/*! 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 25 Find the density \(\rho\) of a f... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 25

Find the density \(\rho\) of a fluid at a depth \(h\) in terms of its density \(\rho_{0}\) at the surface. If a mass \(m\) of fluid has volume \(V_{0}\) at the surface, then it will have volume \(V_{0}-\Delta V\) at a depth \(h\). The density at depth \(h\) is then $$ \rho=\frac{m}{V_{0}-\Delta V} \quad \text { while } \quad \rho_{0}=\frac{m}{V_{0}} $$ which gives $$ \frac{\rho}{\rho_{0}}=\frac{V_{0}}{V_{0}-\Delta V}=\frac{1}{1-\left(\Delta V / V_{0}\right)} $$ However, from Chapter 12, the bulk modulus is \(B=P /\left(\Delta V / V_{0}\right)\) and so \(\Delta V / V_{0}=P / B\). Making this substitution, we obtain $$ \frac{\rho}{\rho_{0}}=\frac{1}{1-P / B} $$ If we assume that \(\rho\) is close to \(\rho_{0}\), then the pressure at depth \(h\) is approximately \(\rho_{0} g h\), and so $$ \frac{\rho}{\rho_{0}}=\frac{1}{1-\left(\rho_{0} g h / B\right)} $$

Short Answer

Expert verified
\( \frac{\rho}{\rho_{0}} = \frac{1}{1 - (\rho_{0} g h / B)} \).

Step by step solution

01

Understanding the Problem

We need to find the density \( \rho \) of a fluid at a depth \( h \) in terms of its density \( \rho_{0} \) at the surface. We are given formulas and conditions that describe how volume changes with pressure and depth.
02

Express Density at Depth

The density \( \rho \) at a depth \( h \) is given by the formula \( \rho = \frac{m}{V_{0} - \Delta V} \). By relating this to the surface density equation \( \rho_{0} = \frac{m}{V_{0}} \), we find \( \frac{\rho}{\rho_{0}} = \frac{V_{0}}{V_{0} - \Delta V} \).
03

Relate Change in Volume to Pressure

Use the definition of bulk modulus \( B \), where \( B = P / (\Delta V / V_{0}) \), which implies \( \Delta V / V_{0} = P / B \). Substitute \( \Delta V / V_{0} \) in the density expression to get \( \frac{\rho}{\rho_{0}} = \frac{1}{1 - (P / B)} \).
04

Substitute Pressure at Depth

Assuming \( \rho \approx \rho_{0} \), the pressure at depth \( h \) is \( P = \rho_{0} g h \). Substituting into the equation gives \( \frac{\rho}{\rho_{0}} = \frac{1}{1 - (\rho_{0} g h / B)} \).
05

Conclusion

We have expressed the density \( \rho \) at a depth \( h \) as \( \frac{\rho}{\rho_{0}} = \frac{1}{1 - (\rho_{0} g h / B)} \), by integrating density relationships, volume change under pressure, and hydrostatic pressure equation.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Bulk Modulus
Bulk modulus is an essential concept in fluid mechanics, particularly when studying fluids under pressure. It measures a material's resistance to uniform compression. In simpler terms, the bulk modulus helps us understand how compressible or incompressible a fluid is when pressure is applied.
The mathematical expression for bulk modulus \( B \) is given by:\[B = \frac{P}{\Delta V / V_{0}}\]where:
  • \( P \) is the pressure applied to the fluid.
  • \( \Delta V \) is the change in volume.
  • \( V_{0} \) is the original volume.
A higher bulk modulus means the fluid is less compressible. For example, water has a high bulk modulus compared to air, making it relatively incompressible. Understanding the bulk modulus is crucial when analyzing volume changes in fluids as they experience different pressures.
Fluid Mechanics
Fluid mechanics is the study of fluids (liquids and gases) and the forces acting on them. It explores the way fluids move and interact with their environment. There are two main branches: fluid statics, which deals with fluids at rest, and fluid dynamics, which involves fluids in motion.
In our context, we're focusing on how the density of a liquid changes with depth, a common topic within fluid statics. Key aspects of fluid mechanics include:
  • Analyzing how pressure in a fluid varies with depth.
  • Understanding how fluid densities can change under various conditions.
  • Considering factors like temperature and fluid properties (e.g., bulk modulus).
By applying principles of fluid mechanics, we can predict how fluids behave in natural settings and engineered systems, such as pipes, and help in designing hydraulic structures.
Hydrostatic Pressure
Hydrostatic pressure is the pressure exerted by a fluid at equilibrium, due to the gravitational pull, on any point in that fluid. It is directly related to the depth of the fluid.
For a fluid at rest, this pressure increases with depth because of the weight of the fluid above pushing down.
The equation for hydrostatic pressure \( P \) at depth \( h \) is:\[P = \rho_{0} g h\]where:
  • \( \rho_{0} \) is the fluid density at the surface.
  • \( g \) is the acceleration due to gravity.
  • \( h \) is the depth within the fluid.
Hydrostatic pressure is fundamental in calculating how fluid density changes with depth, as the increased pressure at greater depths leads to changes in volume, thereby affecting density.
Volume Change under Pressure
Volume change under pressure is a key concept to understand how fluids behave when subjected to different pressures, especially at varying depths. As pressure increases, usually due to increasing depth, the volume of a fluid will decrease if the fluid is compressible.
The relationship between pressure change and volume change is explained through the bulk modulus, which provides a measure of this compressibility. This concept is particularly relevant in the study of how density at different depths can be derived from changes in fluid volume, expressed mathematically as:\[\Delta V / V_{0} = P / B\]This equation shows the fractional change in volume \( \Delta V / V_{0} \) directly relates to the pressure \( P \) applied divided by the bulk modulus \( B \).
Understanding volume change under pressure helps explain why the density of a fluid can increase with depth, as the same mass occupies a smaller volume under higher pressure conditions, leading to a higher density.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A cube of wood floating in water supports a 200-g mass resting on the center of its top face. When the mass is removed, the cube rises \(2.00 \mathrm{~cm}\). Determine the volume of the cube.

