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Some high-speed military aircraft like the SR-71 had real heating problems, but airliners also travel almost at the speed of sound, use most of their fuel to make up for frictional losses, so I would presume that their hulls heat up. They are also cooled by the airflow, but at what temperature does equilibrium set in during cruise? I remember that airliners don't seem to be very hot when you touch them after landing, but they have had time to cool down in the slow winds during descent.

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    $\begingroup$ use most of their fuel to make up for frictional losses - what gave you this idea? Since efficiency is < 50%, most of the fuel is simply burnt for no return. The majority of what is left is burnt to overcome drag which is a consequence of creating lift. I don't have a number, but the overall amount of fuel used to overcome friction is going to be a small fraction. $\endgroup$ – Simon Feb 27 '16 at 9:38
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There are two primary factors that affect the skin temperature of an aircraft in flight: the air temperature, and the speed of the aircraft.

The air temperature where airliners cruise is relatively cold, around -54 °C at 35,000 feet.

As a body like an aircraft moves through air, it compresses the air, which causes the air temperature to rise. The maximum temperature rise will be if the air is completely stopped, such as at a leading edge. This is called the total air temperature, and the amount that the temperature rises is called the ram rise.

Using a simple formula to find the ram rise:

$$RR = \frac{V^2}{87^2}$$

… where $RR$ is in Kelvin, and $V$ is the true airspeed in knots.

Using a typical airliner cruising speed of 500 knots gives a temperature of 33 degrees. This brings the total air temperature to -22 °C, which is still quite cold. At places other than the leading edge, the temperature rise will be less. This is why cargo holds will need heaters to be safe for live animals, even being insulated and pressurized. Airliners just don't fly fast enough to produce a significant amount of heating.

On the other hand, the SR-71 could fly at over 1,910 kts, which gives a ram rise of 482 °C. The air doesn't get much colder as you climb to the altitudes where the SR-71 flew, so this gives a total air temperature of over 400 °C. Speed makes a huge difference.

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    $\begingroup$ Where does the 87 come from in that equation? $\endgroup$ – Holloway Dec 18 '15 at 11:35
  • $\begingroup$ @Holloway If you follow the linked reference you'll see it is an empirical approximation (for the specific system we're discussing) lumping in the heat capacity and recovery factors in the analytical equation. The = should probably be an in this instance. $\endgroup$ – J... Dec 18 '15 at 11:55
  • $\begingroup$ @J... Thanks, I imagined it was a mix of constants but wasn't sure which. $\endgroup$ – Holloway Dec 18 '15 at 11:56
  • $\begingroup$ alternately, you can use the isentropic flow relations and get $T \approx T_\infty(1+\frac{v^2}{531.6\cdot T_\infty})$, where $T_\infty$ is the air temperature in Kelvins and $v$ is the speed in knots $\endgroup$ – costrom Dec 18 '15 at 16:15
  • $\begingroup$ This answer calculates temperature rise due to compression, however it does not address the friction between the air and the surface of the aircraft. I'm assuming the frictional heating is negligible, but someone might want to elaborate on it. $\endgroup$ – Ian Dec 18 '15 at 17:24
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Local air temperature

In fast aircraft, the maximum heating is at the stagnation point. Here the kinetic energy of the flow is completely converted into pressure, which heats the air and, consequently, the structure. Due to the low local speed and the high pressure at and near the stagnation point, the rate of heat transfer is high, too, adding to the heat load.

The formula for the stagnation point temperature $T_s$ of an ideal gas of the temperature $T_{\infty}$ hitting an object with the Mach number Ma is $$T_s = T_{\infty} + T_{\infty}\cdot\frac{(\kappa-1)\cdot Ma^2}{2}$$ For air the ratio of specific heats $\kappa$ is 1.4. The tip of the fuselage nose of an airliner flying at Mach 0.85 will see air temperature to rise by 14.45%. If the air at altitude has a temperature of 220°K (-53.15°C), the air temperature at the stagnation point will be 251.8°K (-21.36°C).

But past the stagnation point the air will accelerate and become faster than flight speed. Now pressure and, consequently, temperature need to drop sufficiently to encourage the flow to stay attached and follow the curvature of the forward fuselage. This acceleration will cool the air, so the flow right above the windshield will be cooler than the ambient air.

Along the cylindrical portion of the fuselage, we find roughly flight speed again, but now friction will change the temperature close to the wall. Again the kinetic energy is converted, but the heating is caused by friction. See the boundary layer plots below:

Frictional and thermal boundary layer

Frictional and thermal boundary layer (picture source)

The temperature close to the wall is now called recovery temperature and is different from the stagnation point temperature because there is a small speed component normal to the surface which carries away some of the heat. The air temperature depends on the ratio between viscous diffusion and thermal diffusion, which is expressed by the Prandtl number Pr. If Pr>1, the air temperature at the wall is higher than the stagnation temperature and for Pr<1, it is colder. The Prandtl number of air is 0.72, so the air surrounding the fuselage is slightly colder than the stagnation temperature.

Fuselage temperature

The fuselage temperature is determined by the equilibrium between thermal conductivity, radiation and convection.

  • Conductivity: Here it is important how much the internal temperature of the fuselage can heat the skin. Cabin temperature is likely around 20°C, so some heating can be expected. However, since most airliners have isolation mats between the outer skin and the internal wall panels, conductivity from the inside is not dominant and will likely raise the skin temperature by a few degrees or less. The low heat conductivity of air (0.0204 W per m² and Kelvin) means that the heating from the inside dominates conductivity.

  • Radiation: Since the top of the fuselage is pointing into space, its far-field radiation budget is negative at night and where it points away from the sun, so radiation will cool it. The lower fuselage, however, is facing either the ground or clouds below, which both are likely hotter than the ambient air. Radiation will not cool it much and is more likely to heat it up. The part of the fuselage in direct sunlight will be significantly hotter again, depending on its color.

  • Convection: This is the dominant factor due to the high speed of the air around the fuselage. Here the air and the fuselage exchange heat by near-field radiation, and since the layer of air is replenished quickly and continuously, the air temperature is impressed on the fuselage.

I did not go to the effort of calculating the end result, but tried to list the main contributors and their magnitude. In general, the fuselage temperature is slightly below the stagnation temperature, and a dark fuselage in bright sunlight or one with little insulation and a hot interior will be several degrees hotter than the stagnation temperature.

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  • $\begingroup$ You distinguish between the kinetic energy of the flow converted into pressure at the stagnation point and 'friction' caused by the airflow against the hull - isn't the first friction too? In both cases, the body of the aircraft transforms the uniform speed of air molecules into heat. $\endgroup$ – yippy_yay Dec 19 '15 at 12:11
  • $\begingroup$ @yippy_yay: No, in the first case it is the pressure rise which heats the flow reversibly, and friction heating is irreversible and isobaric. $\endgroup$ – Peter Kämpf Dec 19 '15 at 12:22
  • $\begingroup$ Okay, but once the heat inside the stagnation point flows into the hull, the process is irreversible. You would still distinguish between friction and this process? $\endgroup$ – yippy_yay Dec 19 '15 at 12:29
  • $\begingroup$ @yippy_yay: Yes, because there is no friction involved. Compression heat occurs in an ideal gas, too. $\endgroup$ – Peter Kämpf Dec 19 '15 at 16:34

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