/*! 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 8 The velocity field for a plane s... [FREE SOLUTION] | 91Ó°ÊÓ

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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?

Short Answer

Expert verified
For a particle moving along the wall, its velocity and acceleration equations are derived using the given velocity field. By verifying, it is concluded that both velocity and acceleration normal to wall are zero. The pressure gradient along the wall is obtained using Bernoulli's equation. The nature of the pressure gradient is then determined based on whether it is positive or negative.

Step by step solution

01

Task 1: Derive the velocity along the wall

Let's start by calculating the velocity formula for fluid particle moving along the wall. With the walls aligned along the x axis, the y coordinate for a particle on the wall becomes 0. So, substitute \(y=0\) into the given velocity field equation to get the velocity. Likewise, because \(\hat{j}\) component refers to y-direction, it will be zero for a particle moving along the x-axis. So ignore \(\hat{j}\) part of the equation. Also, know that \(h=1\), \(x=\) any distance between 0 and \(+10h\)
02

Task 2: Derive acceleration along the wall

Acceleration is the derivative of velocity with respect to time. For a steady flow, the convective acceleration can be represented as \(\vec{A}= \vec{V}. \nabla \vec{V}\). Here \(\nabla \vec{V}\) is the gradient of velocity and can be determined by deriving the velocity equation with respect to position \(x\) and \(\vec{V}. \nabla \vec{V}\) is dot product of the velocity and its gradient.
03

Task 3: Verifying velocity and acceleration normal to the wall

To ensure the correctness of the previous steps, verify that velocity and acceleration normal to the wall are zero. Simply, if a particle moves along the wall (x-axis), its velocity and acceleration in the y-direction (\(\hat{j}\) component) should be zero.
04

Task 4: Plotting the pressure gradient along the wall

Now that we have the velocity and acceleration, we can go further and derive the pressure gradient \(\partial p/\partial x\) along the wall. Using Bernoulli's equation for steady, incompressible flow: \( p+ \frac{1}{2} \rho V^2 = constant\), where \(p\) is pressure, \(\rho\) is fluid density and \(V\) is velocity, derive \(\partial p/\partial x\) by differentiating the equation with respect to position \(x\).
05

Task 5: Determine the type of pressure gradient

It's time to interpret our results. If the pressure gradient \(\partial p/\partial x\) is positive, it means the pressure increases in the flow direction (adverse gradient). If it's negative, the pressure decreases in the flow direction (favorable gradient). Thus, the nature of pressure gradient can be decided.

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Most popular questions from this chapter

An incompressible liquid with negligible viscosity and density \(\rho=1250 \mathrm{kg} / \mathrm{m}^{3}\) flows steadily through a \(5-\mathrm{m}\) -long convergent-divergent section of pipe for which the area varies as \\[ A(x)=A_{0}\left(1+e^{-x / a}-e^{-x / 2 a}\right) \\] where \(A_{0}=0.25 \mathrm{m}^{2}\) and \(a=1.5 \mathrm{m} .\) Plot the area for the first \(5 \mathrm{m} .\) Develop an expression for and plot the pressure gradient and pressure versus position along the pipe, for the first \(5 \mathrm{m}\) if the inlet centerline velocity is \(10 \mathrm{m} / \mathrm{s}\) and inlet pressure is 300 kPa. Hint: Use relation \\[ u \frac{\partial u}{\partial x}=\frac{1}{2} \frac{\partial}{\partial x}\left(u^{2}\right) \\]

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