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#### DynaSoar

##### Well-Known Member
Anyone got a calc or formula handy for aerodynamic heating at a given speed and altitude?

This is not a simple topic and may folks have spent their careers measuring heat transfer.

I haven't found a simple calculator however you can get a rough idea by using the following calculators.

https://www.aerospaceweb.org/design/scripts/atmos.shtml

This is an atmospheric calculator that determines quite a lot of parameters for a given velocity, altitude, and reference length:

Ambient temperature, pressure and desnity, mach number, dynamic pressure, Equivalent Air Speed (the important one), and the Cf in both laminar and turbulent flow. Even though SpaceShipOne exceeded Mach 3, its EAS was only 240 knots so relatively little heat transfer occurred compared to the SR-71 which travels at the same speed but at much lower altitudes.

https://www.aerospaceweb.org/design/scripts/compress.shtml

This flow calculator calculates the pressure, density and temperature ratios across normal and oblique shocks.

The temperatures are the maximum temperatures you could achieve, however in practice, they will be lower due to thermal conductivity and other heat transfer mechanisms.

Bob Krech

Originally posted by bobkrech
This is not a simple topic and may folks have spent their careers measuring heat transfer.

I haven't found a simple calculator however you can get a rough idea by using the following calculators.

https://www.aerospaceweb.org/design/scripts/atmos.shtml

This is an atmospheric calculator that determines quite a lot of parameters for a given velocity, altitude, and reference length:

Ambient temperature, pressure and desnity, mach number, dynamic pressure, Equivalent Air Speed (the important one), and the Cf in both laminar and turbulent flow. Even though SpaceShipOne exceeded Mach 3, its EAS was only 240 knots so relatively little heat transfer occurred compared to the SR-71 which travels at the same speed but at much lower altitudes.

https://www.aerospaceweb.org/design/scripts/compress.shtml

This flow calculator calculates the pressure, density and temperature ratios across normal and oblique shocks.

The temperatures are the maximum temperatures you could achieve, however in practice, they will be lower due to thermal conductivity and other heat transfer mechanisms.

Bob Krech

There's quite a bit more output than I need for the simple estimate, and what's being given on the second page isn't clear because I'm not familiar with the variable names. Would you please run these numbers as an example and show me how to follow them through to get the answer?

One the first page:

alt. 37919 ft.
vel. 5796 ft/sec.
lng. 14.2 ft.
gives
EAS = 1791.18 kts (3044.7 ft/sec.)
among other results.

Now, what do I put in where on the second page to get maximum heating produced and where do I read that from?

Also, on the second page, it gives Critical Cp and vacuum Cp. Are these related to the Cp we usually use in stability calculations, or are these something different?

Originally posted by DynaSoar
There's quite a bit more output than I need for the simple estimate, and what's being given on the second page isn't clear because I'm not familiar with the variable names. Would you please run these numbers as an example and show me how to follow them through to get the answer?

One the first page:

alt. 37919 ft.
vel. 5796 ft/sec.
lng. 14.2 ft.
gives
EAS = 1791.18 kts (3044.7 ft/sec.)
among other results.

Now, what do I put in where on the second page to get maximum heating produced and where do I read that from?

Also, on the second page, it gives Critical Cp and vacuum Cp. Are these related to the Cp we usually use in stability calculations, or are these something different?

Let's begin with your conditions. The important parameters that we need to calculate the maximum gas temperatures at your Mach Number (5.98) and the ambient gas temperature (390 R) which are obtained from the first calculator.

We can use the second calculator to determine the gas temperatures by inputting the M=5.98 into the isentropic flow calculator and dividing the ambient temperature (390 R) by T/To = 0.1227 to obtain the stagnation temperature To = 3080 R = 2620 F = 1440 C. That's the temperature you get if you stop the gas flow at the nose tip of the vehicle. For comparison, steel melts at 1370 C = 2500 F.

We can also use the second calculator to determine the gas temperature behind a normal shock by inputting the M=5.98 into the normal shock calculator and multiplying the ambient temperature (390 R) by T2/T1 = 7.894 to obtain the gas temperature right behind the shock front T2 = 3078 R. That's the temperature you get if you don't stop the gas flow at the nose tip of the vehicle. Not much of a difference at these velocities whether you stop the gas or not.

This is the maximum temperature that the nose tip could obtain if there was perfect heat transfer from the airflow and no thermal conductivity within the nose tip, however this is rarely the case for short term heating. The heat transfer fromn the hot gas to the NC is not 100% efficient, and the NC has a finite heat capacity and a non-zero thermal conductivity, so it will take a finite amout of time before it gets this hot, but I'm guessing it's seconds, not minutes at these velocities at such a low altitude.

Also at velocities over Mach 5, you really can't use ideal gas relationships and gammas of 1.4, since the oxygen in the air dissociates. This reduces the temperature as compared with the ideal gas calculation and increases the pressure, however the density increase remains approximately the same.

Non NC tip temperatures are much lower, but again at this very high velocity, the calculation is not that easy. You can use the oblique shock calculator to find the peak temperatures behind oblique shocks, and make a guesstimate at what the peak temperatures could be.

