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Chapter 4: Problem 17

A tornado has the following velocity components in polar coordinates: \\[ V_{r}=-\frac{C_{1}}{r} \quad \text { and } \quad V_{\theta}=-\frac{C_{2}}{r} \\] Note that the air is spiraling inward. Find an equation for the streamlines. \(r\) and \(\theta\) are polar coordinates.

Short Answer

Expert verified
The equation of the streamlines for a tornado given the velocity components in polar coordinates is \( r = C_{1} \theta C_{2} + C_{1}C \)

Step by step solution

01

Understand the basic definitions

First, it's important to note that a streamline is a curve that is everywhere tangent to the velocity vector. In other words, the fluid particles move along the streamline curve. In polar coordinates, the streamline is defined by the equation: \[ \frac{dr}{V_{r}} = \frac{rd\theta}{V_{\theta}} \] We are given that the radial component of velocity \( V_{r} \) is -\(\frac{C_{1}}{r} \) and the angular component of velocity \( V_{\theta} \) is -\(\frac{C_{2}}{r} \). So we can substitute them into the streamline equation.
02

Substituting the given values

After substituting the given velocity components in the equation, we get: \[ \frac{dr}{-C_{1}/r} = \frac{rd\theta}{-C_{2}/r} \] Simplify the fractions, we get: \[ -C_{1} dr = -C_{2} d\theta \] Divide through by -C_{1} and -C_{2}, we get: \[ \frac{dr}{C_{1}} = \frac{d\theta}{C_{2}} \]
03

Integrate both sides

Now it's time to integrate both sides of the equation: \[ \int \frac{dr}{C_{1}} = \int \frac{d\theta}{C_{2}} \] which gives: \[ \frac{r}{C_{1}} = \frac{\theta}{C_{2}} + C \] where C is the constant of integration.
04

Rearranging the equation

Finally, rearranging the above equation to form the equation of streamlines, we will get: \[ r = C_{1} \theta C_{2} + C{1}C \] This is the final equation for the streamlines of a tornado.

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Most popular questions from this chapter

A constant-density fluid flows in the converging, twodimensional channel shown in Fig. P4.20. The width perpendicular to the paper is quite large compared to the channel height. The velocity in the \(z\) direction is zero. The channel half-height, \(Y\), and the fluid \(x\) velocity, \(u,\) are given by \\[ Y=\frac{Y_{0}}{1+x / \ell} \quad \text { and } \quad u=u_{0}\left(1+\frac{x}{\ell}\right)\left[1-\left(\frac{y}{Y}\right)^{2}\right] \\] Where \(x, y, Y,\) and \(\ell\) are in meters, \(u\) is in \(\mathrm{m} / \mathrm{s}, u_{0}=1.0 \mathrm{m} / \mathrm{s},\) and \(Y_{0}=1.0 \mathrm{m} .\) (a) Is this flow steady or unsteady? Is it one-dimensional, two-dimensional, or three- dimensional? (b) Plot the velocity distribution \(u(y)\) at \(x / \ell=0,0.5,\) and \(1.0 .\) Use \(y / Y\) values of 0 \\[ \pm 0.2,\pm 0.4,\pm 0.6,\pm 0.8, \text { and }\pm 1.0 \\]

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A constant-density fluid flows through a converging section having an area \(A\) given by \\[ A=\frac{A_{0}}{1+(x / \ell)} \\] where \(A_{0}\) is the area at \(x=0 .\) Determine the velocity and acceleration of the fluid in Eulerian form and then the velocity and acceleration of a fluid particle in Lagrangian form. The velocity is \(V_{0}\) at \(x=0\) when \(t=0\)

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