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Velocity of air thru a tube - 4th grade science fair project

  1. May 8, 2008 #1
    Hi.....can anyone help out explain the relationship of diameter to velocity of air flow thru a tube for the following science experiment. A photo is attached.

    The question they are trying to answer thru the attached experiment is:
    Does the diameter of the tube affect how high a ping pong ball will be raised?

    Three rigid clear plastic tubes attached to a board. A small balloon (with both ends cut off) is stretched over the bottom end and a ping pong ball is put into it and it rests on the bottom. To the other end of the balloon (the narrower neck of the balloon) a flexible tube/hose is attached and thru which air is pumped into with an inexpensive balloon pump (see photo - only one tube is shown for set up, but they are testing each of diameters 1/2", 5/8" and 7/8" of flexible tubing). The pump of air causes the ping pong ball to rise a certain level and they record this. Assuming that the same person uses the balloon pump for each of the three trials, they are assuming the "pressure" to be the same.

    The question is .....is there a simple relationship between diameter and velocity if air flow thru the tube which can explain why the larger diameter raises the ball higher. Initially, my son thought intuitively that the pressure would be greater in a narrower tube - which I believe is correct - but is there a relationship between pressure, diameter and velocity that can easily explain this.

    any help would be appreciated.

    Attached Files:

  2. jcsd
  3. May 8, 2008 #2


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    Welcome to PF, Twoxtwo.
    I don't see a balloon in your picture, so that part is confusing.
    In any event, it seems to me that the tube closest in diameter to the ball (such as the one in the photo) will give the most lift. In larger ones, air will leak around the ball and thus reduce the pressure. That's off the top of my head, though; wait for a real scientist to answer.
  4. May 8, 2008 #3


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    Pressure multiplied by an area equals a force. Ideally, if the pump is putting out the same pressure and it's applied to the same area, they should all rise the same. However, they will not rise because there will be too much leakage on the larger diameter tubes.

    I think a rather clever experiment which would be counter-intuitive (for elementary school I think), would be to have the three tubes. Inside the three tubes are three disks, each just slightly smaller than the tube diameter. Have them weighted down each equally, then pump.

    Since the pressure is the same in all three tubes, but the area is larger in the larger tube. There will be a larger force on the larger disk. It should go higher.

    I just think with the current setup, there will be too many variables changing in each tube to get a good relationship.
    Last edited: May 8, 2008
  5. May 8, 2008 #4
    the balloon is not used as the air source but rather to attach the vertical stationary tube (the one the ping pong ball sits in) to the lower flexible tube. The balloon is the white stuff at the bottom of the vertical tube. The balloon air pump is used to as the source of air pumped into the lower tube. The photo doesn't show all three trials but they tested two lower flexible tubes (1/2" and 5/8") and when one pump of air was pressed into each of these, the ping pong ball rose higher - by about an inch - up the vertical tube. We decreased the air flow around the ping pong ball by wrapping them in three cutoff balloons.

    does this explain it any better?
  6. May 8, 2008 #5


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    Which one rose higher, and can you explain the wrapping them a little better?
  7. May 8, 2008 #6
    sorry - i didn't finish that sentence....the 5/8" rose higher - which agrees with the formula you just gave....

    the first trial was with a flex tube of 3/8" and a ping pong ball placed in the vertical tube. It did not rise. We decided there was too much air flow around the ping pong ball in the vertical tube and since we had a bag of small balloons, we figured we could increase the diameter of the PPBall slightly by cutting the top off of a balloon and stretched it open to fit the PPball inside. One wasn't enough so we kept adding a layer of balloon until the ball fit just enough that the ball goes up and comes down again. three balloons did the trick.
  8. May 8, 2008 #7


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    Oh whoops, so the upright tubes stay the same; there's only one? What's changing is the connecting tubes?

    If that's the case, then the answer is simple. Ideally, it shouldn't matter, because as I said, the pump is putting out a certain mass flow of air. When the air goes from whatever sized tube to the larger upright tube, it's velocity and pressure will be whatever, according to that tube.

    However, in real-life tube/pipe flows, there are frictional losses. These losses are proportionate to the velocity SQUARED in the tube. So, when you use a small tube to connect the pump to the upright tube, the velocity goes high, because as I said, there is an easy relationship between velocity and area.

    Since mass flow is equal
    [tex](\rho A V)_1 = (\rho A V)_ 2[/tex]
    Where [tex]\rho[/tex] is the density. Since changes in density are negligible, then area and velocity simple have a linear relationship to area (area, not diameter, so it will be an exponential relationship to diameter).

    So, decreasing the area of the tube, increases the velocity, which exponentially increases the frictional loses. Therefore, there is less pressure actually pushing on the ping pong ball when it finally gets there.

