/*! 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 2 An incompressible frictionless f... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An incompressible frictionless flow field is given by \(\vec{V}=(A x+B y) \hat{i}+(B x-A y) \hat{j},\) where \(A=2 \mathrm{s}^{-1}\) and \(B=2 \mathrm{s}^{-1}\) and the coordinates are measured in meters. Find the magnitude and direction of the acceleration of a fluid particle at point \((x, y)=(2,2) .\) Find the pressure gradient at the same point, if \(\vec{g}=-g j\) and the fluid is water.

Short Answer

Expert verified
The acceleration of the fluid particle is \(0 \ ms^-2\). The pressure gradient at the point (2, 2) is -9800 Pa/m in the x-direction and 0 Pa/m in the y-direction.

Step by step solution

01

Find the velocity

First, plug the given coordinates into the velocity vector. With \( (x, y) = (2,2) \) and \( A = B = 2 s^{-1} \), the velocity vector, \( \vec{V} \), becomes \( \vec{V} = ((2 s^{-1} * 2 m) + (2 s^{-1} * 2 m)) \hat{i} + ((2 s^{-1} * 2 m) - (2 s^{-1} * 2 m))\hat{j} = 8 \hat{i} ms^{-1}. \)
02

Calculate the acceleration

The acceleration of the fluid particle is calculated as the derivative of its velocity with respect to time. In this case, since the flow is steady, the Lagrangian and Eulerian accelerations are the same, and hence, the time derivative of velocity is zero. So, acceleration \( A = 0 \, ms^{-2} \).
03

Compute the pressure gradient

To determine the pressure gradient, we apply the momentum equation, also known as Euler’s equation, for incompressible and frictionless fluid: \( \frac{dp}{dx} = -\rho ( ag + \frac{du^2}{dx}; \ \frac{dp}{dy} = -\rho ( \frac{dv^2}{dy})\). Here \( \frac{dp}{dx} = -\rho ( ag + \frac{du^2}{dx} ) \) and \( \frac{dp}{dy} = -\rho ( \frac{dv^2}{dy} )\), where \( \rho \) is the fluid density, \( a \) acceleration and \( g \) gravity's effect. With acceleration \( a = 0 m/s^{2} \), and velocity \( u = v = 8 m/s \), we find \( \frac{dp}{dx} = -\rho g \) and \( \frac{dp}{dy} = 0 \). For water as fluid, \( \rho = 1000 kg/m^3 \) and \( g = 9.8 m/s^2 \). Hence, the pressure gradient \( \frac{dp}{dx} = -9800\n, Pa/m \) and \( \frac{dp}{dy} = 0\n Pa/m \).

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 91Ó°ÊÓ!

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.

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.

Steady, frictionless, and incompressible flow from left to right over a stationary circular cylinder, of radius \(a,\) is represented by the velocity field \\[ \vec{V}=U\left[1-\left(\frac{a}{r}\right)^{2}\right] \cos \theta \hat{e}_{r}-U\left[1+\left(\frac{a}{r}\right)^{2}\right] \sin \theta \hat{e}_{\theta} \\] Obtain an expression for the pressure distribution along the streamline forming the cylinder surface, \(r=a\). Determine the locations where the static pressure on the cylinder is equal to the freestream static pressure.

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 \(1.2 \mathrm{kg} / \mathrm{m}^{3}\)

The inlet contraction and test section of a laboratory wind tunnel are shown. The air speed in the test section is \(U=50 \mathrm{m} / \mathrm{s} .\) A total- head tube pointed upstream indicates that the stagnation pressure on the test section centerline is \(10 \mathrm{mm}\) of water below atmospheric. The laboratory is maintained at atmospheric pressure and a temperature of \(-5^{\circ} \mathrm{C}\) Evaluate the dynamic pressure on the centerline of the wind tunnel test section. Compute the static pressure at the same point. Qualitatively compare the static pressure at the tunnel wall with that at the centerline. Explain why the two may not be identical.

See all solutions

Recommended explanations on Physics Textbooks

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.