/*! 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 24 The heart is supplied with \(5 \... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 24

The heart is supplied with \(5 \mathrm{~L} / \mathrm{mine}\) of blood from the vena cava (assume one blood vessel with a diameter of \(20 \mathrm{~cm}\) ). The aorta discharges the heart and has a diameter of \(26 \mathrm{~cm}\). Determine the pressure drop across the heart if it generates \(4 \mu \mathrm{W}\) during the process into the system. Assume that there is no internal energy in the fluid and that there is a negligible elevation difference between the outflow of the vena cava and the inflow of the aorta. The density of blood is \(1050 \mathrm{~kg} / \mathrm{m}^{3}\).

Short Answer

Expert verified
The pressure drop across the heart is approximately -8.58 Pa.

Step by step solution

01

Calculate the Flow Rates

First, convert the diameters from centimeters to meters. The diameter of the vena cava is 0.2 m and for the aorta is 0.26 m. Now calculate the cross-sectional areas using the formula for the area of a circle,\[A = \pi \left(\frac{d}{2}\right)^2\]For the vena cava:\[A_{vc} = \pi \left(\frac{0.2}{2}\right)^2 = 0.0314 \, m^2\]For the aorta:\[A_{ao} = \pi \left(\frac{0.26}{2}\right)^2 = 0.0531 \, m^2\]Given flow rate in the vena cava: 5 L/min = 5 x 10^-3 m鲁/sCalculate velocity (v) using the flow rate (Q) and area (A):\[v = \frac{Q}{A}\]For the vena cava:\[v_{vc} = \frac{5 \times 10^{-3}}{0.0314} = 0.159 \text{ m/s}\]For the aorta (flow rate remains the same):\[v_{ao} = \frac{5 \times 10^{-3}}{0.0531} = 0.094 \text{ m/s}\]
02

Use Bernoulli's Equation

Assume that there is no change in elevation and that there's no internal energy (head) due to the non-ideal nature of the flow, we only consider energy conversion between pressure and kinetic energy. According to Bernoulli's principle,\[\frac{P_1}{\rho g} + \frac{v_1^2}{2g} = \frac{P_2}{\rho g} + \frac{v_2^2}{2g}\]Rearranging to solve for pressure drop (\( P_1 - P_2 \)), we multiply through by \( g \rho \):\[P_1 - P_2 = \rho \left( \frac{v_2^2 - v_1^2}{2} \right)\]Substitute \( \rho = 1050 \, kg/m^3\), \(v_1 = 0.159 \, m/s\), and \(v_2 = 0.094 \, m/s\):\[P_1 - P_2 = 1050 \left( \frac{(0.094)^2 - (0.159)^2}{2} \right)\]
03

Calculate Power Contribution to Pressure Drop

The heart generates \(4 \mu W = 4 \times 10^{-6} \text{ W}\). This power contributes to the energy of flowing blood. From power, we calculate the equivalent pressure energy, using \(P = \dot{Q} \cdot (P_1 - P_2)\):\[4 \times 10^{-6} = 5 \times 10^{-3} \cdot (P_1 - P_2)\]Isolating \(P_1 - P_2\), we find:\[P_1 - P_2 = \frac{4 \times 10^{-6}}{5 \times 10^{-3}} = 8 \times 10^{-4} \, Pa\]
04

Summing Contributions for Total Pressure Drop

Add the results from steps 2 and 3 together to find the total pressure drop across the heart:\[P_1 - P_2 = 1050 \left( \frac{0.094^2 - 0.159^2}{2} \right) + 8 \times 10^{-4}\] Evaluating this gives:\[P_1 - P_2 = 1050 \left( \frac{0.0088 - 0.0253}{2} \right) + 8 \times 10^{-4}\]\[P_1 - P_2 = 1050 \times -0.00825 + 8 \times 10^{-4}\]\[P_1 - P_2 = -8.6625 + 8 \times 10^{-4} = -8.5825 \, Pa\]
05

Conclusion: Combining Results

The negative pressure result indicates the flowing blood loses more pressure due to velocity changes, but the heart's power (albeit limited) compensates slightly. Therefore, the net pressure drop is the calculated value from step 4, noting the system specifics where elevation and internal energy changes are ignored.

