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There is a fascinating factoid in this answer to MMRTG- thermal and electrical output:

The electric power of the thermocouples depends on cooling influenced by the environment. On a cold day on Mars it may be 120 W, on a warm day only 110 W. (At the beginning of mission).

I'd never though about specifically environmental effects on RTGs. Answers to Why are RTGs different colors? address incident sunlight, but that does not seem to be what's going on here, since there is a contrast between hot days and cold days, and of course except for sandstorms like this $\tau > 10$ blizzard that blot out the Sun, almost every day is sunny.

Question: Why exactly does curiosity's RTG make more electricity on cold days? Is the temperature of the atmosphere, or something else?

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    $\begingroup$ thermocouples work on temperature differences $\endgroup$ – Organic Marble Aug 9 at 15:43
  • $\begingroup$ @OrganicMarble "Is the temperature of the atmosphere, or something else?" There are several temperatures that the RTG is exposed to in this problem; what does "temperature of the environment" mean exactly? (cf. How cold is the Martian sky at night? Or the day for that matter?) $\endgroup$ – uhoh Aug 9 at 22:43
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    $\begingroup$ @OrganicMarble Imagine turning on the AC in a hot room with stone walls. The air can cool down quickly, but the hot walls are still radiating almost 500 W/m^2 towards you. What would be the "temperature of the environment" in that case? $\endgroup$ – uhoh Aug 9 at 23:10
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    $\begingroup$ Usually I'd say the atmosphere, since radiation is a lot less efficient than conduction and convection, but Mars' atmosphere is thin enough that radiation might actually be significant. $\endgroup$ – Aetol Aug 9 at 23:52
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    $\begingroup$ @OrganicMarble only half an answer, because the radioactive heater actually has constant heat power and not a constant temperature. If the cold side of the RTG gets hotter, the hot side will get hotter, too. $\endgroup$ – fraxinus Aug 11 at 12:48
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There is Carnot's theorem for the theoretical maximum efficiency of heat engines. It is valid not only for mechanical engines like steam engines or Stirling engines but also for solid state devices like the thermocouples used in RTGs.

The Carnot efficiency depends on the upper and lower working temperature.

$$ \eta = 1 - \frac {T_c}{T_h} $$

Tc is the cold temperature in Kelvin, Th the hot temperature of the process. To improve $\eta$, Tc should be small and Th large.

Real thermocouples are much less efficent than the theoretical limit, but they do deliver more voltage when the temperature diffference between hot and cold side is higher.

So an RTG works better if the cooling fins are exposed to a cold environment on Mars. The surface of Mars should be cold and the Sun below the horizon, so the fins could radiate the heat away more efficiently and thus the fins as well as the cold sides of the thermocouples gets colder. Heat transport by the very thin atmosphere would be small but cooling by radiation works better if there is no other hot radiating surface than the fins of the RTG itself.

enter image description here

from 40_ICT_NextGen_Presentation_August_2017_final.pdf

But even next generation thermocouples are far away from the theoretical limit. $\eta$ for Tc = 150 K and Th = 1275 K is 88 %. Even the best thermocouples efficiency is still less than 25 %. But their efficiency is better for a lower temperature of the cold side.

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    $\begingroup$ "Is the temperature of the atmosphere, or something else?" There are several temperatures that the RTG is exposed to in this problem; what does "temperature of the environment" mean exactly? (cf. How cold is the Martian sky at night? Or the day for that matter?) $\endgroup$ – uhoh Aug 9 at 22:39
  • $\begingroup$ Imagine turning on the AC in a hot room with stone walls. The air can cool down quickly, but the hot walls are still radiating almost 500 W/m^2 towards you. What would be the "temperature of the environment" in that case? $\endgroup$ – uhoh Aug 9 at 23:10
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    $\begingroup$ @uhoh whatever the RTG fins see. Some combination of air temperature, ground temperature, and sunlight depending on exact material properties and physical orientation. :) $\endgroup$ – hobbs Aug 9 at 23:32
  • $\begingroup$ @hobbs that's right and it is necessary to address that to at least some extent in an answer to "Why exactly does curiosity's RTG make more electricity on cold days? Is the temperature of the atmosphere, or something else?" $\endgroup$ – uhoh Aug 9 at 23:35
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    $\begingroup$ Glad to see basic thermodynamic principles still apply on Mars too. Carnot's theorem is something I didn't really grasp the significance of at first, but which keeps coming back. $\endgroup$ – Mast Aug 10 at 8:46
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All heat engines, whether mechanical or solid state, produce work based on heat flow across a temperature difference.

The maximum efficiency of a heat engine depends on how large that difference is.

