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Chapter 6: Problem 1

Consider the flow field with velocity given by \(\vec{V}=\left[A\left(y^{2}-x^{2}\right)-B x | \hat{i}+[2 A x y+B y] \hat{j}, A=1 \mathrm{ft}^{-1} \cdot \mathrm{s}^{-1}, B=1\right.\) \(\mathrm{ft}^{-1} \cdot \mathrm{s}^{-1} ;\) the coordinates are measured in feet. The density is \(2 \operatorname{slug} / \mathrm{ft}^{3},\) and gravity acts in the negative \(y\) direction. Calculate the acceleration of a fluid particle and the pressure gradient at point \((x, y)=(1,1)\)

Short Answer

Expert verified
The acceleration of a fluid particle at the point (1,1) is \(-4 \hat{i} -2 \hat{j} \, \text{ft/s}^2\) and the pressure gradient at the same point is \(-8 \hat{i} -96.4 \hat{j}\, \text{lb/ft}^2/\text{ft}\).

Step by step solution

01

Calculate the Acceleration of the fluid particle

The acceleration of a fluid particle in a flow field is given by the material derivative of the velocity vector \(\vec{V}\). The material derivative of the velocity is defined as \(D\vec{V}/Dt = \frac{\partial \vec{V}}{\partial t} + \vec{V} \cdot \nabla \vec{V}\). Since the given velocity field doesn't depend on time, our formula simplifies to \(D\vec{V}/Dt = \vec{V} \cdot \nabla \vec{V}\). We need to calculate the gradient of the velocity field and multiply it with the given velocity field at the point (1, 1). After calculating, we find that the acceleration at a point (1,1) is \(\vec{a} = \nabla \vec{V} \cdot \vec{V} = -4 \hat{i} -2 \hat{j} \, \text{ft/s}^2.\)
02

Calculate Pressure Gradient

Using Euler's equation for fluid motion, we can find an expression to calculate the pressure gradient. The equation is expressed as: \(\rho \frac{D\vec{V}}{Dt}= -\nabla P+\rho \vec{g}\). By approximating the gravity force as a constant \(\vec{g}=-g\hat{j}\), where \(g=32.2 \, \text{ft/s}^2\) is the acceleration due to gravity, we re-arrange Euler's equation to solve for the pressure gradient: \(\nabla P= \rho \frac{D\vec{V}}{Dt} - \rho \vec{g}\). Substituting the given values, we get the pressure gradient to be: \(\nabla P= -8 \hat{i} -96.4 \hat{j}\, \text{lb/ft}^2/\text{ft}\) at the point (1,1).

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Most popular questions from this chapter

An incompressible, inviscid fluid flows into a horizontal round tube through its porous wall. The tube is closed at the left end and the flow discharges from the tube to the atmosphere at the right end. For simplicity, consider the \(x\) component of velocity in the tube uniform across any cross section. The density of the fluid is \(\rho,\) the tube diameter and length are \(D\) and \(L,\) respectively, and the uniform inflow velocity is \(v_{0}\) The flow is steady. Obtain an algebraic expression for the \(x\) component of acceleration of a fluid particle located at position \(x,\) in terms of \(v_{0}, x,\) and \(D .\) Find an expression for the pressure gradient, \(\partial p / \partial x,\) at position \(x .\) Integrate to obtain an expression for the gage pressure at \(x=0\)

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