The heat transfer is proportional to the delta temperature from the skin to the average gas temperature in the boundary layer of the rocket at any point times the gas density times the heat capacity of the gas x the heat transfer coefficient. The boundary layer is probably turbulent so you can use the turbulent Cf in the first calculation as the relevant transfer coefficient. The density is the ambient density from calculator 1 and the density ratio from calculator 2.

I've referenced the following links before.

https://aerodyn.org/Atm-flight/tlimit.html

https://aerodyn.org/Atm-flight/table-skin-t.html

https://www.sierrafoot.org/x-15/pirep4.html

This table of skin temperatures is representative of the skin temperatures obtained at the stagnation points of real vehicles at operational altitudes. The X-15 at M=6 has a skin temperature of 900 K = 1620 R = 1160 F at altitude. This is substantially lower than the what if might expect in your test case, however your flight time woule be shorter. That's because the X-15 is operating at a much higher altitude than the rocket, so the density is at least 1 and up to 2 orders of magnitude lower than the rocket and the rate of heat transfer is that much lower, and it never reaches the maxiumum possible temperature.

You may want to use a castable ceramic NC, probably Al2O3, or a glass or carbon phenolic resin NC for this rocket, and glass or carbon phenolic resin for the fins.

https://naca.larc.nasa.gov/index.cg...imple&order=DESC&keywords=aerodynamic+heating

This report is very close to your proposed rocket flight.

https://naca.larc.nasa.gov/reports/1955/naca-rm-l55a14a/

The temperature probes failed at Mach 5 and ~30 kft in this flight.

Bob Krech

Originally posted by bobkrech
Let's begin with your conditions. The important parameters that we need to calculate the maximum gas temperatures at your Mach Number (5.98) and the ambient gas temperature (390 R) which are obtained from the first calculator.

We can use the second calculator to determine the gas temperatures by inputting the M=5.98 into the isentropic flow calculator and dividing the ambient temperature (390 R) by T/To = 0.1227 to obtain the stagnation temperature To = 3080 R = 2620 F = 1440 C. That's the temperature you get if you stop the gas flow at the nose tip of the vehicle. For comparison, steel melts at 1370 C = 2500 F.

We can also use the second calculator to determine the gas temperature behind a normal shock by inputting the M=5.98 into the normal shock calculator and multiplying the ambient temperature (390 R) by T2/T1 = 7.894 to obtain the gas temperature right behind the shock front T2 = 3078 R. That's the temperature you get if you don't stop the gas flow at the nose tip of the vehicle. Not much of a difference at these velocities whether you stop the gas or not.

This is the maximum temperature that the nose tip could obtain if there was perfect heat transfer from the airflow and no thermal conductivity within the nose tip, however this is rarely the case for short term heating. The heat transfer fromn the hot gas to the NC is not 100% efficient, and the NC has a finite heat capacity and a non-zero thermal conductivity, so it will take a finite amout of time before it gets this hot, but I'm guessing it's seconds, not minutes at these velocities at such a low altitude.

Also at velocities over Mach 5, you really can't use ideal gas relationships and gammas of 1.4, since the oxygen in the air dissociates. This reduces the temperature as compared with the ideal gas calculation and increases the pressure, however the density increase remains approximately the same.

Non NC tip temperatures are much lower, but again at this very high velocity, the calculation is not that easy. You can use the oblique shock calculator to find the peak temperatures behind oblique shocks, and make a guesstimate at what the peak temperatures could be.

The heat transfer is proportional to the delta temperature from the skin to the average gas temperature in the boundary layer of the rocket at any point times the gas density times the heat capacity of the gas x the heat transfer coefficient. The boundary layer is probably turbulent so you can use the turbulent Cf in the first calculation as the relevant transfer coefficient. The density is the ambient density from calculator 1 and the density ratio from calculator 2.

I've referenced the following links before.

https://aerodyn.org/Atm-flight/tlimit.html

https://aerodyn.org/Atm-flight/table-skin-t.html

https://www.sierrafoot.org/x-15/pirep4.html

This table of skin temperatures is representative of the skin temperatures obtained at the stagnation points of real vehicles at operational altitudes. The X-15 at M=6 has a skin temperature of 900 K = 1620 R = 1160 F at altitude. This is substantially lower than the what if might expect in your test case, however your flight time woule be shorter. That's because the X-15 is operating at a much higher altitude than the rocket, so the density is at least 1 and up to 2 orders of magnitude lower than the rocket and the rate of heat transfer is that much lower, and it never reaches the maxiumum possible temperature.

You may want to use a castable ceramic NC, probably Al2O3, or a glass or carbon phenolic resin NC for this rocket, and glass or carbon phenolic resin for the fins.

https://naca.larc.nasa.gov/index.cg...imple&order=DESC&keywords=aerodynamic+heating

This report is very close to your proposed rocket flight.

https://naca.larc.nasa.gov/reports/1955/naca-rm-l55a14a/

The temperature probes failed at Mach 5 and ~30 kft in this flight.

