How high can an airship rise?

[askingask]

So what I know, is that as the airship rises, air density decreases because of altitude, which means the airship becomes as heavy as the air it displaces, where it won‘t generate any lift.

But what I also know, is that at that point the pressure inside the airship is higher then atmospheric pressure.

Now to my question:

Is the pressure inside the airship the main factor which makes the whole airship heavier compared to the low pressure atmosphere?
And would releasing gas from the airship, to neutralize pressure differences, make the airship lighter again compared to the air it displaces?

Could you gradually adjust the pressure of the lifting gas, so it keeps the same pressure relative to the atmosphere while still retaining its lower density to lift the airship all the way to the edge of the atmosphere, or atleast close to it?

 

[Hornbein]

So what I know, is that as the airship rises, air density decreases because of altitude, which means the airship becomes as heavy as the air it displaces, where it won‘t generate any lift.

But what I also know, is that at that point the pressure inside the airship is higher then atmospheric pressure.

Now to my question:

Is the pressure inside the airship the main factor which makes the whole airship heavier compared to the low pressure atmosphere?
And would releasing gas from the airship, to neutralize pressure differences, make the airship lighter again compared to the air it displaces?

Could you gradually adjust the pressure of the lifting gas, so it keeps the same pressure relative to the atmosphere while still retaining its lower density to lift the airship all the way to the edge of the atmosphere, or atleast close to it?

No. Releasing gas reduces the volume of the ship so it sinks.

 

[askingask]

No. Releasing gas reduces the volume of the ship so it sinks.

How, if the pressure outside is equal to the pressure inside the airship?

 

[Ibix]

Is the pressure inside the airship the main factor which makes the whole airship heavier compared to the low pressure atmosphere?
And would releasing gas from the airship, to neutralize pressure differences, make the airship lighter again compared to the air it displaces?

As you point out, the airship will stop rising when its average density is equal to that of the medium it's in. If you can reduce the density of the airship then you can rise higher. That means releasing gas such that the mass decrease is greater than any volume decrease due to the reduced pressure. So the answer would depend on the design of the gas bags - are they more or less rigid, or do they expand so that the internal pressure more or less matches the exterior pressure anyway?

If you were going for altitude I'd think you'd design with the latter plan. Rigid bags would probably weigh more than elastic ones, and releasing gas would make the descent tricky unless you carried spare gas - which would add weight and limit your altitude. But there's definitely an element of "depends on your materials technology" here.

 

[askingask]

That means releasing gas such that the mass decrease is greater than any volume decrease due to the reduced pressure.

If the pressure inside the bags are equal to the pressure outside the bags, then the volume should stay constant, right?

 

[askingask]

or do they expand so that the internal pressure more or less matches the exterior pressure anyway?

I mean if they expand, they would at some point just burst. So you have to release pressure anyway.

 

[Baluncore]

If the pressure inside the bags are equal to the pressure outside the bags, then the volume should stay constant, right?

Then the bags would not lift the airship. It is the difference in hydrostatic pressure, especially the higher pressure inside the top of the bag, that lifts the bag. The pressure at the base of the lifting bag is atmospheric, or with a hot air balloon, is open to the atmosphere.

As you rise from the base, inside the bag, the hydrostatic pressure falls slowly because of the lower density lifting gas. The hydrostatic pressure outside falls faster because of the more dense atmosphere. That results in a higher pressure inside the top of the bag than the atmosphere outside the top of the bag. That is the source of airship lift. The airship hangs from the lift bags that are inside the airship.

Is the pressure inside the airship the main factor which makes the whole airship heavier compared to the low pressure atmosphere?

No. The bags used in airships are not elastic balloons. The bags progressively expand as the airship rises. The bags start off partly filled, with folded envelopes, and are designed to expand to fill the available space inside the airship, only when it reaches maximum altitude.

 

[Lnewqban]

Is the pressure inside the airship the main factor which makes the whole airship heavier compared to the low pressure atmosphere?

What type of airship are you referring to?
Please, see:
https://eaglepubs.erau.edu/introductiontoaerospaceflightvehicles/chapter/lth/

You have the density of the airship and the volume of atmospheric air that is displaced as main factors in buoyancy.

For flexible airbags only, there is a relation between in-out delta pressure and displaced volum of air to be considered.

