/*! 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 93 Consider the steady adiabatic fl... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 93

Consider the steady adiabatic flow of air through a long straight pipe with \(0.05 \mathrm{m}^{2}\) cross-sectional area. At the inlet, the air is at \(200 \mathrm{kPa}\) (gage), \(60^{\circ} \mathrm{C}\), and has a velocity of \(150 \mathrm{m} / \mathrm{s}\). At the exit, the air is at \(80 \mathrm{kPa}\) and has a velocity of \(300 \mathrm{m} / \mathrm{s}\). Calculate the axial force of the air on the pipe. (Be sure to make the direction clear.)

Short Answer

Expert verified
The axial force of the air on the pipe is given by the change in momentum flux. The direction of the force is opposite to the direction of change in velocity.

Step by step solution

01

Identify Given Data

We are given the cross-sectional area of the pipe (0.05 m^2), the initial pressure (200 kPa), the final pressure (80 kPa), the initial velocity (150 m/s), and the final velocity (300 m/s)
02

Determine Initial and Final Momentum

First, you need to understand that the momentum is the mass of the object multiplied by its velocity. But, given that we only have the velocity, we can only calculate the initial and final momentum flux rates (mass flow rate * velocity) or \( \dot{m}v \), where \( \dot{m} \) is the mass flow rate.To find the mass flow rate, we can use the following relationship:\n\( \dot{m} \) = \(\frac{pAV}{RT}\)\nWhere p is the pressure, A is the cross-sectional area, V is the velocity, R is the specific gas constant for air, and T is the Temperature. Find initial and final momentum flux rate by multiplying the initial and final velocities by the mass flow rate.
03

Calculate Force

Now, knowing the initial and final momenta, you can find the force that the air exerts on the pipe by using the conservation of momentum equation: \n\(\Delta F = \Delta (m*v)\)\nSubstitute the change in momentum flux for \(\Delta (m*v)\). You will obtain the axial force on the pipe.
04

Identify Direction of Force

Lastly, remember that force is a vector and therefore has a direction. The direction of the force is opposite to the change in velocity. This gives you the final step of identifying the direction of the applied force.

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

When a plane liquid jet strikes an inclined flat plate, it splits into two streams of equal speed but unequal thickness. For frictionless flow there can be no tangential force on the plate surface. Use this assumption to develop an expression for \(h_{2} / h\) as a function of plate angle, \(\theta\). Plot your results and comment on the limiting cases, \(\theta=0\) and \(\theta=90\).

Obtain expressions for the rate of change in mass of the control volume shown, as well as the horizontal and vertical forces required to hold it in place, in terms of \(p_{1}, A_{1}, V_{1}, p_{2}\) \(A_{2}, V_{2}, p_{3}, A_{3}, V_{3}, p_{4}, A_{4}, V_{4},\) and the constant density \(\rho\).

Considering that in the fully developed region of a pipe, the integral of the axial momentum is the same at all cross sections, explain the reason for the pressure drop along the pipe.

Incompressible liquid of negligible viscosity is pumped, at total volume flow rate \(Q,\) through two small holes into the narrow gap between closely spaced parallel disks as shown. The liquid flowing away from the holes has only radial motion. Assume uniform flow across any vertical section and discharge to atmospheric pressure at \(r=R\) Obtain an expression for the pressure variation and plot as a function of radius. Hint: Apply conservation of mass and the momentum equation to a differential control volume of thickness \(d r\) located at radius \(r\)

A porous round tube with \(D=60 \mathrm{mm}\) carries water. The inlet velocity is uniform with \(V_{1}=7.0 \mathrm{m} / \mathrm{s}\). Water flows radially and axisymmetrically outward through the porous walls with velocity distribution \\[ v=V_{0}\left[1-\left(\frac{x}{L}\right)^{2}\right] \\] where \(V_{0}=0.03 \mathrm{m} / \mathrm{s}\) and \(L=0.950 \mathrm{m} .\) Calculate the mass flow rate inside the tube at \(x=L\)
See all solutions

Recommended explanations on Physics Textbooks

Energy Physics

Read Explanation

Waves Physics

Read Explanation

Nuclear Physics

Read Explanation

Magnetism

Read Explanation

Turning Points in Physics

Read Explanation

Solid State 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.