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Chapter 14: Problem 5

The maximum depth \(d_{\max }\) that a diver can snorkel is set by the density of the water and the fact that human lungs can function against a maximum pressure difference (between inside and outside the chest cavity) of \(0.050 \mathrm{~atm}\). What is the difference in \(d_{\max }\) for fresh water and the water of the Dead Sea (the saltiest natural water in the world, with a density of \(1.5 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\) )?

Short Answer

Expert verified
The difference in snorkeling depth is approximately 0.172 meters.

Step by step solution

01

Understand the problem

The question asks for the difference in maximum snorkeling depth in fresh water compared to the Dead Sea, given a pressure limitation of 0.050 atm. This relates to pressure and depth in fluids of different densities.
02

Recall the relationship between pressure, depth, and density

Use the formula for pressure at depth in a fluid: \[ P = P_0 + \rho g h \] where \( P \) is the pressure at depth, \( P_0 \) is the atmospheric pressure, \( \rho \) is the density, \( g \) is the acceleration due to gravity (9.81 m/s虏), and \( h \) is the depth.
03

Set up the equation for the maximum depth

The diver can withstand a maximum pressure difference of 0.050 atm, we need to calculate \( d_{\max} \) using this pressure difference. Considering that 1 atm = 101,325 Pa:\[ 0.050 \, \text{atm} = 0.050 \times 101325 \, \text{Pa} \] Now, solve for \( h \) from the equation: \[ \rho g h = 0.050 \times 101325 \] This becomes \[ h = \frac{0.050 \times 101325}{\rho g} \]
04

Calculate maximum depth for fresh water

For fresh water, the density \( \rho \) is approximately 1000 kg/m鲁. Substitute \( \rho = 1000 \) kg/m鲁 into the formula to find the depth:\[ h_{fresh} = \frac{0.050 \times 101325}{1000 \times 9.81} \approx 0.515 \, \text{m} \]
05

Calculate maximum depth for Dead Sea water

For Dead Sea water, the density \( \rho \) is given as 1500 kg/m鲁. Substitute \( \rho = 1500 \) kg/m鲁 into the formula to find the depth:\[ h_{Dead Sea} = \frac{0.050 \times 101325}{1500 \times 9.81} \approx 0.343 \, \text{m} \]
06

Determine the difference in maximum depths

Subtract the maximum depth in Dead Sea water from the maximum depth in fresh water to find the difference:\[ \Delta h = h_{fresh} - h_{Dead Sea} = 0.515 \, \text{m} - 0.343 \, \text{m} \approx 0.172 \, \text{m} \]
07

Conclude the calculation

The difference in the maximum depth a diver can snorkel between fresh water and the Dead Sea is approximately 0.172 meters.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pressure and Depth Relationship
Pressure in a fluid is directly related to the depth at which it's measured. Here, as you go deeper into a fluid like water, the pressure increases. This happens because the weight of the fluid above you pushes down on the fluid below. We can calculate this pressure increase using the formula:
  • \( P = P_0 + \rho g h \)
where:
  • \( P_0 \) is the atmospheric pressure at the surface,
  • \( \rho \) is the fluid's density,
  • \( g \) is the acceleration due to gravity (approximately 9.81 m/s虏), and
  • \( h \) is the depth in the fluid.

It鈥檚 essential to remember that pressure doesn't just depend on the depth alone. It's also influenced by the density of the fluid. This explains why different fluids at the same depth can exert different pressures. In the exercise, how pressure changes with depth is crucial because it determines how deep a snorkeler can dive without exceeding the pressure they can tolerate, which is set at 0.050 atm here.
Density in Fluids
Density, symbolized as \( \rho \), is a core characteristic of fluids that determines how much mass exists in a certain volume. It's crucial for calculating pressure in a fluid. The formula to express density is: \( \rho = \frac{m}{V} \), where \( m \) is the mass, and \( V \) is the volume.

