/*! 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 1 The propagation speed of small-a... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 1

The propagation speed of small-amplitude surface waves in a region of uniform depth is given by $$c^{2}=\left(\frac{\sigma}{\rho} \frac{2 \pi}{\lambda}+\frac{g \lambda}{2 \pi}\right) \tanh \frac{2 \pi h}{\lambda}$$ where \(h\) is depth of the undisturbed liquid and \(\lambda\) is wavelength. Using \(L\) as a characteristic length and \(V_{0}\) as a characteristic velocity, obtain the dimensionless groups that characterize the equation.

Short Answer

Expert verified
The dimensionless groups characterizing the wave speed equation are: \(\Pi_1 = \frac{c}{V_0}\), \(\Pi_2 = \frac{h}{L}\), \(\Pi_3 = \frac{\lambda}{L}\), \(\Pi_4 = \frac{\sigma}{\rho V_0^2 L}\) and \(\Pi_5 = \frac{g L}{V_0^2}\).

Step by step solution

01

Understanding Variables and Their Units

In the given equation, \(\sigma\) is surface tension with unit ML/T\(^2\), \(\rho\) is fluid density with unit ML\(^{-3}\), \(h\) is depth of undisturbed liquid with unit L and \(\lambda\) is wavelength with unit L. \(g\) is acceleration due to gravity with unit LT\(^{-2}\) while \(c\) is wave speed with unit LT\(^{-1}\). Additionally, \(V_0\) is a characteristic velocity with unit LT\(^{-1}\) and \(L\) is a characteristic length with unit L.
02

Buckingham鈥檚 Pi Theorem

Using Buckingham鈥檚 Pi theorem, these variables can be combined into dimensionless parameters. From the theorem, if we have n variables in an equation and they form m fundamental dimensions, we can form n-m independent, dimensionless groups. Here, n=8 (including \(c\), \(h\), \(\lambda\), \(\sigma\), \(\rho\), \(g\), \(L\), and \(V_0\)) and m=3 (M, L, T), so the equation will have n-m=5 dimensionless groups.
03

Formation of Dimensionless Groups

The dimensionless groups can be formed as follows: \(\Pi_1 = \frac{c}{V_0}\), \(\Pi_2 = \frac{h}{L}\), \(\Pi_3 = \frac{\lambda}{L}\), \(\Pi_4 = \frac{\sigma}{\rho V_0^2 L}\) and \(\Pi_5 = \frac{g L}{V_0^2}\). Each group is unitless, thus fulfilling the requirement of the dimensionless group, these can be used to analyze the given equation.
04

Substituting Group Into Equation

Substituting these dimensionless groups back into our original equation, the equation becomes \(\Pi_1^{2}=\left(\Pi_4 \cdot \frac{2 \pi}{\Pi_3}+\Pi_5 \cdot \frac{\Pi_3}{2 \pi}\right) \tanh \frac{2 \pi \cdot \Pi_2}{\Pi_3}\). The left-hand side of the equation is now a function of the right-hand side, in accordance with Buckingham鈥檚 Pi theorem.

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Most popular questions from this chapter

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