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How a Hot Air Balloon Works

  1. Jun 9, 2016 #1
    1. The problem statement, all variables and given/known data
    To research the physics of how a hot air balloon works.

    2. Relevant equations


    3. The attempt at a solution
    I just would like someone to check the main points I am going to include. The air inside the balloon is less dense because it is heated so that particles of air are further apart than the particles outside the balloon. The pressure inside is equal to the pressure outside otherwise the balloon would collapse or explode. To maintain the same pressure inside there are less particles but they have more kinetic energy which gives the same pressure as outside. The main point I am confused with are the forces acting on the balloon, particularly the buoyant force. I would have said that the buoyant force is constant since the volume of the balloon is constant. So the balloon rises purely because the weight force is reduced i.e. the weight of the air inside the balloon + balloon + passengers etc is less than the buoyant force. However the first article that comes up in google (http://www.real-world-physics-problems.com/hot-air-balloon-physics.html) talks about manipulating the buoyant force and I don't see how that is possible as the balloon's volume doesn't change? A word for word quote from this site:

    "If the balloon operator wishes to lower the hot air balloon, he can either stop firing the burner, which causes the hot air in the envelope to cool (decreasing the buoyant force), or he opens a small vent at the top of the balloon envelope (via a control line). This releases some of the hot air, which decreases the buoyant force, which also causes the balloon to descend."

    Is this correct?

    Thanks for any help offered!
     
  2. jcsd
  3. Jun 9, 2016 #2

    SteamKing

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    Well, think about it in a logical manner.

    While the balloon is rising, sure, the buoyant force is constant, but you don't want to keep rising, so what must you do to change that?

    Knowing that the buoyant force on the balloon must be greater than the weight of the balloon and all its contents to keep going upward, what could you do to keep the balloon from rising?

    Why is the balloon buoyant in the first place? Remember, Archimedes Law applies to balloons just like it does to floating vessels, like boats.
     
  4. Jun 9, 2016 #3
    Thanks. So is the buoyant force constant all the time or not? Can't you stop it rising by increasing the weight force by letting in cold air (and stop heating the air)? The only way to change the buoyant force is to change the volume of displaced air i.e. the volume of the balloon, do they actually do this in a hot air balloon?
     
  5. Jun 9, 2016 #4

    SteamKing

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    The point is, in order to stop rising, you must change the buoyant force which is causing the balloon to rise. The two methods outlined in the Wiki article do just that.

    If you stop the burner, the hot air in the balloon envelope cools, the envelope shrinks, there is less buoyant force created by the smaller envelope, and the balloon stops rising and eventually begins to descend as the envelope keeps shrinking as it cools.

    If you open a vent and let some of the hot air escape from the balloon envelope, the envelope shrinks, ...
     
  6. Jun 9, 2016 #5
    Some people call the buoyant force the upward force exerted by the surrounding fluid on the object (so it doesn't change if the volume of the balloon doesn't change). Others call it the net upward force on the object. I prefer the first definition, as does Jimmy87. But the article that Jimmy87 is reading and the interpretation that SteamKing is using employ the second definition.

    Chet
     
  7. Jun 9, 2016 #6

    David Lewis

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    When buoyant force > gross weight, the balloon accelerates upward. When the balloon is climbing with constant velocity then buoyant force = gross weight + drag.
     
  8. Jun 9, 2016 #7
    Thanks for the info guys. So to be clear the reason it can lift off is because the weight of the passengers + hot air etc is less than the weight of the displaced air. To stop the balloon rising you change the buoyant force by changing the volume. Would this video I found be wrong:





    It says at 4mins 30s about the volume never changing?
     
  9. Jun 9, 2016 #8
    Could you not stop it rising by increasing the weight force though by letting cold air in to replace the hot air instead?
     
  10. Jun 9, 2016 #9

    SteamKing

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    That's what happens when you open the vent.

    The balloon envelope is obviously not airtight - that's how the hot air from the burner gets inside in the first place. By opening the vent, as the warmer air escapes from the envelope, it is replaced by colder air from the bottom.
     
