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What is a similarity variable, and what is it used for? For what kinds of functions can we expect a similarity solution for a set of partial differential equations to exist?

Short Answer

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#tag_title# Example of Similarity Variables in a PDE #tag_content# Let's consider the heat equation - a parabolic PDE - as an example to illustrate the use of similarity variables: ∂u(x,t)/∂t = α * ∂²u(x,t)/∂x² Here, u(x,t) is the temperature distribution in a one-dimensional rod at position x and time t, and α is the thermal diffusivity. We can assume a similarity solution for the heat equation in the form: u(x,t) = t^(-n) * F(ξ) where ξ = x/t^m represents the similarity variable, F(ξ) is a function of ξ only, and n and m are constants. By substituting this assumed similarity solution into the heat equation, we obtain an ODE for the function F(ξ): F''(ξ) + (2m + 1 - 2n) * ξ * F'(ξ) = 0 Now, by solving this ODE, we can find the similarity solution for the heat equation. In this case, the heat equation, which is a PDE with scaling and symmetry properties, can be transformed into an ODE using similarity variables. This allows us to describe the behavior of the temperature distribution in a rod more easily.

Step by step solution

01

Definition of Similarity Variables

Similarity variables are combinations of independent variables that simplify partial differential equations (PDEs) into ordinary differential equations (ODEs). This technique is known as the method of similarity solutions. They are used to relate the variables in the problem and study the qualitative behavior of the solution.
02

Usage of Similarity Variables

Similarity variables are an essential tool in solving PDEs, as they allow us to find self-similar solutions or invariant solutions for complex problems. They are especially useful in problems with symmetry or repeating structures since the solution pattern can be assumed to have a specific scaling form. In these cases, the similarity method allows us to transform the original PDE into an ODE, which are often easier to solve.
03

Conditions for Similarity Solutions

The method of similarity solutions can be applied to various types of PDEs, such as parabolic, elliptic, and hyperbolic equations, as long as some specific conditions are met. First, the PDE needs to have a certain degree of symmetry or scaling properties. For example, if the PDE describes a physical phenomenon that is invariant under a specific scaling transformation or can be defined by self-similar processes. Second, the PDE must have a well-defined type of invariance group. Depending on the problem, it could be a Lie Group, scaling group, or even symmetry groups. This is important because the similarity method is based on the reduction of the variables by exploiting their invariance properties. In summary, similarity variables can be used for functions that exhibit inherent symmetry or a particular scaling behavior and whose invariance properties can be well described by a suitable group.

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Most popular questions from this chapter

Consider fluid flow over a surface with a velocity profile given as $$ u(y)=100\left(y+2 y^{2}-0.5 y^{3}\right) \mathrm{m} / \mathrm{s} $$ Determine the shear stress at the wall surface, if the fluid is (a) air at $1 \mathrm{~atm}\( and (b) liquid water, both at \)20^{\circ} \mathrm{C}$. Also calculate the wall shear stress ratio for the two fluids, and interpret the result.

A \(5-\mathrm{m} \times 5-\mathrm{m}\) flat plate maintained at a constant temperature of \(80^{\circ} \mathrm{C}\) is subjected to parallel flow of air at \(1 \mathrm{~atm}, 20^{\circ} \mathrm{C}\), and \(10 \mathrm{~m} / \mathrm{s}\). The total drag force acting on the upper surface of the plate is measured to be \(2.4 \mathrm{~N}\). Using the momentum-heat transfer analogy, determine the average convection heat transfer coefficient and the rate of heat transfer between the upper surface of the plate and the air. Evaluate the air properties at \(50^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\).

A metal plate $\left(k=180 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=2800 \mathrm{~kg} / \mathrm{m}^{3}\right.\(, and \)c_{p}=880 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\( ) with a thickness of \)1 \mathrm{~cm}$ is being cooled by air at \(5^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the initial temperature of the plate is \(300^{\circ} \mathrm{C}\), determine the plate temperature gradient at the surface after 2 minutes of cooling. Hint: Use the lumped system analysis to calculate the plate surface temperature. Make sure to verify the application of this method to this problem.

What is the no-slip condition? What causes it?

Air at \(1 \mathrm{~atm}\) is flowing over a flat plate with a free stream velocity of \(70 \mathrm{~m} / \mathrm{s}\). If the convection heat transfer coefficient can be correlated by $\mathrm{Nu}_{x}=0.03 \operatorname{Re}_{x}^{08} \operatorname{Pr}^{1 / 3}$, determine the friction coefficient and wall shear stress at a location \(2 \mathrm{~m}\) from the leading edge. Evaluate air properties at \(20^{\circ} \mathrm{C}\) and $1 \mathrm{~atm}$.

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