/*! 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 30 A sphere of diameter \(D\) that ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 30

A sphere of diameter \(D\) that falls in a fluid ultimately attains a constant velocity, called the terminal velocity, when the drag force exerted by the fluid is equal to the net weight of the sphere in the fluid. The net weight of a sphere can be represented by \(\Delta \gamma \mathcal{V},\) where \(\Delta \gamma\) is the difference between the specific weight of the sphere and the specific weight of the fluid and \(\mathcal{V}\) is the volume of the sphere. In cases where the sphere falls slowly in a viscous fluid, the terminal velocity, \(V,\) depends only on the size of the sphere, \(D,\) the specific-weight difference \(\Delta \gamma,\) and the viscosity of the fluid, \(\mu\). Use dimensional analysis to determine the functional relationship between the terminal velocity and the other relevant variables.

Short Answer

Expert verified
The terminal velocity is proportional to \((D^2 \Delta \gamma) / \mu\).

Step by step solution

01

Identify the Relevant Variables

The task requires a functional relationship involving terminal velocity \( V \), diameter of the sphere \( D \), specific weight difference \( \Delta \gamma \), and fluid viscosity \( \mu \). These variables will be used for dimensional analysis.
02

List the Dimensions of Each Variable

- Terminal velocity \( V \) has dimensions \([L][T]^{-1}\).- Diameter \( D \) has dimensions \([L]\).- The specific weight difference \( \Delta \gamma \) has dimensions \([M][L]^{-2}[T]^{-2}\).- Viscosity of the fluid \( \mu \) has dimensions \([M][L]^{-1}[T]^{-1}\).
03

Set Up the Dimensional Equation

Assume a relationship of the form \( V = k D^a (\Delta \gamma)^b \mu^c \), where \( k \) is a dimensionless constant and \( a, b, \) and \( c \) are exponents to be determined. Equate dimensions on both sides:\[ [L][T]^{-1} = [L]^a [M]^b [L]^{-2b} [T]^{-2b} [M]^c [L]^{-c} [T]^{-c} \]
04

Equate Powers of Each Dimension

Match the powers of \([L]\), \([M]\), and \([T]\) on both sides:1. For \([L]\): \( 1 = a - 2b - c \)2. For \([M]\): \( 0 = b + c \)3. For \([T]^{-1}\): \(-1 = -2b - c \)
05

Solve the System of Equations

Solve these equations simultaneously:1. From \(0 = b + c\), we get \( c = -b \).2. Substitute \(c = -b\) in \(-1 = -2b - c\) to get \(-1 = -2b + b\), so \(b = 1\).3. Substitute \(b = 1\) in \(c = -b\), yielding \(c = -1\).4. Substitute \(b = 1, c = -1\) in \(1 = a - 2b - c\) to give \(1 = a - 2(1) + 1\), so \(a = 2\).
06

Write the Dimensionless Relationship

The relationship is \( V = k D^2 (\Delta \gamma)^1 \mu^{-1} \) or \( V = k \frac{D^2 \Delta \gamma}{\mu} \). This represents how terminal velocity depends on these variables.

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.

Terminal Velocity
Terminal velocity is a fascinating concept in fluid mechanics, especially when considering objects like spheres moving through a fluid. It occurs when the sphere's falling speed becomes constant, meaning it does not accelerate anymore. At this stage, all forces acting on the sphere balance each other out. For a falling sphere, terminal velocity is reached when the downward force due to gravity, or more specifically the net weight, is equal to the upward drag force exerted by the fluid.

Why does terminal velocity become constant? It's simple, really. Initially, as the sphere starts falling, it accelerates because of gravity. However, as it picks up speed, the resistance from the fluid, or the drag force, increases as well. Eventually, these forces reach equilibrium, and the sphere moves at a steady rate.
  • Factors influencing terminal velocity include the object's size, the density difference between the sphere and fluid, and the fluid's viscosity.
  • The formula derived from dimensional analysis shows how each of these factors contributes.
Understanding terminal velocity helps in predicting the behavior of objects moving through fluids, which is crucial in fields like meteorology and engineering.
Drag Force
Drag force is the force exerted by a fluid on an object moving through it. It tends to oppose the object's motion, slowing it down. In the context of a sphere falling through a fluid, drag force arises because parts of the fluid get displaced, creating resistance.

