Kvant Physics Problem 28

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Problem

Two parallel plates are separated by a distance small compared to their dimensions. Several thin, well thermally conducting partitions—screens—are placed between the plates. How does this affect the thermal conductivity between the plates if:

1. the mean free path of the gas molecules filling the space between the plates, i.e. the distance traveled by gas molecules between two collisions, is small compared to the distance between the screens;
2. the mean free path of the gas molecules is large compared to the distance between the plates?

Setup and Assumptions

Two large parallel plates of area $S$ are separated by distance $L$, with $L$ much smaller than the lateral dimensions so edge effects are neglected. The plates are maintained at temperatures $T_1$ and $T_2$, with a small temperature difference $|T_1 - T_2| \ll T$ so that linear transport theory applies.

Between the plates lies a rarefied gas. $N$ thin partitions (screens) are inserted parallel to the plates. Each screen is assumed to be a perfect thermal conductor, so its temperature is uniform over its surface and equal to the local equilibrium temperature imposed by energy exchange with the gas on both sides. The screens are assumed not to exchange heat directly between different positions along the plane and not to obstruct molecular motion except by subdividing the volume.

The gas is characterized by molecular mean free path $\lambda$. Two regimes are considered: $\lambda \ll d$, where $d$ is the spacing between adjacent screens, and $\lambda \gg L$.

The objective is to determine how the effective thermal conductivity between the plates depends on the presence of the screens in both regimes.

Physical Principles

Fourier’s law of heat conduction in the continuum regime is

$q = -\kappa \frac{dT}{dx},$

where $q$ is the heat flux density and $\kappa$ is the thermal conductivity of the gas.

In the diffusive regime, thermal resistances of layers in series add according to

$R = \frac{L}{\kappa S}.$

In the free molecular (Knudsen) regime, heat transfer between two surfaces is governed by molecular transport between wall collisions. The heat flux between two parallel surfaces at temperatures $T_a$ and $T_b$ takes the form

$q = h_K (T_a - T_b),$

where $h_K$ is the Knudsen heat transfer coefficient, independent of separation distance when $\lambda \gg L$, and determined by molecular flux and energy exchange at surfaces.

Thermal resistances in series still add in the free molecular regime when intermediate isothermal surfaces are present, since each gap exchanges energy independently through molecular flights between adjacent boundaries.

Derivation

The total separation $L$ is divided by $N$ screens into $N+1$ identical gas layers, each of thickness

$d = \frac{L}{N+1}.$

Regime 1: $\lambda \ll d$

In this regime, molecular collisions dominate over wall-to-wall transport within each layer, so Fourier’s law applies locally in each gas segment. Each segment behaves as a homogeneous conductor with conductivity $\kappa$.

The thermal resistance of one layer is

$R_1 = \frac{d}{\kappa S} = \frac{L}{(N+1)\kappa S}.$

Since the layers are in series, the total resistance is

$R_{\text{tot}} = (N+1) R_1 = (N+1)\frac{L}{(N+1)\kappa S} = \frac{L}{\kappa S}.$

The screens do not contribute additional resistance because their temperature is continuous across each surface and they are assumed perfectly conducting, enforcing only local equilibrium without adding thermal drop.

The effective conductivity satisfies

$R_{\text{tot}} = \frac{L}{\kappa_{\text{eff}} S},$

hence

$\kappa_{\text{eff}} = \kappa.$

Regime 2: $\lambda \gg L$

In this regime, heat is transported by molecules traveling ballistically between surfaces. Each gas region between adjacent boundaries behaves as a Knudsen layer with heat flux

$q = h_K (T_i - T_{i+1}).$

For one gap,

$R_K = \frac{1}{h_K S}.$

With $N+1$ gaps in series,

$R_{\text{tot}} = \frac{N+1}{h_K S}.$

The total heat flux between the plates is then

$q = \frac{T_1 - T_2}{R_{\text{tot}} S} = \frac{h_K}{N+1}(T_1 - T_2).$

Defining an effective conductivity over distance $L$ via

$q = \kappa_{\text{eff}} \frac{T_1 - T_2}{L},$

gives

$\kappa_{\text{eff}} = \frac{L h_K}{N+1}.$

For the case without screens,

$\kappa_0 = L h_K.$

Therefore,

$\kappa_{\text{eff}} = \frac{\kappa_0}{N+1}.$

Result

For $\lambda \ll d$,

$\boxed{\kappa_{\text{eff}} = \kappa}.$

For $\lambda \gg L$,

$\kappa_0 = L h_K,\qquad \kappa_{\text{eff}} = \frac{\kappa_0}{N+1} = \frac{L h_K}{N+1}.$

Sanity Checks

Dimensional consistency follows since $\kappa$ has units $\mathrm{W,m^{-1},K^{-1}}$, while $L h_K$ also has units $\mathrm{W,m^{-1},K^{-1}}$ because $h_K$ has units $\mathrm{W,m^{-2},K^{-1}}$.

In the diffusive limit, cancellation of $(N+1)$ reflects that subdividing a homogeneous Fourier conductor into identical segments does not change total resistance, since resistances scale linearly with length.

In the free molecular limit, each additional screen introduces an extra independent thermalization boundary, producing a linear increase in total thermal resistance proportional to $N+1$, so conductivity decreases inversely with the number of gaps.

A sign or unit error would most easily occur in converting between conductance per area $h_K$ and bulk conductivity via multiplication by $L$, since one involves per-area flux and the other volumetric transport over a finite distance.