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Suppose that a hollow truncated cone is placed in a wind tunnel with a steady wind speed $V$. The cone is placed in such a way that it's base of area $A_1$ faces the wind (rather than the other side of area $A_2$ for which $A_1>A_2$). Here's a contradiction:

Supposing that $V_2$ is the speed for which wind appears from that end of the cone of area $A_2$, using the conservation of mass law, we get $A_1V=A_2V_2$ and hence $V_2>V$. Now using the relation $\sum F=\dot{m}\Delta V$ (which results from the more general relation $\sum F=\frac{\partial}{\partial t}\int_{C.V}V.\rho (dv)+\int_{C.S}V.\rho (VdA)$) we see that a net thrust is generated which pushes the cone forward and against the wind, but this is impossible.

What's wrong with this argument? I've been giving it some thought but I can't figure out a way to solve it.

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  • $\begingroup$ What do you mean with "hollow"? Does only the lateral surface exist? Are A1 and A2 open or closed? $\endgroup$
    – Holger
    Commented Jan 6, 2014 at 9:15
  • $\begingroup$ Maybe I should have explained a little more. By hollow I mean that the cylinder has a small thickness $t$ which is negligible compared to the dimensions of the cylinder. Also, $A_1$ and $A_2$ are the cross-sectional areas that the air flow can pass through them. $\endgroup$
    – Goodarz Mehr
    Commented Jan 6, 2014 at 10:31
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    $\begingroup$ Even if the conic surface is frictionless the air will be compressed as the diameter decreases. Shouldn't your "conservation of mass" include the local density as well as the velocity and cross-sectional area? (and I would hope :-) that the forces on the conic surface have a net downstream component) $\endgroup$
    – Carl Witthoft
    Commented Jan 6, 2014 at 14:11
  • $\begingroup$ You're right, but is that difference large enough to affect the velocity at the exit? I think the problem may be with the velocity profile that is assumed uniform here, but I can't prove it. $\endgroup$
    – Goodarz Mehr
    Commented Jan 6, 2014 at 14:45

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Air entering the cone at $A_1$ is not traveling at speed $V$, but slower, due to the pressure within the cone.

If you want to see why, let $A_2$ shrink to a very small size.

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  • $\begingroup$ You're right, but still as it travels with a speed $V_1$ lower than $V$ (and hence lower than $V_2$) a forward net thrust is produced. $\endgroup$
    – Goodarz Mehr
    Commented Jan 6, 2014 at 19:31
  • $\begingroup$ @GoodarzMehr: At low mach number (ignoring compresibility) conservation of mass says $\rho A_1V_1 = \rho A_2V_2$, so momentum/second in equals momentum/second out, so no net thrust. (Ignoring drag on the cone.) Jet engines work by increasing $V_2$ by expanding the gas. $\endgroup$
    – Mike Dunlavey
    Commented Jan 6, 2014 at 22:04
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About moving against the wind. This is not entirely impossible. Put a windmill on a boat and let it power the propeller. This worked in an experiment done on a lake in the Netherlands. An explanation was given by a professor in theoretical physics, based on the (non-linear) so-called windmill formula. The wind speed has to be in a certain range.

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  • $\begingroup$ Would you please give a reference for that? $\endgroup$
    – Goodarz Mehr
    Commented Jan 6, 2014 at 10:32
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    $\begingroup$ Wind-powered boats can indeed sail into the wind but this isn't a useful answer to the question. $\endgroup$
    – RedGrittyBrick
    Commented Jan 6, 2014 at 11:10
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    $\begingroup$ @Goodarz Mehr. Unfortunately I do not remember further details. It was 20 years ago or more. The experiment took place in the "Kralingse plassen" near Rotterdam in the Netherlands. It was reported in the Dutch newspaper NRC, together with a photo. Somewhat surprisingly I could not find the windmill formula, in Dutch "molen formule" (molen=mill) with Google. This formula was qubic in the wind speed. If the speed was too low or too high the phenomenon did not occur. $\endgroup$
    – Urgje
    Commented Jan 7, 2014 at 10:41
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    $\begingroup$ @RedGrittyBrick. Although boats can sail into the wind there must always be some angle with the wind speed. This is not so with the arrangement I described above. $\endgroup$
    – Urgje
    Commented Jan 7, 2014 at 10:46
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Your first assumption ($V_2 > V$) is correct, but you neglect the base areas when cutting out a slice of the tunnel for your integration. The forward-facing base sees the pressure of stagnating air at it's center, accelerating towards the edge to whatever speed is reached when air flows around the corner from the forward base to the mantle of the cone. The rear-facing and bigger area sees the pressure of separated flow (weak suction) over the full area. You need to account for the pressure difference as well, not just momentum, because your control areas include the bases of the cone.

If you move the control areas away from the cone, air expands behind the cone and the speed drops to it's old value (neglecting any viscous effects).

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