 

[russ_watters]

But what I also know, is that at that point the pressure inside the airship is higher then atmospheric pressure.

Now to my question:

Is the pressure inside the airship the main factor which makes the whole airship heavier compared to the low pressure atmosphere?
And would releasing gas from the airship, to neutralize pressure differences, make the airship lighter again compared to the air it displaces?

Could you gradually adjust the pressure of the lifting gas, so it keeps the same pressure relative to the atmosphere while still retaining its lower density to lift the airship all the way to the edge of the atmosphere, or atleast close to it?

There's two main types of lifting-gas airships: rigid (Zeppelins) and inflatable (balloons). In neither case is the lifting gas significantly pressurized (except maybe a little for structural rigidity in a Zeppelin?). All that would do is make it heavier/less buoyant. In a balloon the pressure is always exactly equal to atmospheric pressure, so the volume increases as the balloon rises. That's why high altitude balloons look deflated when they take off and inflated when at altitude. The buoyancy of these balloons is roughly constant and mass/weight of lifting gas is constant, except it can be varied for the purpose of control by pumping it out of the envelope and into tanks or venting it. Maximum altitude is reached when the balloon is fully inflated.

https://idstch.com/space/ascending-...eric-balloons-and-high-altitude-surveillance/

Zeppelins and blimps are roughly neutrally buoyant. They are powered and fly with the use of engines/fans, so they aren't as much at the mercy of their buoyancy, but the control techniques are similar: https://en.wikipedia.org/wiki/Buoyancy_compensator_(aviation)#Buoyancy_compensation

 

[askingask]

All that would do is make it heavier/less buoyant. In a balloon the pressure is always exactly equal to atmospheric pressure, so the volume increases as the balloon rises. That's why high altitude balloons look deflated when they take off and inflated when at altitude. The buoyancy of these balloons is roughly constant and mass/weight of lifting gas is constant, except it can be varied for the purpose of control by pumping it out of the envelope and into tanks or venting it. Maximum altitude is reached when the balloon is fully inflated.

That comes closest to the answer I was looking for. Now let us say we have some form of non elastic gasbag filled with hydrogen. Just for simplicity sake. I'm not asking for the material weight of that bag or whatever. All I'm asking for, is that if the bag rises to an altitude where it can't go any higher. Would venting the gas of the bag to equalize pressure, compared to the atmosphere at that altitude, cause the bag to gain altitude again?

 

[askingask]

Then the bags would not lift the airship.

How does this make sense. I thought heavier then air aircraft work on the basis of lower density gases displacing the air. But the pressure of the lifting gas still has to have equal pressure to the atmosphere so that the Gas bag can keep its volume.

 

[Baluncore]

How does this make sense. ... But the pressure of the lifting gas still has to have equal pressure to the atmosphere so that the Gas bag can keep its volume.

The pressure of the lifting gas, compared to the pressure of the atmosphere, should only be equal at the bottom of the bag. As you rise, the hydrostatic pressure of the gas and of the atmosphere fall. The rate that pressure falls is dependent on the density of the gas mixture. Hot air balloons are open at the bottom, which gives them maximum lift at the top.

You must understand that the internal pressure at the top of the bag is greater than the external atmospheric pressure at that height. That upwards pressure on the bag is countered by tension in the bag envelope material, that attaches the bag to the airship framework, so the airship frame, or gondola, hangs from the bag envelope.

All I'm asking for, is that if the bag rises to an altitude where it can't go any higher. Would venting the gas of the bag to equalize pressure, compared to the atmosphere at that altitude, cause the bag to gain altitude again?

To have any lift, there must be positive differential pressure in the top of the bag. The differential pressure at the bottom should be close to zero.
If the bag is under tension due to the expanded volume of the lifting gas, then the density of the lifting gas will be greater because of the internal pressure. Releasing gas until the pressure at the bottom of the bag does not tension the envelope there, will ensure a minimum lifting gas density, so maximum lift, with the minimum mass of gas. It might be better to get a bigger bag.

 

[askingask]

should only be equal at the bottom of the bag

Are you assuming that the bag is open at the bottom? I’m talking about a sealed non elastic bag.

 

[Baluncore]

Are you assuming that the bag is open at the bottom?

No, but it may be considered to be open. The envelope at the bottom should have zero differential pressure.

I’m talking about a sealed non elastic bag.

