modeling buoyancy of a gas envelope in the atmosphere 1

[JoeFrickinFriday]

Greetings -

My background: I have a Ph.D. in mechanical engineering, so I'm well-versed in fluid mechanics, thermodynamics, and all that jazz.

I was recently involved in an online discussion regarding the extraction of mechanical work from the buoyancy of a hydrogen balloon. I took a stab at modeling atmospheric buoyancy, but I'm pretty sure I came up short.

Suppose a vessel is filled with hydrogen gas. It will be buoyant in the atmosphere, provided that the hydrogen is not compressed to the point that it's actually more dense than the surrounding atmosphere. This may happen if the vessel is rigid, or even elastic (such as a weather balloon).

In the case of a rigid vessel, the buoyancy varies as a function of altitude. The work done by the buoyant force can be calculated; if the vessel is allowed to rise (from sea level) to extreme heights (>100 km), the mechanical work done by the buoyant force is equal to the volume of the vessel times the sea-level atmospheric pressure. This can easily be shown in an Excel spreadsheet containing a model of atmospheric properties versus altitude: you can calculate the buoyancy at any altitude, the incremental buoyant work done over each step of altitude, and integrate up to any desired height.

Suppose instead that the vessel is NOT of fixed volume, and is not even elastic. Instead, suppose the vessel is an untensed gas envelope, like one of those super-high altitude balloons. At sea level it's only partially filled, leaving lots of room for expansion. With no tension in the vessel material, the hydrogen inside is allowed to expand so that it's always at ambient pressure. In this situation, how does the buoyancy of the balloon vary with altitude?

 

[MintJulep]

When an object is submerged in a fluid there exits an upward force on the object that is equal to the weight of the fluid that is displaced by the object.

That, along with PV=nRT and a table of the standard atmosphere will allow you to calculate the buoyant force at any altitude.

I don't understand why you are bringing work into the picture.

 

[JoeFrickinFriday]

> don't understand why you are bringing work into the picture.

See paragraph 2 in my OP: the purpose of the original discussion was to identify how much mechanical work could be extracted from a buoyant balloon rising from sea level to an arbitrary altitude.

 

[IRstuff]

So why not make the gas weightless; that would be the absolute best you could do, since that also zeros out the gravity field work component?

TTFN

FAQ731-376

 

[MintJulep]

Work is force x distance.

The buoyant force at any altitude is the maximum weight that can be lifted to that altitude. Therefore the maximum work that could be done is the buoyant force x altitude.

 

[btrueblood]

Er, gave you a star for the first post, Mint. A blue-ish star for that last post, though.

"Work = force x altitude" is not correct:

Work done = integral of: {F X d(altitude)} from zero to final altitude.

Force varies (can vary) over altitude, as you pointed out in earlier post, due mainly to variations in atmospheric density with altitude. Also, some variation will occur due to changes in local "g", which becomes significant if altitude change is big enough.

 

[JoeFrickinFriday]

The root of my mental struggle is understanding the sources and sinks for the energies involved.

Suppose we create a 1 cubic-meter bubble of hydrogen at sea level. In pushing back (lifting up) the atmosphere to make room for this bubble of hydrogen, we have done 101.325 KJ of work. If we hold the volume of this bubble constant and let it rise through the atmosphere, by the time it reaches the "top" (~100,000 meters altitude), the buoyant force has done very nearly 101.325 KJ of work. This seemed reasonable to me: when creating that bubble of H2 at sea level, we did work lifting the atmosphere to make room for it; when we let the atmosphere back down (by letting the H2 bubble rise), we recovered that energy through buoyant work.

Of course there is still some energy left in this 101.325 KPa bubble of hydrogen that up there at the edge of space; if you then adiabatically expand that 1-cubic-meter bubble of H2 to near zero pressure, you can then recover another ~247 KJ.

Where is all this energy coming from?

Things get even more puzzling (to me) if one assumes that the hydrogen gas bag is allowed to expand so as to always be at ambient pressure, regardless of altitude. If we start off with 1 cubic meter of hydrogen at sea level, the buoyant force is 10.88 N. I ran a quick, crude model in EES: 100 runs, starting pressure and temperature 101.325 KPa and 20C respectively, ending pressure and temperature 1.325 KPa and -40C. What I found was that the buoyant force remained pretty much constant; this means that the buoyant work done as our constant-pressure gas envelope heads for the top of the atmosphere (~100,000 m) is an incredible 1.088 MJ.

Again, where is all this energy coming from??? (actually the more relevant question is probably: what am I doing wrong here?) AFAICT, we've done only 101 KJ of work to create this sea-level gas bubble, and suddenly it looks like we can recover nearly ten times that much. Right now the first law is screaming "tu stultus es" over and over again in my head, but I can't see what's wrong. Help...

 

[prex]

The answer to JoeFrickinFriday's question should be relatively simple, if I'm not grossly mistaken.
Assuming the gas inside the balloon and air outside are in equilibrium at any altitude, the two volumes of gas, one inside the balloon, the other one is the same volume filled with air, must contain the same number of moles, as p, v and T are the same. Hence their respective masses are proportional to the respective molecular weights and of course these masses remain constant at different altitudes (as evidently helium's mass is constant).
The conclusion is that, if the change of gravitational constant with altitude can be neglected, the buoyant force remains constant and the work is that force times the altitude change.

prex

http://www.xcalcs.com

: Online engineering calculations

http://www.megamag.it

: Magnetic brakes and launchers for fun rides

http://www.levitans.com

: Air bearing pads

 

[rb1957]

well the density (of the H2) will increase as the atmospheric temperature falls, which will reduce it's bouyancy.

i think you're only looking at part of the problem ...
you've got a volume of H2 at ambient pressure and temperature. this bouyancy will cause the balloon to rise, force causing velocity, and rising causes the balloon to increase it's potential energy. and, of course, you have to account for the weight of the balloon. and can the H2 expand without limit at altitude, or is it limited by the size of the balloon, and once the volume is constrained then the bouyancy won't increase as much; in fact at some stage i think you'll find that the balloon has no bouyancy, that the weight of the H2 = the weight of the atmosphere for the given volume.

