Question

The Hindenburg, the German zeppelin that caught fire in 1937 while docking in Lakehurst, New Jersey, was a rigid duralumin-frame balloon filled with \(2.000 \cdot 10^{5} \mathrm{~m}^{3}\) of hydrogen. The Hindenburg's useful lift (beyond the weight of the zeppelin structure itself) is reported to have been \(1.099 \cdot 10^{6} \mathrm{~N}(\) or \(247,000 \mathrm{lb}) .\) Use \(\rho_{\text {air }}=1.205 \mathrm{~kg} / \mathrm{m}^{3}, \rho_{\mathrm{H}}=\) \(0.08988 \mathrm{~kg} / \mathrm{m}^{3}\) and \(\rho_{\mathrm{He}}=0.1786 \mathrm{~kg} / \mathrm{m}^{3}\) a) Calculate the weight of the zeppelin structure (without the hydrogen gas). b) Compare the useful lift of the (highly flammable) hydrogen-filled Hindenburg with the useful lift the Hindenburg would have had had it been filled with (nonflammable) helium, as originally planned.

Short answer

Question: Compare the useful lifts of hydrogen-filled and helium-filled Hindenburg. Answer: The useful lift with hydrogen gas is \(1.099 \cdot 10^6\mathrm{~N}\). The useful lift with helium gas can be calculated using the formula \(U_{\text{He}} = W_{\text{structure}} + F_{\text{B,He}}\). By comparing the useful lifts of both gases, we can determine the difference in lifting capabilities between hydrogen and helium.

Step-by-step solution

Calculate the weight of air displaced by the hydrogen gas

We'll first find the weight of the air that has been displaced by the hydrogen gas using the given volume of gas, \(\mathrm{V} = 2.000 \cdot 10^5 \mathrm{~m}^3\), and the density of air (\(\rho_{\text{air}}\)): Weight of air displaced (\(W_{\text{air}}\)) = Volume × Density of air × Gravity \(W_{\text{air}} = (2.000 \cdot 10^5 \mathrm{~m}^3)(1.205 \mathrm{~kg/m^3})(9.81 \mathrm{~m/s^2})\)

Calculate the weight of hydrogen gas

Now, we will calculate the weight of the hydrogen gas using the given volume and density of hydrogen (\(\rho_{\mathrm{H}}\)): Weight of hydrogen gas (\(W_{\mathrm{H}}\)) = Volume × Density of hydrogen × Gravity \(W_{\mathrm{H}} = (2.000 \cdot 10^5 \mathrm{~m}^3)(0.08988 \mathrm{~kg/m^3})(9.81 \mathrm{~m/s^2})\)

Calculate the buoyant force with hydrogen gas

To find the buoyant force (\(F_B\)) acting on the hydrogen gas, we will subtract the weight of hydrogen gas from the weight of the displaced air: \(F_B = W_{\text{air}} - W_{\mathrm{H}}\)

Calculate the weight of the zeppelin structure

The given useful lift of the Hindenburg is \(1.099 \cdot 10^6\mathrm{~N}\). To find the weight of the zeppelin structure, we will simply subtract the buoyant force from the useful lift: Weight of zeppelin structure = Useful lift - Buoyant force \(W_{\text{structure}} = (1.099 \cdot 10^6\mathrm{~N}) - F_B\) Now we move to part (b) of the exercise. #b) Comparing useful lifts of hydrogen-filled and helium-filled Hindenburg#

Calculate the weight of helium gas

Using the given volume of gas and the density of helium (\(\rho_{\mathrm{He}}\)), we will calculate the weight of the helium gas: Weight of helium gas (\(W_{\mathrm{He}}\)) = Volume × Density of helium × Gravity \(W_{\mathrm{He}} = (2.000 \cdot 10^5 \mathrm{~m}^3)(0.1786 \mathrm{~kg/m^3})(9.81 \mathrm{~m/s^2})\)

