/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 50 An air-filled, hemispherical she... [FREE SOLUTION] | 91Ó°ÊÓ

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An air-filled, hemispherical shell is attached to the ocean floor at a depth of 10 in. as shown in Fig. P2.50. A mercury barometer located inside the shell reads \(765 \mathrm{mm}\) Hg, and a mercury U-tube manometer designed to give the outside water pressure indicates differential reading of \(735 \mathrm{mm}\) Hg as illustrated. Based on these data, what is the atmospheric pressure at the ocean surface?

Short Answer

Expert verified
The atmospheric pressure at the ocean surface is approximately 1.95 atm.

Step by step solution

01

Understanding the Given Information

We have a hemispherical shell at a depth of 10 inches underwater, with an internal barometer reading and an external water pressure reading. The internal barometer reads a pressure of 765 mm Hg. There is also a differential pressure reading from a U-tube manometer of 735 mm Hg showing the pressure difference between inside and outside the shell.
02

Identify Unknown and Known Variables

The unknown variable is the atmospheric pressure at the ocean surface. The known variable is the internal pressure of 765 mm Hg and the differential manometer reading of 735 mm Hg representing the pressure difference.
03

Calculate Absolute External Water Pressure

To find the absolute pressure outside the shell, add the internal pressure to the differential reading. Thus, the absolute pressure is \(765 \mathrm{mm} \space Hg + 735 \mathrm{mm} \space Hg = 1500 \mathrm{mm} \space Hg\).
04

Convert mm Hg to Atmospheric Pressure

Considering 1 atmosphere is equivalent to 760 mm Hg, convert the absolute pressure into terms of atmospheric pressure: \[\frac{1500 \mathrm{mm} \space Hg}{760 \mathrm{mm} \space Hg/\text{atm}} = \frac{1500}{760} \approx 1.974 \text{ atm}.\]
05

Compute Atmospheric Pressure

Since the pressure difference is due to the pressure underwater, subtract the hydrostatic pressure component (1000 mm water depth) converted to atmospheric pressure from the calculated total absolute pressure. Convert the 10 in depth to mm Hg: \[ 10 \text{ in} \times \frac{25.4 \text{ mm}}{1 \text{ in}} \times \frac{1 \text{ mm Hg}}{13.6 \text{ mm water}} \approx 18.75 \text{ mm Hg}, \] then \[1.974 \text{ atm} - \frac{18.75 \text{ mm Hg}}{760 \text{ mm Hg/atm}} = 1.95 \text{ atm}.\]
06

Final Result Interpretation

The atmospheric pressure at the ocean surface, considering the barometer and depth factors, is approximately 1.95 atm.

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Key Concepts

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

Pressure Measurement
Pressure measurement is a fundamental concept in fluid mechanics and is important in various practical applications such as engineering and meteorology. It involves quantifying the force exerted by a fluid (liquid or gas) on a surface per unit area. This is crucial since understanding pressure distribution can help in designing fluid systems, predicting behavior under different conditions, and ensuring safety in structures and vehicles.

Pressure is typically measured using tools like barometers, manometers, and gauge sensors. Each of these tools has unique functionalities best suited for different applications. Knowing the variable you're measuring helps you choose the correct device. Generally, pressure is categorized into:
  • Absolute Pressure: The total pressure measured relative to a perfect vacuum.
  • Gauge Pressure: The pressure measured relative to the ambient atmospheric pressure.
  • Differential Pressure: The difference in pressure between two points.
In problem-solving, identifying which type of pressure you're dealing with is an essential first step to ensure accurate calculations and conclusions.
Barometers
Barometers are instruments used specifically to measure atmospheric pressure. They are crucial in meteorology for forecasting weather by detecting pressure changes. A commonly used type of barometer is the mercury barometer. Here, mercury rises or falls in a glass tube in response to pressure changes; hence, giving an accurate measurement.

