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When driving a car while smoking with the window open (safety and legal issues aside), I've noticed that the smoke tends to go outside the window.

  1. Why does the smoke go outside?
  2. If the car is standing still and there is wind blowing at the same velocity the car was going - will the smoke behave the same?
dmckee --- ex-moderator kitten
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Joe
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4 Answers4

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It is called Venturi Effect.

The increase in speed of the air surrounding your vehicle comes with a decrease in pressure. That explains too why a chimney works better in windy days.

The Venturi effect is explained by applying the Bernoulli Equation (say, the conservation of energy of a small piece of fluid that moves within the flow) between two points along a streamline (in this case, we would follow a piece of air in a tunnel wind)

$\frac{1}{2} \rho v^2 + \rho g h + p = \text{constant}$

The increase in the first summand when the flow gains speed to adapt itself to the shape of the car, is compensated by a decrease in the pressure $p$. Look what happens in this picture (wikipedia) when the flow changes speed to adapt to the shape of the tube:

(Image from wikipedia) $ $

Note the similarity with the high school equation for the conservation of mechanical energy of a particle:

$\frac{1}{2} m v^2 + m g h = \text{constant}$

(Just change the mass of the particle for the mass of a fluid volume unit, i.e. density, and add an additional summand to accout for the pressure, and you have Bernouilli's equation)

Bernouilli's equation is meant for an incompressible flow (water) which here means that the numerical results would be approximate, but qualitatively the same effect happens.

A related, interesting fact, is that submarine propellers must be carefully designed, in order to avoid points in which water suffers much too rapid a speed increase. When that happens, pressure becomes so low in that points that vacuum bubbles appear. The power released by the implosion of that bubbles against the surface of the propeller, not only is noisy, but also may damage the propeller itself. The phenomenon is called cavitation.

enter image description here

(Image from wikipedia)

