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Chapter 11: Problem 70

In the human body, blood vessels can dilate, or increase their radii, in response to various stimuli, so that the volume flow rate of the blood increases. Assume that the pressure at either end of a blood vessel, the length of the vessel, and the viscosity of the blood remain the same, and determine the factor \(R_{\text {dilated }} / R_{\text {normal }}\) by which the radius of a vessel must change in order to double the volume flow rate of the blood through the vessel.

Short Answer

Expert verified
The radius must increase by a factor of \(\sqrt[4]{2} \approx 1.1892\).

Step by step solution

01

Understanding the Relationship

The volume flow rate of blood flowing through a vessel is determined by Poiseuille's Law, which is given by the formula: \[Q = \frac{\pi \, \Delta P \, R^4}{8 \, \eta \, L}\]where \(Q\) is the volume flow rate, \(\Delta P\) is the pressure difference, \(R\) is the radius of the vessel, \(\eta\) is the viscosity of the blood, and \(L\) is the length of the vessel. Since \(\Delta P\), \(\eta\), and \(L\) are constant, the flow rate \(Q\) is proportional to the fourth power of the radius \(R\). This implies \(Q \propto R^4\).
02

Solving for the New Radius

We want the volume flow rate \(Q\) to double, i.e., \(Q_{\text{new}} = 2Q_{\text{normal}}\). Using the proportionality \(Q \propto R^4\), we can set up the following equation for the normal and dilated radii: \[R_{\text{dilated}}^4 = 2 \, R_{\text{normal}}^4\]
03

Calculating the Radius Ratio

To find the factor by which the radius changes, take the fourth root of both sides of the equation:\[R_{\text{dilated}} = \sqrt[4]{2} \, R_{\text{normal}}\]Thus, the ratio of the dilated radius to the normal radius is:\[\frac{R_{\text{dilated}}}{R_{\text{normal}}} = \sqrt[4]{2}\]
04

Conclusion

The factor by which the radius of the vessel must increase to double the volume flow rate is \(\sqrt[4]{2}\). Numerically, \(\sqrt[4]{2} \approx 1.1892\), indicating a roughly 18.92% increase in the radius.

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Key Concepts

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

Blood Flow
Blood flow is a critical concept in the study of the cardiovascular system. It refers to the movement of blood through the vessels, which is essential for delivering nutrients and oxygen to, and removing waste products from, tissues throughout the body. The efficiency of blood circulation is influenced by several factors, especially the diameter or radius of the blood vessels. According to Poiseuille's Law, the volume flow rate of blood can be expressed as:
  • \( Q = \frac{\pi \, \Delta P \, R^4}{8 \, \eta \, L} \)
Here, \( Q \) is the volume flow rate, \( \Delta P \) represents the pressure difference, \( R \) is the radius of the vessel, \( \eta \) is the blood viscosity, and \( L \) is the length of the vessel.

This equation highlights that blood flow is highly sensitive to changes in the radius of the vessel. A small increase in radius leads to a markedly greater blood flow due to the fourth power relationship. This means that if a vessel's radius is increased, blood flow increases significantly, which is vital for regulating how much blood reaches different parts of the body.
Vessel Dilation
Vessel dilation is the process by which blood vessels enlarge, increasing their radius. This physiological adjustment can occur in response to various stimuli, such as exercise, heat, or changes in metabolic demand. Within the context of Poiseuille's Law, vessel dilation plays a crucial role in enhancing blood flow. If other factors like pressure difference and blood viscosity are held constant, increasing the vessel's radius is the most effective way to boost blood flow without increasing heart workload.
  • When a vessel dilates, its radius \( R \) is multiplied by a factor, which in turn increases the blood flow without a proportional increase in pressure.
  • This process is particularly important during physical exertion when muscles require more oxygen.
The equation given in the exercise, \( \frac{R_{\text{dilated}}}{R_{\text{normal}}} = \sqrt[4]{2} \), shows that a small increase of approximately 18.92% in vessel radius can double blood flow. This illustrates the body's efficient response to increased demand for oxygen and nutrients.
Viscosity
Viscosity refers to the thickness or internal resistance to flow within the blood. It is a key factor in determining how easily blood can move through the circulatory system. Higher viscosity implies greater resistance to flow, making it harder for the heart to pump blood efficiently. Conversely, lower viscosity means blood moves more readily, facilitating improved flow dynamics.

