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At lower speeds (below Mach 5-ish), stagnation temperature (TAT) is a very accurate proxy for skin temperature. But at mid/high hypersonic speeds (especially in the thin upper atmosphere where mass flow is low), thermal radiation bleeds off a significant amount of heat, especially as temperatures climb into the thousands of Kelvin.

I've come up with a very crude formula to estimate the skin temperature. It assumes that the power $P_{absorbed}$ to stagnate the oncoming air [intercepted by an area equivalent to the drag area $C_DA_{ref}$] is radiated away over the entire vehicle skin surface area $A_{rad}$:

$$P_{absorbed} = P_{radiated}$$

$$\frac{1}{2} \dot{m} v^2 = A_{rad} \sigma \epsilon T^4$$

$$\frac{1}{2} \left(\rho v C_D A_{ref} \right) v^2 = A_{rad} \sigma \epsilon T^4$$

$$T = \left( \frac{\rho v^3 C_D A_{ref}}{2 A_{rad} \sigma \epsilon} \right)^\frac{1}{4}$$

($\sigma$ is the Stefan-Boltzmann constant, emissivity $\epsilon$ is estimated at unity, $T$ is skin temperature)

How far off am I? How do you actually estimate hypersonic skin temperatures (without CFD)?


Here are a couple examples:

#   | Airframe | Mach       | Speed     | Altitude | Drag area | T (DATA) | T (stag) | T (rad)
----|----------|------------|-----------|----------|-----------|----------|--------------------
4.  | HTV-2    | Mach 20.   | 5,812 m/s | 125k ft? |  0.05 m^2 | 2,200  K | 21,000 K |  2,771 K  
8.  | X-43A #3 | Mach  9.6  | 3,000 m/s | 109k ft  |  0.10 m^2 | 2,255  K |  4,900 K |  2,143 K  
8.  | X-43A #2 | Mach  6.83 | 2,123 m/s | 109k ft? |  0.10 m^2 | 1,700  K |  2,514 K |  1,650 K  
3.  | X-51A    | Mach  5.1  | 1,500 m/s |  64k ft  |  0.10 m^2 | 2,200  K |  1,355 K |  2,058 K   
13. | SR-71    | Mach  3.2  |   930 m/s |  79k ft  |       m^2 |  ,640  K |   ,651 K |   ,    K  

(Data are actual temperatures, where available. Stag is stagnation temperature. Rad is predicted temp using the above formula. Drag areas are pure guesses. Surface areas were 10-12 m^2 for the X-43A, X-51A, and HTV-2. Mass flows were 20-40 kg/s/m^2), except for the X-51A, which encountered 140 kg/s/m^2).

For predicting skin temp, stagnation temperature seems more accurate at lower mach, formula temperatures at higher mach, as expected. Admittedly, I'm pleased (and surprised) that the formula even yields ballpark figures. However, it's a bit sensitive to drag area and radiating surface area, and these are the only data for which I have estimated surface area, so I can't be confident this formula works well for other aircraft.


Some background: I'm taking on a fun project (nothing serious), so a first approximation (say, to within 100 K) is good enough.


I've tried to follow the etiquette as best as I can, but I'm pretty new to Stack Exchange, so let me know if I should change anything :) Thanks!

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    $\begingroup$ This dissertation has some hypersonic heating formulas which may be of use, as well as a lot of other relevant information. See page 44 for the formulae. $\endgroup$ Mar 6, 2015 at 15:45
  • $\begingroup$ Argh, I have an equation for heating based on the radius of the leading edge somewhere. I just need to find it in all of my textbooks! $\endgroup$
    – tpg2114
    Mar 6, 2015 at 15:48

2 Answers 2

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For viscous hypersonic flows, the heating takes a form:

$$ q_w = \rho_\infty^N V_\infty^M C $$

where the parameters $N$, $M$, and $C$ depend on the configuration and $q_w$ is the heating in $W/cm^2$ (this is all from Hypersonic and High Temperature Gas Dynamics and I highly recommend this book).

