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Air Pressure - Kinetic Energy

  1. Dec 28, 2009 #1
    Hi,

    We learned that, air pressure in a closed container is actually related to the quantity and speed of gas molecules hitting on the container's walls. So kinetic energy of the molecules determines the air pressure.

    However, isn't the fact that air pressure decreases when the speed of gas increases contradicting with the statement above?

    For example, let's think of two balloons hung side by side. When we blow air between them, they get closer to each other due to the fact that the air pressure decreases when the speed of air increases. But, when we blew the air, doesn't also the speed of air molecules increase, in that the kinetic energy of the molecules increase? If so, we would expect the air pressure to increase when we blow, so the balloons should move away from each other.

    Why doesn't it happen like this?

    Thanks in advance.
     
  2. jcsd
  3. Dec 28, 2009 #2

    rcgldr

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    When you (or a common fan) creates a stream of moving air, initially pressure of the affected exhaust air is increased, in a non Bernoulli like interaction where the total energy of that air is increased. Then that air will accelerate from the created higher pressure zone to a lower ambient pressure zone down stream of the source of the higher pressure, increasing in speed and decreasing in pressure in a Bernoulli like fashion (ignoring factors like turbulence). As the speed increases, in order to preserve mass flow, the cross sectional area of the accelerating air decreases in size. The surrounding air and/or in this case, that pair of balloons is drawn into what would otherwise be a void. The result is a circulation of the surrounding air as well as a change in pressure, but I'm not sure which is the dominating factor that draws the balloons together. This link to a Nasa article about propeller's shows what an idealized stream would look like, ignoring the effect on the surrounding air.

    http://www.grc.nasa.gov/WWW/K-12/airplane/propanl.html

    Note also the statement about the interaction across the 'prop disk':

    But at the exit, the velocity is greater than free stream because the propeller does work on the airflow. We can apply Bernoulli'sequation to the air in front of the propeller and to the air behind the propeller. But we cannot apply Bernoulli's equation across the propeller disk because the work performed by the engine (by the propeller) violates an assumption used to derive the equation.

    The so called 'exit velocity' where the affected stream's pressure returns to ambient is non-zero. Due to momentum and viscosity, beyond the point where 'exit velocity' occurs, that stream will end up increasing the pressure of the surrounding air and itself as it slows down. For example, in the case of an aircraft that isn't accelerating vertically, the downwards force of gravity exerted onto the aircraft is transmitted by the aircraft to the air and then through the air by momentum, eventually reaching the ground.
     
    Last edited: Dec 28, 2009
  4. Jan 11, 2010 #3
    Hi,

    Sorry for my late reply and thank you very much for your detailed answer. Since English is not my native language and my knowledge of math does not go beyond high school level, I had difficulty in understanding your explanation. However I have spent some time reading your answer and the link and I think I got some ideas from them.

    Let's think of a wind tunnel and a cube inside it. Face of the cube that is perpendicular to the wind direction would be subject to more collisions by gas molecules and the kinetic energy of molecules would be higher due to speed of the wind. So I expect this side of the cube would be subject to a higher pressure.

    However, the upper face of the cube (that is the face parallel to the wind) would receive less number of molecule collisions since the possibility of collisions is less when the air moves faster (the same amount of air molecules would spend less time on that surface when the wind speed is higher). So I would expect the pressure (and the temperature) on the upper side decrease.

    Thanks again for your time..
     
  5. Jan 11, 2010 #4

    rcgldr

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    The idea behind "faster moving air has less pressure" assumes that no work is done, and that the total energy remains constant. Basically this is a description of what happens when air accelerates from a higher pressure zone to a lower pressure zone, and ignores how those pressure zones were created and are being maintained.

    Regardless of how a pressure differntial is created within a region of air, it should be obvious that air is going to accelerate away from the higher pressure zone towards the lower pressure zone. During this acceleration, the air's speed increases, it's static pressure decreases (in all directions), and Bernoulli equation approximates the relationship between speed and pressure during this acceleration, ignoring issues like turbulence.
    This is not what happens, the relative speeds do not matter. Imagine you observe rain falling onto a 1 meter^2 area section of a treadmill. No matter how fast the treadmill is moving, the amount of rain falling onto that 1 meter^2 area of treadmill is constant. If you switch to using a 1 meter^2 area section of ground, with rain everywhere having some fixed horizontal speed relative to the ground, then again the relative speed between rain and ground doesn't matter, the number of drops of rain per second onto a 1 meter^2 area of the ground remains constant. You get a different set of drops of rain depending on the relative speed, but the number of drops per second remains the same.

    Getting back to your cube example, it's not the relative speed between the cube and air that matters. What matters is the fact that the air accelerated without any external force involved, so that total energy remains constant. For a given mass of air, the total energy, which is a constant in this case, is the sum of 1/2 m v^2 for all the molecules in that mass of air. The static pressure of that air is related to the rate and impulse (force) of collisions of the molecules in that air, which is related to the randomness of those velocities. If that air is accelerated in a particular direction without any change in total energy, then it turns out that the differences between velocities is decreased, the rate and forces of collisions between the molecules is decreased, the static pressure is decreased in all directions, and the dynamic pressure is increased in the direction of flow.
     
    Last edited: Jan 11, 2010
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