From: vince@offshore.ai (Vincent Cate)
Newsgroups: sci.space.tech
Subject: Re: Heat Sink Heat Shields
References: <5dcb47db.0310011151.51d744ce@posting.google.com> <9186edb5.0310012120.758e3a2a@posting.google.com> <5dcb47db.0310030723.3b2688dc@posting.google.com>
NNTP-Posting-Host: 209.88.68.33
Message-ID: <9186edb5.0310042111.41efb20c@posting.google.com>

cray74@hotmail.com (Mike Miller) wrote in message 
news:<5dcb47db.0310030723.3b2688dc@posting.google.com>...
> This link discusses early use of copper heat shields for ICBMs:
> 
> http://www.centennialofflight.gov/essay/Evolution_of_Technology/reentry/Tech19.htm

Thanks!

> > For this a heatsink of around 5% of the mass would be enough for 
> > suborbital and around 15% for orbital.

I tried to post a correction but the post does not seem to have
worked.  Im my simulation the mass for orbital heatsink is 
about 25%, not 15%.  See sample inputs 51 and 51 in my simulator at:

   http://spacetethers.com/spacetethers.html

This simulator has not been tested against any experimental 
heatsink data because I have none (I will be reading that link 
next).  So there is the very real chance that it has bugs.  
If anyone has any real numbers please let me know what they are.

In particular, if Beryllium has to be 25% of the mass, then
copper would have to be more than 100%.  Part of this is
that in my simulations I had a L/D of 0.4 and the ICBMs
had 0.  I have not re-run the simulation with 0 because my
home computer is a text only Linux box.  Plan to tomorrow.
But this could explain the ICBMs.  Also, I don't know how
close to orbital speed the copper heatsink ICBMs got.
It is far easier at 5 km/sec than at 7.7 km/sec.

For a capsule with humans you really do want some lift
though.  It reduces the peak G load.  You can see this
in my simulator.

> But brittle, brittle, brittle. Worse than refractory metals,
> and much worse than some alloys like W-27Re. Suddenly, I
> like Mr. Spencer's idea for a beryllium-copper composite.

It does seem like there should be some way to reenforce it.

> So, a beryllium heat sink would represent 15% of the mass
> of an orbital capsule. Questions:

25% for my simulation (sorry).
 
> For the same re-entry velocity, would the mass of the heat sink
> heat shield vary for different capsule shapes (biconic vs raked
> cone vs Soyuz) and/or different different re-entry paths
> (capsule-type lifting vs. ballistic)?

The more lift you have the more total heat.  You can look
at this as the longer trajectory means more total heat.
Also, if you are getting lift then you are angling your 
heatshield to the flow air and the shockwave is not going 
to be as far away from the capsule, so more heat.  

The capsule has a certain amount of energy given by 1/2 MV^2  
that is going to turn into heat.  The big question is how much 
of that heat goes into the air and how much goes into the 
capsule.  The "stanton number" tells you what portion goes 
into the capsule.  For blunt bodies it is really very small,
numbers like 0.1%.  There is a formula on page 256 of 
Hypersonic Aerotherodynamics to estimate how much heat the 
capsule will get and I use it in my simulator (if you don't 
specify a stanton number in the input).  Here is that part
of the code from mass.java: 

blackBodyRadius = Math.sqrt(blackBodyArea/k.pi);
heatRatePerCC = 18300.0 * Math.pow(ourAir.density, 0.5)
   * Math.pow(airRelativeVelocity.magnitude()/10000.0, 3.05) 
   / Math.sqrt(blackBodyRadius);
heatFromAir = k.timePerDisplay * heatRatePerCC
    * k.SqCMinSqMeter * blackBodyArea;

All of the code is available at http://spacetethers.com/source/

> If I read this correctly, could you replace 5.4kg of Be with
> 1kg of water with transpiration cooling (neglecting the mass
> of the metallic portion of a transpiration heat shield)?
> 
> If so, would that mean a water transpiration heat shield would
> be about 2-4% of an orbital capsule's mass?

Yes except that since I should have said 25%, so it is more like 5%,
neglecting the metallic portion.  

Another note.  The specific heat of materials changes over
temperature, so just taking the value at 300 K and 
multiplying it by the melting point like I did is only
an approximation.  I am not sure how bad it is.

  -- Vince