How to determine heat transfer for survival on Venus?

Question

I'm trying to determine how long a probe could survive on Venus' surface. Assuming its just a sphere of titanium with room temperature air inside, what would be the best method for determining the heat transfer between the atmosphere (superheated carbon dioxide) and the inside of the sphere? How long will it take to reach equivalent temperature?

Answer 1

Summary: The time will depend on how the inside of the spacecraft is insulated, but if we assume that you are in contact with the metal shell of a spacecraft similar to the lunar module (and make a lot of approximations regarding convection in the Venusian atmosphere), you will get serious burns within 15 minutes. The assumptions I make break down as the temperature of the spacecraft approaches the outside temperature (convective transfer slows), but you can probably expect it to reach oven temperature (350 °F, 450 K) within two hours.

First we need to understand what kind of convective regime we are in. The atmosphere of Venus is mostly carbon dioxide (96.5% by volume?) in the supercritical phase, so I'm going to rely a lot on its properties.

Let's first compute the Rayleigh number, which tells us whether the natural convection will be turbulent.

$$\textsf{Ra} = \frac{\rho g \beta}{\eta \alpha} (T-T_\mathrm{env}) D^3$$

Here is the data I'm using.
Atmospheric density: $\rho = 65~\mathrm{kg/m^3}$ (source)
Acceleration of gravity: $g = 8.87~\mathrm{m/s^2}$ (source)
Temperature of environment: $T_\mathrm{env} = 737~\mathrm{K}$ (source)
Temperature of spacecraft surface: $T = 294~\mathrm{K}$
Film temperature (used in calculating the quantities below): $T_f = (T + T_\mathrm{env})/2 = 515.5~\mathrm{K}$
Thermal expansion coefficient at $515.5~\mathrm{K}$: $\beta = 0.00194~\mathrm{K}^{-1}$ (ideal gas law)
Atmospheric pressure: $p = 92~\mathrm{bar}$ (source)
Dynamic viscosity of CO₂ at 515.5 K and 92 bar: $\eta = 2.50 \times 10^{-5}~\mathrm{N \cdot s/m^2}$ (source)
Thermal conductivity of CO₂ at 100 bar and 450 K: $k = 0.03392~\mathrm{W/(m \cdot K)}$ (source)
Isobaric specific heat of CO₂ at 500 K: $c_p = 1014~\mathrm{J/(kg \cdot K)}$ (source)
Thermal diffusivity of atmosphere: $\alpha = k/(\rho c_p) = 5.15 \times 10^{-7}~\mathrm{m^2/s}$
Characteristic size: $D = 6~\mathrm{m}$

I end up with $\mathsf{Ra} = 8.33\times 10^{15}$, which is more than sufficient for Rayleigh–Bénard convection cells to form.

I can't find relations for heat transfer coefficients for a sphere, so I'm just going to treat the spacecraft as a horizontal plate. Our Rayleigh number is also out of the range of validity for this equation (supposed to be $<3\times10^{10}$), but I'm going to ignore that too.

$$h = \frac{0.27 k}{D}\,\mathsf{Ra}^{1/4} \sim 1.5~\mathrm{W/(K \cdot m^2)}$$

How long this takes for this heat transfer coefficient to kill you will depend a lot on the interior structure of the spaceship and possibly convection inside the spaceship. You will survive a lot longer if you are in the center of the spaceship surrounded by foam insulation. I don't really want to try to address that, so let's just assume your body is in contact with the titanium shell. I'm also going to ignore the fact that the convection field outside the spacecraft will dissipate as it cools off (you will probably die before that effect is significant anyway). The specific heat of titanium also changes as it warms up, but let's just use a constant value.

Specific heat of Ti: $c = 0.52~\mathrm{J/(g \cdot K)}$
Spacecraft area: $A= 113~\mathrm{m}^2$
Spacecraft thickness: $L = 1.5~\mathrm{cm}$
Density of Ti: $L = 4.5~\mathrm{g/cm^3}$
Spacecraft mass: $AL\rho_\mathrm{Ti}$ = $2545~\mathrm{kg}$ (similar to dry mass of the lunar module ascent stage)

The characteristic time for Newton's law of cooling is: $$\tau = \frac{mc}{hA} \approx 130.1~\mathrm{minutes}$$

The temperature of the spacecraft shell as a function of time is: $$T(t) = T_\mathrm{env} + \left( T(0) - T_\mathrm{env} \right) \mathrm{e}^{-t/\tau}$$

The time at which the spacecraft shell reaches a temperature $T$ in kelvin is: $$t_T = -\tau \ln \left( \frac{737-T}{443}\right)$$

Burns develop on skin within a second for temperatures above 343 K, so you will start getting severe burns on any portion of your body in contact with the spacecraft shell within 15 minutes.

Edit: I'm also ignoring wind. Winds at the surface of Venus are supposed be pretty still ($<2~\mathrm{m/s}$). But if there are high winds, expect to roast more quickly.

Answer 2

The question seems primarily interested in the rate of heat transfer from Venus to the spherical cow spacecraft via conduction and convection types of transfer. That includes both inelastic collisions of atmospheric molecules with the sphere and also adhesion of hot particulates and/or droplets of any aerosol if that happens.

The high density of the atmosphere means that atomic collisions will happen about 160x more frequently than in Earth's atmosphere; 100x more due to density and 1.6x more due to speed (flux ~ areal density x velocity).

This kind of calculation is central to engineering any aerocraft or lander that may go to bodies with atmospheres. In addition to hot ones like Venus there are cold ones like Titan; either way there's heat transfer from the craft to the environment and it's very important to have a realistic estimate.

That being said, I don't know!

However, the Stefan–Boltzmann constant $\sigma$ is 5.67E-8 W/m²/K⁴ It's the only constant besides pi that I can remember four digits of off the top of my head, because it's got 5, 6, 7, 8

and the Stefan–Boltzmann law says that a sphere of radius $R$ will receive a radiant power of something of the order of

$$4\pi R^2 \sigma T^4$$

A sphere with a volume of 1 cubic meter has a diameter of 1.24 meters. That's about the same size as a one ton spherical cow.

It has an area of about 4.84 square meters. Thinking of the Venusian surface and atmosphere as a blackbody cavity, there will be 73 kilowatts of radiant power incident on our cow sphere!

If it's fairly highly reflective over most of its area, maybe one can get that down to 10 kW (a surface-averaged reflectivity of thermal IR of 85%). It's going to have to be smooth, shiny metal, to do that, and the metal will have to withstand Venus' chemical environment.

If our sphere were indeed a spherical cow with the heat capacity of water needing 4200 Joules for each kg to be raised by one degree, then our spacecraft would warm one degree every 7 minutes, or 100 degrees per every 11.7 hours.

Make all of that 7 times shorter if the spacecraft is dark/absorbing for thermal IR.

While not the central theme of the question, the other answers that will hopefully be posted can be compared to this.