## 13 September 2013

### Space reentry vehicles, part 7

Part 7: Pressure

At subsonic speed, below Mach 0.3, the pressure of the air coming into the hull of an aircraft is roughly 1/2ρv2 where ρ is the density of the fluid medium; for air at sea level = 1.225 kg/m3.  The equation for compressible flow given for above Mach 1 speeds goes as the power of 3.5 rather than 2 and rises much more quickly.

Comparing pressure in compressible (red line) and incompressible (blue line) Mach regimes.  Greater than Mach 1 (red line) has much more pressure (no adjustment for density) than equation for below Mach 0.3 (blue line)(Gamma = 1.4). (My excel worksheet here.)

To be conservative I have used the subsonic speed 1/2 ρv2 estimate in which case the pressure at 11,000 km/s at 10km altitude with air density 0.26 kg/m3 (where most of the capsule slowing takes place) would be 155 atmospheres:

(from here)

11,000 m/s reentry speed at 10km altitude yields a pressure of 155 atmospheres stagnation pressure, which is still a very conservative estimate compared to compressible flow estimates.

A pressure of 155 atmospheres spread evenly on the leading face would give a total weight of 155atm x 101,000Pa x 11.9m2 area of Apollo ÷ 9.8 Newtons to kg = 1,908,000 kg.

This is equivalent to the weight of more than 4 fully loaded 747s:

Against this awesome force a relatively flimsy inner and outer hull is provided.  From Apollo Saturn:
COMMAND MODULE
The basic command module structure consists of a non-pressurized outer shell (the heat shield) and a pressure-tight inner shell for the crew compartment. The inner compartment is formed of aluminium honeycomb sandwich while the heat shield is formed of stainless steel honeycomb sandwich. The space between the inner and outer structures is filled with a special fibrous insulation (Q felt).

This is the aluminium inner compartment of the Apollo reentry capsule:

The Apollo Command Module inner structure was easily able to with stand the 1/3 atmospheric pressure it was pressurised to, not much more though.

The sides of space age reentry capsules such as the Apollo Command Module were even weaker than the base which was supposed to take the brunt of the reentry pressure; though neither were really up to the task. Like meteoroids that are crushed to pieces upon atmospheric entry, these reentry capsules would have cracked open like a walnut.

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This concludes the 7-part blog series Space Reentry Vehicles.  Keep an eye out for the movie version in a couple of months, update: movies take longer than I expected, could be a few months, especially with this much material and my slow rate of production.

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Update 2 Jan 2014:

As a result of Manfred Arcane's astute comment (below) I am updating this post to reflect his concerns. Manfred writes:

...The two figures you present I have seen before, and am convinced are really screwy. The only way to get those altitudes for maximum deceleration is for a ballistic coefficient to be extremely high, much, much higher than for even a reentering warhead. Don't know why they made that choice in their document..

The 100's of G's shown in those two plots clearly does not correspond to manned reentry.

This is correct. I think you're referring to these kinds of diagrams following:

These diagrams are for non-lifting bodies. They come from the NASA document Returning From Space, which is a teaching aid and not particularly rigorous.  As the document explains these reentry profiles are used for simplification purposes, stating:

### Vehicle Shape

The re-entry vehicle’s size and shape help determine the ballistic
coefficient (BC) and the amount of lift it will generate. Because adding lift to the re-entry problem greatly complicates the analysis, we’ll continue to assume we’re dealing only with non-lifting vehicles. In the next section, we’ll discuss how lift affects the re-entry problem.

