Managing Lift when Unloading Zeppelins

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Overview

The idea is how to quickly and with precise control reduce the lift on an airship at the moment it is unloading cargo. This is a major problem with getting light-than-air transportation widely adopted.

The idea is two use two effects:

1. The fact that heating the lifting gas creates more lift with less gas

2. The injection of cold lifting gas has the effect of decreasing lift

Gas cannot be quickly evacuated to reduce lift, since venting gas to the atmosphere is wasteful and impractical, and recompressing the gas cannot be accomplished quickly or reliably (for example, loss of power or such).

The idea is to pump out and compress part of the gas from the gasbag(s) and heat the remainder such that the lift is the same as ambient pressure and temperature. This is done as the airship approaches its unloading point. As each piece of cargo in unloaded (1) the airship crew can decrease the lift by injecting cold lifting gas into the gasbags. 

By evacuating gas from the gas bag and compressing it and heating the remaining gas, the bag can be made to yield the same lift as a larger amount of gas at ambient temperature and atmospheric pressure. This means that when cold lifting gas is nozzled in, the gas bag may be kept at the same volume, brought to a lower pressure, and a lower temperature, resulting in less lift, and this should be able to be accomplished very quickly, with precise control and with minimal need for failure-prone systems such as motors and pumps.

My exchange with ChatGPT below overs this idea, and it seems that the heating, the injection of cold lifting gas, and so on, and the need to decrease lift on the order of 10% of the total mass of the airship, i.e., the mass of the cargo, may all accomplished within reasonable ranges of the amount of the gas that must be removed from a gas bag (perhaps 75%), how much it must be heated (maybe only 50 degrees), and how cold the input gas would need to be (perhaps ambient). In short, it seems all can remain within the operational ranges of all of these systems and materials.

I would still like to check over ChatGPT's calculations again. Comments welcome!

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(1) The cargo winches should have a spring system built in such that they can hang onto the cargo until the lift is decreased, i.e., the cable tension drops to zero, and then release.

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Talk with ChatGPT

does the principle of partial pressures apply to injecting cool hydrogen into a vessel, gasbag of warmer higher pressure hydrogen and how would that look?

Yes, the principle of partial pressures (Dalton's Law) applies when injecting cooler hydrogen into a vessel or gasbag containing warmer, higher-pressure hydrogen. Dalton's Law states that in a mixture of gases, each gas behaves independently, and its contribution to the total pressure is its partial pressure.

What Happens When Cool Hydrogen is Injected?

1. Initial Mixing:

   • The cooler hydrogen, having lower thermal energy, enters the vessel at a lower temperature and pressure.
   • If injected at a different pressure than the existing hydrogen, a temporary pressure gradient is created.
   • The cooler gas may initially sink due to higher density, but diffusion and convection will lead to mixing.

2. Temperature and Pressure Effects:

   • The overall temperature of the system will begin to equilibrate through thermal conduction and convective mixing.
   • If the system is insulated, energy redistribution occurs between the cooler and warmer hydrogen, and the temperature will stabilize at an intermediate value.
   • The total pressure in the vessel will increase by the amount of the added hydrogen’s partial pressure, assuming the volume is constant.

3. Visual Representation:

   • If visualized with a density gradient, at the beginning, the cooler hydrogen might form a distinct layer or plume before fully mixing.
   • Over time, the injected hydrogen will diffuse, eventually leading to a homogeneous mixture.

4. Real-World Considerations:

   • If the gasbag is flexible, its volume might expand instead of increasing pressure.
   • If the system is open or vented, some hydrogen could escape, affecting final conditions.

Would you like a graphical representation or further breakdown of pressure equations?

Yes! We can use the ideal gas law $PV = nRT$ along with Dalton’s Law of Partial Pressures to analyze what happens when cooler hydrogen is injected into a vessel or gasbag containing warmer hydrogen.

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Step 1: Define Initial and Final States

Let’s define:

• $P_1, V_1, T_1$ as the initial pressure, volume, and temperature of the warm hydrogen already in the vessel.
• $P_2, V_2, T_2$ as the initial pressure, volume, and temperature of the injected cool hydrogen.
• $P_f, V_f, T_f$ as the final pressure, volume, and temperature after mixing.

Since both gases are hydrogen, they follow the same gas constant $R$, so we can apply the ideal gas law separately for each gas.

