tinokoloski Posted November 20, 2017 Report Share Posted November 20, 2017 (edited) Hi!I am making an aeolipile for a school project and I was wondering how the air pressure behaves in a situation like this. Is this drawing correct? I thought there would be more air pressure in the ball (leading to air exiting the nozzles quicker) if you have twice the amount of air streaming in the ball. But then the air would push back the air on both sides of the ball limiting the air coming in, but then that would of course again increase the air pressure.Would it work better if there was only one pipe supplying air to the ball? Will the air get out of the ball faster (because it is the only place it can go to?) I know this all seems a bit vague because i'm kinda confused by this myself. I hope you can clear this up for me! Greetings, Tino P.S. here is a schematic of an aeolipile to make it a bit more clear! Edited November 20, 2017 by tinokoloski Quote Link to comment Share on other sites More sharing options...
exchemist Posted November 20, 2017 Report Share Posted November 20, 2017 (edited) Hi!I am making an aeolipile for a school project and I was wondering how the air pressure behaves in a situation like this. Is this drawing correct? I thought there would be more air pressure in the ball (leading to air exiting the nozzles quicker) if you have twice the amount of air streaming in the ball. But then the air would push back the air on both sides of the ball limiting the air coming in, but then that would of course again increase the air pressure.Would it work better if there was only one pipe supplying air to the ball? Will the air get out of the ball faster (because it is the only place it can go to?) I know this all seems a bit vague because i'm kinda confused by this myself. I hope you can clear this up for me! Greetings, Tino P.S. here is a schematic of an aeolipile to make it a bit more clear! I think it is steam, not air, isn't it? The rotation comes from the change of momentum of the steam as it goes through the right-angled bends in the exit tubes. This creates a force ( d/dt (mv) = ma = F ) on the elbows of the exit tubes, in the opposite direction to that in which the steam finally escapes. The rate of exit of steam will determine the rate of change of momentum and thus the turning effect. The higher the pressure in the ball above atmospheric, the faster the rate of exit will be. The pressure and rate of steam entering the ball from the feeder tubes has no effect because the forces are equal and opposite and cancel each other. Edited November 20, 2017 by exchemist Quote Link to comment Share on other sites More sharing options...
Maine farmer Posted November 20, 2017 Report Share Posted November 20, 2017 If you want to improve on the project, you could figure out a way to attach a chain sprocket or belt pulley to the axis and transmit some power to a small generator or mechanical device and see how much useful work you can get out of it. It looks like a fun project anyway! tinokoloski 1 Quote Link to comment Share on other sites More sharing options...
Turtle Posted November 22, 2017 Report Share Posted November 22, 2017 Hi!I am making an aeolipile for a school project and I was wondering how the air pressure behaves in a situation like this....Greetings,Tino Hi!I am making an aeolipile for a school project and I was wondering how the air pressure behaves in a situation like this.... Greetings, Tino You better add a safety valve as in a pressure cooker or you may well cause an explosion. :ebomb: Maine farmer 1 Quote Link to comment Share on other sites More sharing options...
tinokoloski Posted November 22, 2017 Author Report Share Posted November 22, 2017 I think it is steam, not air, isn't it? The rotation comes from the change of momentum of the steam as it goes through the right-angled bends in the exit tubes. This creates a force ( d/dt (mv) = ma = F ) on the elbows of the exit tubes, in the opposite direction to that in which the steam finally escapes. The rate of exit of steam will determine the rate of change of momentum and thus the turning effect. The higher the pressure in the ball above atmospheric, the faster the rate of exit will be. The pressure and rate of steam entering the ball from the feeder tubes has no effect because the forces are equal and opposite and cancel each other. Ah yes i understand now thank you! Yes it is steam, I always confuse them because of the word “air pressure” ;) You better add a safety valve as in a pressure cooker or you may well cause an explosion. :ebomb:Yeah that’s probably necessary! Quote Link to comment Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.