# Shuttle/satellite re-entry temperature, simple question!

1. Jul 17, 2011

### elegysix

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

$E = \int F dx = \int -cv^2 dx$

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

$\Delta$E=nR$\Delta$T

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

dE/dt=dn/dt*R$\Delta$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?

2. Jul 18, 2011

### Andy Resnick

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3. Jul 18, 2011

### cjl

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. Jul 18, 2011

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. Jul 18, 2011

### cjl

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. Jul 18, 2011

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. Jul 18, 2011

### cjl

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?

Last edited: Jul 18, 2011
8. Jul 18, 2011

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. Jul 18, 2011

### sophiecentaur

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. Jul 18, 2011

### cjl

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. Jul 18, 2011

### elegysix

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. Jul 18, 2011

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.

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:

$$\frac{T_{aw}}{T_{e}} = 1 + r \frac{\gamma-1}{2}M^{2}_{e}$$

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

The recovery factor, $r$, is approximated very closely by $\textrm{Pr}^{1/2}$ for laminar flow and $\textrm{Pr}^{1/3}$ for turbulent flow. Here, $\textrm{Pr}$ 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. Jul 19, 2011

### elegysix

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. Jul 19, 2011

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. Jul 19, 2011

### sophiecentaur

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. Jul 19, 2011

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. Jul 19, 2011

### sophiecentaur

Yes, That makes sense. The sacrificial tiles are not a bad solution, I guess.

18. Jul 19, 2011

Don't get me wrong, the tiles are a terrible solution, but they are the best we have at the moment.

19. Jul 19, 2011

### sophiecentaur

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. Jul 19, 2011