/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 32 A balloon \(20 \mathrm{m}\) in d... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 32

A balloon \(20 \mathrm{m}\) in diameter is filled with helium at a gauge pressure of 2.0 atm. A man is standing in a basket suspended from the bottom of the balloon. A restraining cable attached to the basket kecps the balloon from rising. The balloon (not including the gas it contains), the basket, and the man have a combined mass of \(150 \mathrm{kg}\). The temperature is \(24^{\circ} \mathrm{C}\) that day, and the barometer reads \(760 \mathrm{mm} \mathrm{Hg}\) (a) Calculate the mass (kg) and weight (N) of the helium in the balloon. (b) How much force is exerted on the balloon by the restraining cable? (Recall: The buoyant force on a submerged object equals the weight of the fluid- -in this case, the air- -displaced by the object. Neglect the volume of the basket and its contents.) (c) Calculate the initial acceleration of the balloon when the restraining cable is released. (d) Why does the balloon eventually stop rising? What would you need to know to calculate the altitude at which it stops? (e) Suppose at its point of suspension in midair the balloon is heated, raising the temperature of the helium. What happens and why?

Short Answer

Expert verified
Answers are numerical in nature and would vary depending on the values obtained at each step. For (d), the balloon stops rising due to decrease in air density and the subsequent decrease in buoyant force at higher altitudes. For (e), heating the helium causes an increase in the volume of the balloon, lowering the density of helium inside and causing the balloon to rise again due to increased buoyant force.

Step by step solution

01

Part (a) - Calculate the mass and weight of helium

1. The volume of the balloon can be calculated using the formula for volume of a sphere which is given by \(V = 4/3 * \pi * (D/2)^3\) where \(D = 20 m\) is the diameter of the balloon. 2. The total pressure on the helium inside the balloon is the external pressure (atmospheric pressure) plus the gauge pressure inside the balloon. Convert all pressures in atm to Pascal's (Pa) for consistent units. 3. Use the ideal gas law \(PV = nRT\) to find the number of moles of helium in the balloon, where \(R = 8.314 \, J/(mol \cdot K)\) is the universal gas constant and \(T\) is the absolute temperature in Kelvin (which can be found by adding 273.15 to the Celsius temperature).4. The mass (kg) of helium can be calculated from the number of moles using the formula \(mass = n \cdot molar \, mass\), given the molar mass of helium is 4 g/mol (but remember to convert this to kg/mol for consistency). 5. The weight of the helium can be calculated by multiplying its mass by the acceleration due to gravity \( 9.8 \, m/s^2\).
02

Part (b) - Force on the balloon by the restraining cable

1. The buoyant force can be calculated using Archimedes principle. It is equal to the weight of the air displaced by the balloon. 2. The density of air at sea level and at \(24^{\circ} \mathrm{C}\) can be approximated to be \(1.2 \, kg/m^3\). The mass of air displaced by the balloon is density of air times the volume of the balloon.3. The weight of the air is the mass times acceleration due to gravity.4. The total weight of the balloon system (balloon + helium + man and basket) is the sum of the weight of the balloon, man, and basket and the weight of the helium from part (a).5. The force on the balloon by the restraining cable is equal to the total weight of the balloon system subtracted from the buoyant force.
03

Part (c) - Initial acceleration of the balloon

1. When the restraining cable is released, the net force on the balloon is the buoyant force subtracted from the total weight of the balloon system.2. Using Newton's second law, the acceleration can be calculated as \(a = F_{net}/ m_{total}\) where \(F_{net}\) is the net force and \(m_{total}\) is the total mass of the balloon system.
04

Part (d) - Reason the balloon stops rising and altitude calculation

1. The balloon eventually stops rising because as the balloon rises, the atmosphere gets thinner (density decreases) and the buoyant force decreases. At some point, the buoyant force becomes equal to the weight of the balloon system, creating a condition of equilibrium and the balloon stops rising.2. To calculate the altitude at which it stops, information about how air density varies with altitude is required 鈥� this could be obtained from standard atmosphere tables.
05

