Search results
Results from the WOW.Com Content Network
The methods used for solving two dimensional Diffusion problems are similar to those used for one dimensional problems. The general equation for steady diffusion can be easily derived from the general transport equation for property Φ by deleting transient and convective terms [1]
The following steps comprise the finite volume method for one-dimensional steady state diffusion - STEP 1 Grid Generation. Divide the domain into equal parts of small domain. Place nodal points at the center of each small domain. Dividing small domains and assigning nodal points (Figure 1) Create control volumes using these nodal points.
The unsteady convection–diffusion problem is considered, at first the known temperature T is expanded into a Taylor series with respect to time taking into account its three components. Next, using the convection diffusion equation an equation is obtained from the differentiation of this equation.
The diffusion equation is a parabolic partial differential equation.In physics, it describes the macroscopic behavior of many micro-particles in Brownian motion, resulting from the random movements and collisions of the particles (see Fick's laws of diffusion).
The name arises for two reasons. First, the method relies on computing the solution in small steps, and treating the linear and the nonlinear steps separately (see below). Second, it is necessary to Fourier transform back and forth because the linear step is made in the frequency domain while the nonlinear step is made in the time domain.
The Crank–Nicolson stencil for a 1D problem. The Crank–Nicolson method is based on the trapezoidal rule, giving second-order convergence in time.For linear equations, the trapezoidal rule is equivalent to the implicit midpoint method [citation needed] —the simplest example of a Gauss–Legendre implicit Runge–Kutta method—which also has the property of being a geometric integrator.
For premium support please call: 800-290-4726 more ways to reach us
Cnoidal wave solution to the Korteweg–De Vries equation, in terms of the square of the Jacobi elliptic function cn (and with value of the parameter m = 0.9). Numerical solution of the KdV equation u t + uu x + δ 2 u xxx = 0 (δ = 0.022) with an initial condition u(x, 0) = cos(πx).