# How much temperature control would an EVA suit on Mars need?

The average surface temperature on Mars is -55°C and it can go down to -130°C. These are temperatures which sound deadly at first. But on the other hand, the atmospheric pressure on Mars is just 0.63% of that on Earth.

Wouldn't that mean that the thermal conductivity of the atmosphere would be very low? If yes, wouldn't this mean that the space suit wouldn't actually lose much thermal energy to the environment? So could an astronaut stay warm just with their own body heat and maybe a passive insulation layer, or would an active heating system be required?

• ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20140003958.pdf - EVA suit discussion for the whole gamut of missions. – Deer Hunter Dec 5 '14 at 23:58
• @DeerHunter Would you like to look for the relevant information in that document and formulate it as an answer? – Philipp Dec 6 '14 at 3:21
• It is relatively easy to greatly reduce radiative heat loss through a reflective layer (like a space blanket), essentially reflecting the infrared radiation back in instead of letting it escape into the environment. – Blake Walsh Jan 7 '17 at 0:26

You could certainly design an EVA suit that has a low enough thermal conductivity (or effective thermal conductivity) such that it has a loss equal to the typical thermal output from a human body.

We can estimate heat output from the human body based on the typical (suggested) calorific intake.

$$2,000\text{ cal} = 8.37\times10^6\text{ J}$$

If we assume all this is turned into heat energy, then we can design a suit to match this energy input. According to Wikipedia the average surface area of the human body is $1.73\ m^2$.

So we would want to design a suit with a thermal flux of $4.84\times10^6\text{ J}/m^2/\text{day}$.

Joules/day is kind of a messy unit system. We can tidy that up and turn into joules / second, or watts.

$$4.84\times10^6\text{ J}/m^2/\text{day} = 4.84\times10^6\times(1/(60\times60\times24))\mathrm{\ W}/m^2$$

That leaves us with a value of almost exactly $56\mathrm{\ W}/m^2$.

So to design for the worst case scenario (coldest surroundings) you have both conduction from the wall to the atmosphere, and radiation.

The convection from the outside wall of the suit is (derived from Fourier's law):

$$\mathrm{Q} = κA(T_{\mathrm{hot}}-T_{\mathrm{cold}})/d$$

Where:

• $\mathrm{Q}$ = rate of heat transfer (in watts)
• $κ$ = thermal conductivity of the wall (in watts per meter kelvin)
• $A$ = heat transfer surface area (in square meters)
• $T_{\mathrm{hot}}$ = hot temperature (in kelvins)
• $T_{\mathrm{cold}}$ = cold temperature (in kelvins)
• $d$ = distance to ambient temperature (in meters)

The radiation can be calculated with the following equation (derived from the Stefan-Boltzmann law):

$$q = \sigma \times A \times \Delta T^4$$

Where:

• $q$ = heat transfer per unit time (in watts)
• $\sigma$ = Stefan–Boltzmann constant ($5.670373 \times 10^{-8}\, \mathrm{J\, s^{-1}m^{-2}K^{-4}}$)
• $A$ = total surface area of the emitting body (in square meters)
• $\Delta T$ = $T_{\mathrm{hot}}-T_{\mathrm{cold}}$ (in kelvins)

Depending on the design of your suit the internal heat transfer is going to happen in one of two ways, either conduction though material or radiation, if the design is MLI based.

So is it possible to build an EVA suit that doesn't require an external heat source? Yes. It's a question of the mass of the resulting suit and whether the human inside could move around in the potentially heavy suit.

• "It's a question of the mass of the resulting suit and whether the human inside could move around in the potentially heavy suit." - Do you have an answer to this question? Because you are not touching the core of my question: What's the different in heat loss between the environments of Earth and Mars? – Philipp Dec 26 '14 at 4:09
• @Philipp the conductive heat loss will be lower than in earth, the radiated heat loss will be higher. Outside of that it really depends on the design of the suit. – ThePlanMan Dec 26 '14 at 11:10
• The design of a suit could not be based on the mean heat production for a whole day only. A human at rest produces 70 to 100 W of heat energy. But at work the heat production increases. When a human produces a mechanical power of 150 W, the heat production is about 450 W. If the suit does not require an external heat source, it would require an adaptive cooling system to remove 0 to about 500 W of heat from the suit. – Uwe Jan 9 '17 at 13:03
• @Uwe I have one simple response to your comment - heat sinks. All materials have a specific heat capacity (the amount of energy required to raise their temperature by one degree). It would be reasonable to assume you can sink a couple of hundred if not thousand watts into the suit itself. Add to that a few copper (for example) plates (on the back for example) to facilitate heat transfer and you're golden (or copper colored I guess). :D – ThePlanMan Jan 11 '17 at 0:54
• @ThePlanMan: But no heat sinks were used for the Apollo suits on the moon. The cooling was done by evaporating water. If the heat sinks temperature rises by solar radiation above some 10 to 30 °C they are useless to remove heat from the suit. You would need a heavy heat pump to transfer the heat from the cooling garment in the suit to the heat sinks and to overcome the temperature difference between both. But the heat pump would need extra energy for operation. – Uwe Jan 11 '17 at 12:21