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Key Concepts

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

Bernoulli's Equation
In the world of biofluid mechanics, Bernoulli's Equation plays a pivotal role in understanding blood flow within the cardiovascular system. Bernoulli's Equation provides a relationship among pressure, velocity, and elevation in a fluid. While the heart circulates blood, these parameters are of utmost importance. In the context of the exercise, where blood moves from the vena cava to the aorta, Bernoulli's principle primarily showcases the conservation of energy in a flowing fluid. For this scenario, since elevation differences are negligible and there are no changes to internal energy, the equation simplifies to relate only pressure and velocity changes:
  • Initial and final pressures (P1 and P2), influenced by blood velocity changes.
  • Velocity component represented as kinetic energy of blood flow.
Therefore, according to Bernoulli's principle:\[\frac{P_1}{\rho g} + \frac{v_1^2}{2g} = \frac{P_2}{\rho g} + \frac{v_2^2}{2g}\]This equation helps calculate the pressure difference, which is crucial for ensuring efficient blood circulation.
Pressure Drop
The concept of pressure drop is essential in understanding fluid mechanics and, more specifically, biofluid mechanics within the circulatory system. A pressure drop occurs when blood moves through the cardiovascular system, going from a high-pressure area to a lower pressure one. In the exercise, we calculate the pressure drop (P1 - P2) as blood flows from the vena cava to the aorta:
  • The velocity difference between these two vessels impacts the pressure drop.
  • For biological systems, understanding pressure changes enhances insights into heart function and vascular resistance.
The small calculated negative pressure value (\[-8.5825 \, Pa\]) suggests a loss in pressure due to changes in velocity, heavily confirming the vital role of accompanying heart-generated power aiding blood flow.
Blood Flow Rate
Blood flow rate serves as a crucial parameter in understanding the dynamics of the cardiovascular system. It defines the volume of blood passing a certain point within a specific period, usually expressed in liters per minute (L/min). In our scenario, a flow rate of 5 L/min was provided, representing how much blood the vena cava supplies to the heart.
  • Converting flow rate into m鲁/s is essential for calculation purposes.
  • Flow rates remain consistent through connected parts of the system, e.g., vena cava and aorta.
Given the constant flow rate, understanding velocity in these different-sized vessels helps us calculate key parameters such as pressure drop. Thus, ensuring accurate calculations is critical for understanding circulation efficiency and any potential irregularities in blood distribution.
Cardiovascular System
The cardiovascular system is a complex network responsible for maintaining blood circulation, delivering oxygen, and nutrients to tissues, and removing wastes. It's composed of the heart and a vast array of blood vessels. In this exercise's context, the heart generates power pushing blood from the vena cava through the aorta:
  • Heart function includes volume generation (flow rate) and energy provision (power output).
  • Blood vessels' sizes and characteristics influence pressure and flow within this system.
Understanding fluid dynamics, such as pressure drops across these vessels using biofluid mechanics, can provide insights into heart health and system performance. Analyzing changes, as shown in this problem, can also highlight cardiovascular system efficiency amidst different physiological conditions.

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Most popular questions from this chapter

Air at standard atmosphere conditions \(\left(1 \mathrm{~atm}\right.\) and \(\left.25^{\circ} \mathrm{C}\right)\) enters the lungs at \(50 \mathrm{~cm} / \mathrm{s}\) and leaves at a pressure of \(1.1 \mathrm{~atm}, 37^{\circ} \mathrm{C}\), and a velocity of \(60 \mathrm{~cm} / \mathrm{s}\) (with a constant mass flow rate of \(1.2 \mathrm{~g} / \mathrm{s}\), e.g., density changes). The body removes heat from the lungs at a rate of \(15 \mathrm{~J} / \mathrm{g}\). Calculate the power required by the lungs (this would be energy put gained by the system). Assume that there is no internal energy associated with this system.

The left common coronary artery has an axisymmetric constriction because of a plaque buildup (Fig. 3.30). Given the upstream conditions of a velocity of \(20 \mathrm{~cm} / \mathrm{s}\) (systole) and \(12 \mathrm{~cm} / \mathrm{s}\) (diastole), calculate the velocity at the stenosis throat and the pressure difference between the stenosis throat and the inlet during systole and diastole. Assume that the Bernoulli principle can be used and that the density of blood is \(1050 \mathrm{~kg} / \mathrm{m}^{3}\). Assume that there is no difference in height under these conditions.

A \(1.5-\mathrm{m}\) tall person is standing underwater. First, calculate the amount of force acting on the person, assuming that the pressure at the surface of the water is \(1 \mathrm{~atm}\) and that the person's head is just below the water surface. Second, repeat the calculation but assume that the person's head is \(2 \mathrm{~m}\) below the surface. Approximate the volume of the person as a rectangular prism with width of \(50 \mathrm{~cm}\) and depth of \(20 \mathrm{~cm}\). The density of water is \(1000 \mathrm{~kg} / \mathrm{m}^{3}\).

During systole, blood is ejected from the left ventricle at a velocity of \(125 \mathrm{~cm} / \mathrm{s}\). The diameter of the aortic valve is \(24 \mathrm{~mm}\), and there is no heat transfer or temperature change within the system. Assume that systole lasts for \(0.25 \mathrm{~s}\), that the height difference is \(5 \mathrm{~cm}\), and that there is no change in area within this distance. Determine the amount of work performed by the heart during systole and the power that the heart generates. The density of blood is \(1050 \mathrm{~kg} / \mathrm{m}^{3}\), and we will assume there are no internal energy changes associated with blood over this short time interval.

NASA is planning a mission to a newly found planet and will monitor the density of the new planet's atmosphere. Assume that NASA knows that atmosphere behaves as an ideal gas and that the planet's gravitational force is a function of altitude \(\left(g(z)=\frac{18.7 m}{s^{2}\left(1-\frac{z}{10,000 m}\right)}\right)\), where \(z\) is in m). The temperature of the atmosphere is constant at \(250 \mathrm{~K}\), and the gas constant is \(340 \mathrm{Nm} / \mathrm{kg} \mathrm{K}\). Assume that the pressure at the planet's surface is \(2 \mathrm{~atm}\). Calculate the pressure and density at an altitude of 1,5 , and \(9 \mathrm{~km}\).
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