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Elastic Collisions and Drag

  1. Apr 6, 2005 #1
    Some people argue that a non-viscous gas could not result in a drag for moving objects because the kinetic energy could not be dissipated to the gas. However, this view neglects the fact that even if the gas molecules do not interact with each other (i.e. if the gas is inviscid), they still collide with the object which results in a change of momentum and energy.

    Consider a plate moving frontally through a gas with velocity +v. A molecule moving vertically towards the plate with velocity -u in the lab frame has a velocity -u-v in the reference frame of the plate and hence bounces back from the front side with the velocity u+v (as the plate is much heavier than the molecule), i.e. in the lab frame the velocity of the molecule is now u+2v. Correspondingly, at the backside of the plate, a molecule with velocity u in the lab frame has a velocity u-v in the reference frame of the plate and thus after reflection a velocity -u+v, i.e. in the lab frame the velocity is now -u+2v.
    This means in the lab frame the kinetic energy of the molecules being reflected from the front has increased from m/2*u² to m/2*(u+2v)² =m/2*u² +2*m*u*v + 2*m*v², i.e. an increase of +2*m*u*v + 2*m*v².
    Molecules reflected from the back side on the other hand have changed their energy from m/2*u² to m/2*(-u+2v)² = m/2*u² -2*m*u*v + 2*m*v² i.e. a change of -2*m*u*v + 2*m*v².
    Adding and averaging the two contributions one obtains therefore the average increase of the kinetic energy of the molecules with mass m hitting the plate with velocity v as ΔK = 2*m*v². (Note: this value will actually be smaller by about factor 1/2 as the molecules won't be reflected straight back but will be scattered into the whole half-space due to the roughness of the plate's surface).
    (One can derive this result also from the energy and momentum conservation equations applied an elastic collision, with the same result if one assumes the plate mass as large compared to the molecular mass).

    Let's apply this result to air (although the latter is not strictly inviscid in the above defined sense):
    if the plate has a cross section of 1m² and a velocity of 10 m/sec it will collide with 3*10^26 molecules per second, which, assuming a molecule mass of 4.5*10^-26 kg, amounts to a total mass of M=13 kg per second. Replacing m with M in the above result for the energy gain of a molecule, one sees that the kinetic energy gained by the air molecules per second corresponds to the kinetic energy of a plate of about 26 kg (taking the above mentioned correction into account), i.e. a plate of with a mass of 26 kg, a cross section of 1m² and a velocity of 10m/sec would lose its energy within about one second. This result looks actually quite realistic and shows that the assumption of a non-viscous gas can account for observed drag in air, and this obviously suggests also that the same may hold for the aerodynamic lift.
     
  2. jcsd
  3. Apr 6, 2005 #2

    russ_watters

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    You're neglecting the concept of pressure again, Thomas2. If there was no interaction at all between air molecules (that's not what "inviscid" means, btw), then as they hit an object they'd bounce away only on the leading surfaces and there would be a corresponding change in energy, just as you say. But pressure intervenes to make these particles bounce back onto the trailing surfaces of the object, giving back whatever momentum they gained.

    edit: once again, there is real data to back this up. No, you can't ever have zero viscocity, but you can have an extremely low viscocity relative to the other components of the Reynolds number. Simply put, you can measure that as viscocity goes down, drag goes down. Once again, you are arguing against reality.
     
    Last edited: Apr 6, 2005
  4. Apr 6, 2005 #3

    arildno

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    russ:
    One of the major troubles in Thomas2's conception lies here I think:
    That is, he essentially regards an inviscid fluid as a sort of vacuum.
    I don't know how to respond to this..
     
    Last edited: Apr 6, 2005
  5. Apr 6, 2005 #4

    russ_watters

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    I mentioned that. It seems his misunderstands always come back to dealing with pressure. The only way to deal with that in this case is to explain to him that an inviscid fluid still has pressure, which I did. Beyond that.... [shrug]
     
  6. Apr 6, 2005 #5

    arildno

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    Oops, I see you did mention it after all, right at the start.Sorry about that..:redface:
     
    Last edited: Apr 6, 2005
  7. Apr 6, 2005 #6
    Even if you take the molecular interaction into account, the molecules will lose only some of the velocity 2v that they gained on collision with the plate: if you consider that half of the molecules were actually initially not flying towards the plate and average the velocity gained with these, you still have a velocity gain of +v, i.e. the initially (on average) resting molecules will be imparted a velocity such that they co-move with the plate. Interestingly this is just what is observed in form of the 'stagnation' at the rear and front.

    But this was not actually my main point. I merely wanted to show that there is drag in a gas even if the molecules don't interact with each other (whether you call this 'inviscid' or whatever).
     
  8. Apr 6, 2005 #7

    russ_watters

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    Please start using the word "pressure". When you say "molecular interaction" it makes it sound like you are purposefully avoiding taking pressure into account.
    And you base that on what, exactly? Experimental evidence? Mathematical proof? Just saying it doesn't make it true (and the fact is, it isn't true, experimentally or mathematically).
    Well, if you take away all "molecular interactions", (pressure and viscocity) you are no longer talking about a "fluid" and fluid dynamics/aerodynamics no longer applies. What you are saying may be true of a spacecraft encountering a single hydrogen atom every few kilometers, but it has nothing to do with drag, fluid dynamics, or aerodynamics.

    I'm a little incredulous here, Thomas2 - you claim a lot for someone who doesn't even understand what a "gas" is. I am again having doubts about whether you're even serious about this.
     
  9. Apr 9, 2005 #8
    The 'pressure' that the molecules exert on each other is not what causes the drag on an object. It is the one that the molecules exert on the object. If molecules would be point masses, the former would actually be zero as molecules could not collide with each other, but the pressure on macroscopic objects would still be the same.

    The point here is that 'pressure' is actually something different for gases on the on hand and fluids or solids on the other. For gases much of a volume is actually empty space: the volume of a molecule is about 10^-24 cm^3 and with 3*10^19 molecules/cm^3 at atmospheric pressure, this means that only a fraction of 3*10^-5 of a given air volume consists of matter, the rest of empty space. In fluids and solids on the other hand, the molecules or atoms are however tightly packed without any empty space between them. The pressure here is is not associated with the kinetic energy of the molecules but with the static potential energy between the atoms (analogously to compressing a spring for instance).

    In view of this, it is actually the fluid dynamics approach to gases that needs justification and not the 'particle kinematics' approach. Fact is that within a certain distance from an object in a gas there are no collisions between molecules and hence the latter hit the object in a free flight manner as described in my opening post above.
     
  10. Apr 9, 2005 #9

    arildno

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    No. This is simply false, and you know it.
    Stop posting nonsense.
    It seems you are making this up as you go along.
    Where are you? At an asylum?
     
    Last edited: Apr 9, 2005
  11. Apr 9, 2005 #10

    enigma

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    Actually, pressure produces the vast majority (an order of magnitude greater than viscous drag at least) of most ordinary (read: aircraft) flows.

    If you pick up an introductory aerodynamics textbook such as:

    "Fundamentals of Aerodynamics" by John D. Anderson

    You will find that out in chapters 1 and 2 where the basic forces and interactions are derived and explained.

    The situation you are describing is called a "Newtonian Flow" and is only applicable for spacecraft and other applications which are in hard vacuum where the molecular mean free path is of the order of magnitude of the cross section of the craft.
     
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