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Chapter 11: Problem 65

The blood speed in a normal segment of a horizontal artery is \(0.11 \mathrm{m} / \mathrm{s} .\) An abnormal segment of the artery is narrowed down by an arteriosclerotic plaque to one-fourth the normal cross-sectional area. What is the difference in blood pressures between the normal and constricted segments of the artery?

Short Answer

Expert verified
The pressure difference is approximately 96.05 Pa.

Step by step solution

01

Identify Given Values and Conditions

The speed of blood in the normal segment of the artery is given as \( v_1 = 0.11 \, \text{m/s} \). The cross-sectional area of the constricted segment is one-fourth of the normal area, so if the normal area is \( A_1 \), then the constricted area \( A_2 = \frac{1}{4} A_1 \).
02

Relate Velocities Using Continuity Equation

Using the Continuity Equation for fluids: \( A_1 v_1 = A_2 v_2 \). Since \( A_2 = \frac{1}{4} A_1 \), this simplifies to \( A_1 \times 0.11 = \frac{1}{4} A_1 \times v_2 \). Solving for \( v_2 \), we get \( v_2 = 4 \times 0.11 = 0.44 \, \text{m/s} \).
03

Apply Bernoulli's Equation

Bernoulli's equation states: \( P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2 \), where \( \rho \) is the fluid density, and \( P_1 \) and \( P_2 \) are pressures at the normal and constricted segments respectively. We need to find the pressure difference: \( P_1 - P_2 \).
04

Calculate Pressure Difference

Rewriting Bernoulli's equation for the pressure difference: \( P_1 - P_2 = \frac{1}{2} \rho (v_2^2 - v_1^2) \). Substituting \( v_1 = 0.11 \, \text{m/s} \) and \( v_2 = 0.44 \, \text{m/s} \), we compute \( v_2^2 - v_1^2 = 0.44^2 - 0.11^2 = 0.1936 - 0.0121 = 0.1815 \). Thus, \( P_1 - P_2 = \frac{1}{2} \rho \times 0.1815 \).
05

Assume Average Blood Density

Assume the average blood density \( \rho \) is approximately \( 1060 \, \text{kg/m}^3 \). Substitute it into the pressure difference equation to get \( P_1 - P_2 = \frac{1}{2} \times 1060 \times 0.1815 \approx 96.045 \, \text{Pa} \).

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Key Concepts

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

Continuity Equation
The Continuity Equation is a fundamental principle in fluid dynamics, particularly applicable to the study of blood flow in arteries. It states that the mass flow rate of a fluid must remain constant from one cross-section to another in a steady flow. This means that for any segment of a fluid system, the product of the cross-sectional area (A) and the fluid velocity (v) is a constant: \[ A_1 v_1 = A_2 v_2 \] Here, the subscript denotes different segments of a fluid stream.
  • Conservation of Mass: This law ensures that as the area decreases, the velocity of the fluid must increase, assuming the density remains constant.
  • Application to Arteries: In the context of the given problem, as blood flows through an artery that narrows due to a plaque, the velocity must increase in the constricted section (following the continuity equation).
  • This relationship is essential for calculating changes in velocity when the cross-sectional area changes, as we converted in the exercise.
For practical scenarios, this can help predict how changes in arterial sections affect blood flow dynamics.
Fluid Dynamics
Fluid dynamics is the branch of physics concerned with the behavior of liquids and gases in motion. It plays a crucial role in understanding how blood moves through arteries, especially when normal flow is disrupted by abnormalities. Some key points include:
  • Steady vs. Unsteady Flow: Blood flow in arteries is typically steady, meaning it doesn鈥檛 change with time. This is crucial for applying equations like Bernoulli鈥檚 principle.
  • Incompressible Flow: Blood is often treated as an incompressible fluid. This assumption simplifies the mathematics involved as its density remains constant throughout flow.
  • Viscosity: Blood is a viscous fluid, which means it has internal friction when it travels through vessels. This friction must be considered when analyzing flow resistance and pressure changes.
These principles form the foundation for solving complex medical and engineering problems related to the circulatory system.
Pressure Difference in Arteries
Understanding the pressure difference across different segments of an artery is vital for diagnosing cardiovascular issues. Pressure differences can indicate blockages or other circulatory problems. The exercise helpful illustrates this using Bernoulli鈥檚 equation in a practical context.
  • Bernoulli's Principle: It relates pressure, kinetic energy per unit volume, and potential energy per unit volume of a fluid, often simplified in medical physics as \( P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2 \).
  • Application: Using the principle, the problem calculated the pressure difference created by a constriction in an artery.
  • Significance: Such calculations help in assessing cardiac risk and planning treatments by understanding how much additional pressure the heart needs to exert to maintain blood flow.
Analyzing pressure differences allows for a deeper understanding of how various factors affect blood flow, which is crucial in medical diagnosis and treatment.

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Most popular questions from this chapter

To verify her suspicion that a rock specimen is hollow, a geologist weighs the specimen in air and in water. She finds that the specimen weighs twice as much in air as it does in water. The density of the solid part of the specimen is \(5.0 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3} .\) What fraction of the specimen's apparent volume is solid?

The human lungs can function satisfactorily up to a limit where the pressure difference between the outside and inside of the lungs is one-twentieth of an atmosphere. If a diver uses a snorkel for breathing, how far below the water can she swim? Assume the diver is in salt water whose density is \(1025 \mathrm{kg} / \mathrm{m}^{3}.\)

Multiple-Concept Example 8 presents an approach to problems of this kind. The hydraulic oil in a car lift has a density of \(8.30 \times 10^{2} \mathrm{kg} / \mathrm{m}^{3}\). The weight of the input piston is negligible. The radii of the input piston and output plunger are \(7.70 \times 10^{-3} \mathrm{m}\) and \(0.125 \mathrm{m},\) respectively. What input force \(F\) is needed to support the \(24500-\mathrm{N}\) combined weight of a car and the output plunger, when (a) the bottom surfaces of the piston and plunger are at the same level, and (b) the bottom surface of the output plunger is \(1.30 \mathrm{m}\) above that of the input piston?

A lost shipping container is found resting on the ocean floor and completely submerged. The container is \(6.1 \mathrm{m}\) long, \(2.4 \mathrm{m}\) wide, and \(2.6 \mathrm{m}\) high. Salvage experts attach a spherical balloon to the top of the container and inflate it with air pumped down from the surface. When the balloon's radius is \(1.5 \mathrm{m},\) the shipping container just begins to rise toward the surface. What is the mass of the container? Ignore the mass of the balloon and the air within it. Do not neglect the buoyant force exerted on the shipping container by the water. The density of seawater is \(1025 \mathrm{kg} / \mathrm{m}^{3}\)

A solid concrete block weighs 169 N and is resting on the ground. Its dimensions are \(0.400 \mathrm{m} \times 0.200 \mathrm{m} \times 0.100 \mathrm{m} .\) A number of identical blocks are stacked on top of this one. What is the smallest number of whole blocks (including the one on the ground) that can be stacked so that their weight creates a pressure of at least two atmospheres on the ground beneath the first block?
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