Part (e) - Effect of heating the balloon

1. When the helium is heated, it expands causing the volume of the balloon to increase rather than rise in height.2. The density of helium inside the balloon decreases because the mass remains constant while volume increases.3. If the total weight doesn't change significantly, the buoyant force compared to weight increases due to the larger volume, causing the balloon to rise again.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Ideal Gas Law
The ideal gas law is a cornerstone in chemical engineering education, providing insight into the behavior of gases under various conditions. It is commonly presented as the equation \(PV = nRT\), where P represents pressure, V stands for volume, n is the number of moles of gas, R is the universal gas constant, and T is the absolute temperature.

The law allows engineers to predict how a gas will change when subjected to pressure, temperature, or volume changes, and is crucial when calculating the mass and weight of the helium in a balloon, as seen in the textbook exercise. By rearranging the equation to solve for n, we find the number of moles of gas, which can then be multiplied by the gas's molar mass to obtain its mass. This is critical for understanding not only the balloon's behavior but also various industrial processes involving gases.

It's important to remember, however, that the ideal gas law assumes a perfect scenario where gas molecules don't interact with each other. In reality, gases can deviate from this 'ideal' behavior, particularly at high pressures or low temperatures.
Buoyant Force in Action
Buoyant force is the upward force exerted on an object that is immersed in a fluid, whether it鈥檚 a liquid or a gas. This force is crucial for the operation of buoyant objects, from hot air balloons to submarines.

As applied in the textbook problem, the buoyant force is what keeps the balloon afloat, as it counteracts the pull of gravity on the weight of the displaced fluid, which in this case is air. The buoyant force can be calculated by determining the volume of the object in the fluid and multiplying it by the fluid's density and the acceleration due to gravity (\( g \)).

Through understanding this concept, students can appreciate the delicate balance between the weight of the balloon and the buoyant force, which is essential for maintaining the balloon's altitude, or for analyzing its ascent or descent when the balance is disrupted.
Archimedes Principle Explained
Archimedes' principle is a principle of buoyancy that states that the upward buoyant force exerted on a body immersed in a fluid is equal to the weight of the fluid that the body displaces. In terms of formula, it can be expressed as \(F_{\text{buoyant}} = \rho_{fluid} \times V_{\text{displaced}} \times g\), where \(F_{\text{buoyant}}\) is the buoyant force, \(\rho_{fluid}\) is the density of the fluid, \(V_{\text{displaced}}\) is the displaced fluid's volume, and \(g\) is the acceleration due to gravity.

In the context of our balloon scenario, the air displaced by the balloon can be considered the 'fluid'. Once we calculate the volume of the balloon and know the density of air, we can determine the buoyant force on the balloon. Students can use this principle to solve for unknowns such as force or displaced volume in a variety of applications beyond just balloons, further emphasizing its importance in the understanding of fluid mechanics.
Newton's Second Law in Everyday Applications
Newton's second law connects force, mass, and acceleration in the eloquent formula \(F = ma\), where F stands for force, m is mass, and a is acceleration. This law is fundamental in physics and chemical engineering as it describes the motion of objects when an unbalanced force is applied.

In the scenario presented in the exercise, when the restraining cable of the balloon is released, we are observing Newton's second law in action. The net force on the balloon (buoyant force minus the total weight of the balloon system) determines the balloon's initial acceleration as it starts to rise. This law helps students understand how objects in motion behave and allows engineers to design systems that must adhere to these universal principles.

Newton's second law also underpins why the balloon stops rising. As elevation increases, atmospheric density decreases, reducing the buoyant force until it equals the weight of the balloon system, reaching equilibrium and halting ascent. To calculate the altitude where this occurs would require knowledge about the change in air density with altitude.