 

[MintJulep]

Work done = integral of: {F X d(altitude)} from zero to final altitude is correct of course.

But, the recoverable work has to equal the final potential energy of the system, which is mgh.

mg of course is weight, and assuming you reach an equilibrium altitude the buoyant force has balanced the weight.

So don't they kind of have to work out to the same number?

 

[Compositepro]

As Prex said the buoyant force remains constant with altitude and temperature because air and hydrogen are affected the same by temperature and pressure. High altitude balloons will rise until the burst (that is how the payloads are recovered)unless the envelope is strong enough to contain the pressure and increase the density of the hydrogen until the net buoyancy is zero. The physics of bouyancy is just like dealing with weight except the vector is in the opposite direction.

 

[btrueblood]

"But, the recoverable work has to equal the final potential energy of the system, which is mgh."

Erm, maybe, but no, the recoverable work would be the energy you have to put into the winch to pull the balloon back down. Dunno, haven't run the numbers, but it wouldn't necessarily be equal to mgh...you are changing the "system" to recover mgh:

To get the mgh back, you run a compressor, packing all the H2 back into (say) a K-bottle at 10,000 psi (very quickly)...then fold up the balloon nice and small...and then let the whole thing come whistling back down, and smack into the ground at Mach whatever...at least, that converts mgh into 1/2 mV^2...

There was a proposal I remembered reading about years back that would use solar heating variations on a tethered balloon (I think it had one side coated black, the other reflective) to change the buouyancy of the balloon, caused the equilibrium altitude of the balloon to change. Dunno the mechanical details, but net result was that during a warming cycle, the tether would generate net work (free energy from the sun!). How one maintains the balloon's orientation to the sun is left as an excersize for the reader...

 

[zekeman]

Since we are talking about a fixed mass,M of hydrogen with a theoretical massless tensionless envelope we have 2 events going on as it lifts off the ground to atmosphere
1) the mass expands adiabatically to a pressure that is the same as its envronment, p(x). So V(x)=V(0)*(p(0)/p(x))^1/gamma
p(x)=pressure at altitude x
2) the work done on the mass is integral[V(x)*rho(x) dx] limits 0 to x
rho(x)= density of air at height x

The change in energy of the mass is in 2 parts, namely the value of the integral and the the work done by the masss in expanding adiabatically from V(0) to V(x) . The first part is the increase, has potential energy Mx + kinetic energy, M V^2/2g and the second term is the loss of internal energy u(0) -u(x) from the adiabatic work done on the atmosphere.
So the algebraic increase in energy of the mass is
the integral - adiabatic work(Or the loss of internal energy of the mass)



D

 

[zekeman]

Correction Mgx is the potential energy gain and Mg[u0)-ux)] equal the loss of internal energy.

 

[ione]

JoeFrickinFriday,

Further top your post dated 13 Apr 10 13:11 (gain after adiabatic expansion), consider that in an adiabatic transformation all the state variables (pressure volume temperature) change. With an adiabatic expansion the gas cools down and so its density increases. Consequently the equilibrium reached before the adiabatic expansion is lost and this is done at the expanses of altitude (that is at the expense of potential energy).

 

[prex]

JoeFrickinFriday, the answer to your question in the second post (that crossed mine), is also simple.
Your question is analogous to this one: if I let a mass come down from the top of the Everest I can recover energy from it and do work, where is it coming from? The answer is of course obvious: someone spent the same energy before to take that mass on top of the mountain.
For helium it's the same: as it is lighter than air and if there was no helium trapped into earth's crust or generated by radioactive decay and if there were no winds mixing up the atmosphere (caused by solar energy), we would need to go to the upper limit of the atmosphere to get some helium for your experiment.

prex

http://www.xcalcs.com

: Online engineering calculations

http://www.megamag.it

: Magnetic brakes and launchers for fun rides

http://www.levitans.com

: Air bearing pads

 

[zekeman]

"consider that in an adiabatic transformation all the state variables (pressure volume temperature) change. With an adiabatic expansion the gas cools down and so its density increases."

Oh,really?

 

[hydtools]

Joe,
What have you done with the air that moved from above the balloon to under the balloon as the balloon rises? You can not have a trail of nothing under the balloon. Does that represent the work that went into raising the balloon?

Ted

 

[ione]

Zekeman,

Badly need your help to understand your last comment.

Do the equations below work for adiabatic transformation and perfect gas?

P*V^gamma = const
n*R*T*V^(gamma-1) = const

n = number of moles
R = gas constant
gamma = ratio of specific heat

If so, during the transformation variable states change (and remain correlated by equations above) and specifically during an adiabatic expansion temperature drops. Where am I wrong (‘cause certainly am I)?

 

[zekeman]

Ione,

Your equations are OK which show that V goes up when T goes down, so density which is the inverse of V should go down, not up, as you said.

My guess was you meant specific volume, not density.