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Chapter 2: Problem 9

Bathyscaphes are capable of submerging to great depths in the ocean. What is the pressure at a depth of \(5 \mathrm{km}\), assuming that seawater has a constant specific weight of \(10.1 \mathrm{kN} / \mathrm{m}^{3} ?\) Express your answer in pascals and psi.

Short Answer

Expert verified
The pressure at a depth of 5 km in the ocean is \[5.05 \times 10^7 \, \mathrm{Pa}\] or approximately 7320 psi.

Step by step solution

01

Identify the Given Values

The given values are the depth at which the pressure should be calculated, and the specific weight of seawater. The depth is \(5 \, \mathrm{km}\) and the specific weight of seawater is \(10.1 \, \mathrm{kN} / \mathrm{m}^3\). However, it should be noted that depth needs to be in the same unit as in specific weight. Thus, the depth is \(5000 \, \mathrm{m}\).
02

Apply the Pressure Formula

Pressure is calculated by multiplying the specific weight by the depth. So, applying the formula, we get: \[ P = \gamma \times h = 10.1 \, \mathrm{kN/m^3} \times 5000 \, \mathrm{m} \] where \(\gamma\) is the specific weight and \(h\) is the depth.
03

Calculate the Pressure in Pascals

Performing the multiplication gives the pressure in kilopascals (kPa), since \(1 \, \mathrm{kN/m^2} = 1 \, \mathrm{kPa}\). However, the problem asks for the pressure in pascals (Pa), and \(1 \, \mathrm{kPa} = 1000 \, \mathrm{Pa}\). Therefore, we need to multiply the calculated pressure by 1000 to convert it to pascals (Pa).
04

Convert the Pressure to PSI

The problem also asks for the pressure in pounds per square inch (psi). We know that \(1 \, \mathrm{Pa} = 0.000145 \, \mathrm{psi}\). Thus, the pressure in psi can be calculated by multiplying the pressure in pascals by 0.000145.

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Most popular questions from this chapter

A barge is 40 ft wide by 120 ft long. The weight of the barge and its cargo is denoted by \(W\). When in salt-free riverwater. it floats 0.25 ft deeper than when in seawater \(\left(\gamma=64 \mathrm{lb} / \mathrm{ft}^{3}\right) .\) Find the weight \(W\).

A not-too-honest citizen is thinking of making bogus gold bars by first making a hollow iridium \((S=22.5)\) ingot and plating it with a thin layer of gold \((S=19.3)\) of negligible weight and volume. The bogus bar is to have a mass of 100 lbm. What must be the volumes of the bogus bar and of the air space inside the iridium so that an inspector would conclude it was real gold after weighing it in air and water to determine its density? Could lead \((S=11.35)\) or platinum \((S=21.45)\) be used instead of iridium? Would either be a good idea?

A 2 -ft-diameter hemispherical plexiglass "bubble" is to be used as a special window on the side of an above-ground swimming pool. The window is to be bolted onto the vertical wall of the pool and faces outward, covering a 2 -ft- diameter opening in the wall. The center of the opening is 4 ft below the surface. Determine the horizontal and vertical components of the force of the water on the hemisphere.

A submarine submerges by admitting seawater \((S=1.03)\) into its ballast tanks. The amount of water admitted is controlled by air pressure, because seawater will cease to flow into the tank when the internal pressure (at the hull penetration) is equal to the hydrostatic pressure at the depth of the submarine. Consider a ballast tank, which can be modeled as a vertical half- cylinder \((R=8 \mathrm{ft}\) \(L=20 \mathrm{ft}\) ) for which the air pressure control valve has failed shut. The failure occurred at the beginning of a dive from 60 ft to 1000 ft. The tank was initially filled with seawater to a depth of \(2 \mathrm{ft}\) and the air was at a temperature of \(40^{\circ} \mathrm{F}\). As the weight of water in the \(\operatorname{tank}\) is important in maintaining the boat's attitude, determine the weight of water in the tank as a function of depth during the dive. You may assume that tank internal pressure is always in equilibrium with the ocean's hydrostatic pressure and that the inlet pipe to the tank is at the bottom of the tank and penetrates the hull at the "depth" of the submarine.

A barometric pressure of 29.4 in. Hg corresponds to what value of atmospheric pressure in psia, and in pascals?
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