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  • $\begingroup$ "Is the temperature of the atmosphere, or something else?" There are several temperatures that the RTG is exposed to in this problem; what does "temperature of the environment" mean exactly? (cf. How cold is the Martian sky at night? Or the day for that matter?) $\endgroup$ – uhoh Aug 9 at 22:39
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    $\begingroup$ Imagine turning on the AC in a hot room with stone walls. The air can cool down quickly, but the hot walls are still radiating almost 500 W/m^2 towards you. What would be the "temperature of the environment" in that case? $\endgroup$ – uhoh Aug 9 at 23:10
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Solar Irradiance

Mars has seasons, just like Earth. While Mars weather is significantly less interesting than Earth's, due to the thinner atmosphere, the average temperature difference between summer and winter can be more than 50 C. The difference between daytime high and low can be more than 120 C.

If we consider the three forms of heat transfer, we can probably eliminate conduction as a significant factor, given that Curiosity sits high off the ground on six wheels. Although the wheels and suspension are all metal, along with the body, the small contact surface area and relative isolation of the RTG suggest to me that the thermal path to ground is pretty poor.

Furthermore, NASA itself says that the Mars atmosphere acts more like a thermal insulator than a conductor, because of the low density and thermal conductivity of CO2. Therefore, we can eliminate convection as a significant source of thermal variation.

Thus, we are left with the varying solar irradiance impinging on the RTG itself, as well as the portion of the rover body which can transmit a meaningful amount of heat to it as the only significant sources of thermal variation. Both the day/night and seasonal cycles will produce large swings in temperature, as cited above.

The "fin-root temperature" of the RTG is only 157 C. If we look at the average high temps over the year as measured by Curiosity itself, we see a variation of almost 30 C. At a daytime high of 4 C, we are looking at a maximum Carnot efficiency of about: 1 - (277/430) ~= 36% vs. -23 C, which gives: 1 - (250/430) ~= 42%. Obviously, the colder weather improves the theoretical efficiency by up to 17%. Thus, it is reasonable that the actual efficiency is affected by closer to 10%.

Thermodynamics (Added via EDIT)

Curiosity has two temperature sensors: one which measures air temp, and one which measures IR from the ground as a proxy for ground temp. As you can see from the Mars weather report, the ground temp is about 5 C higher, on average, than the air temp. Thus, the difference in air vs. ground does not seem to be a huge contributor (given that the surface rock of Mars serves as an enormous thermal mass, it makes sense that it retains heat better than the atmosphere, leading to the higher average temp).

Calculations have been conducted by Randall Osczevski to estimate the "Earth Equivalent Temperature" (EET) on Mars, which is basically a "wind chill factor" to compare what a human might perceive the Martian weather to be like. Due to the thin atmosphere, Mars would "feel" much warmer than the -60 C nominal temperature, which means that the atmosphere would carry away far less heat by convection. In fact, solar + ground radiation dominates the effective temperature. Transiently, the ground and air temps may differ by up to 20 C, just as air on earth can vary in temperature by a considerable amount just a few tens of meters above the surface.

The atmospheric pressure near the surface of the Earth is on the order of 1000 mbar, whereas, the comparable pressure on Mars is more like 6-8 mbar (more than 100x lighter). I presumed that the lower pressure would result in a lower thermal conductivity, but CO2 appears to be fairly insensitive to pressure after all. Rather, the chemical composition accounts for a 2x difference between Earth (18.5 mW/K.m) and Mars (9.6 mW/K.m).

Conclusion

It's hard to say what the exact contributions of the air temp, ground temp, and direct irradiance are to the fin temperatures of the RTG, but it seems reasonable to consider the ground as a kind of "solar concentrator" in that IR emitted from the ground due to solar heating will impinge upon the RTG, raising its temp somewhat. Any IR coming from the air is surely negligible compared to the ground, leaving the air contribution to be the convection value. Given the EET calculations mentioned above, it seems that even this effect is relatively small.

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  • $\begingroup$ +1 For the "Mars temperatures" mentioned in the last paragraph, are they the temperature of the atmosphere (which you've argued may not present much heat transfer) or are they the temperature of the surface regolith? Lose soil has poor conductivity and and will cool down quickly. My hunch is that on a "warm day" the fins on the underside of the RTG are exposed to a warm surface, and on a "cold day" they are exposed to a cooler surface. What makes the surface warm or cool could be primarily the seasons but there must be at least a small improvement in heat removal (from both) for strong winds $\endgroup$ – uhoh Aug 10 at 2:05
  • $\begingroup$ Your answer omits the background temperature of the surroundings. About half the surroundings are going to be at the temperature of the ground, the other half will be influenced by the temperature of the air (except for the small bit of sky occupied by the sun). Variations in the temperatures of these will affect the equilibrium temperature of the RTG radiators, even under identical sunlight conditions. $\endgroup$ – Christopher James Huff Aug 10 at 2:09
  • $\begingroup$ Thank you for the edit! $\endgroup$ – uhoh Aug 10 at 8:22

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