The problem with your sealed non-elastic bag is that it is too small for the gas it contains at that altitude.

 

[Baluncore]

I thought heavier then air aircraft work on the basis of lower density gases displacing the air.

"Heavier than air" craft fly, usually because they gain lift from an airfoil, but only while they expend energy to hold their airspeed and altitude.

"Lighter than air" craft fly, because they have an average density less than air, so they have positive buoyancy in the air.

That brings up the interesting question of a hot air balloon, a compromise that expends energy to maintain a positive buoyancy.

 

[askingask]

"Heavier than air" craft fly, usually because they gain lift from an airfoil, but only while they expend energy to hold their airspeed and altitude.

"Lighter than air" craft fly, because they have an average density less than air, so they have positive buoyancy in the air.

That brings up the interesting question of a hot air balloon, a compromise that expends energy to maintain a positive buoyancy.

Yea I did a mistake there

 

[askingask]

The problem with your sealed non-elastic bag is that it is too small for the gas it contains at that altitude.

what do you mean by too small? You mean the density inside becomes greater then outside?

 

[Baluncore]

what do you mean by too small?

The bag needs to have sufficient capacity to hold all the lifting gas, without the internal pressure at the lowest point, increasing above the external air pressure there. The envelope would then be under elastic tension, the internal pressure at the bottom would be greater, and so the internal density would be greater than needed for maximum buoyancy.

If the differential pressure does begin to rise at the bottom, you should reduce the excess mass of lifting gas, by venting some from the bottom. There will always be positive internal differential pressure at the top. If you vent that, you will begin to rise momentarily, but then will sink rapidly because it is the differential pressure at the top, that keeps the bag inflated and off the ground.

You mean the density inside becomes greater than outside?

No, I mean the pressure inside becomes greater than that outside, then the density of the lifting gas increases, but not necessarily above the external atmospheric pressure at that point.

 

[askingask]

If the differential pressure does begin to rise at the bottom, you should reduce the excess mass of lifting gas, by venting some from the bottom. There will always be positive internal differential pressure at the top. If you vent that, you will begin to rise momentarily, but then will sink rapidly because it is the differential pressure at the top, that keeps the bag inflated and off the ground.

Now this is interesting, its my first time hearing about this concept of different pressures inside the gas bag. So if you have a gas bag filled only with hydrogen at 1 bar you are telling me that inside that bag at the top the pressure would be higher then at the bottom. Now i don't intuitively understand how that should work. Could you elaborate on this.

No, I mean the pressure inside becomes greater than that outside, then the density of the lifting gas increases, but not necessarily above the external atmospheric pressure at that point.

Yes that fits what I mean, as the pressure rises the density rises. The mass of the bag is given by Density/Volume right? And so for a given volume the the average density of bag and lifting gas = density of atmosphere. So it stops rising.

 

[Baluncore]

So if you have a gas bag filled only with hydrogen at 1 bar you are telling me that inside that bag at the top the pressure would be higher then at the bottom. Now i don't intuitively understand how that should work.

Not quite, and backwards, it is doubly counterintuitive. The hydrostatic pressure of the atmosphere falls as you rise, due to the total mass of the atmosphere above becoming less with height. The same thing happens within a bag of hydrogen, higher up, the pressure gets less. But the density of hydrogen is less than air, so the hydrostatic pressure falls slower with height in hydrogen than in air. That is why, inside the bag of hydrogen, the envelope at the top is pushed upwards. The air pressure has fallen faster with height outside the bag, than inside, with the hydrogen.

Hot air is less dense than cold, so the same hydrostatic mechanism operates inside hot air balloons to give them lift. They are open at the bottom where the burners inject the hot air, so the differential pressure there is zero.

It is easiest to evaluate lift based on the differential density * the volume of bag, than it is by integrating the differential hydrostatic pressure, across the horizontal area component of an entire bag or balloon envelope.
Understanding how a hot air balloon operates, requires an understanding of the differential hydrostatic pressure. The same is true of an optimised bag of lifting gas, the differential pressure is zero at the bottom.

The mass of the bag is given by Density/Volume right?

Mass of gas in bag = density * volume.
Density = mass / volume.

 

[askingask]

The hydrostatic pressure of the atmosphere falls as you rise, due to the total mass of the atmosphere above becoming less with height. The same thing happens within a bag of hydrogen, higher up, the pressure gets less. But the density of hydrogen is less than air, so the hydrostatic pressure falls slower with height in hydrogen than in air.