    I think I finally understand the setup, and you should see a higher travel with a larger connecting tube.
  9. May 8, 2008 #8


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    Y'know, I feel like a total idiot now. I somehow misread the OP and was thinking that it was the vertical tube whose diameter was altered. :redface:
  10. May 8, 2008 #9


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    Danger, I was thinking the exact same thing.
  11. May 8, 2008 #10


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    Two great minds think alike (or fools seldom differ :uhh:). Let's go have a beer and think on it a bit. :biggrin:
  12. May 8, 2008 #11
    Got it....I actually understand it now....(thanks to that physics and math in college.....only it's been a while since I've actually had to think about it)

    There are three vertical tubes and each has a different diameter lower tube. That's what's changing, assuming constant air pressure in pump. So even w/o knowing what that pressure is they can state in their write up, the F=PxA relationship. Because it is a bit complicated, their teacher has suggested to do their report on basics of force.....newtons laws etc.....and their visual - this experiment - just demonstrates an "application" of force.

    Thanks so much for your help.....

    I have four kids under 12 and this website was a great find for future reference.....thanks again!
  13. May 8, 2008 #12


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    Good grief! Do you not know what's causing it?
  14. May 8, 2008 #13


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    Ah ha, there are three vertical tubes, I knew it!! So, the only difference is the lower connecting tubes. Honestly, the initial pressure is the same, and the applied area is the same; ideally the force is the same, and they should raise the same amount.

    You can think of it this way. Pressure and velocity can be thought of similar to potential and kinetic energy. A roller coaster is given potential energy in the same way a fluid is given pressure by a pump. Neglecting friction, the coaster can loop around, speed up by going downhill and slow down by going uphill, but at the end, you can convert all the kinetic energy to potential and climb that same hill (neglecting friction of course).

    In a similar fashion, you can start with the fluid, speed it up by passing it through nozzles, or slow it down by going through diffusers, but at the end, you should be able to get the same pressure that you started with by slowing it's velocity all the way down.

    If you want to display the F = PA relationship, then you need to change either the pressure or the area, and use the SAME size connecting tube. With the setup right now, what you are showing is pressure losses due to friction in a pipe flow. It's definitely a more advanced concept, but if you wish to learn more, do a search for head losses in pipe flow.

    Generally, the losses are something like:

    [tex] h_{loss} = f \frac{L}{D} \frac{V^2}{2} [/tex]

    Where h is proportionate to pressure and density, f is a friction factor based on pipe smoothness and Reynolds number, L is the length, D is the diameter, and V is velocity.
  15. May 8, 2008 #14
  16. May 8, 2008 #15

    This explanation seems to address the fact that the vertical tube is the same - yes there are three vertical tubes but they are identical in diameter and length....and only the lower tube is different - so I'm confused, a little, by your subsequent explanation....that i responded to just before this....

    thanks and sorry for the confusion
  17. May 12, 2008 #16


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    Ideally, the pressure is the same once the fluid reaches the three vertical tubes. However, as (I tried) described, the friction in the lower tubes is changing the final pressure.

    As far as why the ball rose higher in the larger tube, the answer is because there are less frictional losses in the larger tube. As you make the connecting tube smaller, it's velocity goes up, and therefore it's frictional losses go up exponentially.

    Hopefully I explain this right and don't throw you off more. The velocity "hits" the ball, but in your case where there is a stagnant air chamber with a ball, velocity is not pushing the ball up, pressure is. The original "burst" of air stops when it hits the ball. At that point, the only velocity is air leakage around the ball, which is negligible. However, the kinetic energy contained in the moving air is converted into pressure (look up Bernoulli's famous equation). Pressure over an area causes a force.

    Hope this helps a little
  18. May 12, 2008 #17


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    Another way to explain this -- though perhaps not to a 4th grader! -- is by analogy with electricity. Electrical engineers are always saying "voltage and current behave just like fluids flowing in a pipe!", and now we can use that analogy in reverse.

    We have a voltage source (the pump) connected to two resistors (the two tubes), one of which doubles as a current meter (the ball in the larger tube).

    Tubes with smaller diameters have a larger resistance to airflow. This is explained by the frictional losses that others have mentioned. By making the tube diameter smaller (i.e. larger resistance), the total airflow is reduced just as current is reduced for larger resistors connected to a voltage source.

    The reduced airflow is measurable by the lower ball height when using the narrower tube.

    p.s., some commercially available flowmeters are made using this principle. See animated picture at right side of this page:
    Last edited by a moderator: Apr 23, 2017
  19. May 13, 2008 #18
    OK....now I think we have it.......at least well enough that I can explain what's going on in the experiment to my son and his science fair partner (his mom is the one who came up with this idea). He addresses friction in his report and understands it enough so that I think he will be able to explain the phenomenon that is occurring in their tube experiment....

    many many thanks ....
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