Fresh water has a standard density of about 1000 kg/m鲁, which means it is lighter than many other natural bodies of water. On the other hand, the Dead Sea has a notably higher density of 1500 kg/m鲁 due to its high salt content.
  • Higher density means more mass per unit volume.
  • This added weight in turn leads to a higher pressure increase with depth.

      • In the original exercise, this distinction in density between fresh water and Dead Sea water explains the difference in pressure felt at the same depths, affecting the calculation of maximum snorkeling depths.
Buoyancy in Different Water Densities
Buoyancy is the upward force exerted by a fluid, counteracting the weight of an object submerged in it. It's what allows objects to float or sink depending on their density compared to the fluid. The primary factor influencing buoyancy is the density of the fluid.

The relationship between fluid density and buoyancy affects how submerged objects behave. In our scenario, because the density of the Dead Sea is higher than that of fresh water, the fluid itself exerts more buoyant force even at lesser depths. This greater buoyant force contributes to the challenges in snorkeling deeper in such dense water. Understanding buoyancy helps clarify why divers find their maximum depth more limited in the dense waters of the Dead Sea. With the higher density causing a greater increase in pressure per meter of depth, it limits how much deeper one can safely snorkel compared to in less dense fresh water.

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Most popular questions from this chapter

A hollow spherical iron shell floats almost completely submerged in water. The outer diameter is \(50.0 \mathrm{~cm}\), and the density of iron is \(7.87 \mathrm{~g} / \mathrm{cm}^{3}\). Find the inner diameter.

A cylindrical tank with a large diameter is filled with water to a depth \(D=0.30 \mathrm{~m}\). A hole of cross-sectional area \(A=6.2 \mathrm{~cm}^{2}\) in the bottom of the tank allows water to drain out. (a) What is the drainage rate in cubic meters per second? (b) At what distance below the bottom of the tank is the cross-sectional area of the stream equal to one-half the area of the hole?

A venturi meter is used to measure the flow speed of a fluid in a pipe. The meter is connected between two sections of the pipe (Fig. 14-43); the cross- sectional area \(A\) of the entrance and exit of the meter matches the pipe's cross-sectional area. Between the entrance and exit, the fluid flows from the pipe with speed \(V\) and then through a narrow "throat" of crosssectional area \(a\) with speed \(v\). A manometer connects the wider portion of the meter to the narrower portion. The change in the fluid's speed is accompanied by a change \(\Delta p\) in the fluid's pressure, which causes a height difference \(h\) of the liquid in the two arms of the manometer. (Here \(\Delta p\) means pressure in the throat minus pressure in the pipe.) (a) By applying Bernoulli's equation and the equation of continuity to points 1 and 2 in Fig. 14-43, show that $$ V=\sqrt{\frac{2 a^{2} \Delta p}{\rho\left(a^{2}-A^{2}\right)}} $$ where \(\rho\) is the density of the fluid. (b) Suppose that the fluid is fresh water, that the cross-sectional areas are \(60 \mathrm{~cm}^{2}\) in the pipe and \(32 \mathrm{~cm}^{2}\) in the throat, and that the pressure is \(55 \mathrm{kPa}\) in the pipe and \(41 \mathrm{kPa}\) in the throat. What is the rate of water flow in cubic meters per second?

Shows two sections of an old pipe system that runs through a hill, with distances \(d_{A}=d_{B}=40 \mathrm{~m}\) and \(D=110 \mathrm{~m}\). On each side of the hill, the pipe radius is \(2.00 \mathrm{~cm}\). However, the radius of the pipe inside the hill is no longer known. To determine it, hydraulic engineers first establish that water flows through the left and right sections at \(2.50 \mathrm{~m} / \mathrm{s}\). Then they release a dye in the water at point \(A\) and find that it takes \(88.8 \mathrm{~s}\) to reach point \(B\). What is the average radius of the pipe within the hill?

What gauge pressure must a machine produce in order to suck mud of density \(1800 \mathrm{~kg} / \mathrm{m}^{3}\) up a tube by a height of \(2.0 \mathrm{~m} ?\)
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