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Step 2: Apply the Ideal Gas Law

For the warm hydrogen already in the vessel:

$$P_1 V = n_1 R T_1$$

For the injected cool hydrogen:

$$P_2 V = n_2 R T_2$$

After mixing, assuming the system reaches a uniform final temperature $T_f$ and volume remains constant, the final total pressure is given by Dalton’s Law of Partial Pressures:

$$P_f = P_1' + P_2'$$

where $P_1'$ and $P_2'$ are the new partial pressures of each hydrogen component at $T_f$, using the ideal gas law again:

$$P_1' V = n_1 R T_f$$ $$P_2' V = n_2 R T_f$$

So the final pressure is:

$$P_f = \frac{n_1 R T_f}{V} + \frac{n_2 R T_f}{V}$$

Using $n = \frac{P V}{R T}$, we express the initial moles in terms of pressure and temperature:

$$n_1 = \frac{P_1 V}{R T_1}, \quad n_2 = \frac{P_2 V}{R T_2}$$

Substituting:

$$P_f = \left( \frac{P_1 V}{R T_1} + \frac{P_2 V}{R T_2} \right) \frac{R T_f}{V}$$

Canceling $V$ and $R$:

$$P_f = P_1 \frac{T_f}{T_1} + P_2 \frac{T_f}{T_2}$$

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Step 3: Solve for Final Temperature $T_f$

If no heat is lost to the environment, the system follows energy conservation. The total internal energy before mixing equals the total internal energy after mixing:

$$n_1 C_v T_1 + n_2 C_v T_2 = (n_1 + n_2) C_v T_f$$

Since hydrogen behaves as an ideal diatomic gas, its molar heat capacity at constant volume is $C_v = \frac{5}{2} R$, which cancels out:

$$n_1 T_1 + n_2 T_2 = (n_1 + n_2) T_f$$

Solving for $T_f$:

$$T_f = \frac{n_1 T_1 + n_2 T_2}{n_1 + n_2}$$

Substituting $n_1$ and $n_2$:

$$T_f = \frac{\frac{P_1 V}{R T_1} T_1 + \frac{P_2 V}{R T_2} T_2}{\frac{P_1 V}{R T_1} + \frac{P_2 V}{R T_2}}$$

Canceling $V/R$:

$$T_f = \frac{P_1 + P_2 \frac{T_1}{T_2}}{1 + \frac{P_2}{P_1} \frac{T_1}{T_2}} T_1$$

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Step 4: Interpret the Results

• $P_f$ depends on how much cooler hydrogen is added and how its temperature compares to the existing gas.
• $T_f$ is a weighted average of the two initial temperatures, weighted by their initial pressures.
• If significantly cooler hydrogen is injected, $T_f$ will be lower, reducing the final pressure.
• If the gasbag is flexible, the volume might increase instead of pressure.

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Would you like a worked-out example with numbers?

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I want to work out a system where the gas inside a gasbag on an airship is heated so as to provide more lift for a lesser amount of hydrogen ...so less hydrogen but enough to provide something like the same amount of lift with ambient temperature at atmospheric pressure ....but then inject cooler hydrogen in some quantity so as to reduce the lift by cooling the total gas in the gasbag so as to reduce the lift of the gasbag by some percentage (like 10% or less) The idea is to be able to rapidly reduce the lift of an airship quickly and reliably when unloading cargo

That’s a clever approach—using heated hydrogen to increase lift and then injecting cooler hydrogen to reduce lift in a controlled way. Let’s break it down mathematically and conceptually.

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Step 1: Understanding Lift from Hydrogen

Lift comes from the buoyant force, which depends on the density difference between the air and the hydrogen inside the gasbag:

$$F_B = \rho_{\text{air}} V g - \rho_{\text{H}_2} V g$$

where:

• $\rho_{\text{air}}$ = density of ambient air (depends on temperature & pressure),
• $\rho_{\text{H}_2}$ = density of hydrogen inside the gasbag (depends on temperature & pressure),
• $V$ = volume of the gasbag,
• $g$ = gravitational acceleration.

Using the ideal gas law:

$$\rho = \frac{P M}{R T}$$

where:

• $M$ = molar mass (2.016 g/mol for $H_2$, 28.97 g/mol for air),
• $R$ = universal gas constant,
• $T$ = absolute temperature (Kelvin),
• $P$ = pressure (assuming atmospheric, $P \approx 101.3$ kPa).

