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Velocity of a particle exposed to a shock wave?

  1. Oct 29, 2013 #1

    M_1

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    I need to understand how much kinetic energy a particle can absorb which initially is at rest and suddently is exposed to a shock wave.

    Detonation velocities are very high, on the order of 6000 m/s, but I assume this is the velocity
    of the shock front and not of the gas molecules themself, and I assume it is the gas molecules, the air, that drags the particle and accelerates it.

    So I assume I need a velocity vs time function for a typical shock front, or in the best of worlds some values of the final velocity of particles of differents sizes that has been exposed to a some typical shock wave.
     
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  3. Oct 29, 2013 #2

    Simon Bridge

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    You want a pressure profile for the shock front ... and a model for how a particle responds to pressure changes.
     
  4. Oct 29, 2013 #3

    boneh3ad

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    Are you familiar at all with normal shock theory? If you know the velocity of the shock and the conditions of the air, you can get the Mach number, and from that you can fairly easily determine the Mach number of the still air with respect to the shock wave, which will give you exactly the Mach number of the air behind the shock relative to the moving shock. All you have to do then is convert that Mach number back to velocity and back into a stationary reference frame. It is actually a really basic gas dynamics problem, but if you haven't had any gas dynamics, it may be a bit confusing.
     
  5. Oct 29, 2013 #4

    sophiecentaur

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    I guess it must depend upon the mass of the particle, relative to the density of the medium. It must be a 'matching' problem, to get the amount of energy transferred.
     
  6. Oct 29, 2013 #5

    M_1

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    Thanks boneh3ad,
    I'm unfamiliar with shock theory. If I manage to do what you described my understanding is that I will have the air velocity in a stationary reference frame, which is what I want, I think. But I guess that that velocity is very short-lived at a fixed position in space. So I guess that I need the velocity as function of time at a fixed position in space, and then I can calculated the final velocity of the particle due to the time-dependent drag force.

    To complicate things the particle will partly follow the velocity so in fact I would need the velocity as function of both time and space if I should be able to correctly calculate the drag as the particle is accelerated.

    Is this really so basic?
     
  7. Oct 29, 2013 #6

    boneh3ad

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    There is a well-known algebraic relationship between the Mach number before and after a shock.
    [tex]M_2^2 = \dfrac{\left( \gamma - 1 \right)M_1^2 + 2}{2\gamma M_1^2 - \left( \gamma -1 \right)}[/tex]
    where [itex]M_1[/itex] is the Mach number relative to an equivalent stationary shock wave of the flow into which the shock is advancing, [itex]M_2[/itex] is the Mach number relative to an equivalent stationary shock wave of the flow after the shock has passed, and [itex]\gamma[/itex] is the ratio of specific heats, which is 1.4 for diatomic gases like air. So, if you have simply the speed of your shock wave already and the temperature of the air into which it is advancing, then you already know [itex]M_1 = u_1/a_1 = V_{\text{shock}}/\sqrt{\gamma R T}[/itex] where [itex]R[/itex] is the specific gas constant (≈287.1 J kg-1 K-1 for air) and [itex]T[/itex] is the tempetature in Kelvin.

    From there you just use the above equation To get [itex]M_2[/itex]. There exists a relationship for the temperature change across a shock as well
    [tex]\dfrac{T_2}{T_1} = \left[ 1 + \dfrac{2\gamma}{\gamma + 1}\left( M_1^2 -1 \right) \right]\left[ \dfrac{2 + \left( \gamma-1 \right)M_1^2}{\left( \gamma + 1\right)M_1^2} \right].[/tex]
    This will give you [itex]T_2[/itex], which, when combined with [itex]M_2[/itex] will give you the velocity after the shock, [itex]u_2[/itex]. Then it's just a matter of converting that back into a stationary reference frame.

    Essentially what the shock does when it passes through a stationary medium is drag the air behind it along with it at some velocity, and it is that velocity that you are solving for here.

    This works if the shock is of constant strength (constant Mach number) and that velocity of the air behind the shock absolutely will be constant. However, the shock strength won't likely be constant if we are talking about a detonation, as it will weaken as it expands. For that you would have to use knowledge of the pressures to track the changing strength of the shock as it expands, and that would be difficult if not impossible to do analytically, but you are welcome to try. As you may be able to guess though, there is a relationship between the pressures across the shock and Mach number, which is
    [tex]\dfrac{p_2}{p_1} = 1 + \dfrac{2\gamma}{\gamma + 1}\left( M_1^2 -1\right)[/tex]
    and a density relationship, too,
    [tex]\dfrac{\rho_2}{\rho_1} = \dfrac{(\gamma+1)M_1^2}{(\gamma-1)M_1^2 + 2}.[/tex]

    You can find a lot of extra related equations in a report called NACA 1135 that you should be able to find for free pretty easily online, but it may be difficult to use if you don't know what you are doing.

    At any rate, if your detonation that you are specifically talking about is in something like a tube, then this will work wonderfully because the velocity after the shock passes will be constant and you just need to estimate the drag on your particle in your favorite way. If it is a detonation out in the open, It is a more transient problem but you can likely get a decent estimate by assuming constant shock strength, especially if the particle is small and therefore likely to accelerate almost impulsively.
     
  8. Oct 29, 2013 #7

    Simon Bridge

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    That's how I'd have done, yeah: producing a model specifically for what I needed to find out - if the particle is part of the medium or suspended in it (dust in air say) then it would work just to have it match the medium's motion... following the wave.

    boneh3ad has produced the modelling that is more usual though - I'm gonna leave him to it :)
     
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