Good source Jon,
I was also going to suggest the relevant sections of Anderson's
introduction to flight, which has re-entry starting on page 436.
http://ae.sharif.edu/~iae/Download/Introduction%20to%20flight.pdf
Thanks,
Lars Osborne
On Thu, Jan 21, 2016 at 2:20 PM, Jonathan Goff <jongoff@xxxxxxxxx> wrote:
David, Lars,
The FAA had a paper a while back that went into reentry heating
math/analysis. I don't know if it used the same relations or not, but it
looked like a useful source...
I think this was it:
https://www.faa.gov/other_visit/aviation_industry/designees_delegations/designee_types/ame/media/Section%20III.4.1.7%20Returning%20from%20Space.pdf
~Jon
On Thu, Jan 21, 2016 at 3:12 PM, Lars Osborne <lars.osborne@xxxxxxxxx>
wrote:
This is off-topic, but pretty interesting anyway, and I like discussions
that have math in them. There are probably better forums for this.
I am confident assumptions 3,4,6 are invalid. Especially, #6. In direct
sunlight heat flux is closer to 1400 W/m^2, and earth albedo can be up to
1/2 of that (max) , and earth IR is about 218-244 W/m^2. However, in terms
of entry heating it can probably be ignored for first-order analysis.
I know there are some good references out there for this but I don't have
anything good for a first approximation on hand. Maybe this NACA paper will
be helpful.
Thanks,
Lars Osborne
On Thu, Jan 21, 2016 at 11:02 AM, 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
