# Space Shuttle Tiles

D H
Staff Emeritus
I was having trouble imagining many materials of type 2.
Helium II.

Q_Goest
Homework Helper
Gold Member
Hi DH,

• Low thermal conductivity, high heat capacity. This is just delaying the inevitable. The vehicle still fries, it just happens later when the heat pulse finally reaches the aluminum skin. The heat capacity needs to be low, too.
• Low thermal conductivity, low heat capacity. This is the only winning combination. The low conductivity delays and spreads out the heat pulse. The low heat capacity means that the aluminum skin of the vehicle won't heat up all that much when the heat pulse finally does reach the skin.

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.

For a steady state condtion, the kinetic energy of the shuttle is converted to heat energy in the air at the outer surface of the tiles, and this heat flows into the tiles. Let’s call this heating Qk due to it being the conversion of kinetic energy to heat. Under steady state conditions, this heat flows away from the surface of the tiles in 2 ways.
Radiation heat transfer: Qr = e C (Th4 – Tc4) A
Where e is emissivity of the hot surface, C is the Stefan-Bolzmann constant, A is area and Th and Tc are the temperatures of the hot tile surface and the cold surface respectively, the cold surface being the environment. Note that regardless of what thermal conductivity or heat capacity the tiles have, the heat rejected from the surface of the tile is only a function of the temperature.
Conductive heat transfer: Qc = k A dT / s
Where k is the thermal conductivity of the tile, A is area, dT is the difference in temperature between the hot outer surface and the inner surface of the tile and s is thickness of material. This heat is the heat entering the airframe, so the airframe must absorb that heat. More on this in a moment. Note again that heat capacity of the insulation doesn’t enter this heat transfer equation.

We can neglect convective heat transfer away from the tiles because that’s actually Qk which is the heat going into the tiles.

We can also neglect any change in the temperature of the airframe if we assume that the insulation system is designed such that the increase in the airframe temperature is negligable. In other words, the airframe is a heat reservoir or lump mass who’s temperature doesn’t change substantially during the shuttle’s decent. The heat influx from the thermal conductivity through the tile insulation certainly adds to the thermal energy the airframe has as a function of the airframe’s heat capacity. But the overall heat capacity of the airframe is large and if we also assume the conduction of heat away from the inner surface of the tile insulation is high, we can safely model the airframe as a lump mass.

From the above, we can apply conservation of energy to the heat transfer. In order for there to exist a steady state:
Qk = Qr + Qc

So for steady state heat transfer, the heat capacity of the material does not concern us. What we can glean from this analysis is that for two materials with identical thermal conductivity and emissivity but different heat capacity, the temperature Th will be the same and by default, the airframe temperature will be the same also since that is our heat sink. Note also that for a steady state condition, the thermal gradient between the hot outer surface of the tile and the airframe is linear, which again is independent of the heat capacity of the material. The next step is to examine what changes due to there being a transient.

We can break up the transient into the heating up phase and the cooling down phase. During the heating up phase, the difference is that there exists a nonlinear temperature gradient between the hot, outer surface and the cool airframe. The inside of the tile heats up at a rate depending on the heat capacity. During the heating up of the tile, as we look at the temperature on the outer surface, the temperature will drop quickly and flatten out to a temperature near the initial temperature of the tile. As time goes by, heat added by thermal conductivity raises each ‘layer’ (dx) of the tile as a function of its heat capacity, the lower the heat capacity, the more quickly the temperature profile reaches steady state. The higher the heat capacity, the slower the temperature profile reaches steady state. But at no time does the thermal gradient exceed the steady state profile. Each layer inside the insulation is at or below the temperature reached during steady state, the only difference being how quickly it reaches steady state. Remember, we’re comparing two tiles with equal emissivity and equal thermal conductivity, one having a higher heat capacity than the other.

The cooling down phase is slightly different. It starts with the linear temperature profile and as the outer surface cools down, the inner layers remain at the steady state profile until the heat can be removed from the outer surface. Heat still flows into the airframe from the tile and this heat flux is a linear function of the change in temperature dT at the inner surface of the insulation. In other words, the higher the rate of change of temperature at the inner surface of the insulation (dT), the higher the heat transfer rate at that surface. Note this is true throughout the transient and the steady state condition. The slope of the line, being the temperature profile, at the inner surface is proportional to the heat being transferred into our lump mass airframe.

