Kvant Physics Problem 28

Two large parallel plates of area $A$ are separated by a distance $L$, with $L$ much smaller than the plate dimensions so edge effects are neglected.

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Problem

Two parallel plates are separated by a distance small compared to their size. Several thin, thermally well-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 $A$ are separated by a distance $L$, with $L$ much smaller than the plate dimensions so edge effects are neglected. The space between the plates is filled with a gas at uniform pressure $p$ and temperature field varying only in the direction perpendicular to the plates.

A set of $N$ thin partitions (screens) is inserted parallel to the plates. Each screen is assumed to be mechanically thin and in perfect thermal equilibrium internally due to its very high thermal conductivity, so each screen has a well-defined uniform temperature. The screens do not store significant heat compared to the gas and do not introduce additional thermal resistance inside themselves. The gas fills the $N+1$ gaps between adjacent surfaces.

The quantity of interest is the effective thermal conductance between the two plates, or equivalently the heat flux density $q$ for a given temperature difference $\Delta T = T_1 - T_2$ between the plates.

Two regimes are considered for the gas:

First, the hydrodynamic regime where the mean free path $\lambda$ satisfies $\lambda \ll d$, where $d$ is the characteristic spacing between adjacent screens.

Second, the free molecular regime where $\lambda \gg L$.

Physical Principles

In the hydrodynamic regime the gas obeys Fourier’s law of heat conduction

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

where $\kappa$ is the thermal conductivity of the gas.

Thermal resistance of a layer of thickness $\ell$ and area $A$ is

$R = \frac{\ell}{\kappa A}.$

When layers are connected in series, thermal resistances add,

$R_{\text{tot}} = \sum_i R_i.$

In the free molecular regime the heat transfer between two surfaces separated by a distance smaller than the mean free path is determined by ballistic transport. The heat flux between two isothermal surfaces at temperatures $T_a$ and $T_b$ has the form

$q = G (T_a - T_b),$

where $G$ depends on gas properties and pressure but is independent of separation distance as long as collisions in the bulk gas are negligible. The corresponding thermal resistance per gap is

$R_0 = \frac{1}{G}.$

For a sequence of ballistic gaps separated by thermally equilibrating screens, the resistances of each gap add in series.

Derivation

Regime 1: $\lambda \ll d$

In this regime the gas in each sublayer between adjacent surfaces is locally in equilibrium and heat transfer is diffusive. Each gap between neighboring surfaces behaves as a plane-parallel layer of thickness $d_i$.

Let the total distance between plates be $L$. With $N$ screens, the space is divided into $N+1$ layers. If the screens are evenly spaced, each layer has thickness

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

The thermal resistance of one layer is

$R_i = \frac{d_i}{\kappa A} = \frac{L}{(N+1)\kappa A}.$

The total resistance is the sum over all layers,

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

The effective thermal conductance is therefore

$K_{\text{eff}} = \frac{1}{R_{\text{tot}}} = \frac{\kappa A}{L}.$

The presence of the screens does not change the effective thermal conductivity in this regime.

Regime 2: $\lambda \gg L$

In this regime gas molecules travel ballistically between collisions with boundaries. Each gap between adjacent surfaces acts as an independent ballistic heat-transfer element.

For one gap, the heat flux is

$q = G (T_a - T_b),$

so the thermal resistance of one gap is

$R_0 = \frac{1}{G}.$

With $N$ screens, there are $N+1$ gaps. Each screen is in internal thermal equilibrium, so it enforces a well-defined intermediate temperature. The system becomes a series of $N+1$ ballistic conductors.

The total resistance is

$R_{\text{tot}} = (N+1)R_0.$

The effective conductance is

$K_{\text{eff}} = \frac{1}{(N+1)R_0} = \frac{G}{N+1}.$

Thus the thermal conductance decreases in proportion to the number of segments.

Result

For $\lambda \ll d$:

$K_{\text{eff}} = \frac{\kappa A}{L}.$

Substitution introduces no additional parameters since the result is identical to the case without screens. The thermal conductivity of the gas remains $\kappa$ and the effective conductance is

$\boxed{K_{\text{eff}} = \frac{\kappa A}{L} \ \text{W/K}}.$

For $\lambda \gg L$:

$K_{\text{eff}} = \frac{G}{N+1}.$

Expressed relative to the single-gap conductance $G$,

$\boxed{K_{\text{eff}} = \frac{G}{N+1} \ \text{W/K}}.$

Sanity Checks

In the diffusive regime the total thermal resistance depends only on the sum of layer thicknesses. The cancellation of $(N+1)$ confirms dimensional consistency since $\kappa A / L$ has units $\mathrm{W/K}$.

In the ballistic regime each screen introduces a new thermalization boundary, increasing the number of temperature discontinuities. The linear growth of resistance with $N+1$ matches the rule for series-connected conductances.

A potential error would arise from assuming that subdividing a diffusive medium changes the total resistance; this would incorrectly leave a factor of $N+1$ in the denominator. Another common mistake would be treating ballistic transport as distance-dependent, which would incorrectly introduce $L$ into $G$.