# Space Shuttle Tiles

Staff Emeritus
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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."

Small thermal conductivity makes sense. You want something that can insulate the orbiter (which is apparently made of aluminium, at least in part) from the heat from the shock front of compressed gas during re-entry. But I'm confused about the small heat capacity. My first instinct is that you would want a material with a large heat capacity so that you could dump a lot of energy into it without it heating up too much. I'm not sure where that reasoning goes wrong. The only other thought I've had is that maybe you want the tiles to heat up quickly to the temperature of the gas, so that there won't be any subsequent heat transfer to the tiles, which are already in thermal equilibrium with the gas.

Can anyone explain the true reasoning to me?

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IF you are going to dump lots of heat into it, that heat will get radiated somewhere ...
guess where a lot will go ?

yeah out the other side of the tile and into the orbiter framework ... that can only end badly !

cheers
Dave

Staff Emeritus
Gold Member
IF you are going to dump lots of heat into it, that heat will get radiated somewhere ...
guess where a lot will go ?

yeah out the other side of the tile and into the orbiter framework ... that can only end badly !

cheers
Dave

I don't get it. If it has a large heat capacity, that means that dumping a lot of heat into it does not raise its temperature too much. The amount by which it radiates depends on T^4. So what am I missing?

Staff Emeritus
What you're missing is that it is the atmosphere, not the Shuttle, that is dumping the heat. If the tiles had a high heat capacity, all of that thermal energy from the atmosphere streaming past would be transferred through the tiles to the underlying structure. Thanks to the low heat capacity, there was very little transfer of energy from the atmosphere to the Shuttle's structure.

Gold Member
I don't get it. If it has a large heat capacity, that means that dumping a lot of heat into it does not raise its temperature too much. The amount by which it radiates depends on T^4. So what am I missing?

As Davenn said you do not want them conducting inwards to the shuttle. With a low heat capacity them heat up fast as the shuttle enters the atmosphere. After the entry they cool down quickly.

Staff Emeritus
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What you're missing is that it is the atmosphere, not the Shuttle, that is dumping the heat.

No, I do get that that is the situation. I just am not very good at applying physics to it, I guess.

If the tiles had a high heat capacity, all of that thermal energy from the atmosphere streaming past would be transferred through the tiles to the underlying structure. Thanks to the low heat capacity, there was very little transfer of energy from the atmosphere to the Shuttle's structure.

Why not? I.e. why would the energy not be transferred anyway?

As Davenn said you do not want them conducting inwards to the shuttle. With a low heat capacity them heat up fast as the shuttle enters the atmosphere. After the entry they cool down quickly.

He was worried about them radiating energy into the shuttle, not conducting it. In any case we know they won't conduct, because they have been designed to be poor thermal conductors.

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Staff Emeritus
Gold Member
Okay, I've thought about it some more, and I think I've found the flaw in my reasoning: for the high heat capacity case, I was assuming that the tiles would magically remain at a temperature lower than the air. But I think that's wrong. I think that regardless of the heat capacity, the tiles will eventually warm up to the same temperature as the surrounding gas, at which point heating will stop. If that's true, then in the high heat capacity case, it will take a tremendous amount of energy going into the tiles before they heat up to be the same temperature as the surroundings. All of that heat will then start to be conducted (albeit poorly) from the tiles to the body of the shuttle, which would be bad.

In the low heat capacity case, the tiles will heat up to the same temperature as the surroundings much more quickly, and much less heat will be transferred in doing so. This means that much less heat will be conducted to the body of the shuttle through the tiles.

Am I correct?

EDIT: simplified version: you want the tiles to have a low heat capacity so that their capacity to take in heat from the atmosphere is lower.

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Staff Emeritus
Correct. In a material with a high heat capacity, heat conduction dominates over radiative transfer.

Those Shuttle tiles also took advantage of emissivity. The tiles on the bottom of the Shuttle had a black coating (high emissivity) but the bulk of the tile was white (low emissivity).

Gold Member
EDIT: simplified version: you want the tiles to have a low heat capacity so that their capacity to take in heat from the atmosphere is lower.

Right. An air layer is an excellent insulator because air has a low heat capacity.

Staff Emeritus
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Correct. In a material with a high heat capacity, heat conduction dominates over radiative transfer.

Hmm. So what is the significance of that? Are you saying that even though the tile with high C doesn't warm up as fast (and hence won't be radiating as much), it will conduct all of the heat that's transferred to it?

Those Shuttle tiles also took advantage of emissivity. The tiles on the bottom of the Shuttle had a black coating (high emissivity) but the bulk of the tile was white (low emissivity).

Now is the high emissivity so that they can cool radiatively as much as possible? (Which presumably helps lessens the amount of heat they conduct to the interior).

