Off hand sounds like a pretty bad estimate; skin temperature should be much
higher.
1. most of the heat is supplied convectively, not radiatively, radiation is
maybe 20% of the heat load, and that radiative heat flux predominately goes
the other way.
2. The energy transfer coefficient of the orbital energy is nearer 1-2%,
98+% ends up in the shockwave and bypasses the vehicle
3. hypersonic lift to drag is normally nearer 0.5, IRC that's more what the
Shuttle got. I don't think it reached 1 until it got to Mach 3, but I might
have that wrong but it's around there, and I think that's fairly typical of
blunt bodies like the Shuttle. If you've got a high fineness ratio, the
heat transfer coefficients go up hugely, because the convection increases
so much and you'll see even higher skin temperatures.
I think Skylon might be a more similar vehicle than the Shuttle and they
are looking at temperatures more like 900K.
On 21 January 2016 at 19:02, David Summers <dvidsum@xxxxxxxxx> wrote:
I've been working on estimating reentry skin temperatures, and I'd like to
run this gross estimation process by everyone:
1) You start at 7.8 km/s
2) Maximum deceleration on orbital reentry is 1/(Hypersonic Lift to Drag)
3) The worst case assumption for maximum energy loss is that maximum
deceleration happens at maximum velocity (not true, but quickly bounds the
maximum energy bleed rate)
4) The vehicle is roughly the same temperature across its surface, and
only looses energy through radiation
5) Blackbody radiation is 5.7 W/m^2 at 100 K, power goes up with the
fourth power of temperature
6) There is 500 W/m^2 of sunlight / earth light incoming
7) Only a small percentage of the lost orbital energy stays in the vehicle
as heat - I think it is less than 10%?
Given that, the vehicle skin temperature is less than
T=
[(10%*0.5*orbiterMass*(7800^2-(7800-1/hypersonicLiftToDrag)^2)/vehicleArea
+ 500) / 5.7] ^0.25 * 100
Assuming 1/hypersonicLiftToDrag is very small compared to 7800, this
simplifies to:
T= [(10%*0.5*orbiterMass*( 2*7800*1/hypersonicLiftToDrag )/vehicleArea +
500) / 5.7] ^0.25 * 100
T= 100 * [ 140*orbiterMass/(vehicleArea*hypersonicLiftToDrag) + 88 ] ^
0.25
So a reentry vehicle with a mass of 400kg, and area of 200m^2, and a
hypersonic lift to drag of 2 would have a skin temperature of 390 K.
How far off is that? Is there a better way of making a quick estimate of
reentry temperatures?
Thanks!
-David Summers
utsspace.com
