Vehicle Nose Heating

I'm still learning about the heating of nosecones. Launch loops will supply very inexpensive kinetic energy to one-time-use cargo vehicles.

Rockets normally rise vertically, and do most of their acceleration in space vacuum, at least 100 km above the surface. However, the heat resistant nose cones needed to exit the atmosphere "horizontally", from 80 km loop altitude to space vacuum, could be far more expensive. What is needed to protect the vehicle nose from thin but extremely energetic gas?

According to the standard atmosphere model, the atmosphere at 80 km is 1.8458e-5 kg/m³, 1.5e-5 as dense as that at the surface (I expect it to be denser at the equator than at the 45.5425 degree latitude of the standard model). For a 10 km/s exit, the enthalpy (thermal and chemical power, more or less) of the free stream flow approaching the nose cone is ½ ρ V³ = 9.23 MW/m².

A seminal paper in space history is "A Study of the Motion and Aerodynamic Heating of Missiles Entering the Earth's Atmosphere at High Supersonic Speeds" by H. Julian Allen and A. J. Eggers, Jr. The revised October 1957 version is NACA (National Advisory Committee for Aeronautics) Technical Note 4047, available as 19930084817.pdf from NTRS, the NASA Technical Reports Server.

This paper teaches that a hypersonic reentry body nose should be rounded, not pointy. A pointed nose will endure higher heating at the point, and cannot conduct heat fast enough to the body behind, so it will burn away. The thermal stresses may shatter the nose cone. A rounded nose will move the shock wave farther in front of the vehicle, and deflecting more of the entry heat into the air, and less to the stagnation point (the center of a symmetric flow) on the nose itself.

The Allen/Eggers paper concerns warhead entry; the idea is to deliver the warhead near the surface without damaging it, before it does it's job and destroys a city. As my late friend Jordin Kare, formerly a laser physicist at Lawrence Livermore National Labs, said:

• "If we don't do our job exactly right, millions of people will live!".

Launch loop, launching into the equatorial plane, won't overfly most strategic cities, so hopefully billions of people will not be threatened by it. On the other hand, more of those billions can travel to space if nosecones are cheap and effective.

PICA - Phenolic Impregnated Carbon Ablator - is the gold standard for reentry heat shields, but those heat shields cost millions.

Heppenheimer's Facing the Heat Barrier: A History of Hypersonics NASA sp4232 offers a fascinating history of nosecone development. Page 50 offers this amusing sentence:

• Indeed, Kantrowitz recalls Von Braun suggesting that it was possible to build a nose cone of lightweight balsa soaked in water and frozen. In Kantrowitz’s words, “That might be a very reasonable ablator.”

Water evaporates too easily, and balsa is expensive and rare. A possible alternative is chemically modified, porous "nanowood" infused with polydimethylsiloxane ... silly putty. Launch Loop cargo exit nosecones need not reenter, just survive an outbound trip from the perigee of a high apogee Kepler orbit.

Back to Eggers ...

Stagnation point heating is described by equation 44, unfortunately in foot pound second units:

{ \Large { { d H_s } \over { d t } } } ~ = { 6.8e-6 } ~~ { \Large \sqrt{ \rho \over \sigma } } ~ V^3

Let's try scaling that to meters kilograms seconds:

slugs/s^3 = 6.8e-6 (units?) ~ { \Large \sqrt{ slugs/ft^3 \over ft } } ~ { ft^3 / s^3 }

\sqrt{ slugs } / ft = 6.8e-6 (units?)

So, the units of the multiplier are \sqrt{ slugs } / ft . To convert the multiplier to mks, multiply 6.8e-6 by 3.28084 / \sqrt{ 0.0685218 } to get 8.5e-5 \sqrt{ kg } / m .

The metric version of equation 44 is thus:

Eq 44_m: ~~~ { \Large { { d H_s } \over { d t } } } ~ = { 8.5e-5 ~( \sqrt{ kg } / m ) } ~ { \Large \sqrt{ \rho \over \sigma } } ~ V^3

For a 20 cm radius nose, V = 10 km/s, \rho = 1.8458e-5 kg/m³ and \sigma = 0.2m, the heating rate is approximately 820 KW/m² .

For a 50 cm radius nose, the heating rate drops to 520 KW/m².

This is a pessimistic view of enthalpy; at these high velocities, the real gas will be dissociated and partly ionized, which will absorb much of the stream entropy rather than heat. On the other hand, the atomic oxygen and ions bombarding the nose will release energy and recombine. So, treat these numbers as a conservative worst case. Both numbers are a LOT better than 9200 KW/m² of the free stream enthalpy.