Oh so you mean the weight of the hydrogen at the top is pressing down on the hydrogen at the bottom.

 

[Baluncore]

Oh so you mean the weight of the hydrogen at the top is pressing down on the hydrogen at the bottom.

Which is why the pressure is lower at the top, but the air pressure is even lower outside at the top.
https://en.wikipedia.org/wiki/Vertical_pressure_variation

 

[askingask]

Which is why the pressure is lower at the top, but the air pressure is even lower outside at the top.
https://en.wikipedia.org/wiki/Vertical_pressure_variation

Now I don‘t quite understand how this is relevant to my question, I mean is this pressure difference along the hight of the bags not quite small?

 

[Baluncore]

Now I don‘t quite understand how this is relevant to my question,

It is important because the pressure in the bag is different at different heights.
To vent the excess pressure, (the excess mass of gas), you must vent lift gas to get a zero pressure differential at the lowest point of the bag.

If there is excess pressure at the bottom, that excess is present throughout the contents of the bag, so the entire mass is greater than necessary.

I mean is this pressure difference along the hight of the bags not quite small?

The small differential pressure, over a sufficiently large area, lifted the 230 tonne airship, Hindenburg, LZ 129.

 

[askingask]

The small differential pressure, over a sufficiently large area, lifted the 230 tonne airship, Hindenburg, LZ 129.

So the main factor of lift in an airship is not just the air the bags displace but especially that difference in pressure inside the bags?

 

[Baluncore]

Yes, it is the differential pressure across the envelope that causes the lift.

 

[askingask]

Yes, it is the differential pressure across the envelope that causes the lift.

Can you reference any video or article to read? Keep in mind I’m just an undergrad.

 

[hutchphd]

So the main factor of lift in an airship is not just the air the bags displace but especially that difference in pressure inside the bags?

Both of these descriptions (given the laws of gas physics) describe the same situation. Ascribing a "real" cause is an excercise in interpretation. It is what it is and calculations either way will give the correct numbers.

 

[askingask]

Both of these descriptions (given the laws of gas physics) describe the same situation. Ascribing a "real" cause is an excercise in interpretation. It is what it is and calculations either way will give the correct numbers.

Are you telling me that baluncore is overcomplicating this entire thing to confuse me …

 

[hutchphd]

Are you telling me that baluncore is overcomplicating this entire thing to confuse me

No that is very far from what I said. Please pay careful attention.
You are assuming that you are "correct" and that he is therefore "wrong" (or confusing0 In fact try to see that you are each using different words to describe the same essrntially correct physics. Actual physics is done using mathematics for this very reason.w

 

[Baluncore]

Can you reference any video or article to read? Keep in mind I’m just an undergrad.

The best way to understand LTA craft is with the case of a hot air balloon.

Are you telling me that baluncore is overcomplicating this entire thing to confuse me

Is the pressure inside the airship the main factor which makes the whole airship heavier compared to the low pressure atmosphere?
And would releasing gas from the airship, to neutralize pressure differences, make the airship lighter again compared to the air it displaces?

You are confusing yourself by talking about the pressure of hydrogen inside the bag, and the external atmosphere, without realising that the pressures are height and density specific, so you need to consider the differential hydrostatic pressures over the whole of the envelope.

An LTA craft, hangs from bags of lifting gas, in way similar to how a bicycle hangs from its wheel spokes.

 

[askingask]

No that is very far from what I said. Please pay careful attention.
You are assuming that you are "correct" and that he is therefore "wrong" (or confusing0 In fact try to see that you are each using different words to describe the same essrntially correct physics. Actual physics is done using mathematics for this very reason.w

No that is in turn not what I said. I thought my whole understanding of buoyancy was wrong, that is why. Though this meta stuff is irrelevant to the Forum I guess.

 

[askingask]

without realising that the pressures are height and density specific, so you need to consider the differential hydrostatic pressures over the whole of the envelope.

I finally understand what you mean, thank you, though I‘ll probably do a bit of research to this, but thanks for the video.

