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A bow shock is a bow-shaped normal shock created around a blunt object flying supersonic.

enter image description here

The shock itself - mach = 1 - can be very hot at hypersonic speeds. But the subsonic air behind is cooler, which is why we use bow shocks on reentry capsules.

But why is this air cooler? Or, if it isn't, then why does the bow shock produce less heating?

edit: a clarification

enter image description here

Is this correct?

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  • $\begingroup$ they are also used on "entry" vehicles. $\endgroup$
    – JEB
    Commented May 27, 2020 at 13:28
  • $\begingroup$ yep. the only difference is (so far) being on another planet. $\endgroup$ Commented May 27, 2020 at 15:10
  • $\begingroup$ The body surface is only 1500 K, so it cools the air around it. $\endgroup$ Commented May 28, 2020 at 14:30
  • $\begingroup$ @PeterKämpf and since it is a slow subsonic flow behind a bow shock (as opposed to the supersonic flow behind an oblique) this cool air sticks around longer? $\endgroup$ Commented May 28, 2020 at 15:37
  • $\begingroup$ @Abdullah: Yes, and it has more time to cool down, isolating the tip from the hot shock. But longer is relative here. Things still move relatively fast, even behind a straight shock. $\endgroup$ Commented May 28, 2020 at 17:47

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Imagine a solid cone with half-angle $\theta$ moving point-first at a velocity $M_1$ with respect to a compressible ideal gas. If possible, this cone will create a conical attached oblique wave with a half-angle $\beta$ that comes to a point at the point of the solid cone. The shock wave angle $\beta$ is a function of the deflection angle $\theta$, the Mach number $M_1$, and characteristics of the gas. The relation between the deflection angle and the shock wave angle is depicted below for a diatomic ideal gas for various values of Mach number.

Graph showing oblique shock deflection angles. See text for details.

Note that for a given Mach number there is a maximum possible deflection angle. An attached oblique shock wave cannot be formed if the deflection angle is larger than this maximum. The shock detaches and forms a bow shock.

The above assumes a cone that comes to a point, which is not physically possible, and assumes an ideal diatomic gas, which also is not physically possible. A physical object, even one with a highly aerodynamic shape, will have some bluntness to its tip. There will always be a tiny portion of the shock that is detached from the deflecting body.

In a real gas, heating in the shock can raise the temperature to such an extent that the gas dissociates. The region behind the shock will contain dissociated elements that recombine and release heat. Most of this recombination occurs very close to the shock. If the gap between the bow shock and the physical body is small, the recombination heating will result in significant heating of the physical body. But if the gap is large enough, that the recombination occurs somewhat remote from the physical body limits the heat transfer to the physical body.

This reduced heating to a blunt body is one of the two key reasons why a blunt body is preferred over a more aerodynamic shape. The other is that the higher drag slows the body down more than would happen with a more aerodynamic shape. This deceleration is a very desired effect for a reentering body.

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  • $\begingroup$ Hey thanks! :) The region behind the shock will contain dissociated elements that recombine and release heat. wouldn't this released heat then remain in the airflow behind the recombination zone, to be transmitted to the body downstream? Or does radiation cool the air gradually beyond the recombination zone or what? $\endgroup$ Commented May 27, 2020 at 17:15
  • $\begingroup$ Or is it that the relatively cool blunt body creates a cool slow moving boundary layer in front of itself, while most of the hotter, faster air from the recombination zone flows right around this layer? $\endgroup$ Commented May 27, 2020 at 17:19
  • $\begingroup$ @Abdullah - The shock wave does not make the air stand still behind the shock. $\endgroup$ Commented May 27, 2020 at 21:02
  • $\begingroup$ I know. But what is preventing this heat release from being convected to the body? $\endgroup$ Commented May 28, 2020 at 7:13
  • $\begingroup$ I mean, the heat is being transferred to the air, which will transfer it to anything downstream. $\endgroup$ Commented May 28, 2020 at 7:15
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A pointed tip will produce an attached, oblique shock.

A blunt tip will result in a separated (detached) shock.

While the peak temperature in the attached shock is lower than that in the center of the detached shock, the temperature gradient* between the hot air and the surface of the tip is much higher in case of the pointed tip, resulting in much higher energy transfer into the tip. The reason is the distance between the shock and the tip: As the name implies, the attached shock sits directly on the tip while the detached shock keeps some distance between the shock and the tip. And a flatter gradient means less heat energy is transferred into the blunt tip. Most of the heat generated in the detached shock remains with the air and is carried away with the flow.

* That a temperature gradient is maintained lies in the limits of structural materials: The temperature in a hypersonic shock exceeds the melting point of all practical materials, so continuously new, cooler material is exposed to the shock as the tip material melts away. In a blunt nose, thermal conduction or ablative cooling keeps the nose cool. The first reentry bodies of intercontinental ballistic missiles contained a copper heat sink while newer designs use a layer of material which vaporizes during reentry.

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  • $\begingroup$ for this gradient to be steeper, the temperature at the hot end would have to be higher. We're saying this happens with oblique shocks. My question is why it doesn't happen with normals $\endgroup$ Commented May 28, 2020 at 9:39
  • $\begingroup$ @Abdullah But it does happen with detached shocks also. It's not the temperature - that is about equal in both types (actually higher in a detached shock). It is the distance between where that temperature occurs and where cooler material is which soaks up that heat. Temperature divided by distance = temperature gradient. And with a flatter gradient there will be less soaking (= heat transfer). $\endgroup$ Commented May 28, 2020 at 11:36
  • $\begingroup$ the distance between where that temperature occurs and where cooler material is doesn't that temperature occur all the way from the shock right up to our body? After all, all the air in between is the same air that was shocked and heated a fraction of a second ago. or is the new diagram i posted what is happening? $\endgroup$ Commented May 28, 2020 at 11:43
  • $\begingroup$ @ doesn't that temperature occur all the way from the shock right up to our body? No, not at all! That air gets cooled by the body, either by melting/vaporizing it or by heating its heat sink. That is why I stress that the temperature gradient is maintained. The tip must be continuously cooled or it melts away. It is much cooler than the air in the shock and that difference produces the gradient. Compare with the drawings in your question: They say the same. $\endgroup$ Commented May 28, 2020 at 14:27

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