What about the white tiles? Are they deliberately painted white because a poor emitter is also a poor absorber (since it reflects most of the incident light)? In other words, is the whiteness more advantageous in space where all of the heating is radiative, and you don't want a black shuttle that will absorb a lot of radiant heat?

I'm sorry to badger you, I'm just trying to make sure I really understand the concepts here.

Right. An air layer is an excellent insulator because air has a low heat capacity.

Still air is also a poor thermal conductor, is it not?

James_Harford
Hmm. So what is the significance of that? Are you saying that even though the tile with high C doesn't warm up as fast (and hence won't be radiating as much), it will conduct all of the heat that's transferred to it?

It makes sense that the tiles have low C so that during the fireball phase of re-entry they quickly begin shedding, or re-radiating, heat almost as quickly as absorbed. The word is "almost" because some of that heat is not re-radiated but diffuses deeper into the tile in at a rate hampered its low conductance.

So it is a race. The fireball phase must end before too much heat penetrates into the tiles. As soon as the fireball phase ends, the tile quickly radiates its heat back into space -- thanks to its low C. Any remnant diffused deep into the tile will reach the shuttle skin, but not by a dangerous amount.

If the tile has high C, much more heat will be stored in the tile during the fireball phase, the tile will take much longer to cool, and during that time, much more heat will reach the shuttle's skin, with perhaps lethal results.

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Maybe an analogy would help. Think of the Shuttle as a thermos bottle.

Gold Member
I think the thermal capacity of the tiles is of secondary interest. When you get down to it, they are there to delay the temperature rise of the shuttle skin until it's down in cool air. I guess that they will cool down quicker, once they are in the cool atmosphere, because they have low heat capacity but the major aspect of their construction is that the surface gets very hot so it radiates loads of friction-generated energy rather than passing it on to the shuttle body.

I asked a question a while ago (now it seems daft) about why they don't 'fly' the shuttle gradually, down through the atmosphere in order to limit the maximum surface temperature reached. Of course, there is a certain amount of energy to dissipate, one way or another and the whole shuttle would actually get much hotter if the braking process were spread over minutes - or even hours, just like supersonic aircraft only worse.
It's clearly better to 'get it over with quickly' because heat is radiated much faster from a white hot surface in a quick burst.

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Still air is also a poor thermal conductor, is it not?

Agreed. I am not sure how one might see a distinction between poor thermal conductor and poor thermal capacity. I mean, I know the difference but I don't know where you might find one without the other.

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Agreed. I am not sure how one might see a distinction between poor thermal conductor and poor thermal capacity. I mean, I know the difference but I don't know where you might find one without the other.

Water?

Homework Helper
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The thermal gradient through the tile at steady state wouldn't be any different (assuming the thermal conductivity is identical) between a tile with a high heat capacity and a tile with a low one. Assuming the heat is rejected via radiation heat transfer from the hot surface first which I understand is how these tiles are supposed to work, and only secondarily to the air frame, the inner surface of the tile under steady state conditions is dictated by the thermal conductivity, not the heat capacity.

The only difference will be during the transient when the tile heats up or cools down. For a tile with higher heat capacity, the thermal gradient will set up more slowly and dissipate more slowly.

I believe the point of having a low heat capacity is that once you heat up the tille to red hot under a bunsen burner for example, then it will cool down much more quickly so someone can pick it up if it has a low heat capacity. So the low heat capacity is a feature that allows for an impressive demonstration in a classroom, but I don't think a low heat capacity has anything to do with the function of the tile on the shuttle.*

As a side note, I've also heard the tiles have a thermal conductivity which is anisotropic, though I don't know how true that is. I think the coating has a relatively high thermal conductivity so it can conduct heat along the outer surface rapidly. In other words, the thermal conductivity perpendicular to the face is low but the thermal conductivity parallel to the face is high. This allows heat to flow out to the edges and be rejected in relatively cool areas. Unfortunately, I couldn't find anything on the net to confirm or deny this though.

* EDIT: The same thing applies to the heat capacity of aluminum foil for example. If you ever cooked a slice of pizza in a hot oven on top of a piece of aluminum foil, you probably have found you can open the oven and pick up the aluminum foil by the edges and remove your pizza. Even though the foil was obviously way too hot to touch in the oven, the foil doesn't have enough heat capacity to feel hot when you touch it because it cools down so quickly and any residual heat can be absorbed by your hands without getting your skin hot.

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Gold Member
I guess a laminated construction could well help to even out the surface temperatures but still insulate the skin from the outside. Smart thinking, actually, imo.

Staff Emeritus
I think the thermal capacity of the tiles is of secondary interest.
Both the volumetric heat capacity and thermal conductivity are of interest, and they go somewhat hand in hand. A low volumetric heat capacity reduces the magnitude of the heat pulse while a low thermal conductivity delays the heat pulse (and also reduces it as a consequence of the delay).

I asked a question a while ago (now it seems daft) about why they don't 'fly' the shuttle gradually, down through the atmosphere in order to limit the maximum surface temperature reached.
If you're talking about powered flight, yeah, that's pretty much daft. If you're talking about a different reentry angle or a different angle of attack, well that's (possibly) not so daft. The Shuttle had a skip entry capability (shallow reentry angle). It was never used, however.

As a side note, I've also heard the tiles have a thermal conductivity which is anisotropic, though I don't know how true that is. I think the coating has a relatively high thermal conductivity so it can conduct heat along the outer surface rapidly.
I alluded to that in post #8. The coating of the tiles on the bottom of the Shuttle had a higher thermal conductivity and a higher emissivity than the main bulk of the tile. The higher emissivity was the key factor here, IIRC.

Gold Member
If you're talking about powered flight, yeah, that's pretty much daft. If you're talking about a different reentry angle or a different angle of attack, well that's (possibly) not so daft. The Shuttle had a skip entry capability (shallow reentry angle). It was never used, however.

Yebbut I think the issue is to do with the total energy that needs to be dissipated. That would be so much that the whole shuttle temperature would rise to an unacceptable value if you were just to rely on 'forced convection' in the atmosphere on the way down. Radiating at 'several kK' is much more effective - although spectacularly dangerous looking.

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Both the volumetric heat capacity and thermal conductivity are of interest, and they go somewhat hand in hand. A low volumetric heat capacity reduces the magnitude of the heat pulse while a low thermal conductivity delays the heat pulse (and also reduces it as a consequence of the delay).
Just to clarify a few things:
• "Volumetric heat capacity" is the heat capacity multiplied by density?
• The heat pulse referred to is the total heat input during reentry?
• The main consideration regards the transient condition and not the steady state one?
Perhaps you can elaborate and also explain why (assuming I've understood correctly) a low heat capacity might be of benefit for the transient condition.

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AIR&SPACE
Correct. In a material with a high heat capacity, heat conduction dominates over radiative transfer.

Those Shuttle tiles also took advantage of emissivity. The tiles on the bottom of the Shuttle had a black coating (high emissivity) but the bulk of the tile was white (low emissivity).

When I first read that, I had an impulse to correct you: "also take" ; "the Shuttle has" ; "are white"

Then it hit me.
Wow.

JaredJames
I was under the impression Aerogel was used in the tile construction (and loads of other space applications):

http://en.wikipedia.org/wiki/Aerogel

One of the best insulators available. Something like 99% air.

To see just how good they are:

Not really explaining anything in particular to your question, but interesting on the topic and how someone could hold them.

Gold Member
When I first read that, I had an impulse to correct you: "also take" ; "the Shuttle has" ; "are white"

Then it hit me.
Wow.

Staff Emeritus
Just to clarify a few things:
• "Volumetric heat capacity" is the heat capacity multiplied by density?
• The heat pulse referred to is the total heat input during reentry?
• The main consideration regards the transient condition and not the steady state one?
Perhaps you can elaborate and also explain why (assuming I've understood correctly) a low heat capacity might be of benefit for the transient condition.
1. Correct.
2. I'm not a thermal engineer, but I have seen their presentations. They seem to use the term in both senses (total heat transferred to the vehicle structure, and worst case heating).
3. It's the transient (worst case heating) that is of primary concern. Heat the aluminum skin of the Shuttle to over 175 or 200 C and it's game over.

Regarding your final question, look at the set product {high conductivity, low conductivity} × {high capacity, low capacity}.
• High thermal conductivity, high heat capacity, aka Hades. The vehicle fries because of the high conductivity and the vehicle fragments that comprise pieces of structure + tile continue to fry after the breakup thanks to the high heat capacity.
• High thermal conductivity, low heat capacity. The vehicle still fries because of the high conductivity. The thermal conductivity needs to be low.
• 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.

Gold Member
• High thermal conductivity, high heat capacity, aka Hades. The vehicle fries because of the high conductivity and the vehicle fragments that comprise pieces of structure + tile continue to fry after the breakup thanks to the high heat capacity.
• High thermal conductivity, low heat capacity. The vehicle still fries because of the high conductivity. The thermal conductivity needs to be low.
• 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 when the heat pulse hits the aluminum skin, the skin won't heat up all that much.

So what kinds of materials fall into category 2 and 3? I was having trouble imagining many materials of type 2. (I suppose a block of granite would satisfy 3).

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

Homework Helper
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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.

James_Harford
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

Homework Helper
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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.

James_Harford
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.

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.

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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.

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Gold Member
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.