I’ve seen the reverse rocket equation derivation in a few places, (given a dV,
initial mass, and final mass, compute an effective Isp), and it inspired me to
take a look at the energetics from a different perspective:
Let’s consider a kilogram of heat shield substance. It starts out at close to
0K, and is heated to ~2000K.
Specific heat of most solids is on the order of 1000 J / kg, so that’s 2E6 J /
kg = 2e6 m2/s2
This is in turn equivalent to a kinetic energy of ~2e3 m/s, which gives an
effective Isp of ~200 s.
This is over an order of magnitude off than the (conservative) reverse-rocket
derivation which is on the order of 1e5 s.
The conclusion I’m drawing is that only a tiny bit of the re-entry energy is
carried off by the ablated particles of the heat shield, and the vast majority
of the re-entry energy is used in heating the surrounding air. A further
conclusion is that a heat shield is more like a jet engine than a rocket. In a
rocket, the propellant mass is the reaction mass. In a heat shield and a jet
engine a small onboard propellant mass does work on a large reaction mass, the
surrounding atmosphere, increasing efficiency.
Am I getting this right? Engineering school was decades ago for me, and I
dropped out anyways.
On Feb 12, 2018, at 3:44 PM, Henry Spencer <hspencer@xxxxxxxxxxxxx> wrote:
On Mon, 12 Feb 2018, William Claybaugh wrote:
Henry:
Obvious but clever; it is not something I think I would have thought.
Thanks.
You're welcome. To be honest, I don't *think* I invented this, but if I
picked it up from somebody else, I no longer remember who.
Henry