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Most popular questions from this chapter

Ammonia is one of the chemical constituents of industrial waste that must be removed in a treatment plant before the waste can safely be discharged into a river or estuary. Ammonia is normally present in wastewater as aqueous ammonium hydroxide \(\left(\mathrm{NH}_{4}^{+} \mathrm{OH}^{-}\right) .\) A two- part process is frequently carried out to accomplish the removal. Lime (CaO) is first added to the wastewater, leading to the reaction $$\mathrm{CaO}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{Ca}^{2+}+2\left(\mathrm{OH}^{-}\right)$$ The hydroxide ions produced in this reaction drive the following reaction to the right, resulting in the conversion of ammonium ions to dissolved ammonia: $$\mathrm{NH}_{4}^{+}+\mathrm{OH}^{-}=\mathrm{NH}_{3}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})$$ Air is then contacted with the wastewater, stripping out the ammonia. (a) One million gallons per day of alkaline wastewater containing 0.03 mole \(\mathrm{NH}_{3} /\) mole ammoniafree \(\mathrm{H}_{2} \mathrm{O}\) is fed to a stripping tower that operates at \(68^{\circ} \mathrm{F}\). Air at \(68^{\circ} \mathrm{F}\) and 21.3 psia contacts the wastewater countercurrently as it passes through the tower. The feed ratio is \(300 \mathrm{ft}^{3}\) air/gal wastewater, and 93\% of the ammonia is stripped from the wastewater. Calculate the volumetric flow rate of the gas leaving the tower and the partial pressure of ammonia in this gas. (b) Briefly explain in terms a first-year chemistry student could understand how this process works. Include the equilibrium constant for the second reaction in your explanation. (c) This problem is an illustration of challenges associated with addressing undesirable releases into the environment; namely, in developing a process to prevent dumping ammonia into a waterway, the release is instead made to the atmosphere. Suppose you are to write an article for a newspaper on the installation of the process described in the beginning of this problem. Explain why the company is installing the two-part process, and then explain the ultimate fate of the ammonia. Take one of two positions - either that the release is harmless or that it jeopardizes the environment in the vicinity of the plant. since this is a newspaper article, it cannot be more than 800 words.

The volume of a dry box (a closed chamber with dry nitrogen flowing through it) is \(2.0 \mathrm{m}^{3}\). The dry box is maintained at a slight positive gauge pressure of \(10 \mathrm{cm} \mathrm{H}_{2} \mathrm{O}\) and room temperature \(\left(25^{\circ} \mathrm{C}\right) .\) If the contents of the box are to be replaced every five minutes, calculate the required mass flow rate of nitrogen in \(g / \min\) by (a) direct solution of the ideal-gas equation of state and (b) conversion from standard conditions. You may assume the gas in the dry box is well mixed.

A natural gas contains 95 wt\% \(\mathrm{CH}_{4}\) and the balance \(\mathrm{C}_{2} \mathrm{H}_{6}\). Five hundred cubic meters per hour of this gas at \(40^{\circ} \mathrm{C}\) and 1.1 bar is to be burned with \(25 \%\) excess air. The air flowmeter is calibrated to read the volumetric flow rate at standard temperature and pressure. What should the meter read (in SCMH) when the flow rate is set to the desired value?

Air in industrial plants is subject to contamination by many different chemicals, and companies must monitor ambient levels of hazardous species to be sure they are below limits specified by the National Institute for Occupational Safety and Health (NIOSH). In personal breathing-zone sampling (as opposed to area sampling), workers wear devices that periodically collect air samples less than 10 inches away from their noses. Breathing-zone sampling and analysis methods for hundreds of species are set forth in the NIOSH Manual of Analytical Methods. \(^{13}\) For benzene, NIOSH specifies a recommended exposure limit (REL) of 0.1 ppm time-weighted average exposure (TWA), and the Occupational Safety and Health Administration (OSHA) permissible exposure limit (PEL) is 1.0ppm TWA. A worker in a petrolcum refinery has a personal breathing-zone sampler for benzenc clipped to her shirt collar. Following the NIOSH prescription, air is pumped through the sampler at a rate of \(0.200 \mathrm{L} / \mathrm{min}\) by a small battery-operated pump attached to the worker's belt. The sampler contains an adsorbent that removes essentially all of the benzene from the air passing through it. After several hours, the sampler is removed and sent to a lab for analysis, and the worker puts on a fresh sampler. On a particular day when the temperature is \(21^{\circ} \mathrm{C}\) and barometric pressure is \(730 \mathrm{mm}\) Hg, samples are collected during a 4-h period before lunch and a 3.5-h period after lunch. The analytical laboratory reports \(0.17 \mathrm{mg}\) of benzene in the first sample and \(0.23 \mathrm{mg}\) in the second. (a) Calculate the average benzene concentration, \(C_{\mathrm{B}}(\mathrm{ppm}),\) in the worker's breathing zone during each sampling period, where 1 ppm = 1 mol C \(_{6} \mathrm{H}_{6} / 10^{6}\) mol air. (b) The worker's TWA is the average concentration of benzene in her breathing zone during the eight hours of her shift. It is calculated by multiplying \(C_{\mathrm{B}}\) in each sampling period by the time of that period, summing the products over all periods during the shift, and dividing by the total time of the shift. Assume that the worker's exposure during the unsampled 30 minutes was zero, and calculate her TWA. (c) If the worker's exposure is above the recommended limits, what actions might the company take?

