Re: Hydrogen as Lifting Gas for Airship

Wayne Throop wrote:
: SolomonW <SolomonW@xxxxxxxxxxxxxxxxx>
: It would follow the ideal gas law is PV = nRT
: : Raising the temperature in a balloon with a release valve P, V , R remain
: the same so it would be K=nT
: : Say a 40 degree increase in temperature from 10 to 50 degrees
: : The formula would be (273+10)/(275 + 50) = .88 or 12% more lift.

The raio of mass inside the envelope before and after heating doesn't
tell you the ratio of lift before and after heating. Consider the case
of a rarified gas inside the envelope, only 1 percent as dense as the
gas outside. (Why it's that rarified doesn't matter; could be because
it's already hotter than outside, or it could have lighter particles, or
whatever you wish.) Now heat the liftgas enouh to make it half as dense.
You didn't increase your lift by half, you only increased it by a half
a percent. And in fact, even if you applied unlimited heat, so as to
approach a vacuum inside the envelope, you can *never* get more than 1
percent increase in lift.

In the case of hydroegen, a quick googgle sez air at stp is about .83 g/L, and hydrogen has about .09 g/L, which seems to imply you can't get much
more than ten percent extra lift over hydrogen at stp, no matter how
much you heat it.

But why are you guys thinking in terms of fixed-size balloons? (Because the initial question concerned dirigibles?) I think the calculation would be different, if the used balloon was allowed to expand (like they do for the stratospheric balloons, which are shrunken on lift-off), the calculation would be how much the volume changes when heated (at the same pressure), and how much the extra displaced air weighs. The weight of the balloon, and its hydrogen, would then of course remain the same.
The formula being pV = nRT, then V = (nR/p)T, thus the volume is linear with T, the pressure being the same, equal to the ambient pressure at whatever level the balloon happens to be. So, doubling the temperature of the gas inside from ~300K to ~600K should displace twice the volume of air, and increase the lifting force twice.
Of course, one may ask why not fill the balloon to the full to begin with. As previous posters calculated, for a fixed-size balloon, the gains from heating are negligible. The answer would be, I think, to gain lift control, and dispense with all the ballast stuff. In addition, if the used gas is really expensive, it might make sense to heat it to 600k and thus halve the used quantity. That reason is not good enough for hydrogen, however.

--
You'd be crazy to e-mail me with the crazy. But leave the div alone.
.

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