Timeline for Air pressure in balloon
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Jul 2, 2013 at 17:08 | answer | added | Alan Rominger | timeline score: 0 | |
| Jun 2, 2013 at 14:55 | answer | added | Maxim Umansky | timeline score: 1 | |
| Jun 2, 2013 at 13:57 | answer | added | L'Unità | timeline score: -1 | |
| May 16, 2013 at 14:22 | comment | added | OSE | @User58220 You are right, I jumped at the answer a little too quickly. I believe the pressure inside the balloon should obey $dp/dy = \rho g$ with the pressure at the mouth equal to the atmospheric pressure. I'm going to remove my comments to hopefully prevent further confusion. | |
| May 16, 2013 at 2:22 | comment | added | DJohnM | @OSE Isn't the case that the pressures inside the hot-air balloon and outside are the same at the open mouth at the bottom of the balloon? At the top of the balloon, the pressure on the inside must be greater than outside. That's where the lift comes from... | |
| May 15, 2013 at 18:24 | history | edited | David Z |
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| May 15, 2013 at 17:52 | comment | added | Jim | I'm not the best on fluid dynamics and ideal gasses. But it seems to me that since the balloon is open to the outside air, the pressure in the balloon equals the pressure outside the balloon. Otherwise the air in would flow out or vice versa. I know the density in is lower, but I thought the pressure is equal | |
| S May 15, 2013 at 17:50 | history | suggested | Jim | CC BY-SA 3.0 |
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| May 15, 2013 at 17:50 | review | Suggested edits | |||
| S May 15, 2013 at 17:50 | |||||
| May 15, 2013 at 17:47 | comment | added | Wuppy29 | I'm sorry if I'm missing the obvious here, but ρ = m/V which means I would need m. The whole reason to get this pressure is to get the m. And I actually meant density * volume. Sorry for that. | |
| May 15, 2013 at 17:17 | comment | added | OSE | "pressure (kg/m3) * the volume." Those are the units of density, not pressure. Pressure has units force/area. Density has units mass/volume. | |
| May 15, 2013 at 17:14 | comment | added | OSE | Use the ideal gas law with the specific gas constant for air instead. $P = \rho R_{specific}T$ For air $R_{specific} = 287$ J/kg/K | |
| May 15, 2013 at 16:51 | comment | added | dmckee --- ex-moderator kitten | "So to get the pressure in the balloon I would have to know n" Well, you could try that, but the balloon is open to the surrounding air and can gain or lose moles as it warms and cools. | |
| May 15, 2013 at 16:42 | history | asked | Wuppy29 | CC BY-SA 3.0 |