/*! 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

During your summer vacation, you plan an epic adventure trip to scale Mt. Kilimanjaro in Tanzania. Dehydration is a great danger on such a climb, and it is essential to drink enough water to make up for the amount you lose by breathing. (a) During your pre-trip physical, your physician measured the average flow rate and composition of the gas you exhaled (expired air) while performing light activity. The results were \(11.36 \mathrm{L} / \mathrm{min}\) at body temperature (37^) C) and 1 atm, 17.08 mole\% oxygen, 3.25\% carbon dioxide, 6.12 mole\% \(\mathrm{H}_{2} \mathrm{O},\) and the balance nitrogen. The ambient (inspired) air contained 1.67 mole\% water and a negligible amount of carbon dioxide. Calculate the rate of mass lost through the breathing process (kg/day) and the volume of water in liters you would have to drink per day just to replace the water lost in respiration. Consider your lungs to be a continuous steady-state system, with input streams being inspired air and water and \(\mathrm{CO}_{2}\) transferred from the blood and output streams being expired air and \(\mathrm{O}_{2}\) transferred to the blood. Assume no nitrogen is transferred to or from the blood. (b) You made the trip to Tanzania and completed the climb to Uhuru Peak, the summit of Kilimanjaro, at an altitude of 5895 meters above sea level. The ambient temperature and pressure there averaged \(-9.4^{\circ} \mathrm{C}\) and \(360 \mathrm{mm} \mathrm{Hg},\) and the air contained \(0.46 \mathrm{mole} \%\) water. The molar flow rate of your expired air was roughly the same as it had been at sea level, and the expired air contained \(14.86 \% \mathrm{O}_{2}\) \(3.80 \% \mathrm{CO}_{2},\) and \(13.20 \% \mathrm{H}_{2} \mathrm{O} .\) Calculate the rate of mass lost (g/day) through breathing and water you would have to drink (L/day) just to replace the water lost in respiration. (c) The equality of the molar flow rates of expired air at sea level and at Uhuru Peak is due to a cancellation of effects, one of which would tend to increase the rate at higher altitudes and the other to decrease it. What are those effects? (Hint: Use the ideal-gas equation of state in your solution, and think about how the oxygen concentration at a high altitude would likely affect your breathing rate.)

The oxidation of nitric oxide $$\mathrm{NO}+\frac{1}{2} \mathrm{O}_{2} \rightleftharpoons \mathrm{NO}_{2}$$ takes place in an isothermal batch reactor. The reactor is charged with a mixture containing 20.0 volume percent NO and the balance air at an initial pressure of \(380 \mathrm{kPa}\) (absolute). (a) Assuming ideal-gas behavior, determine the composition of the mixture (component mole fractions) and the final pressure (kPa) if the conversion of NO is 90\%. (b) Suppose the pressure in the reactor eventually equilibrates (levels out) at \(360 \mathrm{kPa}\). What is the equilibrium percent conversion of NO? Calculate the reaction equilibrium constant at the prevailing temperature, \(K_{p}\left[(\mathrm{atm})^{-0.5}\right]\), defined as $$K_{p}=\frac{\left(p_{\mathrm{NO}_{2}}\right)}{\left(p_{\mathrm{NO}}\right)\left(p_{\mathrm{O}_{2}}\right)^{0.5}}$$ where \(p_{i}(\mathrm{atm})\) is the partial pressure of species \(i\left(\mathrm{NO}_{2}, \mathrm{NO}, \mathrm{O}_{2}\right)\) at equilibrium. (c) Assuming that \(K_{\mathrm{p}}\) depends only on temperature, estimate the final pressure and composition in the reactor if the feed ratio of NO to \(\mathrm{O}_{2}\) and the initial pressure are the same as in \(\operatorname{Part}(\) a), but the feed to the reactor is pure \(\mathrm{O}_{2}\) instead of air. (d) Replace the partial pressures in the expression for \(K_{\mathrm{p}}\), and use the result to explain how reactor pressure influences the conversion of NO to \(\mathrm{NO}_{2}\)

