ChatGPT on Problem of Debarking Cargo from Airships

I've got the idea that keeping cooled lifting gas (hydrogen, helium) in a tank, in a compressed state, and bleeding that into the lifting gas bags when unloading cargo might be a good way to keep an airship at a constant altitude when unloading cargo. How does that sound? What kinds of ship dimensions and volumes and relative temperatures would be required for something like that to work?

Your idea is fundamentally sound—using compressed and cooled lifting gas (hydrogen or helium) as ballast to compensate for weight changes when unloading cargo could help maintain altitude. However, the feasibility depends on several factors, including gas volume, compression ratios, temperature effects, and the size of the airship. Let's break it down:

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Key Considerations

1. Compression and Density Changes:

   • Helium and hydrogen are compressible, and their densities increase significantly under higher pressure.
   • For example, at 300 bar (≈4400 psi), helium’s density is about 50 times greater than at atmospheric pressure.

2. Volume Requirements:

   • The airship needs enough onboard storage to hold a sufficient quantity of compressed gas.
   • A full cargo load’s weight should roughly match the mass of gas needed to be released to maintain equilibrium.

3. Thermodynamic Effects:

   • Compression generates heat, and expanding gas cools. If expansion is too rapid, icing or condensation could occur.
   • Insulation and heating mechanisms may be needed.

4. Size and Mass Scaling:

   • The storage tanks must be lightweight but strong, likely high-strength carbon-fiber-wrapped pressure vessels.
   • The lifting gas volume required is massive, meaning a large airship.

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Rough Calculation Example

Let’s assume:

• You unload 10 metric tons (10,000 kg) of cargo.
• The airship's lifting gas is helium at ambient temperature and pressure (1 atm, 20°C).
• The density of helium at 1 atm and 20°C is 0.1786 kg/m³.
• To offset 10,000 kg, you'd need to vent about 56,000 m³ of helium at ambient pressure.

If you instead store helium in tanks:

• At 300 bar, helium’s density is ≈50 times that of ambient.
• To store 56,000 m³ of ambient helium in compressed form, you'd need about 1,120 m³ of tank space (56,000 ÷ 50).
• Modern high-pressure tanks weigh around 15 kg per m³ of stored gas, so the tanks themselves could weigh 16,800 kg, which is a problem.

For hydrogen, which is much lighter (density 0.0899 kg/m³ at 1 atm, 20°C):

• You’d need 111,000 m³ of hydrogen at ambient pressure.
• At 300 bar, hydrogen’s density is about 23 times ambient, so you'd need around 4,826 m³ of storage.
• Tank weight would be substantial but possibly less than helium due to hydrogen’s lower molecular mass.

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Ship Dimensions

• Modern large airships (e.g., the Zeppelin NT) have volumes in the range of 8,400–25,000 m³.
• A truly heavy-lift airship like Lockheed Martin’s concepts could exceed 100,000 m³.
• A cargo airship using this method would likely need to be on the order of 150,000–500,000 m³ to make practical use of compressed lifting gas.

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Alternative Approach

Instead of storing vast amounts of compressed gas, consider:

1. Water or Liquid Hydrogen Ballast:

   • If the airship carries liquid hydrogen (LH2, density ≈70.8 kg/m³ at -253°C), converting it to gaseous hydrogen for lift could be more efficient than compressing gas.
   • Water ballast could be picked up or released from the environment to manage buoyancy.

2. Hybrid Lift Airship:

   • Some designs use partial aerodynamic lift from wings, reducing reliance on gas compression.

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Conclusion

Your method could work in large-scale airships (100,000+ m³ envelopes), but the tank weight and volume requirements are major challenges. Using liquid hydrogen or water ballast might be a more practical way to achieve the same altitude stability when unloading cargo.

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what is the differential lift if one injected some amount of compressed (cooled) hydrogen into the gasbags of an airship? There'd be more hydrogen, but it would be cooler.

The differential lift from injecting compressed, cooled hydrogen into an airship’s gasbags depends on the tradeoff between two opposing effects:

1. Increased Lift from More Hydrogen:

   • Injecting additional hydrogen increases the number of buoyant gas molecules inside the envelope, which could provide more lift.

2. Decreased Lift from Cooling:

   • A cooler gas is denser, meaning each cubic meter provides less lift.

Step-by-Step Analysis

Let’s compare the initial (ambient) hydrogen in the envelope to the cooled, injected hydrogen using the ideal gas law:

$$PV = nRT$$

where:

• $P$ = pressure (assumed constant at 1 atm inside the gasbag)
• $V$ = volume
• $n$ = number of moles of gas
• $R$ = universal gas constant
• $T$ = absolute temperature (Kelvin)

For buoyancy, lift is determined by the difference in density between air and hydrogen:

$$\text{Lift} = V (\rho_{\text{air}} - \rho_{\text{H}_2})$$

where:

• $\rho_{\text{air}}$ is the density of air
• $\rho_{\text{H}_2}$ is the density of hydrogen

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Example Calculation

Baseline: Hydrogen at Ambient Conditions

• Assume ambient temperature is 20°C (293 K).
• Hydrogen density at 1 atm, 20°C ≈ 0.0899 kg/m³.

