/*! 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 53 Water discharges from a horizont... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 53

Water discharges from a horizontal cylindrical pipe at the rate of 465 \(\mathrm{cm}^{3} / \mathrm{s}\) . At a point in the pipe where the radius is \(2.05 \mathrm{cm},\) the absolute pressure is \(1.60 \times 10^{5} \mathrm{Pa} .\) What is the pipe's radius at a constriction if the pressure there is reduced to \(1.20 \times 10^{5} \mathrm{Pa}\) ?

Short Answer

Expert verified
The radius at the constriction is found using the continuity and Bernoulli's equations.

Step by step solution

01

Apply the Continuity Equation

The continuity equation states that the flow rate must be constant throughout the pipe. The flow rate is given by \[ Q = A_1 v_1 = A_2 v_2, \] where \( A \) is the cross-sectional area and \( v \) is the velocity of the fluid. For a cylindrical pipe, the area \( A \) is \( \pi r^2 \). Given the flow rate, 465 \( \mathrm{cm}^3/\mathrm{s} \), we have: \[ A_1 v_1 = A_2 v_2 = 465. \] The radius \( r_1 \) at point 1 is 2.05 cm, so: \[ A_1 = \pi (2.05)^2 \text{ cm}^2. \] Simplify this to find \( v_1 \): \[ v_1 = \frac{465}{\pi (2.05)^2}. \]
02

Bernoulli's Equation Setup

Bernoulli's equation for incompressible flow states that: \[ P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2, \] where \( P \) is pressure, \( \rho \) is the density of water, \( v \) is the speed of the fluid. Given, \( P_1 = 1.60 \times 10^5 \) Pa and \( P_2 = 1.20 \times 10^5 \) Pa, and we are solving for \( v_2 \) before finding \( r_2 \).
03

Solve for Velocity at Constriction

Use the continuity equation from Step 1 to substitute \( v_1 \) in Bernoulli's equation: \[ \frac{1}{2}\rho v_1^2 - \frac{1}{2}\rho v_2^2 = P_2 - P_1. \] Rearrange to get: \[ \frac{1}{2}\rho (v_2^2 - v_1^2) = P_1 - P_2. \] Input the densities and pressures, solve for \( v_2 \): \[ v_2 = \sqrt{\frac{2(P_1 - P_2)}{\rho} + v_1^2}. \]
04

Find New Radius using Continuity Equation

Once \( v_2 \) is known from Step 3, utilize the continuity equation again to find \( r_2 \): \[ A_2 = \pi r_2^2 = \frac{465}{v_2}. \] Solve for \( r_2 \): \[ r_2 = \sqrt{\frac{465}{\pi v_2}}. \] This gives the radius at the constriction.

