/*! 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 32 Shows two sections of an old pip... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 32

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?

Short Answer

Expert verified
The average radius of the pipe in the hill is approximately 3.03 cm.

Step by step solution

01

Understanding the Problem

We have a pipe system with three sections: left (A), middle (within the hill), and right (B). The radius on both sides of the hill is given as \(2.00 \mathrm{~cm}\), and we need to find the radius inside the hill. We're given the flow speed on the left and right \(2.50 \mathrm{~m/s}\) and the time for the dye to travel from A to B \(88.8 \mathrm{~s}\).
02

Calculate the Flow Rate on Both Sides

The flow rate \(Q\) through a pipe is given by \(Q = A \cdot v\), where \(A\) is the cross-sectional area and \(v\) is the speed of the fluid. For the sections on either side, the area \(A = \pi r^2\) with \(r = 0.02 \mathrm{~m}\). Hence, \(Q = \pi (0.02)^2 \cdot 2.50\).
03

Calculate Total Distance and Use Flow Time

The total distance the dye travels is \(2 \times 40 + 110 = 190 \mathrm{~m}\). We can use the travel time \(88.8\mathrm{~s}\) to relate the average flow speed \(v_{avg}\) by \(190 = v_{avg} \times 88.8\), allowing us to solve for \(v_{avg}\).
04

Calculate Flow Rate Within the Hill

For a given average speed \(v_{avg}\) over the entire system, continuity of flow implies constant flow rate. Thus, \(Q = A_{hill} \cdot v_{hill} = A_{side} \cdot v_{side}\). Compute \(A_{hill}\) (area and not directly the radius) from \(Q\) and \(v_{avg}\).
05

Solve for Radius in Hill

Using \(A_{hill} = \pi r_{hill}^2\), solve for \(r_{hill}\) with \(A_{hill}\) found from Step 4, yielding the average radius in the hill.

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.

Continuity Equation
In fluid dynamics, understanding how fluid moves through a system is crucial, and the continuity equation helps with that. It is based on the principle of conservation of mass. This principle states that, for an incompressible and steady fluid flow in a closed system, the mass flow rate must remain constant.

This is expressed as:
  • \(A_1 \cdot v_1 = A_2 \cdot v_2\)
where:
  • \(A_1\) and \(A_2\) are the cross-sectional areas at two different points along the pipe,
  • \(v_1\) and \(v_2\) are the fluid velocities at those points.

The continuity equation is fundamental in determining the unknown dimensions of a pipe, such as the radius inside a hill, by ensuring that the flow rate calculated at one section of the pipe can be related directly to another section. This ensures that engineers can accurately measure or estimate pipe dimensions without direct physical measurement.
Flow Rate
The flow rate of a fluid through a pipe is a measure of how much fluid passes a given point per unit time. It is a critical parameter in fluid dynamics and can be calculated using the formula:
  • \(Q = A \cdot v\)
where:
  • \(Q\) is the flow rate, often expressed in cubic meters per second (m鲁/s),
  • \(A\) is the cross-sectional area of the pipe,
  • \(v\) is the velocity of the fluid.
The flow rate remains constant throughout an unchanging, closed pipe system as stated by the principle of continuity. This makes it possible to use velocities and areas at known sections of the pipe to determine unknown sections.

In practice, knowing the flow rate allows engineers to diagnose issues and plan maintenance for system efficiency, ensuring that systems run smoothly and without unexpected blockages or leaks.
Cross-Sectional Area
In any given pipe, the cross-sectional area is a key parameter in determining how both the velocity and the flow rate of a fluid will behave. It affects the volume of fluid that can move through a section of the pipe at any given time. The area is determined using the formula for the area of a circle:
  • \(A = \pi r^2\)
where \(r\) is the radius of the pipe.

Understanding the cross-sectional area is essential, especially when dealing with changes in the diameter of a pipe.
  • A decrease in area will result in an increase in fluid velocity to maintain the same flow rate, according to the continuity equation.
  • An increase in radius leads to a lower velocity for the same reason.
In engineering applications, knowing the cross-sectional area allows practitioners to make calculations related to pressure, velocity, and the capacity of piping systems. This is extremely important for designing efficient systems.

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

What fraction of the volume of an iceberg (density \(917 \mathrm{~kg} / \mathrm{m}^{3}\) ) would be visible if the iceberg floats (a) in the ocean (salt water, density \(1024 \mathrm{~kg} / \mathrm{m}^{3}\) ) and (b) in a river (fresh water, density \(1000 \mathrm{~kg} / \mathrm{m}^{3}\) )? (When salt water freezes to form ice, the salt is excluded. So, an iceberg could provide fresh water to a community.)

A spring of spring constant \(3.75 \times 10^{4} \mathrm{~N} / \mathrm{m}\) is between a rigid beam and the output piston of a hydraulic lever. An empty container with negligible mass sits on the input piston. The input piston has area \(A_{i}\), and the output piston has area \(18.0 A_{2}\). Initially the spring is at its rest length. How many kilograms of sand must be (slowly) poured into the container to compress the spring by \(5.00 \mathrm{~cm}\) ?

Three children, each of weight \(356 \mathrm{~N}\), make a log raft by lashing together logs of diameter \(0.30 \mathrm{~m}\) and length \(2.00 \mathrm{~m}\). How many logs will be needed to keep them afloat in fresh water? Take the density of the logs to be \(800 \mathrm{~kg} / \mathrm{m}^{3}\).

A flotation device is in the shape of a right cylinder, with a height of \(0.650 \mathrm{~m}\) and a face area of \(4.00 \mathrm{~m}^{2}\) on top and bottom, and its density is \(0.300\) times that of fresh water. It is initially held fully submerged in fresh water, with its top face at the water surface. Then it is allowed to ascend gradually until it begins to float. How much work does the buoyant force do on the device during the ascent?

A piston of cross-sectional area \(a\) is used in a hydraulic press to exert a small force of magnitude \(f\) on the enclosed liquid. A connecting pipe leads to a larger piston of cross-sectional area \(A\) (Fig. 14-31). (a) What force magnitude \(F\) will the larger piston sustain without moving? (b) If the piston diameters are \(3.50 \mathrm{~cm}\) and \(60.0 \mathrm{~cm}\), what force magnitude on the large piston will balance a \(20.0 \mathrm{~N}\) force on the small piston?
See all solutions

Recommended explanations on Physics Textbooks

Wave Optics

Read Explanation

Rotational Dynamics

Read Explanation

Medical Physics

Read Explanation

Famous Physicists

Read Explanation

Physics of Motion

Read Explanation

Linear Momentum

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.