# What is the relation between airspeed and generated heat at constant drag?

I understand higher speed, and higher drag mean higher adiabatic compression leading to heating up the air, and the plane as result. X-15 needed ablative coating to prevent overheating. On reentry of space vehicles heat shielding is the most critical problem. This is perfectly understandable as higher speed with constant or growing air density means higher drag - and this results in higher speed.

I'd like to know how are these values related though.

[for the purpose of this question, let's neglect the issues of engine efficiency. We have a magical rocket engine that weighs nothing and needs no air nor fuel to produce thrust. This is a purely an aerodynamics question.]

At given pressure, drag is directly proportional to lift (both being quadratically proportional to airspeed and linearly to air density). That means higher airspeed allows raising the altitude, lowering air density, and thus lowering drag until lift and gravity even out.

Let's maintain such a flight: We're constantly accelerating (slowly); and climbing, at such a rate, that lift remains constant, and evens out with gravity (minus a minimal delta to retain climb). The increase of lift coming from rising airspeed is being offset by decreasing air density as the craft climbs.

With constant lift, and drag being directly proportional to lift, drag remains constant too.

Maintaining constant drag through climb like that, air density is inversely proportional to square of airspeed.

How would heat creation behave in this situation though?

• A bit of a topic drift but pet-peeve: "minus a minimal delta to retain climb" - be careful, it's not excess lift that causes a climb in a steady-state climb. In fact ,lift is less than weight in a climb... search this site a bit for more explanations. – Radu094 May 25 '16 at 11:24
• "increase of lift coming from rising airspeed is being offset by decreasing air density" - if you are 'accelerating' as in IAS is increasing, then decreasing air density is already taken into account (IAS vs TAS). What will happen is AOA will decrease to offset the increasing IAS (or TAS & q ) and keep lift constant – Radu094 May 25 '16 at 11:25
• I'll leave a full answer to someone more math oriented, but generally heat (Q?) will be proportional with speed ^ 4. As you mentioned drag & lift is proportional to ^-2 . If you do the math you will see that spending a lot of time at high altitude can become a problem. Which is why most re-entries from space are not done by slowly decelerating at high alt, but violently dropping into the lower atmosphere – Radu094 May 25 '16 at 11:31
• I'm definitely no expert on the topic, but this question/answer has the equation for "ram rise", which by definition relates the quantities that determine that temperature rise. – Steve May 25 '16 at 12:34
• The situation for the X-15 and space re-entry vehicles (both hypersonic velocities) is different than for your example of a slowly accelerating climbing flight. Because of the hypersonic speed, shock waves are formed that have a huge energy loss in the surrounding flow. That energy needs to go somewhere, and most of it is converted into heat which affects the aircraft. At subsonic speeds, I don't believe aerodynamic heating is an issue. – Jordy May 26 '16 at 11:49

You express a desire to simplify things by assuming a weightless rocket engine, so I follow with a weightless vehicle which only creates drag to overcome aerodynamic resistance.

This resistance has two parts: One is adiabatic compression when the air ahead of the approaching vehicle is compressed, and the second is friction. Friction shows up in strong shocks and in a layer of high shear stress around the vehicle which is called the boundary layer. Let's also neglect other heat sources like solar radiation which becomes more intense with altitude.

If we look at small changes, the adiabatic temperature rise ∆T due to a deceleration from a speed v to zero is: $$∆T = \frac{v^2}{2\cdot c_p}$$ where $c_p$ is the specific heat at constant pressure (unit $\frac{J}{kg\cdot K} = \frac{m^2}{s^2\cdot K}$). The amount of heat Q which gets transferred to our vehicle depends on this temperature increase, but also on the air density $\rho$ and the flow speed: $$Q = v\cdot\rho\cdot c_p\cdot∆T$$ Denser air contains more heat energy, so the proportionality with density should make sense. But why speed? If the hot air is replenished at a faster rate, more heat can be transported to the vehicle's surface per unit of time. Now we have a heat flow which grows with the cube of flow speed, but that is not all. We still need to understand how density will be affected. When air is heated, density drops, but the flow at the forward facing parts of the vehicle is compressed as well, so density goes up in the end: $$\frac{\rho}{\rho_{\infty}} = \left(1 + \frac{\kappa}{2}\cdot Ma^2\cdot c_p\right)^{\frac{1}{\kappa}}$$ with $\kappa$ as the ratio of the specific heats and $Ma$ the Mach number of the flow. Now things get messy because all parameters will change with temperature, so let me short-cut to NACA report 1381 which covers the re-entry heating of spheres. Here the rate of heat transfer is proportional to the cube of speed, so flying faster at constant dynamic pressure will cause a quadratic rise in the temperature of the surrounding air and a rate of heat transfer rising with the cube of speed due to convection.

• How do the units work out in $(\frac{v^2}{c_p} - 1)$? $1$ is dimension-less, so $c_p$ would have to be the same as $v^2$, namely $\mathrm{m}^2\mathrm{s}^{-2}$. But if it is specific heat, it is $\mathrm{J}\mathrm{kg}^{-1}\mathrm{K}^{-1}$. The $\mathrm{kg}$ reduces with the $\mathrm{J}$ to $\mathrm{m}^2\mathrm{s}^{-2}$ all right, but what about the Kelvin? – Jan Hudec Jun 27 '16 at 9:37
• @JanHudec: Thank you - of course the equation was wrong. I should have looked it up properly. – Peter Kämpf Jun 27 '16 at 19:14

A given airplane moving through the atmosphere at at given weight and velocity requires a specific and determinate amount of energy. That known energy can be expressed in foot-pounds, calories, btu's, etc. There are simple, straightforward conversions between these metrics. btu's typically represent heat energy.

A simple approach to determine the energy in foot-pounds is to fly at a known HP setting and fly for a specific time frame. The product of that HP times that time represents energy. Simply convert it to btu's.

• Considering I don't have a supersonic airplane to conduct the experiment myself, would you care to provide the expected results curve? – SF. May 26 '16 at 19:04
• Of course, not all of the spent energy will be transformed into heating by friction. A lot (actually, quite most) of the energy will be spent accelerating air downward and forward. – Radu094 May 26 '16 at 21:13