Problem 89 - Fundamentals of Physics

Question

For seawater of density \(1.03 \mathrm{~g} / \mathrm{cm}^{3}\), find the weight of water on top of a submarine at a depth of \(255 \mathrm{~m}\) if the horizontal cross-sectional hull area is \(2200.0 \mathrm{~m}^{2} .(\mathrm{b})\) In atmospheres, what water pressure would a diver experience at this depth?

Step-by-Step Solution

Answer

The weight of the water is approximately 5,650,000,900 N. The water pressure experienced by the diver is approximately 25.35 atm.

Step 1: Find the Pressure on the Submarine

To calculate the pressure at the depth of 255 m, we use the formula for pressure in a fluid: \( P = \rho \cdot g \cdot h \), where \( \rho = 1.03 \text{ g/cm}^3 = 1030 \text{ kg/m}^3 \), \( g = 9.8 \text{ m/s}^2 \), and \( h = 255 \text{ m} \). Calculate the pressure: \( P = 1030 \times 9.8 \times 255 = 2,568,660 \text{ N/m}^2 \).

Step 2: Calculate the Weight of the Water on the Submarine

Weight (force due to pressure) can be determined by multiplying pressure by area: \( F = P \cdot A \). Here, pressure \( P = 2,568,660 \text{ N/m}^2 \) and area \( A = 2200.0 \text{ m}^2 \). So, \( F = 2,568,660 \times 2200.0 = 5,650,000,900 \text{ N} \).

Step 3: Convert Water Pressure into Atmospheres

Given that 1 atmosphere is equivalent to \( 101,325 \text{ N/m}^2 \), convert the pressure from pascals to atmospheres. Pressure at depth is \( 2,568,660 \text{ N/m}^2 \), so pressure in atmospheres is \( \frac{2,568,660}{101,325} \approx 25.35 \text{ atm} \).

Key Concepts

Pressure Calculation

Understanding the concept of pressure is key in fluid mechanics, especially when dealing with underwater activities. Pressure in a fluid, like seawater, is calculated using the formula: \( P = \rho \cdot g \cdot h \). Here, \( \rho \) is the density of the fluid, \( g \) is the acceleration due to gravity (9.8 m/s\(^2\)), and \( h \) is the depth of the fluid above the point in question.
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For the submarine problem, we know:

• Density \( \rho = 1.03 \text{ g/cm}^3 = 1030 \text{ kg/m}^3 \)
• Depth \( h = 255 \text{ m} \)
• Gravity \( g = 9.8 \text{ m/s}^2 \)

The multiplication of these values gives us the pressure \( P \) exerted on the hull at that depth, which calculates to \( 2,568,660 \text{ N/m}^2 \). This force is immense due to the substantial depth, emphasizing the significant role pressure plays in deep-sea environments.

Density of Seawater

In fluid mechanics, density is a fundamental property that indicates how much mass a fluid contains in a particular volume. Seawater, unlike pure water, has a higher density because it contains salts and minerals. Its density is often about 1.03 g/cm\(^3\) (or 1030 kg/m\(^3\) in SI units).
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Knowing the density of seawater is crucial for calculations involving buoyancy, pressure, and force. The extra density compared to freshwater means seawater exerts more force at any given depth, affecting everything from organisms living in the ocean to mechanical structures like submarines.
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This concept helps us evaluate conditions under the sea and design more efficient and resistant marine vessels that can withstand these unique underwater challenges.

Forces on Submarine Hull

The pressure calculated at a certain depth can be translated into force when applied to a surface, such as a submarine's hull. This force is what counters the pull of gravity and keeps submarines buoyant while they navigate the depths. Calculating this involves understanding pressure distribution over a given area.
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The formula to find this force is \( F = P \cdot A \), where \( P \) is the pressure calculated previously \( (2,568,660 \text{ N/m}^2) \) and \( A \) is the area of the hull \( (2200 \text{ m}^2) \).
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The force exerted on the hull by the weight of the water above it is calculated to be 5,650,000,900 N. This force must be considered for structural integrity during the design and operation of submarines to ensure that they can withstand the immense pressure from the surrounding water.

Conversion to Atmospheres

In many situations, especially in underwater diving and marine operations, it's practical to express pressure in atmospheres rather than pascals. An atmosphere is a unit defined as the average atmospheric pressure at sea level and is equivalent to 101,325 N/m\(^2\).
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To convert the water pressure from pascals (Pa) to atmospheres (atm), divide the given pressure by the sea level atmospheric pressure.

• Given depth pressure: \( 2,568,660 \text{ N/m}^2 \)
• Conversion: \( \frac{2,568,660}{101,325} \approx 25.35 \text{ atm} \)

This means that at a depth of 255 meters, a submarine or diver would experience a pressure of approximately 25.35 times the normal atmospheric pressure at sea level. This conversion is essential for understanding how extreme pressures at such depths can affect both machinery and biological organisms.