Problem 15 - Biofluid Mechanics: An Introduction to Fluid Mechanics, Macrocirculation, and Microcirculation

Question

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.

Step-by-Step Solution

Answer

The power required by the lungs is approximately 18 W.

Step 1: Identify the given data

The problem provides the following data: initial pressure \( P_1 = 1 \text{ atm} \), initial temperature \( T_1 = 25^{\circ}C \), initial velocity \( V_1 = 50 \text{ cm/s} \), final pressure \( P_2 = 1.1 \text{ atm} \), final temperature \( T_2 = 37^{\circ}C \), final velocity \( V_2 = 60 \text{ cm/s} \), mass flow rate \( \dot{m} = 1.2 \text{ g/s} \) or \( 0.0012 \text{ kg/s} \), and heat removed \( q = 15 \text{ J/g} \) or \( 15000 \text{ J/kg} \). The goal is to calculate the power required by the lungs.

Step 2: Apply the Steady Flow Energy Equation

The steady flow energy equation for this system is: \[ \dot{Q} - \dot{W} = \dot{m} \left( h_2 - h_1 + \frac{V_2^2 - V_1^2}{2} \right) \] where \( \dot{Q} \) is the heat transfer, \( \dot{W} \) is the work done by the system, \( h \) is the specific enthalpy, and \( V \) is the velocity. Since there is no internal energy, only the terms with heat, work, enthalpy, and kinetic energy need to be considered.

Step 3: Convert velocities to basic units

Convert the velocities from cm/s to m/s for consistency: \[ V_1 = 50 \text{ cm/s} = 0.5 \text{ m/s} \] \[ V_2 = 60 \text{ cm/s} = 0.6 \text{ m/s} \] These are necessary for using the equation in basic SI units.

Step 4: Calculate the kinetic energy change term

Find the kinetic energy change using the velocities: \[ \frac{V_2^2 - V_1^2}{2} = \frac{(0.6)^2 - (0.5)^2}{2} = 0.055 \text{ J/kg} \]

Step 5: Calculate the heat transfer term

The heat removed from the system is given as \( q = 15000 \text{ J/kg} \). Since this is removed, it will be negative in energy balance: \[ \dot{Q} = -\dot{m} \times q = -0.0012 \times 15000 = -18 \text{ W} \]

Step 6: Calculate power required using energy equation

Substituting the known values into the energy equation \( \dot{Q} - \dot{W} = \dot{m} \left( \frac{V_2^2 - V_1^2}{2} \right) \), we solve for \( \dot{W} \): \[ -18 - \dot{W} = 0.0012 \times 0.055 \] \[ - \dot{W} = 18 - 0.000066 = 17.999934 \] \[ \dot{W} = 17.999934 \text{ W} \] The power required by the lungs is approximately \( 18 \text{ W} \).

Key Concepts

Steady Flow Energy Equation

The Steady Flow Energy Equation is fundamental in biofluid mechanics, enabling us to analyze the energy changes in a fluid moving through a control volume. The equation considers various energy forms such as heat transfer, work done by the system, changes in enthalpy, and kinetic energy.
In a closed system like the human lungs, fluid flow can be considered steady, meaning that the properties of the system do not change with time. The equation used in the given problem is:\[ \dot{Q} - \dot{W} = \dot{m} \left( h_2 - h_1 + \frac{V_2^2 - V_1^2}{2} \right) \]Where:

• \( \dot{Q} \) is the rate of heat transfer
• \( \dot{W} \) is the work done by the system
• \( \dot{m} \) is the mass flow rate
• \( h \) represents specific enthalpy
• \( V \) is the velocity of the fluid

In this scenario, internal energy changes are neglected, focusing solely on heat, work, enthalpy, and kinetic energy. Understanding how these components interact helps in calculating the metabolic and mechanical processes within biological systems.

Kinetic Energy Calculation

Kinetic energy is a crucial component in understanding the movement of fluids in biological systems. It is determined by the velocity of the fluid as it enters and exits a specific area.
To calculate the kinetic energy change in the lungs, use the formula for kinetic energy change:\[ \frac{V_2^2 - V_1^2}{2} \]Given the final and initial velocities of the air entering and leaving the lungs, convert them into consistent units (from cm/s to m/s):

• Initial velocity, \( V_1 = 0.5 \text{ m/s} \)
• Final velocity, \( V_2 = 0.6 \text{ m/s} \)

Substitute these values into the kinetic energy equation:\[ \frac{(0.6)^2 - (0.5)^2}{2} = 0.055 \text{ J/kg} \]This result represents the change in kinetic energy due to the velocity difference of the air as it moves through the lungs, illustrating how fluid dynamics principles apply to physiological processes.

Power Calculation in Biological Systems

Power calculation is essential to evaluate how the lungs perform energetically. It represents the rate of energy input required by the system, derived from the difference between heat transfer and kinetic energy transformations.
Using the Steady Flow Energy Equation, we solve for the power needed by the lungs. Given the data:

• Heat removed (negative since it's leaving the system): \( \dot{Q} = -18 \text{ W} \)
• Mass flow rate: \( \dot{m} = 0.0012 \text{ kg/s} \)
• Kinetic energy change: \( 0.055 \text{ J/kg} \)

The power \( \dot{W} \) is calculated by rearranging the energy equation:\[-18 - \dot{W} = 0.0012 \times 0.055\]Therefore:\[-\dot{W} = 18 - 0.000066 = 17.999934 \text{ W}\]This shows that the lungs require approximately \( 18 \text{ W} \) of power. This calculation reflects not only biological efficiency but also how mechanical work equates to physiological function within an organism.