For the stagnation point (like the leading edge of a body):

$$ M = 3;~N=0.5;~C=1.83\times 10^{-8}R^{-1/2}\left(1-\frac{h_w}{h_0}\right) $$

where $R$ is the radius, $h_w$ is the wall enthalpy and $h_0$ is the total enthalpy.

For a laminar flat plate at local angle $\phi$ to the flow at position $x$ meters from the leading edge:

$$ M = 3.2;~N=0.5;~C=2.53\times10^{-9}\left(\cos\phi\right)^{1/2}\left(\sin\phi\right)x^{-1/2}\left(1-\frac{h_w}{h_0}\right)$$

and for a turbulent flat plate:

$$N = 0.8$$

and if $V_\infty \leq 3962 m/s$:

$$M=3.37;~C=3.89\times10^{-8}\left(\cos\phi\right)^{1.78}\left(\sin\phi\right)^{1.6}x_T^{-1/5}\left(\frac{T_w}{556}\right)^{-1/4}\left(1-1.11\frac{h_w}{h_0}\right)$$

where $T_w$ is the wall temperature and $x_T$ is the distance along the body measured from the onset of the turbulent boundary layer.

For $V_\infty > 3962 m/s$:

$$M=3.7;~C=2.2\times10^{-9}\left(\cos\phi\right)^{2.08}\left(\sin\phi\right)^{1.6}x_T^{-1/5}\left(1-1.11\frac{h_w}{h_0}\right)$$

Phew, I believe I typed all those correctly. These are approximations but are really the simplest approach to get answers without requiring simulation or data measurements. Great for an initial estimate.

In the aforementioned book, these expressions are attributed to the paper Aerothermodynamics of Transatmospheric Vehicles. The approach there is to assume that the heating can take a form like mentioned in the first expression and relating that to the wall temperature for a fully catalytic material, then finding the values for $C$, $M$ and $N$ that are roots of the system.

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  • $\begingroup$ Thanks for the very timely (and thorough) response. I'm actually a novice at aerodynamics; what does wall enthalpy and total enthalpy refer to? $\endgroup$ Mar 6, 2015 at 17:08
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    $\begingroup$ @HephaestusAetnaean The enthalpy is related to the temperature and depends on the equation of state. You can actually assume calorically perfect gas (CPG) for hypersonic problems assuming you pick the appropriate values for the constants. So for CPG in hypersonic conditions, $\gamma = c_p/c_v \rightarrow 1$ and enthalpy is $h = c_p T$. So the total enthalpy will be based on the stagnation temperature (post-bow shock stagnation temperature) and the wall enthalpy will be based on the temperature of the wall. $\endgroup$
    – tpg2114
    Mar 6, 2015 at 17:23
  • $\begingroup$ OH it's literally just... yes, haha :) Thank you. And accepted. $\endgroup$ Mar 6, 2015 at 17:33
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    $\begingroup$ @HephaestusAetnaean Happy to help. That's the great thing about the StackExchange network -- there is probably somebody who has seen the answer to your question somewhere. I was lucky I had the book on my desk and could find it quickly. Sorry I don't have more time to dig into the theory/literature more -- both about how these approximations came to be and about other approximations that exist. But hopefully it's enough to get you started. $\endgroup$
    – tpg2114
    Mar 7, 2015 at 3:15
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    $\begingroup$ And I should also point out that these models are based on $T_w = f(q_w, \sigma\epsilon)$ so that the emissivity is included. So you had all the relevant physics in your question, it was just a matter of assuming a functional form under various configurations and iterating for an approximate root to that form. $\endgroup$
    – tpg2114
    Mar 7, 2015 at 3:18
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I found a NASA technical memo for very accurate "Real-Time Aerodynamic Heating and Surface Temperature Calculations for Hypersonic Flight Simulation." The authors also discuss a bit how they obtained their expression.

I have to say, though, Dave's dissertation and tgp2114's answer are wholly sufficient and more straightforward.

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