/*! 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 24 A velocity field is given by \(\... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 24

A velocity field is given by \(\vec{V}=a x t \hat{i}-b y \hat{j},\) where \(a=0.1 \mathrm{s}^{-2}\) and \(b=1 \mathrm{s}^{-1}\). For the particle that passes through the point \((x, y)=(1,1)\) at instant \(t=0\) s, plot the pathline during the interval from \(t=0\) to \(t=3\) s. Compare with the streamlines plotted through the same point at the instants \(t=0,1,\) and \(2 \mathrm{s}\).

Short Answer

Expert verified
The pathline for the particle during \(t=0\) to \(t=3\)s is given by \(x(t) = 1 + 0.05t^2\) and \(y(t) = 1 - t\), which is a parabola moving up in the x-direction and a straight line dropping in y-direction. The streamlines through the point at \(t=0,1,2\)s show similar trajectories, representing the snapshot of fluid flow at those instants of time. The main difference is that the pathline shows the entire path of the fluid particle between 0 and 3s whilst the streamlines are at concrete instants of time.

Step by step solution

01

Write Down Given Information

The velocity vector function is given by \(\vec{V}=a x t \hat{i} - b y \hat{j}\), with \(a=0.1 \, \mathrm{s}^{-2}\) and \(b=1 \, \mathrm{s}^{-1}\). We also know that the particle is at point \((1,1)\) at time \(t=0\).
02

Find the Pathline

The pathline is the trajectory that a fluid particle will take in the field of fluid flow. We can integrate the velocity vector to get the position function. For the x-component, we have \(x(t) = \int a x t \, dt \), and for the y-component, we have \(y(t) = \int -b y \, dt\).
03

Solve Integrals

For the integrals, note that the term \(x(t)\) or \(y(t)\) in the integral complicates things as these are functions of \(t\), not constants. However, at \(t = 0\), \(x = 1\) and \(y = 1\), so we can treat \(x\) and \(y\) as constants while integrating. This will give us an approximation for small \(t\). Doing this, the result yields \(x(t) = 1 + 0.05t^2\) and \(y(t) = 1 - t\).
04

Plot the Pathline

Use the resulting equations to plot the pathline from \(t=0\) to \(t=3\). This is a parabola in the x-direction and a straight line dropping in y-direction. This represents the trajectory of the particle.
05

Find Streamlines

Streamlines at a given instant in time provide a snapshot of the flow field. They are described by \(dx/Vx = dy/Vy\). Substituting for \(Vx\) and \(Vy\) and using \(a\) and \(b\) values, integrate this equation from the point \((1,1)\). Do this for the required time points \(t=0,1,2\).
06

Plot Streamlines

Use the resulting equations to plot the streamlines at \(t=0,1,2\). Compare the streamline plots with the pathline plot. They should match at the respective time points because both streamlines and pathlines represent the trajectory of the particle, but streamlines are time-instant specific whereas a pathline represents the particle's entire journey.

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Most popular questions from this chapter

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