Understanding the Calculation of Pressure Exerted by Submerged Objects

Pressure exerted by submerged objects is the force per unit area caused by fluid displacement. This calculation is essential in fluid mechanics and engineering.

In this article, you will find detailed formulas, variable explanations, tables of common values, and real-world application examples. Mastering these concepts ensures accurate pressure assessments in submerged environments.

• Calculate the pressure exerted by a steel cube submerged 5 meters underwater.
• Determine the pressure on a submarine hull at 200 meters depth in seawater.
• Find the force exerted by a submerged wooden log of known volume and density.
• Compute the pressure difference between two depths in freshwater for a submerged object.

Comprehensive Tables of Common Values for Pressure Calculations

Fundamental Formulas for Calculating Pressure Exerted by Submerged Objects

Pressure exerted by a fluid at a certain depth is governed primarily by hydrostatic principles. The fundamental formula is:

Expressed in HTML-friendly format:

Where:

• pressure (P) is the hydrostatic pressure in Pascals (Pa) or Newtons per square meter (N/m²).
• density (ρ) is the fluid density in kilograms per cubic meter (kg/m³).
• gravity (g) is the acceleration due to gravity, approximately 9.81 meters per second squared (m/s²).
• depth (h) is the vertical depth below the fluid surface in meters (m).

This formula calculates the pressure exerted by the fluid on the submerged object at depth h. It assumes the fluid is incompressible and static.

Calculating Total Pressure on a Submerged Object

When calculating the total pressure on an object submerged in fluid, atmospheric pressure at the surface must be considered:

• atmospheric_pressure (P₀) is the pressure exerted by the atmosphere at the fluid surface, typically 101,325 Pa at sea level.

This total pressure is critical for engineering applications such as submarine hull design, underwater pipelines, and dam walls.

Buoyant Force and Its Relation to Pressure

The buoyant force acting on a submerged object is derived from the pressure difference between the top and bottom surfaces of the object. Archimedes’ principle states:

Where:

• buoyant_force (F_b) is in Newtons (N).
• density_fluid (ρ) is the fluid density (kg/m³).
• gravity (g) is acceleration due to gravity (m/s²).
• volume_submerged (V) is the volume of the object submerged in the fluid (m³).

This force acts upward, opposing the weight of the object, and is a direct consequence of pressure differences.

Pressure Distribution on Submerged Surfaces

Pressure on submerged surfaces varies linearly with depth. For flat surfaces, the average pressure is at the midpoint depth:

For curved or irregular surfaces, integration of pressure over the surface area is required to find total force.

Detailed Explanation of Variables and Their Typical Values

• Density (ρ): Varies by fluid type. Freshwater is approximately 1000 kg/m³, seawater about 1025 kg/m³, and mercury 13,546 kg/m³. Temperature and salinity affect density.
• Gravity (g): Standard value is 9.81 m/s² but can vary slightly with location on Earth.
• Depth (h): Measured from the fluid surface vertically downward. Depth directly influences pressure linearly.
• Atmospheric Pressure (P₀): Standard atmospheric pressure at sea level is 101,325 Pa but varies with altitude and weather conditions.
• Volume (V): Volume of the submerged portion of the object, critical for buoyant force calculations.

Real-World Applications and Case Studies

Case Study 1: Pressure on a Submerged Steel Cube in Freshwater

A steel cube with side length 0.5 m is submerged 5 meters below the surface of freshwater. Calculate the pressure exerted on one face of the cube and the total force on that face.

Given:

• Density of freshwater, ρ = 1000 kg/m³
• Gravity, g = 9.81 m/s²
• Depth, h = 5 m
• Side length of cube, L = 0.5 m

Step 1: Calculate pressure at 5 m depth

Step 2: Calculate area of one face

Step 3: Calculate force on one face

Result: The pressure on the cube face is 49,050 Pa, and the total force exerted is 12,262.5 Newtons.

Case Study 2: Buoyant Force on a Submerged Wooden Log

A wooden log with volume 0.2 m³ and density 700 kg/m³ is fully submerged in freshwater. Calculate the buoyant force and determine if the log will sink or float.

Given:

• Density of wood, ρ_object = 700 kg/m³
• Density of freshwater, ρ_fluid = 1000 kg/m³
• Gravity, g = 9.81 m/s²
• Volume of log, V = 0.2 m³

Step 1: Calculate buoyant force

Step 2: Calculate weight of the log

Step 3: Compare forces

• Buoyant force (1962 N) > Weight (1373.4 N)
• Net upward force = 1962 – 1373.4 = 588.6 N

Result: The log experiences a net upward force and will float when submerged.

Additional Considerations for Accurate Pressure Calculations

• Fluid Compressibility: At great depths, fluid density may increase due to compression, affecting pressure calculations.
• Temperature Effects: Temperature variations alter fluid density and viscosity, influencing pressure and buoyancy.
• Dynamic Pressure: Moving fluids exert additional pressure components; Bernoulli’s equation may be required.
• Surface Tension: For small objects, surface tension can affect pressure distribution.
• Non-Uniform Shapes: Complex geometries require integration of pressure over the surface area for total force.

Authoritative Resources for Further Study

• Engineering Toolbox: Hydrostatic Pressure
• Encyclopedia Britannica: Pressure (Physics)
• NIST: Hydrostatic Pressure Standards
• Woods Hole Oceanographic Institution: Fluid Mechanics

Understanding the calculation of pressure exerted by submerged objects is fundamental for engineers, scientists, and professionals working with fluids. Accurate pressure assessment ensures safety, efficiency, and reliability in underwater structures, vessels, and fluid systems.

By mastering the formulas, variables, and real-world applications detailed here, you can confidently analyze and design systems involving submerged objects under various fluid conditions.