/*! 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 37 When a human takes a breath, the... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

When a human takes a breath, the inhaled air flows through the nostrils and trachea before splitting into two primary bronchial tubes. The primary tubes further split to form smaller tubes, and eventually the air passages end in sacs, called alveoli. In the alveoli, oxygen and carbon dioxide are exchanged with the blood. The typical trachea is \(2 \mathrm{cm}\) in diameter; the right primary bronchial tube has a diameter of \(12 \mathrm{mm}\) and that of the left is \(10.0 \mathrm{mm}\). The average adult takes 12 breaths per minute, with each breath taking in about \(0.5 \mathrm{L}\) of air at ambient conditions, which may be taken to be \(1.0 \mathrm{atm}\) and \(25^{\circ} \mathrm{C}\) The velocities of air in the two bronchial tubes are related by the approximation $$\frac{u_{\mathrm{L}}}{u_{\mathrm{R}}}=\left(\frac{D_{\mathrm{R}}}{D_{\mathrm{L}}}\right)^{0.5}$$ where \(u_{L}\) and \(u_{R}\) are velocities in the left and right bronchial tubes and \(D_{L}\) and \(D_{R}\) are the diameters of the left and right tubes, respectively. The temperature of the air in the bronchial tubes may be assumed to have reached \(37^{\circ} \mathrm{C}\). Recognizing that half of a breathing cycle is exhaling, estimate the mass flow rates and the velocities of air flowing through the trachea and each of the primary bronchial tubes.

Short Answer

Expert verified
The mass flow rate of air is approximately \(116 mg/s\) while the velocities of air flowing through the trachea, left bronchial tube, and right bronchial tube are approximately \(0.32 m/s\), \(0.254 m/s\), and \(0.232 m/s\) respectively.

Step by step solution

01

Determine the Mass Flow Rate of Air

The breathing rate is given as 12 breaths per minute, with each breath taking in about \(0.5L\) of air at ambient conditions. Total air breathed in per minute can be calculated as \(12 \times 0.5L\). This volume has to be converted to \(m^{3}\) because the standard SI unit for volume is cubic meter. Hence, the total air breathed in per minute or the volume flow rate is \(12 \times 0.5L \times (10^{-3}m^{3}/L) = 0.006m^{3}/min\). Now the flow of air per second or the volume flow rate \(Q\) during half the cycle (inhalation) is \(0.006m^{3}/min \times (1min/2 \times 60s) = 0.0001m^{3}/s\). Given the temperature is \(37^{\circ}C\), with the use of Ideal Gas equation(\(PV = nRT\) where \(R = 0.0821 atm.L/mol.K\), \(P = 1 atm\) and \(T = 310K\), the molar volume of air \(V_{m} = RT/P = 0.0821 \times 310 / 1 = 25.4 L/mol = 0.0254 m^{3}/mol\), Hence the molar flow rate is \(Q/V_{m} = 0.0001 / 0.0254 = 0.004 mol/s\). Since air is mainly composed of nitrogen and oxygen and the average molar mass can be considered as \(29g/mol\), the mass flow rate of air is \(0.004 mol/s \times 29 g/mol \times (1kg/1000g) = 0.000116 kg/s or 116 mg/s\).
02

Calculate the Velocity of air in the Trachea

The velocity of air in the trachea can be calculated by considering the trachea as a cylindrical tube. The volume flow rate \(Q\) through the trachea can be related to the cross-sectional area \(A_{T}\) of the trachea and the velocity of air \(u_{T}\) by the equation \(Q = A_{T}u_{T}\). \(A_{T}\) is given by \(\pi (D_{T}/2)^2\), where \(D_{T} = 2cm = 0.02m\) which will give \(A_{T} = \pi(0.02/2)^2 = 0.000314 m^{2}\). Substituting this in the continuity equation gives \(u_{T} = Q/A_{T} = 0.0001 / 0.000314 = 0.32 m/s\).
03