Drag force is complex to calculate because it depends on several factors:
  • The shape and size of the object: A larger surface area means more fluid comes into contact with the object, creating more drag.
  • The speed of the object: As velocity increases, the drag force also increases, often following a quadratic relation.
  • The properties of the fluid, such as viscosity, which we'll explore further.
In balanced conditions, drag equals the weight force minus buoyancy, and the sphere stops accelerating once terminal velocity is achieved. Understanding drag force is vital in a variety of applications from designing vehicles to predicting the behavior of falling objects.
Viscosity
Viscosity is a measure of a fluid's resistance to deformation, often thought of as the "thickness" of the fluid. It's an inherent property that affects how fluids flow and consequently, how objects move through them.

Fluids like honey have high viscosity, which means they flow slowly and resist motion more significantly than low viscosity fluids like water.
  • Viscosity directly affects drag by determining how easily fluid layers move past one another.
  • In our exercise, it acts in the opposite direction of the sphere's motion, slowing it down. This means that the drag force and, consequently, terminal velocity depend heavily on viscosity.
  • The relationship crafted via dimensional analysis beautifully reflects how an increase in viscosity leads to a decrease in terminal velocity.
Grasping the concept of viscosity is pivotal when studying fluid dynamics, demonstrating why certain fluids behave differently and how they affect objects in their flow.

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

Wind blowing at a steady velocity \(V\) toward a vertical column of diameter, \(D,\) generates velocity fluctuations behind the column that have a frequency \(\omega\). The frequency of the generated velocity fluctuations can also depend on the density and viscosity of the air. Determine a nondimensional functional relationship that relates the frequency of the velocity fluctuations to the influencing variables. Identify any named conventional dimensionless groups.

The performance of a torpedo is to be tested using a model in a wind tunnel. The prototype torpedo is designed to move at \(30 \mathrm{~m} / \mathrm{s}\) in water at \(20^{\circ} \mathrm{C}\). If a 1: 4 model is used in a wind tunnel with air at \(20^{\circ} \mathrm{C}\) and an airspeed of \(110 \mathrm{~m} / \mathrm{s}\), what must the air pressure in the wind tunnel be to achieve dynamic similarity? If dynamic similarity is achieved and a drag force of \(600 \mathrm{~N}\) is measured in the model, what is the corresponding drag force in the prototype?

A hydraulic jump is a phenomenon associated with high-velocity flow of a liquid in an open channel in which the flow depth of the liquid suddenly changes from a lower depth, \(h_{1}\), to a higher depth, \(h_{2}\). When the flow is occurring in a horizontal rectangular channel, the relationship between \(h_{1}\) and \(h_{2}\) depends only on the lower-depth velocity of flow, \(V_{1}\), and the gravity constant, \(g\). Use dimensional analysis to determine the functional relationship between \(h_{2}\) and the influencing variables, expressed in terms of dimensionless groups. Identify any named conventional dimensionless groups that occur in this relationship.

A small plane is designed to cruise at \(350 \mathrm{~km} / \mathrm{h}\) at an elevation of \(1 \mathrm{~km}\), where the typical atmospheric pressure is \(90 \mathrm{kPa}\). The drag characteristics of the aircraft are to be studied in a pressurized wind tunnel, where the air in the wind tunnel is the same temperature as the air in the prototype. The model is to be at 1: 6 scale, and the airspeed in the wind tunnel is to be \(260 \mathrm{~km} / \mathrm{h}\). (a) What scaling law should be used in designing the model study? (b) What pressure should be used in the wind tunnel? (c) If the drag on a component of the model aircraft is measured as \(15 \mathrm{~N},\) what is the drag on the corresponding component in the prototype?

An orifice plate is a flat plate with a central opening that is sometimes used to measure the volume flow rate in a pipe. The pressure drop across an orifice plate can be assumed to be a function of the diameter of the opening in the plate, the diameter of the pipe, the velocity of flow in the pipe, and the density and viscosity of the fluid. Use dimensional analysis to determine the functional relationship between the pressure drop across the plate and the influencing variables. Express this functional relationship in both a dimensional and nondimensional form. Identify any named conventional dimensionless groups that appear in the nondimensional relationship.
See all solutions

Recommended explanations on Physics Textbooks

Dynamics

Read Explanation

Solid State Physics

Read Explanation

Fundamentals of Physics

Read Explanation

Kinematics Physics

Read Explanation

Mechanics and Materials

Read Explanation

Electricity

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.