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Say I have a helium balloon that has an envelope of 3800 cubic feet (to lift a payload of 200lbs at sea level atmospheric density and gravitational attraction). What will the maximum altitude (in feet) be for this craft before it achieves neutral bouyancy with the atmosphere? ALSO: At what point will the helium envelope no longer be able to contain the helium (in feet)? What if this envelope was rigid? (in this case I would include the airframe weight in with the overall payload weight)

2007-10-15 06:14:19 · 1 answers · asked by CaptainGarry 1 in Science & Mathematics Physics

1 answers

Not enough info, I think. How much excess lift does it have on the ground? Or does 3800 ft^3 exactly balance 200 lb at STP? (If so, it's not going anywhere.)
The lift of the helium is the mass of the displaced air Ma minus the mass of the helium Mh, multiplied by g. In both cases, mass = density * volume. With a perfectly flexible balloon (whose added pressure due to elasticity is zero), if it can rise at sea level it theoretically can rise without limit, because the balloon expands freely and thus Ma remains constant. The altitude where a real flexible balloon loses lift would be a complicated computation because the pressure it adds with increasing volume must be known. The altitude where it bursts depends on its flexibility and strength.
With a rigid balloon, the max altitude is where the air density decreases to the point where Ma = Mh + Mpayload. The formula for air pressure is
P = P0 * EXP((-mm * g * (h - h0)) / (R * T))
where
P, P0 = pressure at actual, reference altitudes in Pa
mm = mean molecular weight of air = 28.964
h, h0 = actual, reference altitudes in m
R = universal gas constant = 8.3145
T = temperature K
The formula for density D is
D = 1.292 kg/m^3 *P/101325, for dry air at 0 C, 101325 Pa being standard sea-level atmospheric pressure
I hope this is some help.

2007-10-18 17:33:24 · answer #1 · answered by kirchwey 7 · 0 0

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