Explaining the Science of Heat Shield Tiles on the Space Shuttle

[D H]

Hi DH,

I guess I don’t understand how the heat capacity enters into this. Let’s look at the steady state condition first, then consider how that differs from a transient condition.

This analysis is somewhat mistaken. There was no steady state condition. The structure would have failed well before steady state was reached were the Shuttle to be subject to thousands upon thousands of acetylene torches aimed at the Shuttle's belly for hours on end.

The Shuttle was not an infinitely massive heat sink. Every ounce counts in spaceflight. The Shuttle structure did heat up some during reentry. The Shuttle was designed to tolerate realistically worst-case transients, but not your steady state scenario. Peak heating was a short lived phenomenon, about ten minutes or so in duration.

This picture of the last reentry of Atlantis as viewed from space is just too cool:

Click here to embiggen.

The Shuttle appears to be near the start of its first S turn in this photo. Those four S turns were where the Shuttle dumped a good bit of its speed and were also where the Shuttle was subject to peak heating.

 

[Q_Goest]

Ok. We start with the temperature at the surface at T_h, but temperature is still ambient immediately beneath, where it subsequently rises at a rate proportional to the rate of increase in heat density and inversely proportional to heat capacity. This increases the temperature gradient further into the tile, where the above described increase in temp repeats. Hence heat capacity inversely affects the time required for temperature profile reach steady state. Makes sense.

Thank you.

Yes, that’s correct. Note that for tiles that have a relatively low heat capacity such as the Shuttle, steady state will be reached much sooner than for tiles with a higher heat capacity.

Even in the initial transient condition during which the tiles heat up, the Shuttle airframe is subjected to this heat flux much sooner with the tiles it has regaredless of whether or not it comes to steady state. Clearly, suggesting the lower heat capaciity of the shuttle tiles is a benefit to reducing this heat flux is misinformed.

 

[James_Harford]

This analysis is somewhat mistaken. There was no steady state condition. The structure would have failed well before steady state was reached were the Shuttle to be subject to thousands upon thousands of acetylene torches aimed at the Shuttle's belly for hours on end.

The Shuttle was not an infinitely massive heat sink. Every ounce counts in spaceflight. The Shuttle structure did heat up some during reentry. The Shuttle was designed to tolerate realistically worst-case transients, but not your steady state scenario. Peak heating was a short lived phenomenon, about ten minutes or so in duration.

I would agree with all of your points, except the conclusion. I think that by "steady state" Q_Goest is referring to the temperature profile of the tile, and is able to treat the shuttle as an infinitely massive heat sink precisely because peak heating is short lived -- that and the low rate of maximum (i.e. steady state) heat flow into the airframe. If peak heating ends before steady state is reached, so much the better, but as he points out, the (extremely?) low heat capacity of the tile makes that unlikely. So it would seem that low tile conductivity and relatively high heat capacity of the airframe are the key elements that protect the shuttle, with tile capacity having an insignificant role.

 

[sophiecentaur]

There is an excellent electrical analogue to this, in the form of series resistors and shunt capacitors. A capacitor represents heat capacity and series resistors represent the thermal resistance. Applying a voltage pulse of appropriate duration and amplitude at the input can represent the sudden but brief temperature rise on the skin of the tile. The exercise is to limit the maximum voltage across the load resistor.

For a better model, you can take a series of Pi sections of R and C.
Clearly , what happens to the voltage on the output load will depend upon the detailed values of the various components. If you want to limit the output voltage, one way would be to have a huge capacitor and a huge series resistor - but you can't, in the thermal version. You have to go for a compromise.

 

[AlephZero]

There is some possibility of confusion about what "steady state solution" means here. I would take it NOT as meaning uniform temperature everywhere and no heat flow -any form of insulation would delay reaching that state, but would not change the damage caused when it was reached. Rather, I would take it as meaning a uniform heat flux through the thickness of the tile, and therefore a linear temperature gradient through the tile.

The heat flux in that condition is determined only by the conductivity of the tile, not the thermal capacity. Increasing the thermal capacity changes the situation in two ways, one good and one bad, and the question is whether you can live with the bad way.

The good change is delaying the increase in heat flux at the inner surface of the tile, by storing heat in the tile. the bad change is that when the external heat source is removed, the heat stored in the tile is conducted away in both directions - into the shuttle as well as back into the atmosphere.

Inventing some numbers (hopefully plausible numbers, but they are guesses), if the external temperature is 1100C, heating up the middle of the tile to 600C while the inside surface only reaches 100C seems like a good idea - except that if the external temperature then drops back to 100C, only half of the heat stored at 600C is going to diffuse back into the air, and the other half is going to end up in the shuttle skin. If the temeperature limit for the skin is say 200C, you need the thermal capacity of the tile much lower than the thermal capacity of the skin.

 

[James_Harford]

There is some possibility of confusion about what "steady state solution" means here. I would take it NOT as meaning uniform temperature everywhere and no heat flow -any form of insulation would delay reaching that state, but would not change the damage caused when it was reached. Rather, I would take it as meaning a uniform heat flux through the thickness of the tile, and therefore a linear temperature gradient through the tile.

Q_Goest has given the most detailed analysis on this question, which I believe is in good agreement with all of your points, except for the significance of the heat capacity of the tile. I found it very persuasive in spite of my earlier concerns on that exact point.

 

[Q_Goest]

Thanks AlephZero, I agree. Maybe one minor point:

the bad change is that when the external heat source is removed, the heat stored in the tile is conducted away in both directions - into the shuttle as well as back into the atmosphere.

The rate of heat transfer at the inner surface duiring 'steady state' (or any state during which the thermal gradient is aproximately linear as I've mentioned earlier) is always going to be higher than during this cool down period. We can be sure of this because the rate of heat transfer due to thermal conductivity is linearly proportional to dT/s at the inner surface. That's the slope of the temperature gradient at the inner surface. That slope only decreases as we go from a linear temperature gradient to one experienced during the transient. Agreed?

 

[AlephZero]

Agreed. The main point I was making was that the heat stored in the thermal capacity of the tile has to go somewhere, and it doesn't all go back out to the atmosphere just because that would a nice thing to happen.

I don't know enough about the shuttle tile system to estimate the thermal time constant (i.e the time for the system to approach a steady state condition) compared with duration of the heat input, so I don't know how relevant your refinement of the argument is in practice.