/*! 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 60 A mercury barometer is carried i... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 60

A mercury barometer is carried in a car on a day when there is no wind. The temperature is \(20^{\circ} \mathrm{C}\) and the corrected barometer height is \(761 \mathrm{mm}\) of mercury. One window is open slightly as the car travels at \(105 \mathrm{km} / \mathrm{hr}\). The barometer reading in the moving car is \(5 \mathrm{mm}\) lower than when the car is stationary. Explain what is happening. Calculate the local speed of the air flowing past the window, relative to the automobile.

Short Answer

Expert verified
The reduction in the barometer reading when the car is moving is owing to the Bernoulli's principle, which states that an increase in speed of a fluid leads to a decrease in pressure. Hence, as the speed of the car (and thus the air flowing past the window) increases, the pressure inside the car drops, reducing the height of the mercury in the barometer. The calculation of the local speed of the air relative to the car can be achieved using Bernoulli's equation, substituting the pressure difference calculated from the change in barometer height, and the density of air.

Step by step solution

01

Understanding Bernoulli's Principle

According to Bernoulli's principle, an increase in speed of fluid (in this case, air) occurs simultaneously with a decrease in pressure. So as the car speeds up, the air flowing past the open window increases in speed. The faster moving air results in a decrease in pressure inside the car, which in turn reduces the height of the mercury in the barometer.
02

Applying Bernoulli's Principle to the given problem

The Bernoulli's equation can be expressed as: \[ P + \frac{1}{2} \cdot \rho \cdot v^2 = \text{constant} \] where \(P\) is the pressure, \(\rho\) is the density and \(v\) is the speed. This equation states that the sum of the pressure and the kinetic energy per unit volume is a constant for a steady flow of an incompressible, non-viscous fluid. In this scenario, the change in height of the mercury in the barometer (\(\Delta h = 5 \, \text{mm}\)) can signify a change in pressure difference. This pressure difference corresponds to the change in kinetic energy per unit volume due to the speed of the moving car.
03

Calculate the local speed of air

To get the local speed of the air relative to the car, rearrange Bernoulli's equation to solve for \(v\). \[ v = \sqrt{\frac{2 \Delta P}{\rho}} \] where \( \Delta P = \rho_{\text{mercury}} g \Delta h\) is the pressure difference (with \(\rho_{\text{mercury}}\) being the density of mercury and \(g\) gravitational acceleration). Substituting the given values calculates the relative speed of the air.

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!

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

The velocity field for a plane source located distance \(h=1 \mathrm{m}\) above an infinite wall aligned along the \(x\) axis is given by \\[ \begin{aligned} \vec{V} &=\frac{q}{2 \pi\left[x^{2}+(y-h)^{2}\right]}[x \hat{i}+(y-h) \hat{j}] \\\ &+\frac{q}{2 \pi\left[x^{2}+(y+h)^{2}\right]}[x \hat{i}+(y+h) \hat{j}] \end{aligned} \\] where \(q=2 \mathrm{m}^{3} / \mathrm{s} / \mathrm{m} .\) The fluid density is \(1000 \mathrm{kg} / \mathrm{m}^{3}\) and body forces are negligible. Derive expressions for the velocity and acceleration of a fluid particle that moves along the wall, and plot from \(x=0\) to \(x=+10 h .\) Verify that the velocity and acceleration normal to the wall are zero. Plot the pressure gradient \(\partial p / \partial x\) along the wall. Is the pressure gradient along the wall adverse (does it oppose fluid motion) or not?

Water flows steadily up the vertical 1-in.-diameter pipe and out the nozzle, which is 0.5 in. in diameter, discharging to atmospheric pressure. The stream velocity at the nozzle exit must be \(30 \mathrm{ft} / \mathrm{s}\). Calculate the minimum gage pressure required at section (1). If the device were inverted, what would be the required minimum pressure at section (1) to maintain the nozzle exit velocity at \(30 \mathrm{ft} / \mathrm{s} ?\)

Consider the flow field with velocity given by \(\vec{V}=\left[A\left(y^{2}-x^{2}\right)-B x | \hat{i}+[2 A x y+B y] \hat{j}, A=1 \mathrm{ft}^{-1} \cdot \mathrm{s}^{-1}, B=1\right.\) \(\mathrm{ft}^{-1} \cdot \mathrm{s}^{-1} ;\) the coordinates are measured in feet. The density is \(2 \operatorname{slug} / \mathrm{ft}^{3},\) and gravity acts in the negative \(y\) direction. Calculate the acceleration of a fluid particle and the pressure gradient at point \((x, y)=(1,1)\)

The stream function of a flow field is \(\psi=A x^{2} y-B y^{3},\) where \(A=1 \mathrm{m}^{-1} \cdot \mathrm{s}^{-1}, B=\frac{1}{3} \mathrm{m}^{-1} \cdot \mathrm{s}^{-1},\) and the coordinates are measured in meters. Find an expression for the velocity potential.

Consider frictionless, incompressible flow of air over the wing of an airplane flying at \(200 \mathrm{km} / \mathrm{hr}\). The air approaching the wing is at 65 kPa and \(-10^{\circ} \mathrm{C}\). At a certain point in the flow, the pressure is 60 kPa. Calculate the speed of the air relative to the wing at this point and the absolute air speed.
See all solutions

Recommended explanations on Physics Textbooks

Famous Physicists

Read Explanation

Fluids

Read Explanation

Solid State Physics

Read Explanation

Classical Mechanics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Turning Points in Physics

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.