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I When should the drag coefficient of a rocket peak?

  1. Apr 17, 2018 #1
    Hi,

    When considering the drag coefficient of a rocket in transonic/supersonic flight, I always expected the drag coefficient to peak at exactly Mach 1 due to the abrupt increase in pressure that occurs when breaking the sound barrier. However, I have been looking at several websites and articles that predict the drag coefficient of a rocket to peak at slightly above Mach 1 (say Mach 1.2). I.e. Figure 4 on this website:

    http://what-when-how.com/space-science-and-technology/rocket-propulsion-theory/

    Could anyone explain why the drag coefficient peaks at slightly above Mach 1 instead of exactly Mach 1?

    Thanks!

    /krihamm
     
  2. jcsd
  3. Apr 22, 2018 #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
     
  4. Apr 22, 2018 #3

    jack action

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    I think @boneh3ad 's expertise could answer this question ...
     
  5. Apr 22, 2018 #4

    256bits

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    Here'e a picture of a plane flying through the air and increasing its speed. courtesy of Nasa.
    https://history.nasa.gov/SP-367/fig87.jpg
    https://history.nasa.gov/SP-367/chapt5.htm
    as it transitions form subsonic to supersonic speeds, called the transonic range.
    One can see that supersonic flow of the fluid ( air ) occurs over the body not just at Mach 1 for the free stream flow, but at some speed lower.
    Subsonic flow is characterized as the fluid being treated as incompressible and the equations in that region work particularly fine.
    At around say 0.8 of the free stream Mach number, called the drag-divergence Mach number, the local Mach number at certain places over the wing becomes supersonic, as the air has to speed up over the body, and we see the beginnings of shock waves. and increased drag coefficient.
    This is due to the fact of the compressibility of the air, and called wave drag.

    The formation of the bow shock and the implications for the drag coefficient is probably what the question details as the object ( plane ) increase its speed beyond that of free stream Mach 1.
    that's as far as I get.


    fig87.jpg
     
  6. Apr 23, 2018 #5

    boneh3ad

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    There seems to be a lot of incorrect statements in the post by @256bits, even if the general idea is correct. I feel I ought to clarify a few things.

    This is inaccurate. Subsonic flow is characterized by a Mach number less than 1 It says absolutely nothing about whether the flow is compressible or not. In order to treat a flow of gas as incompressible, the Mach number needs to be less than about 0.3, so there is a sizable Mach number range that is subsonic but compressible.

    The Mach number at which the flow becomes locally sonic (and beyond which, supersonic) over an airfoil is called the critical Mach number. There is immediately a small drag increase because you now have to contend with wave drag, but the shocks that form are still so incredibly weak that they don't contribute much. The drag-divergence Mach number is close to but slightly larger than the critical Mach number and represents when the drag on the airfoil makes a sharp rise, generally corresponding to the formation of stronger shocks as that local supersonic region gets greater and greater than Mach 1. The critical Mach number and drag-divergence Mach number are not the same thing. Also, the critical Mach number varies based on the shape of the body, with some airfoils becoming critical around Mach 0.6 and others pushing it nearly to Mach 0.9.

    In some sense, I suppose that yes, wave drag is a result of the compressibility of air. In reality, though, it is more specifically due to flows that are supersonic. You will not see any wave drag in a flow that is, say, Mach 0.999, even though it is highly compressible, because it is still subsonic. There are fundamental changes that occur in flow fields when the Mach number increases beyond 1. Specifically, the behavior becomes more wave-like (mathematically, the solutions transition from elliptic to hyperbolic). It is this change that gives rise to wave drag.

    What large pressure increase are you referencing here?

    Your link does not cite its sources. The figures (and very possibly the equations) are lifted from some other site without attribution. It is difficult to say with certainty what they are even plotting. They say drag coefficient versus Mach number and angle of attack, but over what shape? I am not sure I want to speculate without knowing what shape was plotted there.
     
    Last edited: Apr 23, 2018
  7. Apr 25, 2018 #6

    256bits

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    Well noted. And informative.

    At least the OP has received a few responses, although he/she hasn't returned.
     
  8. Apr 25, 2018 #7

    boneh3ad

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    My original response came off as more combative than I intended. Hopefully I managed to fix it.
     
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