Problem 24
Question
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 \(1050 \mathrm{~kg} / \mathrm{m}^{3}\).
Step-by-Step Solution
Verified Answer
The pressure drop across the heart is approximately -8.58 Pa.
1Step 1: 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 m³/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):\[v_{ao} = \frac{5 \times 10^{-3}}{0.0531} = 0.094 \text{ m/s}\]
2Step 2: 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 \(v_2 = 0.094 \, m/s\):\[P_1 - P_2 = 1050 \left( \frac{(0.094)^2 - (0.159)^2}{2} \right)\]
3Step 3: 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:\[P_1 - P_2 = \frac{4 \times 10^{-6}}{5 \times 10^{-3}} = 8 \times 10^{-4} \, Pa\]
4Step 4: 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:\[P_1 - P_2 = 1050 \left( \frac{0.0088 - 0.0253}{2} \right) + 8 \times 10^{-4}\]\[P_1 - P_2 = 1050 \times -0.00825 + 8 \times 10^{-4}\]\[P_1 - P_2 = -8.6625 + 8 \times 10^{-4} = -8.5825 \, Pa\]
5Step 5: 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.
Key Concepts
Bernoulli's EquationPressure DropBlood Flow RateCardiovascular System
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.
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.
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 m³/s is essential for calculation purposes.
- Flow rates remain consistent through connected parts of the system, e.g., vena cava and aorta.
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.
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