Problem 22 - Biofluid Mechanics: An Introduction to Fluid Mechanics, Macrocirculation, and Microcirculation

Question

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 \(1000 \mathrm{~kg} / \mathrm{m}^{3}\).

Step-by-Step Solution

Answer

Force at surface: 10132.5 N; Force at 2 m depth: 12044.5 N.

Step 1: Understand the question and pressure formula

We need to find the force acting on a person underwater. Force due to water pressure is given by the formula: \( F = P \times A \), where \( P \) is the pressure at a certain depth and \( A \) is the surface area upon which the pressure acts. The pressure at depth \( h \) is given by \( P = P_0 + \rho gh \), where \( P_0 \) is the atmospheric pressure (\(1 \text{ atm} = 101325 \text{ Pa}\)), \( \rho \) is the density of the fluid (\(1000 \text{ kg/m}^3\)), and \( g \) is the gravitational acceleration (\(9.81 \text{ m/s}^2\)).

Step 2: Calculate surface area of the person

Since the person can be approximated as a rectangular prism, the surface area facing upward can be calculated as \( A = \text{width} \times \text{depth} = 0.5 \text{ m} \times 0.2 \text{ m} = 0.1 \text{ m}^2 \).

Step 3: Calculate pressure at the head underwater at first situation

When the person is just below the water surface, the pressure is simply atmospheric pressure: \( P_1 = 101325 \text{ Pa} \).

Step 4: Calculate force when head is just below water surface

Using the pressure we have just calculated, the force on the top surface can be found by \( F_1 = P_1 \times A = 101325 \text{ Pa} \times 0.1 \text{ m}^2 = 10132.5 \text{ N} \).

Step 5: Calculate pressure at 2 m depth

At a depth of 2 m, the pressure is given by \( P_2 = P_0 + \rho gh = 101325 \text{ Pa} + 1000 \text{ kg/m}^3 \times 9.81 \text{ m/s}^2 \times 2 \text{ m} = 120445 \text{ Pa} \).

Step 6: Calculate force when head is 2 m underwater

Using the calculated pressure for 2 m depth, the force is \( F_2 = P_2 \times A = 120445 \text{ Pa} \times 0.1 \text{ m}^2 = 12044.5 \text{ N} \).

Key Concepts

Fluid Mechanics

In the study of Fluid Mechanics, understanding how fluids exert forces on objects is crucial. When we talk about a fluid, it could be anything from water to air, and each behaves differently under various conditions. One key aspect is how pressure changes within the fluid itself.
Pressure in fluids increases with depth due to the weight of the fluid above. At sea level, the atmospheric pressure is approximately one atmosphere (1 atm), equivalent to 101,325 Pascals (Pa). But as you dive deeper into a fluid like water, this pressure increases due to the weight of the water above pushing down. That's why your ears pop when you swim to the bottom of a pool.
In scenarios where objects or individuals are submerged in water, calculating the variations in pressure is vital. This pressure exerted by static fluid forms the foundation of hydrostatics—a branch of Fluid Mechanics focused on fluids at rest. It's applied broadly in engineering and physics, allowing predictions of behaviors of submerged structures or bodies.

Pressure Force Calculations

When calculating the force exerted by fluid pressure, it's essential to know both the pressure acting on an object and the area over which this pressure is applied. From physics, the force exerted by a fluid is determined by the formula: \[ F = P \times A \]Here, \( F \) stands for force, \( P \) for pressure, and \( A \) for area.
Understanding this requires knowledge of how pressure builds underwater. Pressure not only comes from atmospheric pressure on the fluid surface, \( P_0 \), but also from the weight of the fluid above a given depth, represented by:\[ P = P_0 + \rho gh \]Where \( \rho \) is the fluid density and \( g \) is the acceleration due to gravity (9.81 m/s²).
In the original exercise solution, force calculations underneath the water involve these principles. For example, when a person is standing with their head just below the surface, the pressure mainly reflects atmospheric conditions. But at greater depths, the added pressure from the water column influences the total force acting on the person.

Submerged Surface Area

Understanding submerged surface area is critical when assessing the force on an object in fluid. The concept applies when an object like our person in the exercise is approximated to a simple geometric form—in this case, a rectangular prism.
To calculate the area upon which water exerts pressure, we consider the person's surface projected upwards (or the cross-sectional area). For a geometric shape with known dimensions, such as the given rectangle with a width of 0.5 m and a depth of 0.2 m, the surface area \( A \) equals:\[ A = \text{width} \times \text{depth} = 0.5 \text{ m} \times 0.2 \text{ m} = 0.1 \text{ m}^2 \]
The concept of the submerged surface area is fundamental in calculations as force due to fluid pressure acts perpendicular to the surface. It highlights why accurate geometry understanding and measurement are vital in engineering and physics, impacting various fields from civil to mechanical engineering. Properly calculating these forces ensures structures like dams or underwater labs are built to withstand pressures.