/*! 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 6 The \(x\) component of velocity ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 6

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.

Short Answer

Expert verified
The simplest \(y\) component of velocity is \(v = -2y + C\). The fluid acceleration is \(\frac{du}{dt} = 4x - 4y\). The pressure gradient at point (2,1) is 6 Pa/m. The pressure distribution along the positive x axis is given by the equation \(p2 = p0 - \rho g u^2 + \rho g u2^2\).

Step by step solution

01

Finding the simplest y-component of velocity

In incompressible flow, the divergence of the velocity field is zero. Hence, the \(y\) component of the velocity, \(v\), can be found by solving the continuity equation \(\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} = 0\). With \(u = A x = 2x\) and \(\frac{\partial u}{\partial x} = 2\), it is implied \(v\) must be a function of \(y\) only and direct integration of \(\frac{\partial v}{\partial y} = -\frac{\partial u}{\partial x}\) gives \(v = -2y + C\), where \(C\) is the constant of integration.
02

Fluid Acceleration

The acceleration in incompressible flow is determined by the formula \(\frac{du}{dt} = \frac{\partial u}{\partial t} + u \frac{\partial u}{\partial x} + v \frac{\partial u}{\partial y}\). Given that the flow is steady, we have \(\frac{\partial u}{\partial t} = 0\). Also \(\frac{\partial u}{\partial x} = A = 2\). Substituting \(u = 2x\) and \(v = -2y\) gives fluid acceleration as \(\frac{du}{dt} = 4x - 4y\).
03

Pressure Gradient

The pressure gradient at any point can be found by applying the Euler's equation for inviscid flow in \(x\)-direction \(\rho \frac{du}{dt} = - \frac{dp}{dx}\), substituting the variables gives \(-\frac{dp}{dx} = 6\). Hence the pressure gradient is 6 Pa/m.
04

Pressure Distribution Along x-axis

Applying Bernoulli's equation along a streamline from \(x = 0\) to \(x = 2\) gives \(\frac{p0}{\rho g} + \frac{u^2}{2g} = \frac{p2}{\rho g} + \frac{u2^2}{2g} \), where \(p0\) and \(p2\) are the pressures at \(x = 0\) and \(x = 2\) respectively and \(u2 = 2x = 4m/s\). Solving for \(p2 = p0 - \rho g u^2 + \rho g u2^2\), gives the pressure distribution.

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

Consider the flow represented by the stream function \(\psi=A x^{2} y,\) where \(A\) is a dimensional constant equal to 2.5 \(\mathrm{m}^{-1} \cdot \mathrm{s}^{-1}\). The density is \(1200 \mathrm{kg} / \mathrm{m}^{3}\). Is the flow rotational? Can the pressure difference between points \((x, y)=(1,4)\) and (2,1) be evaluated? If so, calculate it, and if not, explain why.

A horizontal axisymmetric jet of air with 0.4 in. diameter strikes a stationary vertical disk of 7.5 in. diameter. The jet speed is \(225 \mathrm{ft} / \mathrm{s}\) at the nozzle exit. A manometer is connected to the center of the disk. Calculate (a) the deflection, if the manometer liquid has \(\mathrm{SG}=1.75,(\mathrm{b})\) the force exerted by the jet on the disk, and (c) the force exerted on the disk if it is assumed that the stagnation pressure acts on the entire forward surface of the disk. Sketch the streamline pattern and plot the distribution of pressure on the face of the disk.

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 the flow field formed by combining a uniform flow in the positive \(x\) direction and a source located at the origin. Let \(U=30 \mathrm{m} / \mathrm{s}\) and \(q=150 \mathrm{m}^{2} / \mathrm{s} .\) Plot the ratio of the local velocity to the freestream velocity as a function of \(\theta\) along the stagnation streamline. Locate the points on the stagnation streamline where the velocity reaches its maximum value. Find the gage pressure there if the fluid density is \(1.2 \mathrm{kg} / \mathrm{m}^{3}\)

\(\mathrm{A}\) crude model of a tornado is formed by combining a \(\operatorname{sink},\) of strength \(q=2800 \mathrm{m}^{2} / \mathrm{s},\) and a free vortex, of strength \(K=5600 \mathrm{m}^{2} / \mathrm{s} .\) Obtain the stream function and velocity potential for this flow field. Estimate the radius beyond which the flow may be treated as incompressible. Find the gage pressure at that radius.
See all solutions

Recommended explanations on Physics Textbooks

Further Mechanics and Thermal Physics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Turning Points in Physics

Read Explanation

Particle Model of Matter

Read Explanation

Astrophysics

Read Explanation

Conservation of Energy and 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.