A wooden cylinder has a mass \(m\) and a base area \(A\). It floats in water with its axis vertical. Show that the cylinder undergoes SHM if given a small vertical displacement. Find the frequency of its motion. When the cylinder is pushed down a distance \(y\), it displaces an additional volume \(A y\) of water. Because this additional displaced volume has mass \(A y \rho_{w}\), an additional buoyant force \(A y \rho_{w} g\) acts on the cylinder, where \(\rho_{w}\) is the density of water. This is an unbalanced force on the cylinder and is a restoring force. In addition, the force is proportional to the displacement and so is a Hooke's Law force. Therefore, the cylinder will undergo SHM, as described in Chapter \(11 .\) Comparing \(F_{B}=A \rho_{w} g y\) with Hooke's Law in the form \(F=k y\), we see that the elastic constant for the motion is \(k=A \rho_{w} g\). This, acting on the cylinder of mass \(m\), causes it to have a vibrational frequency of $$ f=\frac{1}{2 \pi} \sqrt{\frac{k}{m}}=\frac{1}{2 \pi} \sqrt{\frac{A \rho_{w} g}{m}} $$

An \(80-\mathrm{kg}\) metal cylinder, \(2.0 \mathrm{~m}\) long and with each end of area \(25 \mathrm{~cm}^{2}\), stands vertically on one end. What pressure does the cylinder exert on the floor? $$ P=\frac{\text { Normal force }}{\text { Area }}=\frac{(80 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)}{25 \times 10^{-4} \mathrm{~m}^{2}}=3.1 \times 10^{5} \mathrm{~Pa} $$

A balloon and its gondola have a total (empty) mass of \(2.0 \times 10^{2} \mathrm{~kg}\). When filled, the balloon contains \(900 \mathrm{~m}^{3}\) of helium at a density of \(0.183 \mathrm{~kg} / \mathrm{m}^{3}\). Find the added load, in addition to its own weight, that the balloon can lift. The density of air is \(1.29 \mathrm{~kg} / \mathrm{m}^{3}\).

How high would water rise in the essentially open pipes of a building if the water pressure gauge shows the pressure at the ground floor to be \(270 \mathrm{kPa}\) (about \(40 \mathrm{lb} / \mathrm{in.}^{2}\) )? Water pressure gauges read the excess pressure just due to the water, that is, the difference between the absolute pressure in the water and the pressure of the atmosphere. The water pressure at the bottom of the highest column that can be supported is \(270 \mathrm{kPa}\). Therefore, \(P=\rho_{w} g h\) gives $$ h=\frac{P}{\rho_{w} g}=\frac{2.70 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}}{\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)}=27.5 \mathrm{~m} $$
See all solutions

Recommended explanations on Physics Textbooks

Circular Motion and Gravitation

Read Explanation

Waves Physics

Read Explanation

Torque and Rotational Motion

Read Explanation

Engineering Physics

Read Explanation

Oscillations

Read Explanation

Work Energy and Power

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.