/*! 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 121 \(\mathrm{A}\) crude model of a ... [FREE SOLUTION] | 91Ó°ÊÓ

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\(\mathrm{A}\) crude model of a tornado is formed by combining a \(\operatorname{sink},\) of strength \(q=2800 \mathrm{m}^{2} / \mathrm{s},\) and a free vortex, of strength \(K=5600 \mathrm{m}^{2} / \mathrm{s} .\) Obtain the stream function and velocity potential for this flow field. Estimate the radius beyond which the flow may be treated as incompressible. Find the gage pressure at that radius.

Short Answer

Expert verified
The stream function for this flow field is \(1400/ \pi r\) m²/s and the velocity potential is \(-1400 cos\theta/\pi\) m²/s. The specific values of the radius for the flow to be treated as incompressible and the gage pressure at that radius depend on the values of the speed of sound and the fluid density.

Step by step solution

01

Calculate the Stream Function

Stream function (\( \psi \)) for a flow comprising a sink and a free vortex can be given by superposing the stream functions for each component. The stream function for a sink is \(-q/\(2\pi r\), and for a free vortex it's \(K/\(2\pi r\). So, the combined stream function is \((K-q)/2\pi r\). Given that \(q=2800 m^2/s\) and \(K=5600 m^2/s\), we get \( \psi = (5600-2800)/2\pi r = 1400/ \pi r \)
02

Calculate the Velocity Potential

The velocity potential (\(\phi\)) is null for a free vortex and \(-q/2\pi cos\theta\) for a sink. By adding both terms \(\phi = -q/2\pi cos\theta\), given that \(q=2800 m^2/s\), we get \(\phi = -2800/2\pi cos\theta = -1400 cos\theta/\pi \)
03

Calculate Radius for Incompressible Flow

A flow field may be treated as incompressible if the speed of the flow is much less than the speed of sound. For our tornado model, the radial velocity (\(u_r\)) is \( -q/2\pi r\), and the tangential velocity (\(u_\theta\)) is \( K/2\pi r\). The speed \(U\) \(= sqrt(u_r^2 + u_\theta^2)\). Let's equate \(U/c\) (c is speed of sound) to a very small number (like 0.03), and solve this equation for \(r\).
04

Calculate the Gage Pressure at the Radius

The gage pressure at a point in a fluid is given by \(P = P_0 + 1/2 * rho * U^2\), where \(P_0\) is reference pressure (we can take it as atmospheric pressure), rho is fluid density, and \(U\) is speed of the flow. Substitute \(U\) from step 3 into this equation to calculate the gage pressure. The value of \(rho\) should be known or assumed.

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