Waffle's Crazy Peanut
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Eduardo Guerras Valera
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  • Language corrections to my non-native english are highly welcome. – Eduardo Guerras Valera Jan 16 '13 at 03:21
  • Hmm... I haven't seen your answer for the past few weeks. Nice though :-) – Waffle's Crazy Peanut Jan 16 '13 at 04:48
  • Great answer to my first question! But what about my second question? I think that according to the same Venturi effect, the air pressure when there's wind blowing should be lower than the atmospheric pressure, which is the air pressure at some distant point where there is no wind. Therefore in the case of wind the smoke should be leaving the car at an increased rate - am I right? – Joe Jan 16 '13 at 06:20
  • @Joe: yes, it is the same. That is why aerodynamics research can be done with a model of the car fixed to the ground inside a wind tunnel, instead of having a moving car on the road against a static atmosphere. – Eduardo Guerras Valera Jan 16 '13 at 08:36
  • If the car and the velocity of the air inside the wind tunnel are correctly scaled, there is no difference. If the model is not downscaled, there isn't any problem. But you cannot build a wind tunnel for an Airbus, and there is a whole branch of fluid dynamics dedicated to that question, how a smaller model scales to reality. Google for Reynolds Number if you are curious about that. – Eduardo Guerras Valera Jan 16 '13 at 08:46
  • But I'm not asking about a wind tunnel, I'm asking about real-world wind. What I was trying to say is that the air pressure in the area where the wind is blowing should be lower than atmospheric because of the Venturi effect, and then you have the speed gain caused by flowing around the car on top of that, so that altogether the air pressure is lower in the case of wind blowing than in the case of a moving car. So I'm saying the two cases are not the same - does this make sense? (I guess that in wind tunnels they can compensate for that, but I'm not interested in that right now) – Joe Jan 16 '13 at 13:54
  • @Joe, the only thing that matters is the relative difference in the flow (squared)speed between the front part of your car and the side where the window is. Whatever happens in the rest of the atmosphere, whether there is a windy day or not, it doesn't matter. You can achieve the air speed difference between that points due to wind, due to the speed of the car, or both things together. Does that answer your question? – Eduardo Guerras Valera Jan 17 '13 at 01:50
  • @Eduardo Both you and tpg2114 seem to be getting at the same general idea, and I don't disagree, but the astrophysicist in me sees a possibly turbulent boundary layer (Kelvin-Helmholtz? though maybe that's a stretch). Do you think the lateral transport could maybe just be mixing due to turbulent eddies. –  Jan 17 '13 at 03:22
  • @Chris, Happy new year! Interesting topic, but I think it would not apply here, unless he completely opened the window of the car. If he only opens the window a bit (that is the picture I have in mind) the question is completely different: a laminar flow around a solid surface with a small hole. Car designers do big efforts to keep air flow around the vehicle in the laminar regime, and even within a turbulent flow, around solid surfaces there is a so-called Laminar Boundary Layer (here). But if he fully opens the window, you may be right. – Eduardo Guerras Valera Jan 17 '13 at 08:47
  • Car manufacturers always advise not to drive with opened windows, in order to spare fuel. Because a fully opened window gives rise to turbulences and, as you say, lateral transport, with unpredictable interchange of momentum that slows down the car, as if it were friction. – Eduardo Guerras Valera Jan 17 '13 at 08:53
  • @EduardoGuerras - I disagree. Of course a simple change of reference frame should not matter, but there is no such change that relates the two cases since the boundary conditions are different. In the case of wind there is some distant point where there is no wind blowing, and therefore the flowing air is at a lower pressure to begin with. I've written my own answer to try and explain my view. I welcome any criticism or comments. – Joe Jan 19 '13 at 09:53
  • @Joe, to a piece of air inside your car, the distant point doesn't matter. The air that reaches your car has no memory of its past circumstances. And, for Bernouilli equation, two and only two points matter (that shall lie along the same stramline). You can use a very distant one as you like, or one in front of the cat or inside. It is all the same, provided that you give the right values P, v, density. – Eduardo Guerras Valera Jan 19 '13 at 10:10
  • @EduardoGuerras - You are right, I was wrong. Thank you for having this discussion with me. I deleted my answer, in which my main mistake was that I used Bernoulli's principle for a system where the airflow was not steady, and this is a requirement without which the principle does not hold. – Joe Jan 19 '13 at 14:51
  • @EduardoGuerras - I still have one question about your answer though. You say that "the flow gains speed to adapt itself to the shape of the car", but if we ignore this speed gain we still get a pressure difference, don't we? The air outside the car is moving relative to the air inside the car simply because the car is moving, and therefore the pressure outside is lower. – Joe Jan 19 '13 at 14:54
  • @Joe, the essential thing is that the car has at least another "hole" that is in contact with the atmosphere in a point with different conditions. Imagine a floating dust particle that enters the car through the front and follows a streamline until it goes out through the window. If the car were perfectly closed and aisled and there were only a hole (the slit you created by opening the window a bit) then, after a very short time, the pressures inside and outside would be balanced and, no matter how windy, the smoke would not leave through the slit. – Eduardo Guerras Valera Jan 19 '13 at 18:00
  • @It wouldn't work either, if you had two holes in contact with external points with identical conditions in that idealized, perfectly sealed car (e.g. left and right windows). – Eduardo Guerras Valera Jan 19 '13 at 18:03
  • @Look at the first picture. The difference in pressures in the U-shaped part is compensated by the extra weight of the left water column. There is no such mechanism in your car, so the air keeps flowing and the smoke goes out wih it. – Eduardo Guerras Valera Jan 19 '13 at 18:08
  • @EduardoGuerras - thank you for all the explanations in the comments. Maybe you should add some of those to your answer? I Certainly found them helpful, and I think these comments are bound to be cleaned up. – Joe Jan 20 '13 at 10:20
  • By the way, cavitation is an issue for all marine propellers, not just submarines. – David Richerby Mar 15 '14 at 01:23
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I'll answer your questions in reverse order:

There is no difference in the fluid dynamics of this problem (may not always be true as boundary layers may be different, but that's not a factor really here) if the car is moving forward at speed $V$ into quiescent air or the car is stationary with a wind moving towards it at speed $V$. It's just a change of reference frame which in this case isn't an issue.

