test15.m

matlab中PDE工具箱的帮助文件pde.pdf的例子全部用代码实现,花了我一个月的时间,有限元应用程序

pdecirc(0,0,1,'C1')
pdecirc(0,0,0.8,'C2')
pdecirc(0,0,0.6,'C3')
pdecirc(0,0,0.5,'C4')
pdecirc(0,0,0.4,'C5')
pderect([-0.2 0.2 0.2 0.9],'R1')
pderect([-0.1 0.1 0.2 0.9],'R2')
pderect([0 1 0 1],'SQ1')
(C1+C2+C3+C4+C5+R1+R2)*SQ1
% In boundary mode you need to remove a number of subdomain borders. Using
% Shift-click, select borders and remove them using the Remove Subdomain
% Border option from the Boundary menu until the geometry consists of four
% subdomains: the stator, the rotor, the coil, and the air gap. In the plot below,
% the stator is subdomain 1, the rotor is subdomain 2, the coil is subdomain 3,
% and the air gap is subdomain 4. Note that the numbering of your subdomains
% may be different.
% Before moving to the PDE mode, select the boundaries along the x-axis and set
% the boundary condition to a Neumann condition with g = 0 and q = 0. In the
% PDE mode, turn on the labels by selecting the Show Subdomain Labels option
% from the PDE menu. Double-click each subdomain to define the PDE
% parameters:
% ?In the coil both μ and J are 1, so the default values do not need to be changed.
% ?In the stator and the rotor μ is nonlinear and defined by the equation above.
% Enter μ as
% 5000./(1+0.05*(ux.^2+uy.^2))+200
% ux.^2+uy.^2 is equal to |?A |2. J is 0 (no current).
% ?In the air gap μ is 1, and J is 0.
% 52
% Initialize the mesh, and continue by opening the Solve Parameters dialog box
% by selecting Parameters from the Solve menu. Since this is a nonlinear
% problem, the nonlinear solver must be invoked by checking the Use nonlinear
% solver. If you want, you can adjust the tolerance parameter. The adaptive
% solver can be used together with the nonlinear solver. Solve the PDE and plot
% the magnetic flux density B using arrows and the equipotential lines of the
% magnetostatic potential A using a contour plot. The plot clearly shows, as
% expected, that the magnetic flux is parallel to the equipotential lines of the
% magnetostatic potential.