In our original exercise, the internal barometer gives a reading of 765 mm Hg. This value indicates the atmospheric pressure inside the hemispherical shell.

barometers are necessary tools not only for weather prediction but also used in activities like hiking to determine altitude above sea level, as changes in elevation affect atmospheric pressure. Understanding barometric pressure can give valuable insights into potential weather changes, allowing for proactive planning and adjustment in various outdoor activities.
Manometers
Manometers are devices used to measure the pressure of a gas or liquid. U-tube manometers, one specific type, are simple yet effective tools for measuring pressure differences between two points. They consist of a U-shaped tube containing a fluid, often mercury, which moves in response to pressure changes on either side of the tube.

The key function of a manometer is to measure differential pressure. This means it can show the difference between two pressure points. In our exercise, the U-tube manometer shows a difference of 735 mm Hg between the inside and outside of the shell. This indicates how much more pressure is exerted by the external water.

Manometers are used in various industries for applications like HVAC systems, laboratory experiments, and marine engineering. Understanding how to read and interpret manometer readings is important for ensuring proper functioning and maintenance of these systems.
Hydrostatic Pressure
Hydrostatic pressure is the pressure exerted by a fluid at rest due to the force of gravity. It plays a key role in many disciplines of science and engineering, such as hydraulics, oceanography, and civil engineering. At any given depth, this pressure is uniform and acts equally in all directions. It depends on three factors: the density of the fluid, the depth of the fluid, and gravitational acceleration.

In mathematical terms, hydrostatic pressure can be expressed as \( P = \rho gh \), where \( \rho \) is the fluid's density, \( g \) is the acceleration due to gravity, and \( h \) is the depth of fluid.

In the given exercise, the pressure exerted by the ocean water 10 inches above the shell's depth contributes to the total external pressure. It's essential to account for hydrostatic pressure when calculating total pressure or trying to understand how pressure changes at different depths. This concept is crucial in applications like diving, where pressure changes significantly with depth, affecting buoyancy and decompression.

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Most popular questions from this chapter

The water level in an open standpipe is \(80 \mathrm{ft}\) above the ground. What is the static pressure at a fire hydrant that is connected to the standpipe and located at ground level? Express your answer in psi.

Because of elevation differences, the water pressure in the second floor of your house is lower than it is in the first floor. For tall buildings this pressure difference can become unacceptable. Discuss possible ways to design the water distribution system in very tall buildings so that the hydrostatic pressure difference is within acceptable limits.

A tank of cross-sectional area \(A\) is filled with a liquid of specific weight \(\gamma_{1}\) as shown in Fig. \(\mathrm{P} 2.137 a .\) Show that when a cylinder of specific weight \(\gamma_{2}\) and volume \(\psi\) is floated in the liquid (see Fig. \(\mathrm{P} 2.137 b\) ), the liquid level rises by an amount \(\Delta h=\) \(\left(\gamma_{2} / \gamma_{1}\right) \forall / A\).

A 0.3 -m-diameter pipe is connected to a 0.02 -m-diameter pipe, and both are rigidly held in place. Both pipes are horizontal with pistons at each end. If the space between the pistons is filled with water, what force will have to be applied to the larger piston to balance a force of \(90 \mathrm{N}\) applied to the smaller piston? Neglect friction.

An open rectangular settling tank contains a liquid suspension that at a given time has a specific weight that varies approximately with depth according to the following data: $$\begin{array}{cc}\boldsymbol{h}(\mathbf{m}) & \gamma\left(\mathbf{N} / \mathbf{m}^{3}\right) \\\\\hline 0 & 10.0 \\ 0.4 & 10.1 \\\0.8 & 10.2 \\\1.2 & 10.6 \\\1.6 & 11.3 \\\2.0 & 12.3 \\\2.4 & 12.7 \\\2.8 & 12.9 \\\3.2 & 13.0 \\\3.6 & 13.1 \\ \hline\end{array}$$ The depth \(h=0\) corresponds to the free surface. Determine, by means of numerical integration, the magnitude and location of the resultant force that the liquid suspension exerts on a vertical wall of the tank that is \(6 \mathrm{m}\) wide. The depth of fluid in the tank is \(3.6 \mathrm{m}\).

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