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Answer to Captain’s Question





I think I have it Captain!!!! Say you attach a length of pipe (or hose) to a balloon.
You attach a funnel to the other end of the hose.
You hold the funnel out the window of a moving car.



Okay, how fast would you need to go to fully inflate a standard balloon? ( I understand there would be lots of variables in this).

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Well, you ought to be able to make a pretty good guess at the speed required. Ignore the funnel for a minute (which I suspect wont increase the pressure all that much, anyway: see below). You can apply the Bernoulli equation, which in this case simplifies down to:

V2/2 = P/r

where V is the inlet velocity (or velocity of the car, if you prefer), P is the pressure in the balloon, and r is air density. http://www.balloonhq.com/balloon_car/12.html color=#3d0264this page says that "a balloon inflating for the first time needed about 30 Torr for inflation." 30 Torr = 0.579 psi= 83.4 pounds/sq. ft. r for air is 0.00233 slugs/ft3. So the velocity required to inflate a balloon ought to be:

V2 = 2*(83.4pounds/ft2)/0.00233 slugs/ft3 = 71600 ft2/sec2
V = 268 ft/sec = 182 mph.

Testable Hypothesis: Unless you have an Indy car, you aint gonna do it.

Having a funnel may help a little bit, but I suspect that any advantage will be small, because the entire funnel and ballon is effectively a dead-head, where there is no air velocity and constant pressure throughout. If the funnel helps at all I think it would mainly be in capturing and directing the air, regardless of angle of attack.
 


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