ for $t \in I$ , the formula gives $T (t, x) > 0$ . Direct differentiation gives $\partial_{t} T = - λ a e^{- λ t} \cos (k \cdot x)$ and $Δ T = - ∣ k ∣^{2} a e^{- λ t} \cos (k \cdot x)$ . The definition of $λ$ therefore yields $ρ c \partial_{t} T = κ Δ T$ .

1.2givenA1A2algebra

The integral of $\cos (k \cdot x)$ over $Q$ is zero because $m \in Z^{3} ∖ {0}$ , so $\int_{Q} ρ e d x = ρ c T_{*} L_{1} L_{2} L_{3}$ for all $t \geq 0$ . Equivalently, the periodic boundary flux has zero net integral.

2.1givenA1A2A4F1F2step 1.1

The constant density and zero velocity satisfy mass balance; $σ = b = 0$ satisfy momentum balance. Since $D = 0$ , $e = c T$ , $q = - κ \nabla T$ , and $r = 0$ , the internal-energy equation in [F1] reduces to $ρ c \partial_{t} T = κ Δ T$ , which holds by step 1.1. For the material formulation, choose a bounded $C^{1}$ ball $B$ containing $Q$ in its interior, take $U = R^{3}$ , $χ (t, a) = a$ and $ρ_{0} = ρ$ . Every admissible material subregion $V ⋐ B$ is stationary. Constant mass, zero momentum and zero stress give its mass, linear and angular balances. Integrating $ρ c \partial_{t} T + div q = 0$ over $V$ , with differentiation permitted by smoothness on compact closures and the divergence theorem [F2] under [A4], gives the adopted material total-energy balance. Thus the hypotheses of [F1], including $Q \subset B$ for all $t \in I$ , are met. Periodicity is [A2].

3.1givenA1A3A4F2step 1.1step 2.1algebra

Using $s = c \log (T / T_{*})$ and step 1.1, $\partial_{t} (ρ s) = ρ c \partial_{t} T / T = κ Δ T / T$ . Also, $div (q / T) = - κ Δ T / T + κ ∣ \nabla T ∣^{2} / T^{2}$ . Their sum is the stated nonnegative entropy-production density; since $r = 0$ , this verifies the local entropy inequality. On each stationary $V$ from step 2.1, integration and [F2] give $\frac{d}{d t} \int_{V} ρ s d x + \int_{\partial V} (q / T) \cdot n d S = \int_{V} κ ∣ \nabla T ∣^{2} / T^{2} d x \geq 0$ , which is exactly the adopted material inequality [A3]. Positivity of $T$ and compactness make all these fields and derivatives bounded on each integration region.

4.1givenA2step 3.1algebra∎

Integrate the local entropy relation in step 3.1 over $Q$ . Periodicity cancels the boundary integral of $q / T$ , as required by [A2], so $\frac{d}{d t} \int_{Q} ρ s d x = \int_{Q} κ ∣ \nabla T ∣^{2} / T^{2} d x$ . Because $a \neq 0$ and $k \neq 0$ , $\nabla T = - a e^{- λ t} \sin (k \cdot x) k$ is nonzero on an open subset of each cell; the integrand is positive there and nonnegative everywhere. The integral is strictly positive for every finite $t \geq 0$ .

Conduction preserves energy while entropy increases

Example

Example references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources