Shuttle/satellite re-entry temperature, simple question!

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Just curious, but I was wondering if anyone knew how to find the maximum temperature of the surface of a shuttle/satellite upon reentering the atmosphere?


My guess would be:
F=-cv^2

[itex]E = \int F dx = \int -cv^2 dx[/itex]

PV=nRT --- I think PV has units of joules, maybe not, but assuming it is, so I'd say

[itex]\Delta[/itex]E=nR[itex]\Delta[/itex]T

But what would n be here? it's usually the moles of gas... so maybe

dE/dt=dn/dt*R[itex]\Delta[/itex]T ?

dn/dt would be some function of the velocity and the density of the atmosphere... I don't know, I'll just stop here for now lol.

what do you guys think?
 

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  • #2
Andy Resnick
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  • #3
cjl
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It's a fairly complex problem, so you really only have one of two choices. You can measure it, or you can do fairly advanced computer simulations. The DSMC method for simulating rarefied gas flows would be a good choice here.
 
  • #4
boneh3ad
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And of course it will vary throughout the re-entry process depending on the altitude and whether or not the boundary layer is laminar or turbulent at a given point in space and time. A turbulent boundary layer transfers heat to the surface an order of magnitude faster than a laminar boundary layer.

It will also depend on the size of the object, the shape of the object, if it is in Earth's atmosphere or that of Mars or another celestial object and many other things. It is an incredibly complex problem.
 
  • #5
cjl
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And of course it will vary throughout the re-entry process depending on the altitude and whether or not the boundary layer is laminar or turbulent at a given point in space and time. A turbulent boundary layer transfers heat to the surface an order of magnitude faster than a laminar boundary layer.
Yep, and it also depends on how rarefied the flow is. The heat transfer for a highly rarefied (Kn > 1) flow is yet another case that is different from both laminar and turbulent boundary layers.

As the object reenters, it'll likely go from a very rarefied, free molecular flow with high thermal accomodation down to a continuum-regime flow (probably turbulent), and during the whole process, the heat transfer and drag will be changing dramatically.

(It's quite an interesting problem though)
 
  • #6
boneh3ad
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Actually, the boundary layer on a blunt body (space capsule, shuttle entering bottom first as it does part of the way down) is typically mostly laminar during re-entry. Of course, that only goes for the blunt side on the flow side, the back side is just crazy and separated but isn't as hot to begin with so it isn't much of a problem. In general, hypersonic flows like to stay laminar much longer than subsonic flows do.
 
  • #7
cjl
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Really? That's fascinating - I've spent some time studying highly rarefied hypersonic flows (rarefied to the degree that direct simulation methods are required), but I haven't really looked at continuum hypersonic flows, so I was just guessing about the turbulence there. That's a really interesting result though. Is there any particular mechanism that keeps them laminar, or is it just a strange property of very high speed flows?
 
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  • #8
boneh3ad
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I would argue that we don't really fully understand the phenomenon yet, but they have done IR thermography on the underside of the shuttle starting with STS-114 and viewed the heat transfer properties behind a boundary layer trip. It clearly showed the turbulent wedge from the trip as well as the greater area of laminar flow around it. There is a wealth of information on this work, for example Berry et al. 2008 (paper).

However, there are a lot of unsolved mysteries and compressible stability theory is incredibly far behind the more traditional and easily approachable incompressible stability theory. We know that Tollmien-Schlichting waves still dominate below Mach 4 on a flat plate or similar surface, but beyond that, the most important mechanism is the second Mack mode, which is a much different instability that only arises in compressible flows. The problem with it is that while T-S waves are suppressed by wall cooling, second mode instabilities are destabilized, so changing wall temperatures is no longer a viable way to affect boundary layer stability. Additionally, with a blunt body or the blunted tip of a cone or wedge, the massive entropy layer that forms behind the bow shock is stabilizing.

Of course, there are many, many more things that go into it, but I don't know even close to all of it yet (and likely no one really does) and there aren't a lot of great sources on high speed transition at this point in time. For basic stability, check out Mack 1984 (abstract), as it is the definitive source for theory, but it only does some work with compressible flows and that is mostly theoretical work, as most of the empirical trends hadn't been observed yet in 1984.
 
  • #9
sophiecentaur
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Something that I always wondered about is the possibility of controlling the descent rate by 'flying' / skimming through the upper atmosphere so that the vehicle loses KE at a lower rate. I appreciate that we are talking supersonic /ultrasonic flight and that control would not be trivial but, if you could extend the actual time in which the energy was dissipated, the power would be lower and so the temperature problem would be a lot less. You could even consider avoiding the radio blackout due to the plasma around the nose of the vehicle.
I think is must 'just' be a matter of flight control. IS that problem insoluble?
 
  • #10
cjl
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Something that I always wondered about is the possibility of controlling the descent rate by 'flying' / skimming through the upper atmosphere so that the vehicle loses KE at a lower rate. I appreciate that we are talking supersonic /ultrasonic flight and that control would not be trivial but, if you could extend the actual time in which the energy was dissipated, the power would be lower and so the temperature problem would be a lot less. You could even consider avoiding the radio blackout due to the plasma around the nose of the vehicle.
I think is must 'just' be a matter of flight control. IS that problem insoluble?
That's how pretty much all manned vehicles reenter actually - the shuttle and even capsules fly a lifting trajectory to decrease heating rates and acceleration loads. Interestingly enough, stretching out the reentry decreases the maximum heating rate, but it actually increases the integrated heat load, so it isn't completely beneficial. It's still used though, as much as anything to minimize the acceleration to which the passengers are submitted.

See this image for an example:

http://www-public.tu-bs.de:8080/~y0021684/pic/apollo11_reentry.png [Broken]
 
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  • #11
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So what I gather is that unless we measure it, or simulate it with a computer, we wouldn't even have an educated guess.

Now, I'm not asking for an exact model or anything... I'm just looking for a ballpark approximation. If you wanted to simulate this with a computer, you'd have to have some equations to start with... what would those be?




In 1957 the soviets did the first human spaceflight, so how did they decide whether or not their shuttle would disintegrate upon reentry?

Was it a Fortran nightmare or something? lol

Or did they melt enough equipment prior to that, that they knew what to make the front of their ship with?
 
  • #12
boneh3ad
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Something that I always wondered about is the possibility of controlling the descent rate by 'flying' / skimming through the upper atmosphere so that the vehicle loses KE at a lower rate. I appreciate that we are talking supersonic /ultrasonic flight and that control would not be trivial but, if you could extend the actual time in which the energy was dissipated, the power would be lower and so the temperature problem would be a lot less. You could even consider avoiding the radio blackout due to the plasma around the nose of the vehicle.
I think is must 'just' be a matter of flight control. IS that problem insoluble?
That's how pretty much all manned vehicles reenter actually - the shuttle and even capsules fly a lifting trajectory to decrease heating rates and acceleration loads. Interestingly enough, stretching out the reentry decreases the maximum heating rate, but it actually increases the integrated heat load, so it isn't completely beneficial. It's still used though, as much as anything to minimize the acceleration to which the passengers are submitted.

See this image for an example:

http://www-public.tu-bs.de:8080/~y0021684/pic/apollo11_reentry.png [Broken]
It may be losing KE at a lower rate, but that means it is spending more time at higher speeds. This means it is exposed to the extreme temperatures for a longer time. The problem is that the temperatures aren't caused by the rate of deceleration, but the rate of motion. The faster the object moves, the hotter it gets. Basically, in the frame of reference of the entry vehicle, the air is coming on at a given atmospheric temperature, pressure, density, etc. and some free stream Mach number. This air stagnates (again, reference frame of the vehicle) against the walls of the vehicle, raising it to an even higher temperature. This temperature, which is typically slightly different from the stagnation temperature, is called either the adiabatic wall temperature or recovery temperature and is typically enormously high compared to the relatively warm free stream temperature in the atmosphere.

So what I gather is that unless we measure it, or simulate it with a computer, we wouldn't even have an educated guess.

Now, I'm not asking for an exact model or anything... I'm just looking for a ballpark approximation. If you wanted to simulate this with a computer, you'd have to have some equations to start with... what would those be?

In 1957 the soviets did the first human spaceflight, so how did they decide whether or not their shuttle would disintegrate upon reentry?

Was it a Fortran nightmare or something? lol

Or did they melt enough equipment prior to that, that they knew what to make the front of their ship with?
The ballpark figure would be the adiabatic wall temperature, since that is the theoretically highest temperature the wall could reach. It is typically defined as:

[tex]\frac{T_{aw}}{T_{e}} = 1 + r \frac{\gamma-1}{2}M^{2}_{e}[/tex]

where
[itex]T_{aw}[/itex] is the adiabatic wall temperature
[itex]T_{e}[/itex] is the total temperature or stagnation temperature of the free stream just outside the boundary layer but inside the shock
[itex]r[/itex] is the recovery factor
[itex]\gamma[/itex] is the ratio of specific heats ([itex]\approx 1.4[/itex] in air)
[itex]M_{e}[/itex] is the edge Mach number just outside the boundary layer but inside the shock layer

The recovery factor, [itex]r[/itex], is approximated very closely by [itex]\textrm{Pr}^{1/2}[/itex] for laminar flow and [itex]\textrm{Pr}^{1/3}[/itex] for turbulent flow. Here, [itex]\textrm{Pr}[/itex] is the Prandtl number.

Edge quantities can be calculated from either normal or oblique shock relations depending on your situation.

This would give an absolute worst-case scenario if you assumed the flow was fully turbulent and that the craft lingered long enough to heat up to that adiabatic wall temperature. Of course this will never actually happen, but early designs would have been extremely conservative like this or else incredibly risky. I would imagine the Soviets probably used some variation of this. They probably melted some equipment along the way as well.

It is interesting to note that we still don't have a good answer to this, so the thermal protection system (TPS) is typically vastly overdesigned. Getting a better grip on this would let us make a much less bulky TPS and greatly increase our payload and/or decrease the amount of fuel required to reach orbit.
 
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  • #13
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thats really cool, thanks for posting boneh3ad, I was looking at your equation, and I was wondering how you'd come up with a value for Te if you wanted to apply this.

Te - "is the total temperature or stagnation temperature of the free stream just outside the boundary layer but inside the shock"

From the looks of it, we could look up the values for all the other constants, given a set of conditions, but I have to ask, how would we get a value for Te?
 
  • #14
boneh3ad
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Well for a given altitude you should be able to get a freestream total temperature. Total temperature is constant across a shock, so it should be nominally the same as the free stream.
 
  • #15
sophiecentaur
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That's how pretty much all manned vehicles reenter actually - the shuttle and even capsules fly a lifting trajectory to decrease heating rates and acceleration loads. Interestingly enough, stretching out the reentry decreases the maximum heating rate, but it actually increases the integrated heat load, so it isn't completely beneficial. It's still used though, as much as anything to minimize the acceleration to which the passengers are submitted.

See this image for an example:

http://www-public.tu-bs.de:8080/~y0021684/pic/apollo11_reentry.png [Broken]
That's an interesting comment (and so are the more detailed ones, later). It partially answers my question in as far as the present re-entry is at least one least worst solution.
However, in the end, you are converting GPE to Thermal energy and, if you convert it slowly enough, then could not the rate of dissipation be arbitrarily low - obviating the cooling problem?
I guess the possible flaw in that argument is that you still need to get rid of this heat somewhere and you have the inside temperature to consider as well as the surface temperature. A short burst of intense temperature on the outside reduces the time for heat to get through to the living quarters of the ship and, once the ship is falling slowly through the atmosphere, its skin can cool down pretty rapidly and before the inside temperature gets unbearable.
How warm is the skin of the Shuttle when it actually lands, I wonder? `Fry an egg`?
 
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  • #16
boneh3ad
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It is more than just converting gravitational potential energy into thermal energy. Entry vehicles such as the shuttle have an enormous amount of kinetic energy as well. They typically re-enter the atmosphere at over 7200 m/s and have to dissipate that energy almost entirely due to drag, which is where the heating comes from upon re-entry. You can't come in too shallow or you stay at the high temperatures for way too long and burn up, and you can't come in too steep or you will get to way too high of a temperature and burn up. You have to hit that sweet spot. No matter how you slice it, you still have a cooling problem that requires a lot of science and engineering forethought. This is still an ongoing area of improvement.
 
  • #17
sophiecentaur
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Yes, That makes sense. The sacrificial tiles are not a bad solution, I guess.
 
  • #18
boneh3ad
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Don't get me wrong, the tiles are a terrible solution, but they are the best we have at the moment.
 
  • #19
sophiecentaur
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What's wrong with 'sacrificial', as a protection process?
The only alternative would be to use a lot of energy in a retro rocket - like the Moon Landings had to - but much much more.
 
  • #20
boneh3ad
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The other alternative is to better understand the physics so that you don't have to sacrifice so much material and simultaneously to develop new materials such as CMCs that can potentially withstand the heat without relying on ablation. Ideally, you would want a completely reusable orbiter. The shuttle is not completely reusable, and as the Columbia incident shows, the tiles are not infallible.
 
  • #21
cjl
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Actually, the tiles on the shuttle are not sacrificial - they are reusable. The shuttle is somewhat unique in that it is (to my knowledge) the only large reentry vehicle which has not used ablative materials as the heat shield. This is part of the reason its thermal protection system is so delicate - the materials to withstand 1200-2000+C and stay intact (with good insulation properties no less) are much more exotic, expensive, and fragile than a simple ablative is.
 
  • #22
cjl
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This would give an absolute worst-case scenario if you assumed the flow was fully turbulent and that the craft lingered long enough to heat up to that adiabatic wall temperature. Of course this will never actually happen, but early designs would have been extremely conservative like this or else incredibly risky. I would imagine the Soviets probably used some variation of this. They probably melted some equipment along the way as well.
The Soviets (and the US) used pretty much exclusively ablative heat shields for the early vehicles, so the limiting factor isn't actually the temperature, but rather the heating rate, the dynamic pressure, and the total heat load. The maximum heating rate and dynamic pressure would determine what materials were suitable (with too high of a dynamic pressure or heating rate, the material would come off in large chunks instead of slowly charring and burning away), and the total heat load would determine the thickness. I imagine there were some initial unmanned tests, along with some laboratory testing with various heat sources to determine what materials would work. Certainly none of them would stand up to the full adiabatic wall temperature, but that wasn't the design goal anyways.
 
  • #23
cjl
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Oh, and here's an interesting paper I came across a while ago that is somewhat related to this. It discusses heating and acceleration for various reentry profiles, including ballistic (with a range of ballistic coefficients), and lifting. It talks a lot about ballistic missile reentry in addition to capsules and the shuttle, but it's the same basic problem for warheads and manned vehicles:

http://exoaviation.webs.com/pdf_files/Atmospheric Re-Entry.pdf
 
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  • #24
cjl
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Hi,

Just for fun, I ran some numbers regarding the possibility of a "Steel Re-entry Sheild."

Assumptions are needed, of course. Here's my work.

http://www.thermospokenhere.com/wp/04_tsh/D318___shuttle/shuttle.html [Broken]

Good Luck, Jbo
That's interesting, but terribly unrealistic of course. The vast majority of the energy of reentry is absorbed by atmospheric heating, rather than heating of the vehicle (which is very good, because if all of the heat went into the vehicle, the amount of insulation would be impractically large, as your numbers show).
 
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  • #25
boneh3ad
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Actually, the tiles on the shuttle are not sacrificial - they are reusable. The shuttle is somewhat unique in that it is (to my knowledge) the only large reentry vehicle which has not used ablative materials as the heat shield. This is part of the reason its thermal protection system is so delicate - the materials to withstand 1200-2000+C and stay intact (with good insulation properties no less) are much more exotic, expensive, and fragile than a simple ablative is.
Fair enough. I probably should have caught that myself. They are still a 30-year old technology that are very damage-prone, expensive and time-consuming to repair and maintain, and are not nearly reliable enough for any industry other than space flight (for example, airlines). New materials are needed for sure, though I doubt you would disagree with that.

The Soviets (and the US) used pretty much exclusively ablative heat shields for the early vehicles, so the limiting factor isn't actually the temperature, but rather the heating rate, the dynamic pressure, and the total heat load. The maximum heating rate and dynamic pressure would determine what materials were suitable (with too high of a dynamic pressure or heating rate, the material would come off in large chunks instead of slowly charring and burning away), and the total heat load would determine the thickness. I imagine there were some initial unmanned tests, along with some laboratory testing with various heat sources to determine what materials would work. Certainly none of them would stand up to the full adiabatic wall temperature, but that wasn't the design goal anyways.
And yet out of all the factors you mentioned, the bottom line comes down to speed, temperature and the time spent at those conditions. Heating rate is determined by the velocity (effectively the convection coefficient) and the temperature of the boundary layer (itself a function of velocity and other smaller factors). Increase the velocity and you increase the temperature gradient and the convection coefficient. Additionally, once you get into high enough temperatures, you not only worry about convection, but the heat absorbed by the craft due to the radiative heat transfer of the surrounding plasma (and of course heat lost by the craft due to radiation if it gets hot enough). Dynamic pressure is of course a function of velocity as well.

Ultimately, the TPS of any spacecraft is vastly overdesigned right now since there is no good way to tell what the actual heat transfer characteristics of the system will be aside from empirical data and correlations, but for those you have to test it first. Earlier spacecraft would have been even more over-designed (though with the Soviets, who knows).

Obviously they wouldn't design it for adiabatic wall temperature, but it wouldn't be very difficult to find [itex]T_{aw}[/itex] as a function of spacecraft altitude and velocity (which are known) and then do a very basic, conservative heat transfer analysis. You could get an okay engineering approximation even by hand with enough work and diligence, which is likely related to what they actually did.
 

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