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Chapter 9: Problem 81

The approximate diameter of the aorta is \(0.50 \mathrm{~cm}\); that of a capillary is \(10 \mu \mathrm{m}\). The approximate average blood flow speed is \(1.0 \mathrm{~m} / \mathrm{s}\) in the aorta and \(1.0 \mathrm{~cm} / \mathrm{s}\) in the capillaries. If all the blood in the aorta eventually flows through the capillaries, estimate the number of capillaries in the circulatory system.

Short Answer

Expert verified
The estimated number of capillaries in the circulatory system is given by the formula \( n = \frac{R_{aorta}}{A_{capillary} \times V_{capillary}} \). Substitute the known values and calculate n. Be sure to use appropriate units before calculating.

Step by step solution

01

Calculate the Area of the Aorta and Capillaries

First, calculate the cross-sectional area of both the aorta and the capillary using the formula for the area of a circle: \( A = \pi r^{2} \). Given the diameters, the radius is half the diameter. So, for the Aorta \( A_{aorta} = \pi (0.25 cm)^{2} \) and for the Capillary \( A_{capillary} = \pi (5 \mu m)^{2} \). But remember to convert the radii to appropriate units before calculating. Here, convert cm to m and 碌m to m.
02

Calculate the Flow Rate in the Aorta

The flow rate is calculated by multiplying the cross-sectional area by the speed of the flow. In this case, the flow rate in the aorta \( R_{aorta} = A_{aorta} \times V_{aorta} \), where \( V_{aorta} = 1.0 m/s \) is the speed of the blood flow in the aorta. Remember to use SI units before multiplying.
03

Calculate the Number of Capillaries

Using the principle of continuity, the total flow rate in the aorta is equal to the total flow rate in the capillaries combined. Hence, \( R_{aorta} = n \times A_{capillary} \times V_{capillary} \), where n is the total number of capillaries we want to find and \( V_{capillary} = 1.0 cm/s \). Simplify this equation to solve for n: \( n = \frac{R_{aorta}}{A_{capillary} \times V_{capillary}} \). Substitute the known values and calculate n to get the estimated number of capillaries in the circulatory system.

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Key Concepts

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

Aorta
The aorta is the largest artery in the human body, playing a crucial role in the circulatory system. It is responsible for carrying oxygen-rich blood from the heart to the rest of the body.
The diameter of the aorta is approximately 0.50 cm, which translates into a radius of 0.25 cm.
  • This large size allows it to handle high pressures as blood is ejected from the heart.
  • The wide diameter reduces resistance, ensuring efficient blood flow.
The aorta's substantial diameter compared to other blood vessels helps distribute blood quickly, making it an essential component in fluid dynamics studies within the human body.
Capillaries
Capillaries are the smallest blood vessels in the body, with a diameter of about 10 micrometers (\(10 \mu \text{m} \)). These tiny vessels are where the exchange of nutrients, gases, and waste products occurs between blood and tissues.
Despite their small size, capillaries are numerous, forming an extensive network throughout the body.
  • This massive network ensures sufficient surface area for exchange processes.
  • Their thin walls facilitate the rapid diffusion of substances.
Since they have a much smaller diameter compared to the aorta, the blood flow speed is also reduced, which allows more time for the exchange of substances.
Blood Flow Rate
Blood flow rate in the circulatory system is a key concept in fluid dynamics, indicating the volume of blood passing through a vessel per unit time. It is influenced by the cross-sectional area of the vessel and the velocity of blood flow.
In the aorta, the blood flow speed is approximately 1.0 m/s, while in the capillaries, it drops to 1.0 cm/s.
  • The flow rate in any section of the circulatory system can be calculated using the formula: \( R = A \times V \), where \( R \) is the flow rate, \( A \) is the cross-sectional area, and \( V \) is the flow speed.
  • A higher speed in the aorta is due to its larger diameter, facilitating rapid transport of blood.
  • Lower speed in the capillaries ensures an adequate time for exchange processes.
Understanding blood flow rate is essential for calculating the number of capillaries that can handle the blood circulating from the aorta.
Continuity Equation
The continuity equation is a fundamental principle in fluid dynamics, stating that the flow rate must remain constant from one cross-section of a vessel to another.This principle is applied in the circulatory system to ensure the same volume of blood that enters the capillaries via the aorta exits without loss.
  • For the aorta and capillaries, it is expressed as \( R_{\text{aorta}} = n \times A_{\text{capillary}} \times V_{\text{capillary}} \).
  • This equation helps determine the number \( n \) of capillaries by rearranging it to solve for \( n \).
  • The continuity equation is crucial for analyzing how changes in vessel size and flow velocity affect the overall blood distribution.
By utilizing the continuity equation, one can effectively estimate complex biological structures like the extensive capillary network.