Now, I happen to think that the lifting, flying-wing theory for Apollo and other reentry vehicles is highly contentious. But according to theory the slow down happened well above the 10km altitude I used for the pressure in the Rho V squared equation. And Returning From Space offers these diagrams reflecting the more conventional view on the reentry profile:

According to the above diagram there's a knee  in the curves for Apollo and Gemini at about 50 km altitude, in which case the atmosphere would be incredibly thin, as the following table shows, from equation from Engineering toolbox.
 Meters alt ATMs 1 0.9998814 1000 0.8869929 5000 0.5331348 10000 0.2609053 15000 0.114075 20000 0.0427153 25000 0.0127493 30000 0.0026444 35000 0.0002773 40000 4.907E-06 45000 #NUM!
At an altitude of 45,000 metres Excel 2007 doesn't even want to calculate the pressure! But where there is drag there will also be significant pressure, no matter how thin the atmosphere.
Manfred writes:

Hence your projection of pressure is off by three orders of magnitude.

This is correct, assuming the slowing occurs much higher than 10km altitude. But I was relying on the huge underestimate of using the equation for incompressible flow (Rho V squared) compared to the equation given for compressible flow which was even more orders of magnitude larger than the discrepancy you point out.
Here is the graph for the (Edit 1/Jun14:) incompressible flow equation, going up to Apollo reentry speed of Mach 33:

120 million atmospheres! I suggest that cancels out the three orders of magnitude you mention.  Thanks for pointing out that flaw Manfred, it improved my post.

1. Sir,
your calculation for dynamic pressure on Apollo makes a colossal error. By the time Apollo is at 10 km, its velocity is less than 1 km/sec (refer to: http://www.aulis.com/images_reentry/apollo-8_re-entry.jpg). It will have slowed down by close to a factor of 30 from 10 km/sec, in fact. Hence your projection of pressure is off by three orders of magnitude.
Regards.
PS. The two figures you present I have seen before, and am convinced are really screwy. The only way to get those altitudes for maximum deceleration is for a ballistic coefficient to be extremely high, much, much higher than for even a reentering warhead. Don't know why they made that choice in their document, but maximum deceleration for a manned spacecraft reentering would be in excess of 30km altitude. The reentry flight path angle would be very shallow, so as to cause reentry deceleration not to exceed about 8 G's, the upper bound on what humans can endure. This drives the pressure down considerably, because it is the pressure that causes the deceleration that must be bounded above to be tolerable. The 100's of G's shown in those two plots clearly does not correspond to manned reentry.

1. Manfred, well spotted. I thought about including that consideration initially in the post but got lazy.

This is a fair criticism, and I have updated the post to reflect that. I appreciate the effort you made to find and point out that flaw, and I hope I have addressed your concerns in the update above.

2. Sir,
It is rather easy to code up Allen and Eggars formulas (which are very simple and elegant, BTW) for reentry deceleration and compute quantities such as the dynamic pressure acting on these sorts of reentry vehicles. I have done so and arrived at values that max out (for ICBM warheads) at multiple (>10X) hundreds of Kilo-Pascals. However, atmospheric pressure at sea level is itself around 100 Kilo-Pascals, so this is not a horribly high value after all. Again, when you uses these same formulas for computing deceleration of warheads, you get generally at most about 120 g's, but for reentry angles and ballistic coefficients that are much higher than employed for manned reentry. Prompted by your line of inquiry here, I lowered both of these quantities in some of my tools for implementing these formula in order to be more representative of manned reentry (even ignoring lift) and was able to lower the maximum level of deceleration g's to around 8-10 g's - suitable for manned flight. -->
Then, if you multiply the 10 g's = 100 m/s^2 by the command module mass of ~5000 Kg, you get a maximum dynamic pressure level of ~500,000 Pascals = 500 kPascals, which is only 5X sea level atmospheric pressure. I would encourage you to investigate Allen's wonderful formulas (If my memory serves correct, I believe the reference you consulted for your plots above has some of the formulas documented.), and you might begin to see how reentry was a dicey proposition, but within the realm of engineering solution. It turns out that heat loads were probably more worrisome than the dynamic loads to NASA engineers, once they turned to blunt RVs (meaning low ballistic coefficient values) reentering at low reentry angles.
Regards,
Ray

1. Thanks for the info. I haven't considered those angles before.