Since hydrogen is much less dense than air, heating it reduces its density further, increasing lift.

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Step 2: How Heating Hydrogen Increases Lift

At ambient conditions (25°C, 1 atm):

• $\rho_{\text{H}_2} \approx 0.0838$ kg/m³.
• $\rho_{\text{air}} \approx 1.184$ kg/m³.

If we heat the hydrogen, its density decreases as:

$$\rho_{\text{H}_2} (T) = \frac{P M}{R T}$$

For example, at 50°C (323 K), hydrogen density drops to $\approx 0.069$ kg/m³, increasing lift per unit volume.

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Step 3: Injecting Cool Hydrogen to Reduce Lift

To reduce lift, we cool the hydrogen by injecting cooler hydrogen at a lower temperature $T_2$. This increases the overall density of the gas inside the gasbag, decreasing buoyancy.

Using a weighted average approach, the final temperature after mixing cool hydrogen ($T_2$) with the existing heated hydrogen ($T_1$) is:

$$T_f = \frac{n_1 T_1 + n_2 T_2}{n_1 + n_2}$$

where $n_1$ and $n_2$ are the respective moles of hot and cool hydrogen.

Since $n = PV/RT$, we can rewrite:

$$T_f = \frac{P V (T_1 / R T_1) + P V (T_2 / R T_2)}{(P V / R T_1) + (P V / R T_2)}$$

Simplifying:

$$T_f = \frac{T_1 + r T_2}{1 + r}$$

where $r = \frac{P_2 T_1}{P_1 T_2}$ (assuming equal volumes for simplicity).

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Step 4: How Much Cooling is Needed to Reduce Lift by 10%?

If you want to reduce lift by 10%, you need to increase hydrogen density by 10%, meaning a lower final temperature.

Since density is inversely proportional to temperature:

$$\rho_{\text{H}_2} (T_f) = \rho_{\text{H}_2} (T_1) \frac{T_1}{T_f}$$

For a 10% increase in density:

$$\frac{T_1}{T_f} = 1.1$$ $$T_f = \frac{T_1}{1.1}$$

Substituting into our temperature mixing equation:

$$\frac{T_1 + r T_2}{1 + r} = \frac{T_1}{1.1}$$

Solving for $r$ (the ratio of cool hydrogen needed):

$$r = \frac{T_1 - \frac{T_1}{1.1}}{\frac{T_1}{1.1} - T_2}$$

For example:

• If initial hydrogen is at 50°C (323 K),
• Inject hydrogen at 10°C (283 K),
• The required ratio $r$ (cool hydrogen / hot hydrogen) is:

$$r = \frac{323 - 293.6}{293.6 - 283} = \frac{29.4}{10.6} \approx 2.77$$

So you would need to inject 2.77 times the amount of existing hydrogen to cool it down enough to reduce lift by 10%.

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Step 5: Practical Implementation

• Injection Mechanism: A controlled release valve or injector system could introduce cooler hydrogen from a separate tank.
• Rate of Mixing: Faster mixing means more immediate lift reduction—could use turbulence-inducing nozzles.
• Emergency Lift Dumping: If a larger reduction is needed quickly, a higher volume of cool gas could be injected.

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Final Thoughts

• You can increase lift efficiency by heating hydrogen in normal operations.
• A precise injection system of cooler hydrogen would allow for rapid and reliable lift reduction without venting gas.
• The system could be tuned for small or large lift adjustments, allowing for smooth cargo unloading.

Would you like to simulate this with different temperatures and amounts of gas?

Simulation Results

• Final Temperature After Mixing: ~293.6 K (~20.6°C)
• Ratio of Cool Hydrogen to Hot Hydrogen Required: ~2.76 (i.e., 2.76 times the amount of existing hydrogen needs to be injected as cool hydrogen)
• Percentage of Final Gasbag Volume That Must Be Cool Hydrogen: ~73.4%

Interpretation

• To reduce lift by 10%, you need to inject a significant amount of cool hydrogen—about 73.4% of the final gas mixture.
• This means that if the gasbag initially contained 100 units of hot hydrogen, after injection it would contain 73.4 units of cool hydrogen and 26.6 units of the original heated hydrogen.

Would you like to tweak the parameters or visualize the effect?