Magnesium sulfate has a number of uses, some of which are related to the ability of the anhydrate form to remove water from air and others based on the high solubility of the heptahydrate \(\left(\mathrm{MgSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}\right)\) form, also known as Epsom salt. The densities of the anhydrate and heptahydrate crystalline forms are 2.66 and \(1.68 \mathrm{g} / \mathrm{mL},\) respectively. Suppose you wish to form a 20.0 wt\% \(\mathrm{MgSO}_{4}\) aqueous solution by simply pouring crystals of one of the forms into a tank of water while the temperature is held constant at \(30^{\circ} \mathrm{C}\). The specific gravity of the 20.0 wt\% solution at \(30^{\circ} \mathrm{C}\) is \(1.22 .\) Answer the following questions for both forms of the \(\mathrm{MgSO}_{4}\) crystals: (a) What volume of water should be in the tank before crystals are added if the final product is to be 1000 kg of the 20 wt\% solution? (b) Suppose the tank diameter is \(0.30 \mathrm{m}\). What is the height of liquid in the tank before the crystals are added? (c) What is the height of the water in the tank after addition of the crystals but before they begin to dissolve? (d) What is the height of liquid in the tank after all the MgSO \(_{4}\) has dissolved?

A fuel cell is an electrochemical device that reacts hydrogen with oxygen from the air to produce water and DC electricity. A proposed application is replacement of the gasoline-fueled internal combustion engine in an automobile with a \(100 \mathrm{kW}\) fuel cell. You are on a summer internship with a gas supplier planning to transport hydrogen to service stations for use in cars powered by fuel cells. The hydrogen is to be transported in tube trailers, each of which has 10 tubes of length \(10.5 \mathrm{m}\) and diameter \(0.56 \mathrm{m}\). Hydrogen in the tubes at 2600 psig and an average temperature of \(298 \mathrm{K}\) is discharged at service stations to a final pressure of 55 psig. Refueling cach fuel-cell-powered automobile is estimated to require 4.0 kg of hydrogen. (a) You and your office-mate- an intern from a different university - have been asked to estimate the number of automobiles that can be refueled by one tube-trailer load of hydrogen. He does a very quick calculation and comes up with a value of 95 cars. Speculate how he did it and provide support for your speculation. What was his mistake? (b) Do the calculation using the SRK equation of state. Instead of using Eqs. \(5.3-11\) and \(5.3-13\) for the parameter \(\alpha,\) use the following correlation developed specifically for hydrogen: \(^{23}\) \(\alpha=1.202 \exp \left(-0.3228 T_{\mathrm{r}}\right)\) (c) Do the calculation using the law of corresponding states. (d) In which of the three estimates would you have the greatest confidence, and why?

The bacteria acetobacter aceti convert ethanol to acetic acid in the presence of oxygen according to the reaction $$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}+\mathrm{O}_{2} \rightarrow \mathrm{CH}_{3} \mathrm{COOH}+\mathrm{H}_{2} \mathrm{O}$$ In a continuous fermentation process, ethanol enters the top of the fermenter at a rate of \(145 \mathrm{kg} / \mathrm{h}\), and the air fed to the bottom of the fermenter is \(25 \%\) in excess of the amount required to consume all of the ethanol. A gas stream containing nitrogen and unreacted oxygen leaves the top of the fermenter, and a liquid stream containing acetic acid, water, and \(10 \%\) of the entering ethanol leaves the bottom. Assume that none of the ethanol, water, and acetic acid in the reactor is vaporized. The fermenter operates at \(30^{\circ} \mathrm{C},\) maintains a liquid \((\mathrm{SG}=0.95)\) height of \(4.5 \mathrm{m},\) and is open to the atmosphere (i.e., the pressure at the top of the fermenter is 1 atm). (a) What is the volumetric flow rate of air as it enters the bottom of the fermenter? What is the volumetric flow rate of gas leaving the top of the fermenter? (b) Assume a linear relationship between the fraction of oxygen reacted and the position of gas bubbles rising through the liquid in the fermenter: for example, half of the oxygen reacted is consumed in the bottom half of the fermenter. At the vertical midpoint of the fermenter, the average bubble diameter is \(1.5 \mathrm{mm}\). What is the average bubble diameter at the entry point of the air and as the gas leaves the liquid at the top of the fermenter?
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