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.