Injected Hydrogen: Cooled to -50°C (223 K)

• Using the ideal gas law, at constant pressure:

  $$\frac{\rho_{\text{cold}}}{\rho_{\text{ambient}}} = \frac{T_{\text{ambient}}}{T_{\text{cold}}}$$ $$\frac{\rho_{\text{cold}}}{0.0899} = \frac{293}{223} \approx 1.314$$
  • New density: 0.0899 × 1.314 ≈ 0.118 kg/m³

• The lift per cubic meter changes from:

  $$\text{Lift}_{\text{ambient}} = 1.225 - 0.0899 = 1.1351 \text{ kg/m³}$$ $$\text{Lift}_{\text{cold}} = 1.225 - 0.118 = 1.107 \text{ kg/m³}$$
  • Loss of lift per cubic meter due to cooling: $$1.1351 - 1.107 = 0.0281 \text{ kg/m³} (≈2.5$$

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Offsetting the Loss with Additional Gas

By injecting more hydrogen, you increase the total number of molecules inside the envelope, but pressure inside the gasbag remains constant (assuming a flexible envelope).

• At 223 K, hydrogen molecules occupy 23.9% less volume than at 293 K.
• So, for every 1 m³ of ambient hydrogen, you could fit 1.314 m³ worth of molecules if you inject hydrogen at -50°C.

Does this compensate for the lift loss?

• Since density increases by 31.4%, but the amount of gas increases by the same factor, the total lift should stay roughly the same.
• However, if the envelope is rigid (not flexible), then the cooling will just increase density and reduce overall lift.

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Conclusion

• If the envelope can expand, injecting cold hydrogen should keep lift nearly constant because the increased number of hydrogen molecules compensates for the density change.
• If the envelope volume is fixed, cooling will reduce lift unless the expansion of injected gas offsets it.
• The net effect depends on how much the gasbag expands and the degree of cooling.

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what about pumping out (and possibly compressing (slightly) the existing gas in the gasbag (the hotter gas) at the same time one pumps in the colder lifting gas?

 If you simultaneously pump out (and slightly compress) the existing warm hydrogen while injecting cooler hydrogen, you create a dynamic lift control system that balances the effects of temperature and density changes. Let’s analyze the potential impact:

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Key Effects of This Process

1. Increased Hydrogen Mass per Unit Volume

   • Injecting cooler hydrogen increases its density.
   • The total number of gas molecules inside the envelope increases (if the volume expands).

2. Maintaining or Enhancing Lift by Removing Warmer, Less Dense Gas

   • Warmer hydrogen is less dense and provides slightly more lift per cubic meter.
   • However, removing it reduces the total gas mass, preventing the overall lifting gas from becoming too heavy.

3. Compression and Storage Considerations

   • The removed gas can be slightly compressed and stored for later use (to help regulate buoyancy).
   • Compression generates heat, so cooling may be needed.

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Step-by-Step Lift Calculation

Baseline: Hydrogen at Ambient Conditions

• Assume ambient temperature: 20°C (293 K).
• Density of hydrogen at 1 atm: 0.0899 kg/m³.
• Air density: 1.225 kg/m³.
• Lift per cubic meter: $$\text{Lift}_{\text{ambient}} = 1.225 - 0.0899 = 1.1351 \text{ kg/m³}$$

New Hydrogen Injected at -50°C (223 K)

• From the ideal gas law, the density of hydrogen increases by a factor of $\frac{293}{223} \approx 1.314$.
• New hydrogen density: $$\rho_{\text{cold}} = 0.0899 \times 1.314 = 0.118 \text{ kg/m³}$$
• New lift per cubic meter: $$\text{Lift}_{\text{cold}} = 1.225 - 0.118 = 1.107 \text{ kg/m³}$$
• Lift loss per cubic meter: 0.0281 kg/m³ (≈2.5% reduction).

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Effect of Pumping Out Warmer Hydrogen

If we remove a portion of the warmer hydrogen while injecting the cooler gas, we can prevent lift loss. The key is to balance the mass and density so that the envelope maintains the same overall buoyancy.

• Say we remove 1 m³ of warm hydrogen (293 K) and replace it with 1 m³ of cold hydrogen (223 K):
  • The density of the gas inside the envelope increases.
  • But the number of hydrogen molecules inside stays constant, preserving the lifting force.

If we compress the removed hydrogen into a storage tank:

• If stored at 3 atm (moderate compression), its density increases by 3× (≈0.27 kg/m³).
• Storing 1 m³ of compressed hydrogen at 3 atm would take up 1/3 of its original volume, freeing up space inside the airship’s envelope.

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Final Considerations

• For a Flexible Envelope:

  • The gasbag expands slightly, offsetting density changes and keeping lift nearly constant.
  • This method is most efficient.

• For a Rigid Envelope:

  • Cooling the gas alone would reduce lift.
  • You’d need to remove gas in a way that keeps the internal pressure stable.
  • Slightly compressing and storing warm gas might allow you to compensate for the cooling effect.

• Energy Costs:

  • Compressing and storing the removed hydrogen requires energy input.
  • If the hydrogen is later warmed and re-expanded, this could be used to recover lift when needed.

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Conclusion

Your method of pumping out warmer hydrogen while injecting cooler hydrogen is a valid way to maintain or regulate lift, provided the system allows for expansion or reintroduction of stored gas later. This could serve as an effective altitude control mechanism, reducing the need for dumping gas overboard.