Problem 10
Question
During peak systole, the heart delivers to the aorta blood that has a velocity of \(100 \mathrm{~cm} / \mathrm{s}\) at a pressure of \(120 \mathrm{mmHg}\). The aortic root has a mean diameter of \(25 \mathrm{~mm}\). Determine the force acting on the aortic arch if the conditions at the outlet are a pressure of \(110 \mathrm{mmHg}\) and a diameter of \(21 \mathrm{~mm}\) (Fig. 3.27). The density of blood is \(1050 \mathrm{~kg} / \mathrm{m}^{3}\). Assume that the aorta is rigid and that blood is incompressible. Ignore the weight of the blood vessel and the weight of blood within the blood vessel. FIG. \(3.27\) Figure for Homework Problem 3.10.
Step-by-Step Solution
Verified Answer
The force acting on the aortic arch is calculated using Bernoulli's equation, considering pressure differences and velocity changes.
1Step 1: Convert Units
First, we need to convert all units to the standard SI units for consistency. - The diameter of the aorta at the root is \(25\, \text{mm} = 0.025\, \text{m}\).- The diameter of the aorta at the outlet is \(21\, \text{mm} = 0.021\, \text{m}\).- The velocity of blood is \(100\, \text{cm/s} = 1\, \text{m/s}\). - Convert pressures from mmHg to Pascals using the conversion factor: \(1 \text{ mmHg} = 133.322 \text{ Pa}\). Thus, \(120 \text{ mmHg} = 15998.64 \text{ Pa}\) and \(110 \text{ mmHg} = 14665.42 \text{ Pa}\).
2Step 2: Calculate Cross-sectional Areas
Calculate the cross-sectional areas of the aortic root and the outlet using the formula for the area of a circle, \(A = \pi r^2\).- For the aortic root, the radius \(r\) is \(0.0125\, \text{m}\), so the area \(A_1 = \pi (0.0125)^2\approx 4.91 \times 10^{-4} \text{ m}^2\).- For the outlet, the radius \(r\) is \(0.0105\, \text{m}\), so the area \(A_2 = \pi (0.0105)^2 \approx 3.46 \times 10^{-4} \text{ m}^2\).
3Step 3: Apply Bernoulli's Equation
Use Bernoulli's equation to find the force acting on the aortic arch:\[P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2\]Given:- \(P_1 = 15998.64 \text{ Pa}\)- \(P_2 = 14665.42 \text{ Pa}\)- \(\rho = 1050 \text{ kg/m}^3\)- \(v_1 = 1 \text{ m/s}\)- Calculate \(v_2\) by noting volume flow rate and equating \(A_1v_1 = A_2v_2\):\[v_2 = \frac{A_1}{A_2}v_1 \approx \frac{4.91 \times 10^{-4}}{3.46 \times 10^{-4}}(1) \approx 1.42 \text{ m/s}\]
4Step 4: Calculate Forces Using Bernoulli’s Equation
Rearrange Bernoulli's equation to solve for the force due to pressure:\[F = P_1A_1 - P_2A_2 = \frac{1}{2}\rho(A_2v_2^2 - A_1v_1^2)\]Substitute the known values:\[F = (P_1A_1 - P_2A_2) + \frac{1}{2}\rho (A_2v_2^2 - A_1v_1^2)\]Substitute and calculate the terms, computing the force exerted due to pressure difference and velocity changes.
Key Concepts
Blood Flow DynamicsAortic Arch ForceVelocity and Pressure in Blood Vessels
Blood Flow Dynamics
Blood flow dynamics form an essential aspect of understanding how blood moves through the circulatory system. This involves analyzing the flow, pressure, and velocity of blood within vessels. These three components are interlinked closely as they determine how effectively blood can deliver oxygen and nutrients to tissues.
One way to analyze blood flow dynamics is through Bernoulli's equation. This equation helps us understand the relationship between blood pressure, kinetic energy, and gravitational potential. For blood, while the potential energy usually remains constant due to short distances in vessels, pressure and velocity are critical.
One way to analyze blood flow dynamics is through Bernoulli's equation. This equation helps us understand the relationship between blood pressure, kinetic energy, and gravitational potential. For blood, while the potential energy usually remains constant due to short distances in vessels, pressure and velocity are critical.
- Blood flow is generally continuous, which allows for the application of the continuity equation: the flow rate remains the same throughout a closed system.
- The change in diameter along the aorta causes variations in velocity and pressure, as blood moves from the heart through to various parts of the body.
Aortic Arch Force
The force acting on the aortic arch is a key concept in cardiovascular physiology. It results from the pressure and velocity changes as blood moves through the aorta from the heart.
To determine this force, one would typically apply Bernoulli's equation, which accounts for energy conservation along a streamline. This approach considers:
This includes calculating the pressure differential force \(P_1A_1 - P_2A_2\) along with the velocity dependent component \(\frac{1}{2}\rho(A_2v_2^2 - A_1v_1^2)\).
By summing these components, the resultant force reveals how the heart's pumping action and constriction or dilation of the vessels affects the aortic arch.
To determine this force, one would typically apply Bernoulli's equation, which accounts for energy conservation along a streamline. This approach considers:
- The kinetic energy term given by \(\frac{1}{2}\rho v^2\) which represents the energy associated with the motion of the blood.
- The pressure energy term \(P\) that accounts for the force exerted by the blood on vessel walls.
This includes calculating the pressure differential force \(P_1A_1 - P_2A_2\) along with the velocity dependent component \(\frac{1}{2}\rho(A_2v_2^2 - A_1v_1^2)\).
By summing these components, the resultant force reveals how the heart's pumping action and constriction or dilation of the vessels affects the aortic arch.
Velocity and Pressure in Blood Vessels
Understanding the relationship between velocity and pressure in blood vessels is fundamental in cardiovascular mechanics. These two parameters are inversely related due to the conservation of energy and the continuity equation.
In a narrowed portion of a vessel, like at the outlet point of the aorta:
In a narrowed portion of a vessel, like at the outlet point of the aorta:
- The area \(A\) decreases, leading to an increase in velocity \(v\) because the flow rate must remain constant (continuity equation).
- As velocity increases, the kinetic energy component in Bernoulli's equation also increases, resulting in a decrease in pressure energy.
- Hence, narrower vessels see higher velocities but lower pressures.
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