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
The continuity equation is a fundamental principle in fluid dynamics. It asserts that the mass flow rate must remain constant from one cross-section of a pipe to another. This means the rate at which fluid enters a section of the pipe is equal to the rate at which it exits - no fluid is lost.
For a cylindrical pipe, like the one in the exercise, the equation is represented as:
  • Flow rate (\(Q\)) = Cross-sectional area (\(A\)) \( \times \) Velocity (\(v\)) of the fluid.
  • Mathematically, \( Q = A_1 v_1 = A_2 v_2 \) , where \( A_1 \) and \( A_2 \) are areas at different points of the pipe, and \( v_1 \) and \( v_2 \) are corresponding velocities.
Consequently, if the pipe narrows (and thus the cross-sectional area decreases), the velocity of the fluid must increase to maintain the same flow rate. This principle is crucial when analyzing fluid behavior in varying pipe diameters.
Fluid Dynamics
Fluid dynamics is the study of how fluids (liquids and gases) move. In the context of the exercise, understanding fluid dynamics involves recognizing how fluid properties like pressure, velocity, and density interact and change within pipelines.
Fluids behave predictably under different conditions due to fundamental laws of physics:
  • Pressure: In a fluid, this is the force exerted per unit area. Changes in pipe diameter affect pressure.
  • Velocity: This is the speed of the fluid at a given point along the pipeline's path. As per fluid dynamics laws, velocity increases when fluid moves through narrower parts of the pipe.
Studying these aspects helps us understand how designs in industries ranging from water supply to aeronautics optimize fluid movement for efficiency and safety.
Cross-Sectional Area
Cross-sectional area in pipes refers to the shape and size of the section cut perpendicular to the flow of fluid in the pipe. For cylindrical pipes, this area is found using the formula \( A = \pi r^2 \), where \( r \) is the radius of the pipe's interior.
This concept is pivotal in calculations involving:
  • Flow rate: Greater cross-sectional area usually means a slower fluid velocity, as more space is available for fluid to flow given a constant flow rate.
  • Pressure: If the cross-sectional area decreases, often through constriction, the fluid's velocity increases, impacting pressure based on Bernoulli's principle.
Understanding this allows for the efficient design of systems where controlling the speed and pressure of fluid is essential, like in pipelines and even blood vessels.
Pipe Constriction
Pipe constriction refers to a reduction in the diameter of a pipe, which affects the flow of fluid within. The change in diameter can cause changes in velocity and pressure, a phenomenon that can be explained using Bernoulli's principle, which relates fluid speed and pressure at different points on a streamline.
When the pipe constricts:
  • The cross-sectional area \( A \) decreases, causing the fluid velocity \( v \) to increase in order to maintain constant flow rate, as per the continuity equation.
  • Pressure \( P \) decreases at the constriction point, based on Bernoulli's principle, which states: \( P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2 \) .
This understanding is crucial for designing systems where maintaining or controlling fluid pressure is important, such as intravenous systems in medicine or fuel lines in vehicles.

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

You are designing a machine for a space exploration vehicle. It contains an enclosed column of oil that is 1.50 m tall, and you need the pressure difference between the top and the bottom of this column to be 0.125 atm. (a) What must be the density of the oil? (b) If the vehicle is taken to Mars, where the acceleration due to gravity is \(0.379 g,\) what will be the pressure difference (in earth atmospheres) between the top and bottom of the oil column?

A cube of compressible material (such as Styrofoam or balsa wood) has a density \(\rho\) and sides of length \(L .\) (a) If you keep its mass the same, but compress each side to half its length, what will be its new density, in terms of \(\rho ?\) (b) If you keep the mass and shape the same, what would the length of each side have to be (in terms of \(L )\) so that the density of the cube was three times its original value?

At \(20^{\circ} \mathrm{C}\) , the surface tension of water is 72.8 dynes/cm and that of carbon tetrachloride \(\left(\mathrm{CCl}_{4}\right)\) is 26.8 dynes/cm. If the gauge pressure is the same in two drops of these liquids, what is the ratio of the volume of the water drop to that of the \(\mathrm{CCl}_{4}\) drop?

A liquid is used to make a mercury-type barometer, as described in Section \(13.2 .\) The barometer is intended for spacefaring astronauts. At the surface of the earth, the column of liquid rises to a height of 2185 \(\mathrm{mm}\) , but on the surface of Planet \(\mathrm{X},\) where the acceleration due to gravity is one-fourth of its value on earth, the column rises to only 725 \(\mathrm{mm} .\) Find (a) the density of the liquid and (b) the atmospheric pressure at the surface of Planet \(\mathrm{X}\) .

A shower head has 20 circular openings, each with radius 1.0 \(\mathrm{mm} .\) The shower head is connected to a pipe with radius 0.80 \(\mathrm{cm} .\) If the speed of water in the pipe is \(3.0 \mathrm{m} / \mathrm{s},\) what is its speed as it exits the shower-head openings?
See all solutions

Recommended explanations on Physics Textbooks

Conservation of Energy and Momentum

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Turning Points in Physics

Read Explanation

Quantum Physics

Read Explanation

Oscillations

Read Explanation

Classical Mechanics

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.