 

[James Demers]

The OP's question, I think, calls for a hypothetical envelope of volume V and mass M that is perfectly rigid and thus maintains constant volume. Fill this vehicle with one atm. of helium at sea level, and turn it loose. As it rises, the external pressure drops. Releasing helium will indeed reduce the mass of the vehicle, and reduce its density, allowing it to rise higher. If the mass of the contained helium at any given altitude is m, the limit on altitude will be the height at which the density of the vehicle, (M+m)/V, equals the density of the atmosphere at that same given altitude. (Scientific high-altitude balloons will get to this height before bursting, where V is the mechanical limit on the volume of the envelope.)
You could get a bit higher if you evacuated this (extremely) theoretical envelope, leaving a high vacuum, before releasing the vehicle. Since m=0, this "vacuum balloon" would rise to the point where the atmosphere has density M/V.

 

[askingask]

The OP's question, I think, calls for a hypothetical envelope of volume V and mass M that is perfectly rigid and thus maintains constant volume. Fill this vehicle with one atm. of helium at sea level, and turn it loose. As it rises, the external pressure drops. Releasing helium will indeed reduce the mass of the vehicle, and reduce its density, allowing it to rise higher. If the mass of the contained helium at any given altitude is m, the limit on altitude will be the height at which the density of the vehicle, (M+m)/V, equals the density of the atmosphere at that same given altitude. (Scientific high-altitude balloons will get to this height before bursting, where V is the mechanical limit on the volume of the envelope.)
You could get a bit higher if you evacuated this (extremely) theoretical envelope, leaving a high vacuum, before releasing the vehicle. Since m=0, this "vacuum balloon" would rise to the point where the atmosphere has density M/V.

This is exactly what I asked for. Thank you.

 

[Baluncore]

There are some changes that need to happen, to take the thought experiment towards a real experiment.

a hypothetical envelope of volume V and mass M that is perfectly rigid and thus maintains constant volume.

Using a sphere of the lightest weight film would weigh less and still limit the volume. Also, the top of the envelope needs thicker film to withstand greater differential pressure and surface tension, than at the bottom, where thinner film can be used. That film mass distribution will tend to capsize the balloon, unless there is a suspended payload that defines the base of the envelope.

Fill this vehicle with one atm. of helium at sea level, and turn it loose.

A film envelope will lift with only about 10% hydrogen or helium. Why waste the limited helium resource, by 100% filling a rigid envelope, when a spherical film envelope is easier to transport and costs 90% less to partially fill?

As it rises, the external pressure drops. Releasing helium will indeed reduce the mass of the vehicle, and reduce its density, allowing it to rise higher.

As it rises, a film envelope will expand with the lift gas, until finally, the base of the film envelope becomes spherical. Then the mode of operation changes. I would put a one-way flap valve at the base to release the excess lift gas, so there is zero differential pressure pushing the base of the envelope downwards from inside.

Scientific high-altitude balloons will get to this height before bursting, where V is the mechanical limit on the volume of the envelope.

Bursting is used to aid recovery of the payload, and to clear the controlled airspace. By continuously venting excess lift gas from an opening in the base, a greater height than the burst height would be reached.

 

[Baluncore]

You could get a bit higher if you evacuated this (extremely) theoretical envelope, leaving a high vacuum, before releasing the vehicle.

Consider a rigid spherical envelope, that you want to fly as a vacuum balloon, to a height where the external air pressure is; Pa.

Initially, on the ground, the envelope is manufactured, filled with air at one bar.
Start pumping out air, until the internal pressure is Pa below local atmospheric.
The balloon will then start to lift, with only a partial vacuum.
The envelope then only needs to withstand an external crush of Pa.
The internal depression of Pa, reduces the density of the internal air.
No special, rare or expensive lifting gas is needed.

As the balloon rises, pump out more air, maintaining a differential pressure of Pa.
When the balloon has reached an altitude where the external air pressure is Pa, the envelope will contain a true vacuum.

At no time during the flight was the external crush greater than Pa.
The (solar powered) pump never needed to develop a pressure difference greater than Pa.

The big mistake with vacuum balloons, is attempting to design them to withstand the crush of a high vacuum at sea level.

 

[askingask]

Consider a rigid spherical envelope, that you want to fly as a vacuum balloon, to a height where the external air pressure is; Pa.

Initially, on the ground, the envelope is manufactured, filled with air at one bar.
Start pumping out air, until the internal pressure is Pa below local atmospheric.
The balloon will then start to lift, with only a partial vacuum.
The envelope then only needs to withstand an external crush of Pa.
The internal depression of Pa, reduces the density of the internal air.
No special, rare or expensive lifting gas is needed.