Bob Krech

Thanks, I'll study this well. Beside showing me how, it'll help me understand the variable names used.

When I feel I'm familiar with it, I'll do the next step, and sim the launch from the 100k' balloon platform, do the calcs and see ask if they look reasonable. The burnout speed will be higher but the burnout altitude will be more so.

Originally posted by DynaSoar
When I feel I'm familiar with it, I'll do the next step, and sim the launch from the 100k' balloon platform, do the calcs and see ask if they look reasonable. The burnout speed will be higher but the burnout altitude will be more so.

Doc

I hope you're not thinking about a baloon launched rocket project. If launching rockets from a baloon from 100 kft was cheap and reliable, NASA would have been doing it for 50 years. The fact is that it's much cheaper and simplier to launch from the ground to a given altitude than from a baloon platform.

here https://www.nsroc.com/front/what/SRHB.pdf

Of particular interest is the Super Arcus which can loft a 10 pound payload to 90 km in an 8 ft long x 4.5" package, and the Viper Dart which achieves 7000 ft/sec in 2.3 seconds.

Below are other classic cheap sounding rockets. Each is capable of high altitude and/or speed.

JPL PWN-1 Loki-Dart
Aerojet General PWN-2 Aerobee-Hi
University of Michigan/NACA PWN-3 Nike-Cajun
University of Michigan PWN-4 Exos
Cooper Development PWN-5 Rocksonde 200
Atlantic Research PWN-6 Kitty
Atlantic Research PWN-7 Rooster
Space Data PWN-8 Loki Datasonde
Aerojet/UTC PWN-9 Kangaroo
Space Data PWN-10 Super Loki Datasonde
Space Data PWN-11 Super Loki Datasonde
Space Data PWN-12 Super Loki ROBIN

https://www.designation-systems.net/dusrm/n-3.html

Bob Krech

Originally posted by bobkrech
Doc

I hope you're not thinking about a baloon launched rocket project. If launching rockets from a baloon from 100 kft was cheap and reliable, NASA would have been doing it for 50 years. The fact is that it's much cheaper and simplier to launch from the ground to a given altitude than from a baloon platform.

That's precisely what I intend to do.

The fact that NASA can't make it cheap and reliable is hardly surprising.

There are high altitude balloon teams all over the country who do it as a hobby. Most of them are also amateur radio groups. There's the lift and the communication & control.

I not only intend to, but I intend to do it twice (at least). What NASA doesn't do is bring them back. I intend to build a reuseable balloon lifted launch tower. I've already figured out I can either buy everything off the shelf or hire someone that does it, for less than the cheapest new car you can find. With it I can win the Stark Draper prize and buy a second set of reloads, which should be all I'd need for a second launch. That's not part of the Stark Draper prize, it's a point I want to make.

It's my IGY+50 project, which gives me between 2.5 and 4 years to pull it off. If I can wait that long.

"Can't" never did nothing. -- Pat Gordzelik

I think where you're going to run into trouble is not with the physics, or the cost of building the project, or even the practicality, it will be the regulators.

To fly large, or high-flying rockets, you have to clear the down-range area where the rocket would land (if anything went wrong). When you are launching from a static location, such as Black Rock, that's not too terribly difficult. Launch a balloon and read the prevailing winds. Given the launch location, prevailing wind and a little math, you can come up with an outer-bounds sort of calculation that tells you where the rocket can come down. Then you can be sure that it is clear. Then you can launch.

With a tethered baloon launch, you may be able to do this as well. Of course, you'll need 100,000' of tether and enough balloon power to lift the rocket and the tether. Your launch site will have to be much larger too, because the launch site will be a few miles away from your starting point.

With a free flight balloon, it's very, very difficult to assure a cleared landing zone. From what I've read, most of the successful balloon launches have been done at see, with ships following the balloon. NASA can't really use this method because of range restrictions and safety rules. So, for them, it's much cheaper to just build a bigger rocket, or add another stage.

The FAA will need to be assured that such a launch will be safe and that you will be able to clear the landing zone. Selling such a launch project to the FAA will be a chore. Hope you have a couple of good lawyers on the team.

urbanek

If the altitudes/locations become an issue, I suggest you contact NASA Wallops. They have let other groups use their range.

The problems Dave described are exactly why the picked that site for both balloons and rockets. Also, don't try to attribute NASA's inability to efficiently manage multi-gazillion dollar programs with their ability to do smaller ones well.

Originally posted by rstaff3
If the altitudes/locations become an issue, I suggest you contact NASA Wallops. They have let other groups use their range.

The problems Dave described are exactly why the picked that site for both balloons and rockets. Also, don't try to attribute NASA's inability to efficiently manage multi-gazillion dollar programs with their ability to do smaller ones well.

I had them in mind. I've been there a few times. A water recovery would be great for the platform, but floatation for the rocket stages would require some refiguring. But then, it might be the only way to get AST clearance, for the vehicle at least. I plan on working with an existing balloon project on the platform launch/recovery. They do these regularly, just not on this scale.

Oh, I know NASA can do things on the cheap if they set their mind to it. I know a guy who does the electronics for their circum-Antarctic balloons.

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