Now for the actual hard part...

Your car is not going fast enough to make the flow compressible so we can assume it isn't. We can also assume that the flow doesn't have heat added/removed and for the most part inviscid. This all means that the flow is adiabatic and reversible, aka isentropic.

The isentropic assumption is a handy one because it allows us to say that total pressure is constant along streamlines. Since we'll assume the atmosphere is uniform (which for this works well enough, it is pretty uniform over the width of the car) this means that the total pressure is the same everywhere upstream where the flow isn't moving. Which means the total pressure is constant in the entire flow when moving.

Now, because it's incompressible and isentropic, we can define the total pressure as:

$$P_0 = P_s + \frac{1}{2}\rho V^2$$

which is Bernoulli's Principle where $P_0$ is the total pressure, $P_s$ is the static pressure, $\rho$ is the air density and $V$ is the air (or car) speed. Since the left hand side is constant, this tells us that as we go faster the $\frac{1}{2}\rho V^2$ term gets bigger which means $P_s$ gets smaller.

The air inside your car is, for our purposes, not moving. Or if it is moving, it's moving far slower than the air outside the car and can be assumed still. This means the static pressure inside your car is equal to the total pressure (at least initially, the pressure inside will decrease with the window open as the flow begins to move). However, because outside the air is moving with speed $V$, we know that the static pressure outside the car is lower.

This difference in the static pressures from inside the car to outside creates flow from inside the car to outside. So the flow brings with it the smoke and that's why it leaves the window.

tpg2114
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  • Makes sense, but see my comment to Eduardo as well. Any thoughts? –  Jan 17 '13 at 03:23
  • @ChrisWhite Yes, the turbulence will certainly alter the process, but you would see it even without turbulence. In fact, it's considerably more complicated when you consider turbulence -- a turbulent spatial mixing layer (which is what this essentially becomes) spreads faster than a laminar one. Depending on how far the front of the door is from the back, the spread could be considerable meaning the thickness of the layer extends into the car window. This would slow the transport of smoke to the outside of the car. However, it could also enhance it by increasing the mixing of clean air and... – tpg2114 Jan 17 '13 at 04:43
  • the smoke. Without (considerably) more complicated analysis than that, it's really tough to say what the exact effects would be. However, the pressure gradient is strong enough that the dominant mechanism will still be the laminar transport due to the gradient while the turbulence would alter it by either enhancing or detracting depending on far more parameters and complicated physics that we could answer theoretically. The turbulence just at the wall before it encounters the open window is zero because there is the viscous sublayer within the boundary layer so it would take time to grow. – tpg2114 Jan 17 '13 at 04:46
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To the first question: no idea, honestly. My intuition says that air should come by the front windows and out from the rear ones. Probably air creates difference of pressures in the neighboorhood of the window and suck air out, just like a vaccum cleaner: it sucks air by making it go out... Making theoretic studies may be very difficult, but if we could see a wind tunnel simulation with windows opened we may see what air is doing there.

To the second question: definitely yes. The two situations are actually the same, that's why they use wind tunnels to test aerodynamics of cars instead of getting them out in tracks (they also do it, but for driving stuff rather than aerodynamics).

MyUserIsThis
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  • Why would theoretic studies by very difficult? – gerrit Jan 15 '13 at 17:50
  • Theoretic predictions on dynamics of fluids are hard because you usually can't solve the differential equations you will get too, but we would have to get there and see what comes out. – MyUserIsThis Jan 15 '13 at 18:28
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The air flowing around the outside of your car near the windows will be at a reduced pressure due to the Bernoulli effect. There will also be some sort of vent letting air into the cabin, the vent intake must be located at a point in the airflow where you have higher stagnation pressure, so you get a net flow going in the vent and out the window, I'm guessing.

user1631
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