Download Tutorial 11 Summary Background

Tutorial 11 Summary In this tutorial we will examine the transient behavior of the quenching process of a two dimensional triangle. The equation that describes this process is known as the diffusion equation. To solve this equation, a Brownian motion scheme will be used to obtain the average temperature as a function of time. Background Diffusion Diffusion describes the spread of particles through random motion from regions of high concentration to regions of low concentration. This concentration gradient is the reason of the mass transfer from one region of a system to another. A classic example is the diffusion of an ink drop in a glass of water. Due to the random nature of diffusion, this process can be described using Brownian motion. Brownian motion is the random motion of particles suspended in a fluid (a liquid or a gas) resulting from their collision with the other atoms or molecules in the gas or liquid. Mass and energy transport Next semester (in your transport phenomena class), you will learn that the equation that governs mass transfer (Fick’s law) has the same mathematical form as the equation that governs energy transfer (Fourier law): ߲‫ܥ‬
߲ଶ‫ܥ‬
= ‫ ܦ‬ଶ ߲‫ݐ‬
߲‫ݔ‬
߲ܶ
߲ଶܶ
= ߙ ଶ ߲‫ݐ‬
߲‫ݔ‬
The two equations above are the unsteady state one dimensional Fick law and Fourier law, respectively. In the Fick’s equation, the letters C, t, D and x stand for concentration, time, diffusivity and distance, respectively. In the Fourier equation, T, t, D and x represent temperature, time, thermal diffusivity and distance. We can see that both physical processes (mass and energy transfer) are governed by the same equations. Therefore, we can think of energy transport as mass transfer! Generalizing for three dimensions, the Fourier law becomes: ߲ܶ
= ߙ‫׏‬ଶ ܶ ߲‫ݐ‬
Where ‫׏‬ଶ represents the Laplacian in the x,y,z directions (or any other coordinate system). (1) The problem In this tutorial we will examine the unsteady state mean temperature evolution of a hot triangle immersed in a cold medium. The initial temperature of the triangle is T1 and the surroundings are held at a constant temperature To throughout the process. Since this is an unsteady state problem, the triangle temperature will be time dependent and we expect that it tends to To at sufficiently long times. The triangle is placed in a Cartesian coordinate system and it is centered at x=0. The sides of the triangle are 2b. The picture is shown below: Y Surroundings at To Triangle initially at T1 X Due to symmetry, we expect that the mean temperature on the right half is the same as that on the left. Therefore, the problem becomes: Y Since this is a two dimensional problem, equation 1 becomes: X ߲ଶܶ ߲ଶܶ
߲ܶ
= ߙ ቈ ଶ + ଶ ቉ ߲‫ݐ‬
߲‫ݔ‬
߲‫ݕ‬
We can now set up the boundary conditions for the system. This is a partial differential equation with three independent variables: time, y and x. We have to write down boundary conditions for each of these independent variables. For time, we recognize that a t=0 the temperature inside the triangle is T1: ܶ|௧ୀ଴ = ܶଵ For y, we note that at y=0 (the base of the triangle) the temperature must be To. Moreover, at the border the same temperature must hold. This can be translated into: ܶ|௬ୀ଴ = ܶ௢ ܶ|௬ୀ௛(௫) = ܶ௢ Note that h(x) is the height of the triangle as a function of x, which has the form of a straight line. The equation for h is therefore: ‫ݔ‬
݄(‫ܾ = )ݔ‬ξ3 ቂ1 െ ቃ ܾ
For x, we recognize that there must be a symmetry at x=0. At this point, the temperature must be maximum (because of symmetry). This can be translated into: ߲ܶ
= 0 ฬ
߲‫ ݔ‬௫ୀ଴
We can render this problem dimensionless by defining the following variables: ‫ݔ‬
‫ = כ ݔ‬ ܾ
‫ݕ‬
‫ = כ ݕ‬ ܾ
‫= כݐ‬
ܶ‫= כ‬
ߙ‫ݐ‬
ܾଶ
ܶ െ ܶ௢
ܶଵ െ ܶ௢
Substituting these definitions in the original equation and the boundary condition, we obtain the following problem: ߲ܶ ‫כ‬
߲ଶܶ‫߲ כ‬ଶܶ ‫כ‬
= ቈ ‫כ‬ଶ + ‫כ‬ଶ ቉ ߲‫כ ݐ‬
߲‫ݔ‬
߲‫ݕ‬
ܶ ‫| כ‬௧ ‫ כ‬ୀ଴ = 1 ܶ ‫| כ‬௬‫ כ‬ୀ଴ = 0 ܶ|௬‫ כ‬ୀ௛(௫ ‫ = ) כ‬0 ߲ܶ ‫כ‬
= 0 ฬ
߲‫ כ ݔ‬௫‫ כ‬ୀ଴
Given that ݄‫ = ) כ ݔ( כ‬ξ3[1 െ ‫ ] כ ݔ‬ As you can already imagine, this is a hard problem to solve. We have a second order partial differential equation with three independent variables. There are many numerical methods that can solve this problem. We will use a Brownian motion scheme to solve for the mean temperature. Solution strategy We will first place a large number of points inside the triangle. Each of this points represent little chunks of energy (remember the analogy to mass transfer – these points would be like molecules!). Once we have the x,y coordinates of each of these points, we will propose a random displacement in each of these coordinates (to simulate the stochastic nature of diffusion!). The magnitude of this random displacement has to be consistent with how easy it is for the particle to diffuse. This information is included in the standard deviation of the position squared of the molecules for a Brownian process, which it can be shown to be <x2> = 2Dt, where D is the diffusion constant. Therefore, we can propose a displacement of (2't)1/2 * randn, where randn is a normally distributed random number of mean zero and standard deviation of unity. To enforce the boundary conditions, we will do the following: Reflection: The symmetry boundary condition (dT/dx = 0 at x=0) will be enforced by reflecting the dots Random displacement yields x<0 Put particle back in by setting x=abs(x) that try to cross the boundary x=0. In other words, if the x coordinate of a molecule is less than zero after proposing its random displacement, this x coordinate will be set to the absolute value of x. Dissipation: The boundary conditions at y=0 and y=h will be enforced by letting each of the particles leave the system if the new coordinates (after the random displacements) are out of the triangle. Physically, this means that energy is being lost through the faces at y=0 and y=h(x). Mean temperature profile: The mean temperature will be represented by how many particles are inside the triangle. We expect that all of the particles will eventually escape from our system (this means that the triangle is in thermal equilibrium with the surroundings!).