/*! 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 12 An incompressible liquid with a ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 12

An incompressible liquid with a density of \(900 \mathrm{kg} / \mathrm{m}^{3}\) and negligible viscosity flows steadily through a horizontal pipe of constant diameter. In a porous section of length \(L=2 \mathrm{m}\) liquid is removed at a variable rate along the length so that the uniform axial velocity in the pipe is \(u(x)=U e^{-x / L},\) where \(U=20 \mathrm{m} / \mathrm{s} .\) Develop expressions for and plot the acceleration of a fluid particle along the centerline of the porous section and the pressure gradient along the centerline. Evaluate the outlet pressure if the pressure at the inlet to the porous section is \(50 \mathrm{kPa}(\text { gage })\)

Short Answer

Expert verified
The expressions for the acceleration and pressure gradient along the centerline of the porous section are given by \(a(x)=U^2*(1/L)*e^{-2x/L}\) and \(dp/dx = -蟻U*(1/L)*e^{-x/L}\) respectively. The outlet pressure for the given conditions can be computed using the formula: \(p(L) = 50kPa - 蟻UL*(1-e^{-1}).\)

Step by step solution

01

Determine the acceleration

We start by calculating the acceleration of the fluid. Acceleration, \(a(x)\), can be determined by taking the derivative of the velocity function with respect to time: \(a(x)=dv/dt\). However, since the flow is steady, we can write it in terms of spatial variation as: \(a(x)=u(du/dx)\). Given \(u(x)=Ue^{-x/L}\), the derivative \(du/dx\) is found as follows: \(du/dx=(-U/L)e^{-x/L}\). Multiply the result by \(u(x)\) to obtain \(a(x)\): \(a(x)=u*(du/dx)=U^2*(1/L)*e^{-2x/L}.\)
02

Derive the pressure gradient

Now we derive the expression for the pressure gradient. We start by noting the momentum equation for steady, frictionless flow: \(dp/dx = -蟻u(du/dx)\) where \(蟻\) is the fluid density. Substituting our known values, we have: \(dp/dx = -蟻U*(1/L)*e^{-x/L}\).
03

Calculate the outlet pressure

To find the outlet pressure, we integrate the pressure gradient from \(0\) (at the inlet) to \(L\) (at the outlet): \(p(x=L) = p(x=0) + 鈭玣rom x=0 to L (dp/dx) dx\). Substituting our expression for \(dp/dx\) gives: \(p(L) = 50kPa - 蟻U*(鈭玣rom x=0 to L e^{-x/L} dx).\) Computing the integral, we obtain: \(p(L) = 50kPa - 蟻UL*(1-e^{-1}),\) where \(蟻=900kg/m^{3}, U=20m/s, L=2m\). Calculate \(p(L)\) to obtain the gage pressure at the outlet of the porous section.

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

To model the velocity distribution in the curved inlet section of a water channel, the radius of curvature of the streamlines is expressed as \(R=L R_{0} / 2 y .\) As an approximation, assume the water speed along each streamline is \(V=10 \mathrm{m} / \mathrm{s}\) Find an expression for and plot the pressure distribution from \(y=0\) to the tunnel wall at \(y=L / 2,\) if the centerline pressure (gage) is \(50 \mathrm{kPa}, L=75 \mathrm{mm},\) and \(R_{0}=0.2 \mathrm{m} .\) Find the value of \(V\) for which the wall static pressure becomes 35 kPa.

The \(x\) component of velocity in an incompressible flow field is given by \(u=A x,\) where \(A=2 \mathrm{s}^{-1}\) and the coordinates are measured in meters. The pressure at point \((x, y)=(0,0)\) is \(p_{0}=190 \mathrm{kPa}\) (gage). The density is \(\rho=1.50 \mathrm{kg} / \mathrm{m}^{3}\) and the \(z\) axis is vertical. Evaluate the simplest possible \(y\) component of velocity. Calculate the fluid acceleration and determine the pressure gradient at point \((x, y)=(2,1) .\) Find the pressure distribution along the positive \(x\) axis.

Imagine a garden hose with a stream of water flowing out through a nozzle. Explain why the end of the hose may be unstable when held a half meter or so from the nozzle end.

Determine whether the Bernoulli equation can be applied between different radii for the vortex flow fields (a) \(\vec{V}=\omega r \hat{e}_{\theta}\) and (b) \(\vec{V}=\hat{e}_{\theta} K / 2 \pi r\)

You (a young person of legal drinking age) are making homemade beer. As part of the process you have to siphon the wort (the fermenting beer with sediment at the bottom) into a clean tank using a \(5-\mathrm{mm}\) ID tubing. Being a young engineer, you're curious about the flow you can produce. Find an expression for and plot the flow rate \(Q\) (liters per minute \()\) versus the differential in height \(h\) (millimeters) between the wort free surface and the location of the hose exit. Find the value of \(h\) for which \(Q=2 \mathrm{L} / \mathrm{min}\)
See all solutions

Recommended explanations on Physics Textbooks

Classical Mechanics

Read Explanation

Electric Charge Field and Potential

Read Explanation

Translational Dynamics

Read Explanation

Scientific Method Physics

Read Explanation

Radiation

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.