As the balloon rises, pump out more air, maintaining a differential pressure of Pa.
When the balloon has reached an altitude where the external air pressure is Pa, the envelope will contain a true vacuum.

At no time during the flight was the external crush greater than Pa.
The (solar powered) pump never needed to develop a pressure difference greater than Pa.

The big mistake with vacuum balloons, is attempting to design them to withstand the crush of a high vacuum at sea level.

That is genius, never thought about that, but based on our discussion here, it makes absolutely sense. The only problem here is that to keep the envelop lightweight the pressure difference has to be small. But if the pressure difference is small net buoyancy will also be very small. Question is where is the optimal point, if one exists even, where the weight of the envelop is small enough and the pressure difference is big enough to ensure the density difference.

 

[askingask]

Consider a rigid spherical envelope, that you want to fly as a vacuum balloon, to a height where the external air pressure is; Pa.

Initially, on the ground, the envelope is manufactured, filled with air at one bar.
Start pumping out air, until the internal pressure is Pa below local atmospheric.
The balloon will then start to lift, with only a partial vacuum.
The envelope then only needs to withstand an external crush of Pa.
The internal depression of Pa, reduces the density of the internal air.
No special, rare or expensive lifting gas is needed.

As the balloon rises, pump out more air, maintaining a differential pressure of Pa.
When the balloon has reached an altitude where the external air pressure is Pa, the envelope will contain a true vacuum.

At no time during the flight was the external crush greater than Pa.
The (solar powered) pump never needed to develop a pressure difference greater than Pa.

The big mistake with vacuum balloons, is attempting to design them to withstand the crush of a high vacuum at sea level.

Perhaps this could be especially useful for missions on mars.

 

[Baluncore]

The only problem here is that to keep the envelop lightweight the pressure difference has to be small. But if the pressure difference is small net buoyancy will also be very small.

A functional vacuum balloon does not yet exist, because a sufficiently low-mass, crush-surviving envelope, has still not been engineered. But, I expect it will happen.

 

[askingask]

A functional vacuum balloon does not yet exist, because a sufficiently low-mass, crush-surviving envelope, has still not been engineered. But, I expect it will happen.

Honestly hydrogen is pretty good already and people over estimate its danger. No need for expensive helium.

 

[Baluncore]

Honestly hydrogen is pretty good already and people over estimate its danger.

If you use hydrogen, the cost of insurance will be greater for historical reasons.
Hydrogen molecules are small, fast, and pass through many envelope materials. Helium diffuses at a lower rate.

 

[askingask]

for historical reasons.

Lol

 

[akhmeteli]

The big mistake with vacuum balloons, is attempting to design them to withstand the crush of a high vacuum at sea level.

I respectfully disagree. If the pressure difference between exterior and interior of the vacuum balloon is small, the shell needs to be very light to provide buoyancy, so it would be more difficult to prevent the balloon failure, in spite of the lesser pressure difference. According to our calculations/computations, it is easier to design a vacuum balloon for sea level and low interior pressure / high pressure difference than, say, for sea level and low pressure difference, or for higher altitudes.

We offered a design of a sea level and high pressure difference vacuum balloon at Eng 2021, 2(4), 480-491

 

[Baluncore]

According to our calculations/computations, it is easier to design a vacuum balloon for sea level and low interior pressure / high pressure difference than, say, for sea level and low pressure difference, or for higher altitudes.

That is true, but to rise to any altitude where the air pressure is Pa, the balloon must also be neutrally buoyant, with a differential envelope pressure of Pa, at sea level. I ask why a high vacuum would be needed at sea level, if that design will later fly at an external pressure of Pa.

A demonstration system, capable of neutral buoyancy at sea level with high vacuum, would be unable to rise above sea level.

That challenge has the same design and material constraints, as if you halve the mass of the carcass, at half an atmosphere differential pressure, with half the crush, and the balloon could rise from sea level to half an atmosphere = 5470 metres.

The limiting factor is the composite materials needed, and the composite structures required for both tension and compression, to prevent the crush.

I predict the first partial-vacuum balloon to fly will be in a hot and dry atmosphere. There should be a few drops of water inside the balloon, so the RH of the internal air, remains close to 100%, or would that be cheating?