Calculate the Velocities of air in the Bronchial Tubes

Applying the same principle to both bronchial tubes, we can use \(Q = A_{i}u_{i}\), where \(i=R, L\) representing the right and left bronchial tubes respectively. Ratio of \(u_{L}\) and \(u_{R}\) is given by: \(u_{L}/u_{R} = (D_{R}/D_{L})^{0.5} = (12/10)^{0.5} = 1.095\). The total volume of air through the bronchial tubes must equal the volume through the trachea i.e. \(u_{L}A_{L} + u_{R}A_{R} = u_{T}A_{T}\). Substituting the earlier relation, we get \(u_{R} = (u_{T}A_{T} / (A_{L} + 1.095A_{R}))\). For \(A_{L} = \pi (D_{L}/2)^2 = \pi(10 \times 10^{-3}/2)^2 = 0.0000785 m^{2}\) and \(A_{R} = \pi(12 \times 10^{-3}/2)^2 = 0.00113 m^{2}\), we find \(u_{R} = 0.232 m/s\) and \(u_{L} = 1.095u_{R} = 0.254 m/s\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Understanding Fluid Mechanics
Fluid mechanics is a branch of physics concerned with the behavior of fluids (liquids, gases, and plasmas) and the forces on them. It encompasses various principles, including how fluids move and interact with their environment. Fluid flow can be characterized by properties such as velocity, pressure, density, and temperature.

When looking at the breathing process, fluid mechanics becomes significant because air, a fluid, moves through the respiratory system. The flow of air through the trachea into the bronchial tubes and eventually into the alveoli, where gas exchange occurs, is a prime example of fluid mechanics at work within the human body. Understanding how the properties of air as a fluid—such as its flow rate, pressure, and temperature—change and interact inside the bronchial tubes is essential for calculating the mass flow rate and velocities in this example.

In practice, these calculations often require applying various equations and laws to predict and determine the behavior of the fluid flow, which will be further elaborated with concepts such as the ideal gas law, volume flow rate, and the continuity equation.
Applying the Ideal Gas Law
The ideal gas law is a fundamental equation that describes the state of an ideal gas. It is often expressed as \(PV = nRT\), where \(P\) represents pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the universal gas constant, and \(T\) is temperature measured in Kelvin.

In fluid mechanics and particularly in our breathing example, the ideal gas law helps us to determine the density and mass of the air being inhaled and exhaled. This is crucial for calculating the mass flow rate of air through the respiratory system. By knowing the ambient conditions, such as temperature and pressure, and the volume of air inhaled, we can relate these variables to find the molar volume of air and subsequently the mass flow rate.

This task requires converting known quantities into appropriate units (such as liters to cubic meters or grams to kilograms) and ensuring that temperature is provided in absolute terms (Kelvin), which helps to ensure accuracy in the calculations and hence, a more educated understanding of the fluid's behavior.
Volume Flow Rate Essentials
Volume flow rate, usually represented by \(Q\), is a measure of the volume of fluid that passes through a given surface per unit time. It's typically expressed in units of cubic meters per second (\(m^3/s\)) in the SI system. The breathing process provides a context for understanding the volume flow rate, as a certain volume of air flows through the respiratory system at a regular pace.

Calculating the volume flow rate is imperative for understanding the quantity of air entering the lungs during each breath, and how quickly it does so. With the given exercise, the number of breaths per minute and the volume inhaled in each breath are used to ascertain the volume flow rate, which is then used to determine the velocity of airflow within the different portions of the bronchial network.

To connect volume flow rate to actual mass flow rate, the ideal gas law is used to first determine the molar flow rate, which, when multiplied by the molar mass of air, provides the mass flow rate. This process underscores the interplay between volume flow rate and mass flow rate, essential for a comprehensive analysis of fluid dynamics in practical applications.
The Continuity Equation Connection
The continuity equation is a fundamental principle in fluid mechanics that describes the conservation of mass in fluid flow. It states that, within a closed system, the mass of fluid entering a portion of the system must equal the mass exiting it. Mathematically, it's often expressed as \(A_1v_1 = A_2v_2\), where \(A\) represents the cross-sectional area and \(v\) denotes the velocity of the fluid at points 1 and 2 along the flow path.