In Poiseuille's Law:
  • Viscosity is represented by \( \eta \), directly influencing the denominator of the formula \( Q = \frac{\pi \, \Delta P \, R^4}{8 \, \eta \, L} \).
  • This means that reductions in blood viscosity can enhance flow by decreasing the resistance blood experiences as it navigates through vessels.
While viscosity was held constant in the original exercise, understanding its role is crucial when considering broader physiological contexts. Modifications in blood viscosity, through changes in temperature or hydration levels, for example, can also significantly affect blood flow efficiency.

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Most popular questions from this chapter

Poiseuille's law remains valid as long as the fluid flow is laminar. For sufficiently high speed, however, the flow becomes turbulent, even if the fluid is moving through a smooth pipe with no restrictions. It is found experimentally that the flow is laminar as long as the Reynolds number Re is less than about 2000: \(\mathrm{Re}=2 \bar{v} \rho R / \eta .\) Here \(\bar{v}, \rho,\) and \(\eta\) are, respectively, the average speed, density, and viscosity of the fluid, and \(R\) is the radius of the pipe. Calculate the highest average speed that blood \(\left(\rho=1060 \mathrm{~kg} / \mathrm{m}^{3}, \eta=4.0 \times 10^{-3} \mathrm{~Pa} \cdot \mathrm{s}\right)\) could have and still remain in laminar flow when it flows through the aorta \(\left(R=8.0 \times 10^{-3} \mathrm{~m}\right)\).

A Venturi meter is a device that is used for measuring the speed of a fluid within a pipe. The drawing shows a gas flowing at speed \(v_{2}\) through a horizontal section of pipe whose cross-sectional area is \(A_{2}=0.0700 \mathrm{~m}^{2} .\) The gas has a density of \(\rho=1.30 \mathrm{~kg} / \mathrm{m}^{3} .\) The Venturi meter has a cross-sectional area of \(A_{1}=0.0500 \mathrm{~m}^{2}\) and has been substituted for a section of the larger pipe. The pressure difference between the two sections is \(P_{2}-P_{1}=120 \mathrm{~Pa}\). Find (a) the speed \(v_{2}\) of the gas in the larger original pipe and (b) the volume flow rate \(Q\) of the gas.

A hydrometer is a device used to measure the density of a liquid. It is a cylindrical tube weighted at one end, so that it floats with the heavier end downward. It is contained inside a large "medicine dropper," into which the liquid is drawn using a squeeze bulb (see the drawing). For use with your car, marks are put on the tube so that the level at which it floats indicates whether the liquid is battery acid (more dense) or antifreeze (less dense). (a) Compared to the weight \(W\) of the tube, how much buoyant force is needed to make the tube float in either battery acid or antifreeze? (b) Is a greater volume of battery acid or a greater volume of antifreeze displaced by the hydrometer to provide the necessary buoyant force? (c) Which mark is farther up from the bottom of the tube? Justify your answers.

A person who weighs \(625 \mathrm{~N}\) is riding a \(98-\mathrm{N}\) mountain bike. Suppose the entire weight of the rider and bike is supported equally by the two tires. If the gauge pressure in each tire is \(7.60 \times 10^{5} \mathrm{~Pa}\), what is the area of contact between each tire and the ground?

One of the concrete pillars that support a house is \(2.2 \mathrm{~m}\) tall and has a radius of \(0.50 \mathrm{~m}\). The density of concrete is about \(2.2 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\). Find the weight of this pillar in pounds \((1 \mathrm{~N}=0.2248 \mathrm{lb})\).
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