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Most popular questions from this chapter

A large balloon of mass \(226 \mathrm{~kg}\) is filled with helium gas until its volume is \(325 \mathrm{~m}^{3}\). Assume the density of air is \(1.29 \mathrm{~kg} / \mathrm{m}^{3}\) and the density of helium is \(0.179 \mathrm{~kg} / \mathrm{m}^{3}\). (a) Draw a force diagram for the balloon. (b) Calculate the buoyant force acting on the balloon. (c) Find the net force on the balloon and determine whether the balloon will rise or fall after it is released. (d) What maximum additional mass can the balloon support in equilibrium? (e) What happens to the balloon if the mass of the load is less than the value calculated in part (d)? (f) What limits the height to which the balloon can rise?

A high-speed lifting mechanism supports an \(800-\mathrm{kg}\) object with a steel cable that is \(25.0 \mathrm{~m}\) long and \(4.00 \mathrm{~cm}^{2}\) in cross-sectional area. (a) Determine the elongation of the cable. (b) By what additional amount does the cable increase in length if the object is accelerated upward at a rate of \(3.0 \mathrm{~m} / \mathrm{s}^{2} ?\) (c) What is the greatest mass that can be accelerated upward at \(3.0 \mathrm{~m} / \mathrm{s}^{2}\) if the stress in the cable is not to exceed the elastic limit of the cable, which is \(2.2 \times 10^{8} \mathrm{~Pa}\) ?

The viscous force on an oil drop is measured to be equal to \(3.0 \times 10^{-13} \mathrm{~N}\) when the drop is falling through air with a speed of \(4.5 \times 10^{-4} \mathrm{~m} / \mathrm{s}\). If the radius of the drop is \(2.5 \times 10^{-6} \mathrm{~m}\), what is the viscosity of air?

Suppose a distant world with surface gravity of \(7.44 \mathrm{~m} / \mathrm{s}^{2}\) has an atmospheric pressure of \(8.04 \times 10^{4} \mathrm{~Pa}\) at the surface. (a) What force is exerted by the atmosphere on a disk-shaped region \(2.00 \mathrm{~m}\) in radius at the surface of a methane ocean? (b) What is the weight of a \(10.0-\mathrm{m}\) deep cylindrical column of methane with radius \(2.00 \mathrm{~m}\) ? (c) Calculate the pressure at a depth of \(10.0 \mathrm{~m}\) in the methane ocean. Note: The density of liquid methane is \(415 \mathrm{~kg} / \mathrm{m}^{3}\).

Old Faithful geyser in Yellowstone Park erupts at approximately 1-hour intervals, and the height of the fountain reaches \(40.0 \mathrm{~m}\) (Fig. P9.57). (a) Consider the rising stream as a series of separate drops. Analyze the free-fall motion of one of the drops to determine the speed at which the water leaves the ground. (b) Treat the rising stream as an ideal fluid in streamline flow. Use Bernoulli's equation to determine the speed of the water as it leaves ground level. (c) What is the pressure (above atmospheric pressure) in the heated underground chamber \(175 \mathrm{~m}\) below the vent? You may assume the chamber is large compared with the geyser vent.
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