So during this cool down phase, the hotter, outer layers of the insulation cool down first. This cooling results in the outer layers of the insulation getting cooler while the inner layers are largely unaffected. You can imagine a bell shaped curve being set up where before, there was a linear one from a high temperature to a low one. The shape of this bell curve will always be below the linear line, so the rate of heat transfer into the airframe at the inner surface will only drop over time as it cools. The rate at which it cools is only a function of the heat capacity of the insulation. The higher the heat capacity, the slower this curve will develop and the more heat energy the insulation retains over time. Although a higher heat capacity will mean that during cool down, the airframe will absorb more heat, this amount of heat is negligable. But also consider that a higher heat capacity will mean that during the heating phase, the amount of heat transfer to the airframe will be less. So assuming these two phases are roughly the same length of time, there is no change in the total amount of energy transmitted to the airframe.

I don’t see any way for heat capacity to significantly alter the total amount of thermal energy transferred to the airframe. The airframe (lump mass) is not significantly heated during reentry so any slight differences between the warm up and cool down transients can’t be noticeably. Getting back to the OP:
I came across an example figure in my first-year physics textbook depicting a tile of the same material used for the heat shield on the space shuttle. The tile is hot enough to be glowing red, and yet a person is holding it by the edges. The caption explains that this is due to the "extremely small thermal conductivity and small heat capacity of the material."
I think what the book is referring to when it talks about the heat capacity, is the ability of someone to hold this tile shortly after we remove the source of heat. I don't think it has anything to do with the function of the tile on the shuttle.

I guess I don’t understand how the heat capacity enters into this.

Thank you for this wonderfully written analysis. My own concerns about tile heat capacity, and of reaching the steady state condition, are resolved. The deciding factor seems to be that the airframe acts as an efficient large capacity heat sink with a capacity which is many times greater than that of tiles of the highest possible heat capacity. So the shuttle is safe, provided Qc is small.

One sentence, however, is not clear:

During the heating up of the tile, as we look at the temperature on the outer surface, the temperature will drop quickly and flatten out to a temperature near the initial temperature of the tile..

That seems to say there is effectively no temperature gradient between the outer surface and interior. Very unintuitive, as it suggests high conductivity at the very least. Could you have instead meant the temperature profile drops quickly and flattens towards the initial temperature as measured from the surface inwards?

But also consider that a higher heat capacity will mean that during the heating phase, the amount of heat transfer to the airframe will be less. So assuming these two phases are roughly the same length of time, there is no change in the total amount of energy transmitted to the airframe.

Very "cool" observation!

Thanks again!

-James

Q_Goest
Homework Helper
Gold Member
Hi James,
Q_Goest said: During the heating up of the tile, as we look at the temperature on the outer surface, the temperature will drop quickly and flatten out to a temperature near the initial temperature of the tile..
That seems to say there is effectively no temperature gradient between the outer surface and interior. Very unintuitive, as it suggests high conductivity at the very least. Could you have instead meant the temperature profile drops quickly and flattens towards the initial temperature as measured from the surface inwards?
I'll try again, it's hard to explain without drawing a graph.

Imagine the temperature gradient of the tile prior to heating. The temperature is at some low, 'ambient' temperature, let's say it's at 70 F. So if we graphed temperature as a function of the tile thickness, we would have a flat line at 70 F. For this graph, the x axis starts at 0 at the outer surface and ends at s which is the thickness of the tile. So the temperature is a flat line from 0 to s at 70 F. Compare this to when the temperature of the tile reaches the steady state value; the outer surface is at Th and the inner surface is at 70 F and between those two points is a straight line. So at x = 0, we have a temperature of Th and at x = s, the temperature is 70 F.

Now imagine what happens during the transient. It starts out with a flat line at 70 F and ends at steady state with a linear drop from Th to 70 F. Between those two times, what happens?. Once we start heating the outer surface, the outer surface (x = 0) immediately goes to Th but just a little bit inside the outer surface is still cool. As time goes on, the temperature at x = 0 stays at Th but inside the tile, the temperature rises till it reaches the linear, steady state temperature profile.

I was just trying to describe what happens to the temperature profile as a function of time. This change in the temperature profile will occur the same way regardless of what the heat capacity is of the insulation, but the change to steady state will be more rapid for a tile with a lower heat capacity.

I was just trying to describe what happens to the temperature profile as a function of time. This change in the temperature profile will occur the same way regardless of what the heat capacity is of the insulation, but the change to steady state will be more rapid for a tile with a lower heat capacity.

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.

D H
Staff Emeritus
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:

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
Homework Helper
Gold Member
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.

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.

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sophiecentaur
Gold Member
2020 Award
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
Homework Helper
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.

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
Homework Helper
Gold Member
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