The current global reliance on fossil fuels for heating, transportation, and electric power generation raises concems regarding the release of \(\mathrm{CO}_{2}\) and \(\mathrm{CH}_{4},\) which are greenhouse gases thought to lead to climate change, and NO, which contributes to smog. One potential solution to these problems is to produce transportation fuels from renewable biomass. You have been asked to evaluate a proposed process for converting forest residues to alcohols that may be used as transportation fuels. In the first stage of the process, steam and dry wood from hybrid poplar trees (which grow between five and eight feet a year and can be harvested roughly every five years) are fed to a gasifier in which the biomass is converted to light gases in the following reactions: $$\begin{aligned} \mathrm{C}+\mathrm{H}_{2} \mathrm{O} & \rightarrow \mathrm{CO}+\mathrm{H}_{2} \\\ \mathrm{CO}+\mathrm{H}_{2} \mathrm{O} & \rightarrow \mathrm{CO}_{2}+\mathrm{H}_{2} \\ \mathrm{C}+\mathrm{CO}_{2} & \rightarrow 2 \mathrm{CO} \\ \mathrm{C}+2 \mathrm{H}_{2} & \rightarrow \mathrm{CH}_{4} \\ \mathrm{CH}_{4}+\mathrm{H}_{2} \mathrm{O} & \rightarrow \mathrm{CO}+3 \mathrm{H}_{2} \end{aligned}$$ The effluents from the reactor are a gas stream containing \(\mathrm{H}_{2}, \mathrm{CO}, \mathrm{CO}_{2}, \mathrm{CH}_{4},\) and \(\mathrm{H}_{2} \mathrm{O},\) and a solid char stream that contains only carbon and hydrogen. The char is discarded and the gases go through additional steps in which the hydrogen and carbon monoxide are converted to mixed alcohols. This problem only concerns the gasifier. \(\cdot\) Elemental composition of biomass: 51.9 mass \(\%\) C \(, 6.3 \%\) H, and \(41.8 \%\) O \(\cdot\) Pressure and temperature of entering steam: \(155^{\circ} \mathrm{C}, 4.4 \mathrm{atm}\) \(\cdot\) Feed ratio of steam to biomass: 1.1 kg steam/kg biomass \(\cdot\) Yield and dry-basis composition of product gas: 1.35 kg dry gas/kg biomass at \(700^{\circ} \mathrm{C}, 1.2\) atm; 50.7 mol\% \(\mathrm{H}_{2}, 23.8 \%\) CO, \(18.0 \% \mathrm{CO}_{2}, 7.5 \% \mathrm{CH}_{4}\) (a) Taking a basis of \(100 \mathrm{kg}\) of biomass fed, draw and completely label a flowchart for the gasifier incorporating the given data, labeling the volumes of the steam fed and the gases produced. Perform a degree-of-freedom analysis. (b) Calculate the mass and mass composition of the char and the volumes of the steam feed and product gas streams. (c) List advantages and possible drawbacks of using biomass rather than petroleum as a fuel source.
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