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Why Doesn't Thermal Radiation Render Thermionic Conversion Unable To Approach Carnot Efficiency?

Physics Asked on January 6, 2022

Thermionic Conversion follows the classic Richardson-Dushmann Equation for thermionic current as a function of temperature squared:

$$J_{RD} = A_0 T^2 expleft(-frac{phi}{k_B T}right)$$
where

  • $J_{RD}$ is the emission current density.
  • $T$ is the emission temperature.
  • $A_0$ is a material specific correction factor
  • $phi$ is the material’s environment-mediated work function, and
  • $k_B$ is is the Boltzmann constant

But thermal power loss is a function of the fourth power of temperature following the Stefan-Boltzmann law:
begin{equation}
P=Asigma T^4
end{equation}

where

  • $P$ is the power lost to thermal radiation
  • $T$ is the emission temperature.
  • $A$ is the area of thermal emission, and
  • $sigma$ is the Stefan-Bolzmann constant

Although I see a lot of talk about space charge buildup being the limiting factor on thermionic conversion efficiency, the power of 2 disadvantage from thermal radiation losses seems to make thermionic conversion a fundamentally impractical approach to approximating Carnot efficiency.

What am I missing?

One Answer

According to this article, "Thermionic Energy Conversion in the Twenty-first Century: Advances and Opportunities for Space and Terrestrial Applications", the thermal radiation actually is a design concern.

You mention two power laws, and yes it does imply that at high temperatures the efficiency must drop, but it also equally implies that at low enough temperatures, the effect of the thermal radiation must become negligible.

The power of the power law is not the only thing, and in evaluating whether a technology sinks or swims, it comes down to a practical question of what are the actual coefficients, etc., and what is the final tradeoff that gets made in engineering.

Answered by Nanite on January 6, 2022

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