Calculate the buoyant force with helium gas

To find the buoyant force (\(F_{\text{B,He}}\)) acting on the helium gas, we will subtract the weight of helium gas from the weight of the displaced air: \(F_{\text{B,He}} = W_{\text{air}} - W_{\mathrm{He}}\)

Calculate the useful lift with helium gas

We can calculate the useful lift of the helium-filled Hindenburg by adding the buoyant force of helium to the weight of the zeppelin structure: Useful lift with helium = Weight of zeppelin structure + Buoyant force with helium \(U_{\text{He}} = W_{\text{structure}} + F_{\text{B,He}}\)

Compare the useful lifts

Finally, we can compare the useful lift with hydrogen gas (\(1.099 \cdot 10^6\mathrm{~N}\)) and the useful lift with helium gas (\(U_{\text{He}}\)) to see the difference.

Key concepts

Density

The concept of density is essential in understanding buoyancy and how gases like hydrogen and helium affect objects such as zeppelins. Density is the mass of a substance per unit volume, expressed as \( \rho = \frac{m}{V} \), where \( \rho \) is the density, \( m \) is the mass, and \( V \) is the volume. In the context of the Hindenburg, we look at the density of air, hydrogen, and helium to understand the buoyant force, which is the ability of the gas to make the zeppelin float.

- **Air density:** The density of air is approximately \( 1.205 \mathrm{~kg/m^3} \). This value is crucial as the air's density helps us determine how much lift a zeppelin can achieve when filled with lighter gases.- **Hydrogen density:** Hydrogen is much lighter, with a density of \( 0.08988 \mathrm{~kg/m^3} \), making it a very attractive option for lifting objects because it displaces a greater volume of air for its mass.- **Helium density:** Helium, another light gas, has a density of \( 0.1786 \mathrm{~kg/m^3} \). Though not as light as hydrogen, helium is non-flammable, offering a safer alternative for zeppelins at the cost of slightly reduced lift.

In the exercise, by knowing these densities, we calculate the weights and buoyant forces necessary to assess how much lift each gas can provide.

Hydrogen vs Helium

Choosing between hydrogen and helium as lifting gases involves a balance between efficiency and safety. Let's explore the characteristics and trade-offs of each gas:

- **Hydrogen:** Despite being highly flammable, hydrogen was favored historically for its low density, giving it a strong lifting power. The Hindenburg utilized hydrogen because it offered a greater useful lift due to displacing more air with less mass compared to helium. This makes it a potent lifting agent, which is why the Hindenburg's lift was approximately \( 1.099 \cdot 10^6 \mathrm{~N} \) using hydrogen.- **Helium:** Although helium's density is about twice that of hydrogen, making it less capable of lifting the same weight, it bears the advantage of being non-flammable. Initially, the Hindenburg was planned to use helium to ensure safety. Helium offers a reliable, albeit less efficient, lift compared to hydrogen. In our exercise, determining the lift capacity with helium helps understand safety versus efficiency trade-offs.

When comparing useful lifts, the choice becomes clear that while hydrogen offers unparalleled lift, helium's safety is crucial for passenger and public safety.

Zeppelin Structure Weight

Understanding the weight of the zeppelin structure is crucial in evaluating how much additional load it can carry. The structure of a zeppelin comprises materials like duralumin, a lightweight yet sturdy metal, critical in determining the overall lifting capacity.

- **Weight Calculation:** In the exercise, the useful lift was stated as \( 1.099 \cdot 10^6 \mathrm{~N} \). By calculating the buoyant force generated by hydrogen, and subtracting it from the given useful lift, we can deduce the weight of the zeppelin itself. This calculation is a vital step as it allows us to understand how much weight the vessel can carry apart from itself.- **Impact on Lift:** The zeppelin structure's weight inherently affects the load it can carry. A heavier structure reduces the payload capacity, making the need for lightweight construction materials paramount to maximize carriage potential.

Accurate assessment of the structure's weight and the comparison between hydrogen and helium's lift capabilities is essential in aeronautics, ensuring that designs meet requirements for both efficiency and safety.