 

[akhmeteli]

That is true, but to rise to any altitude where the air pressure is Pa, the balloon must also be neutrally buoyant, with a differential envelope pressure of Pa, at sea level. I ask why a high vacuum would be needed at sea level, if that design will later fly at an external pressure of Pa.

But vacuum balloons can have some applications at sea level as well, so I don't understand why designing vacuum balloons for sea level would be a "big mistake".

A demonstration system, capable of neutral buoyancy at sea level with high vacuum, would be unable to rise above sea level.

Technically, yes, but, for example, at the altitude of 150 m, the air density is just 1.8% less than that at sea level (and the pressure is about 1.8% less), so a system designed for a range from 0 m to 150 m will not be much different from a purely sea level system. On the other hand, the distance of direct vision at the altitude of 150 m is about 44 km, so this altitude may be of interest, say, for antenna applications.

I predict the first partial-vacuum balloon to fly will be in a hot and dry atmosphere. There should be a few drops of water inside the balloon, so the RH of the internal air, remains close to 100%, or would that be cheating?

I have not thought about "dry", but cold air is better for a vacuum balloon. The pressure being equal, the buoyancy force is higher for cold air as air density is higher.

 

[Baluncore]

The pressure being equal, the buoyancy force is higher for cold air as air density is higher.

The temperature of a partial-vacuum balloon will equilibrate with the environment. The saturated air, inside the balloon, will have a lower density than the dry air outside. That is significantly enhanced at higher temperatures.

 

[akhmeteli]

The temperature of a partial-vacuum balloon will equilibrate with the environment. The saturated air, inside the balloon, will have a lower density than the dry air outside. That is significantly enhanced at higher temperatures.

At high altitudes, there cannot be "hot atmosphere" (unless one considers altitudes unreasonable for balloons). As for sea level, let us compare temperatures of 0 deg C and 50 deg C. The pressure of saturated vapor pressure at 50C is about 12% of the atmospheric pressure, but the air density and, therefore, the buoyancy force is about 15% less at 50C than at 0C (for the same atmospheric pressure).

 

[Baluncore]

On the other hand, the distance of direct vision at the altitude of 150 m is about 44 km, so this altitude may be of interest, say, for antenna applications.

At 150 metres you could use a fixed tower, or a tethered helium balloon. I have flown gyro kites well above 150 m, and they generate their own power.

At high altitudes, there cannot be "hot atmosphere" (unless one considers altitudes unreasonable for balloons).

A high-vacuum balloon at sea level is a demonstration package, of less practical use than a stepladder. It will hover best below sea level, on a cold and dry night, during an anticyclone. Think the Dead Sea, Death Valley, or deep underground in a mine. The problem with mines is that they get hotter with depth, and have RH=100%.

Only a partial-vacuum balloon can benefit from internal saturation, RH=100%. At the depressed boiling point of water, the lift gas remaining would be steam from the air, being a clear cloud chamber.

 

[akhmeteli]

At 150 metres you could use a fixed tower, or a tethered helium balloon. I have flown gyro kites well above 150 m, and they generate their own power.

Fixed towers of this height are (very?) expensive, they require some land to own or rent, they require lighting for aircraft safety, which creates serious maintenance problems.

As for helium balloons, they have their own set of issues (don't get me started:-) ). Furthermore, helium balloons are competitors to stratospheric vacuum balloons too.

I don't know much about gyro kites, but I believe they are wind-dependent.

And tethered balloons require some real estate, and their altitude is wind-dependent

A high-vacuum balloon at sea level is a demonstration package, of less practical use than a stepladder. It will hover best below sea level, on a cold and dry night, during an anticyclone. Think the Dead Sea, Death Valley, or deep underground in a mine. The problem with mines is that they get hotter with depth, and have RH=100%.

So you don't think there is any application for low-altitude vacuum balloons. I believe there are some important applications, such as antennas, freight transportation in remote areas, etc., so let us agree to disagree.

And let us not forget that, while making a sea-level balloon is really hard, making a stratospheric balloon is significantly harder.

Only a partial-vacuum balloon can benefit from internal saturation, RH=100%. At the depressed boiling point of water, the lift gas remaining would be steam from the air, being a clear cloud chamber.

So again, internal saturation does not seem important at low altitudes, and temperature and water vapor content are low at high altitudes.