This principle is crucial when analyzing the airflow through the trachea and the bronchial tubes in the human respiratory process. As air moves through different sections of the bronchial tree, its velocity changes in response to the different diameters of the tubes; however, the overall mass flow rate remains constant. The exercise makes use of this concept to solve for the unknown velocities in the left and right bronchial tubes, based on their diameters and the known velocity in the trachea.

Such an application highlights the versatility of the continuity equation and its essential role in predicting fluid behavior in various systems, ensuring that mass conservation is upheld across diverse scenarios.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Steam reforming is an important technology for converting refined natural gas, which we take here to be methane, into a synthesis gas that can be used to produce a varicty of other chemical compounds. For example, consider a reformer to which natural gas and steam are fed in a ratio of 3.5 moles of steam per mole of methane. The reformer operates at 18 atm, and the reaction products leave the reformer in chemical equilibrium at \(875^{\circ} \mathrm{C}\). The steam reforming reaction is $$\mathrm{CH}_{4}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{CO}+3 \mathrm{H}_{2}$$ and the water-gas shift reaction also occurs in the reformer. $$\mathrm{CO}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{CO}_{2}+\mathrm{H}_{2}$$ The equilibrium constants for these two reactions are given by the expressions At \(875^{\circ} \mathrm{C}, K_{\mathrm{R}}=872.9 \mathrm{atm}^{2}\) and \(K \mathrm{w} \mathrm{G}=0.2482 .\) The process is to produce \(100.0 \mathrm{kmol} / \mathrm{h}\) of hydrogen. Calculate the feed rates (kmol/h) of methane and steam and the volumetric flow rate \(\left(\mathrm{m}^{3} / \mathrm{min}\right)\) of gas leaving the reformer.

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?

In froth flotation, air is bubbled through an aqueous solution or slurry to which a foaming agent (soap) has been added. The air-soap bubbles carry finely dispersed solids and hydrophobic materials such as grease and oil to the surface where they can be skimmed off in the foam. An ore-containing slurry is to be processed in a froth flotation tank at a rate of 300 tons/h. The slurry consists of \(20.0 \mathrm{wt} \%\) solids (the ore, \(\mathrm{SG}=1.2\) ) and the remainder an aqueous solution with a density close to that of water. Air is sparged (blown through a nozzle designed to produce small bubbles) into the slurry at a rate of \(40.0 \mathrm{ft}^{3}\) (STP)/1000 gal of slurry. The entry point of the air is 10 \(\mathrm{ft}\) below the slurry surface. The tank contents are at \(75^{\circ} \mathrm{F}\) and the barometric pressure is 28.3 inches of Hg. The sparger design is such that the average bubble diameter on entry is \(2.0 \mathrm{mm}\). (a) What is the volumetric flow rate of the air at its entering conditions? (b) By what percentage does the average bubble diameter change between the entry point and the slurry surface?

A stream of air enters a \(7.50-\mathrm{cm}\) ID pipe with a velocity of \(60.0 \mathrm{m} / \mathrm{s}\) at \(27^{\circ} \mathrm{C}\) and 1.80 bar (gauge). At a point downstream, the air flows through a \(5.00 \mathrm{cm}\) ID pipe at \(60^{\circ} \mathrm{C}\) and 1.53 bar (gauge). What is the average velocity of the gas at this point.

A gas cylinder filled with nitrogen at standard temperature and pressure has a mass of \(37.289 \mathrm{g}\). The same container filled with carbon dioxide at STP has a mass of 37.440 g. When filled with an unknown gas at STP, the container mass is \(37.062 \mathrm{g}\). Calculate the molecular weight of the unknown gas, and then state its probable identity.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.