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Chapter 13: Problem 47

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

Expert verified
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

01

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 \(W_{\text{air}} = (2.000 \cdot 10^5 \mathrm{~m}^3)(1.205 \mathrm{~kg/m^3})(9.81 \mathrm{~m/s^2})\)
02

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 \(W_{\mathrm{H}} = (2.000 \cdot 10^5 \mathrm{~m}^3)(0.08988 \mathrm{~kg/m^3})(9.81 \mathrm{~m/s^2})\)
03

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: \(F_B = W_{\text{air}} - W_{\mathrm{H}}\)
04

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#
05

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 \(W_{\mathrm{He}} = (2.000 \cdot 10^5 \mathrm{~m}^3)(0.1786 \mathrm{~kg/m^3})(9.81 \mathrm{~m/s^2})\)
06

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: \(F_{\text{B,He}} = W_{\text{air}} - W_{\mathrm{He}}\)
07

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 \(U_{\text{He}} = W_{\text{structure}} + F_{\text{B,He}}\)
08

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.

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Most popular questions from this chapter

In a horizontal water pipe that narrows to a smaller radius, the velocity of the water in the section with the smaller radius will be larger. What happens to the pressure? a) The pressure will be the same in both the wider and narrower sections of the pipe. b) The pressure will be higher in the narrower section of the pipe. c) The pressure will be higher in the wider section of the pipe d) It is impossible to tell.

A water pipe narrows from a radius of \(r_{1}=5.00 \mathrm{~cm}\) to a radius of \(r_{2}=2.00 \mathrm{~cm} .\) If the speed of the water in the wider part of the pipe is \(2.00 \mathrm{~m} / \mathrm{s}\), what is the speed of the water in the narrower part?

The average density of the human body is \(985 \mathrm{~kg} / \mathrm{m}^{3}\) and the typical density of seawater is about \(1020 \mathrm{~kg} / \mathrm{m}^{3}\) a) Draw a free-body diagram of a human body floating in seawater and determine what percentage of the body's volume is submerged. b) The average density of the human body, after maximum inhalation of air, changes to \(945 \mathrm{~kg} / \mathrm{m}^{3}\). As a person floating in seawater inhales and exhales slowly, what percentage of his volume moves up out of and down into the water? c) The Dead Sea (a saltwater lake between Israel and Jordan ) is the world's saltiest large body of water. Its average salt content is more than six times that of typical seawater, which explains why there is no plant and animal life in it. Two-thirds of the volume of the body of a person floating in the Dead Sea is observed to be submerged. Determine the density (in \(\mathrm{kg} / \mathrm{m}^{3}\) ) of the seawater in the Dead Sea.

A very large balloon with mass \(M=10.0 \mathrm{~kg}\) is inflated to a volume of \(20.0 \mathrm{~m}^{3}\) using a gas of density \(\rho_{\text {eas }}=\) \(0.20 \mathrm{~kg} / \mathrm{m}^{3}\). What is the maximum mass \(m\) that can be tied to the balloon using a \(2.00 \mathrm{~kg}\) piece of rope without the balloon falling to the ground? (Assume that the density of air is \(1.30 \mathrm{~kg} / \mathrm{m}^{3}\) and that the volume of the gas is equal to the volume of the inflated balloon).

Water of density \(998.2 \mathrm{~kg} / \mathrm{m}^{3}\) is moving at negligible speed under a pressure of \(101.3 \mathrm{kPa}\) but is then accelerated to high speed by the blades of a spinning propeller. The vapor pressure of the water at the initial temperature of \(20.0^{\circ} \mathrm{C}\) is \(2.3388 \mathrm{kPa}\). At what flow speed will the water begin to boil? This effect, known as cavitation, limits the performance of propellers in water
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