  11. Jun 9, 2016 #10
    Thanks. So if you were to look very closely at the forces involved before and after opening the vent would you see both a reduction in the buoyant force (due to reduced volume/shrinking) and an increase in weight force? For example, when the balloon is rising the buoyant force arrow on your diagram should be longer than the weight force. If you then drew a second diagram after the vent is opened (assuming you opened in such a way to wanted it to descend) would you make the length of the weight arrow bigger than the first diagram (as well as making the buoyant arrow smaller of course since the new force would be down)? Also, is the shrinking of a hot air balloon noticeable when the vent is opened?
     
  12. Jun 9, 2016 #11

    rcgldr

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    The volume in the balloon remains about the same. What changes is the density of the air inside the balloon, hotter air is less dense than cooler air.
     
  13. Jun 9, 2016 #12
    Ahhhh so much confusion. Is the original article write or wrong? It talks about reducing the buoyant force which must mean the volume of the balloon changes. I am more inclined to think it would change because to start with the balloon on a hot air balloon is completely collapsed so if you turn the burner off the volume must shrink?
     
  14. Jun 9, 2016 #13

    rcgldr

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    As posted earlier, the article appears to have two different meanings for buoyant force. Initially it refers to the upwards buoyant force of the displaced air (Archimedes principle), but later uses the term "net buoyant force" which includes the weight of the balloon, air, gondola, passengers, ballast, ... .

    Assuming the volume remains nearly constant, what's changing is the weight of the air inside the balloon. The weight is less when there's hotter and less dense air (so less mass of air for the same volume) trapped inside the balloon. During expansion due to heat, the excess air escapes out of the bottom of the balloon. The air that escapes at the bottom will be the cooler higher density air. If the vent at the top is used, then the hotter lower density air escapes through the top vent. During this process the volume inside the balloon remains about the same.
     
    Last edited: Jun 9, 2016
  15. Jun 9, 2016 #14

    SteamKing

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    There's a difference between being "right" and being "write".

    I have no idea what you're trying to say here.
     
  16. Jun 9, 2016 #15
    Great, thanks that was my original thoughts. How can the balloon not shrink though. If cold air rushes in then surely a mixture of cold and hot air will exert less pressure than just hot air (i.e. before the vent was opened) so the balloon will shrink? SteamKing said something about the balloon shrinking in post #4.
     
  17. Jun 9, 2016 #16
    Whoops my bad sorry I meant "right". I am dyslexic.

    What I am trying to say is that the video (post #7) and rcgldr are saying that the volume is always constant whether you are opening or closing the vent. However the article and yourself are saying that the balloon shrinks when the vent is opened. The bottom line is I want to know whether or not to say in my assignment that the balloon is lowered by letting cold air to increase the weight AND reduce the buoyant force (due to the balloon shrinking) or whether to say the cold air ONLY increases the weight force (by purely increasing the density of the air inside) to enable the balloon to descend.
     
  18. Jun 9, 2016 #17

    rcgldr

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    To clarify the point being made in the video from post #7, the pressure at the bottom of the balloon is the same as the pressure of the surrounding air at the same height as the bottom of the balloon, otherwise air would be flowing into or out of the balloon due to any difference in pressure. The pressure gradient (how much the pressure decreases as height increases) is a bit less inside the balloon than outside the balloon because the density of the hot air inside the balloon is a bit less, so the pressure near the top of the balloon is a bit higher than the pressure of the air outside of the balloon at the same height. However, the materials used in balloons don't stretch much in response to slight differences in pressure, so the volume inside the balloon is nearly (but not exactly) constant. The basic concept just considers the volume to be constant, so it's just the weight of the air inside the balloon that's changing corresponding to the change in density related to temperature.
     
  19. Jun 10, 2016 #18

    David Lewis

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    Yes, unless the balloon is rising slowly at a constant rate of climb. In that situation, neglecting air resistance, buoyant force and weight are equal and opposite.
     
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