Ch 12: Fluid Mechanics

Young & Freedman Calc - University Physics 15th Edition

Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook

Young & Freedman Calc 15th Edition

Ch 12: Fluid Mechanics

Problem 12b

Chapter 12, Problem 12b

You are designing a diving bell to withstand the pressure of seawater at a depth of 250 m. What is the gauge pressure at this depth? (You can ignore changes in the density of the water with depth.) At this depth, what is the net force due to the water outside and the air inside the bell on a circular glass window 30.0 cm in diameter if the pressure inside the diving bell equals the pressure at the surface of the water? (Ignore the small variation of pressure over the surface of the window.)

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To find the gauge pressure at a depth of 250 m, use the formula for gauge pressure: \( P_{\text{gauge}} = \rho g h \), where \( \rho \) is the density of seawater (approximately 1025 kg/m³), \( g \) is the acceleration due to gravity (9.81 m/s²), and \( h \) is the depth (250 m). Substitute these values into the formula to calculate the gauge pressure.

To find the net force on the window, first calculate the area of the circular window. The diameter is given as 30.0 cm, so the radius \( r \) is 15.0 cm or 0.15 m. Use the formula for the area of a circle: \( A = \pi r^2 \). Substitute the radius into this formula to find the area.

The net force on the window is due to the difference in pressure between the outside and inside of the diving bell. The pressure inside the bell is equal to the atmospheric pressure at the surface, \( P_{\text{atm}} \). The pressure outside is the sum of atmospheric pressure and gauge pressure: \( P_{\text{outside}} = P_{\text{atm}} + P_{\text{gauge}} \).

The net force \( F_{\text{net}} \) on the window is given by the difference in pressure times the area: \( F_{\text{net}} = (P_{\text{outside}} - P_{\text{inside}}) \times A \). Since \( P_{\text{inside}} = P_{\text{atm}} \), the net force simplifies to \( F_{\text{net}} = P_{\text{gauge}} \times A \).

Substitute the calculated gauge pressure and the area of the window into the formula for net force to find the net force on the window due to the water pressure at 250 m depth.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Gauge Pressure

Gauge pressure is the pressure relative to atmospheric pressure. It is calculated by subtracting atmospheric pressure from the absolute pressure. In the context of a diving bell, gauge pressure at a certain depth can be determined using the formula P = ρgh, where ρ is the density of the fluid (seawater), g is the acceleration due to gravity, and h is the depth.

Buoyancy and Net Force

The net force on an object submerged in a fluid is influenced by buoyancy, which is the upward force exerted by the fluid. For the diving bell, the net force on the window is the difference between the force due to the water pressure outside and the air pressure inside. This can be calculated using the formula F = P * A, where P is the pressure difference and A is the area of the window.

Pressure and Area Relationship

Pressure is defined as force per unit area. When calculating the force on a surface due to pressure, it is crucial to consider the area over which the pressure is applied. For a circular window, the area can be calculated using A = πr², where r is the radius. This relationship helps determine the total force exerted by the pressure on the window.

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Textbook Question

In intravenous feeding, a needle is inserted in a vein in the patient's arm and a tube leads from the needle to a reservoir of fluid (density 1050 kg/m³) located at height h above the arm. The top of the reservoir is open to the air. If the gauge pressure inside the vein is 5980 Pa, what is the minimum value of h that allows fluid to enter the vein? Assume the needle diameter is large enough that you can ignore the viscosity of the liquid.

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Textbook Question

You are designing a diving bell to withstand the pressure of seawater at a depth of 250 m. (a) What is the gauge pressure at this depth? (You can ignore changes in the density of the water with depth.)

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Textbook Question

The liquid in the open-tube manometer in Fig. 12.8a is mercury, y₁=3.00 cm,and y₂=7.00 cm. Atmospheric pressure is 980 millibars. What is (a) the absolute pressure at the bottom of the U-shaped tube; (b) the absolute pressure in the open tube at a depth of 4.00 cm below the free surface; (c) the absolute pressure of the gas in the container; (d) the gauge pressure of the gas in pascals?

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Textbook Question

BIO. There is a maximum depth at which a diver can breathe through a snorkel tube (Fig. E12.17) because as the depth increases, so does the pressure difference, which tends to collapse the diver's lungs. Since the snorkel connects the air in the lungs to the atmosphere at the surface, the pressure inside the lungs is atmospheric pressure. What is the external– internal pressure difference when the diver's lungs are at a depth of 6.1 m (about 20 ft)? Assume that the diver is in fresh-water. (A scuba diver breathing from compressed air tanks can operate at greater depths than can a snorkeler, since the pressure of the air inside the scuba diver's lungs increases to match the external pressure of the water.)

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Textbook Question

Ear Damage from Diving. If the force on the tympanic membrane (eardrum) increases by about 1.5 N above the force from atmospheric pressure, the membrane can be damaged. When you go scuba diving in the ocean, below what depth could damage to your eardrum start to occur? The eardrum is typically 8.2 mm in diameter. (Consult Table 12.1.)

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Textbook Question

A barrel contains a 0.120-m layer of oil floating on water that is 0.250 m deep. The density of the oil is 600 kg/m³. (a) What is the gauge pressure at the oil–water interface? (b) What is the gauge pressure at the bottom of the barrel?