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如何用matlab通过编程求解油膜压力？ 已有1人参与
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cguoguo2014
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2楼2015-10-31 10:56:48
FMStation
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http://www.mathworks.com/help/pd ... n-a-thin-plate.html == Problem Parameters == k = 400; % thermal conductivity of copper, W/(m-K) rho = 8960; % density of copper, kg/m^3 specificHeat = 386; % specific heat of copper, J/(kg-K) thick = .01; % plate thickness in meters stefanBoltz = 5.670373e-8; % Stefan-Boltzmann constant, W/(m^2-K^4) hCoeff = 1; % Convection coefficient, W/(m^2-K) % The ambient temperature is assumed to be 300 degrees-Kelvin. ta = 300; emiss = .5; % emissivity of the plate surface == Create the PDE Model with a single dependent variable == numberOfPDE = 1; pdem = createpde(numberOfPDE); == Geometry == width = 1; height = 1; % define the square by giving the 4 x-locations followed by the 4 % y-locations of the corners. gdm = [3 4 0 width width 0 0 0 height height]'; g = decsg(gdm, 'S1', ('S1')'); % Convert the DECSG geometry into a geometry object % on doing so it is appended to the PDEModel geometryFromEdges(pdem,g); % Plot the geometry and display the edge labels for use in the boundary % condition definition. figure; pdegplot(pdem,'edgeLabels','on'); axis([-.1 1.1 -.1 1.1]); title 'Geometry With Edge Labels Displayed'; == Definition of PDE Coefficients == c = thick*k; % Because of the radiation boundary condition, the "a" coefficient % is a function of the temperature, u. It is defined as a MATLAB % expression so it can be evaluated for different values of u % during the analysis. a = @(~,state) 2*hCoeff + 2*emiss*stefanBoltz*state.u.^3; f = 2*hCoeff*ta + 2*emiss*stefanBoltz*ta^4; d = thick*rho*specificHeat; specifyCoefficients(pdem,'m',0,'d',0,'c',c,'a',a,'f',f); == Boundary Conditions == applyBoundaryCondition(pdem,'Edge',1,'u',1000); == Initial guess == setInitialConditions(pdem,0); == Mesh == hmax = .1; % element size msh = generateMesh(pdem,'Hmax',hmax); figure; pdeplot(pdem); axis equal title 'Plate With Triangular Element Mesh' xlabel 'X-coordinate, meters' ylabel 'Y-coordinate, meters' == Steady State Solution == R = solvepde(pdem); u = R.NodalSolution; figure; pdeplot(pdem,'xydata',u,'contour','on','colormap','jet'); title 'Temperature In The Plate, Steady State Solution' xlabel 'X-coordinate, meters' ylabel 'Y-coordinate, meters' axis equal p=msh.Nodes; plotAlongY(p,u,0); title 'Temperature As a Function of the Y-Coordinate' xlabel 'X-coordinate, meters' ylabel 'Temperature, degrees-Kelvin' fprintf('Temperature at the top edge of the plate = %5.1f degrees-K\n', ... u(4)); == Transient Solution == specifyCoefficients(pdem,'m',0,'d',d,'c',c,'a',a,'f',f); endTime = 5000; tlist = 0:50:endTime; numNodes = size(p,2); % Set the initial temperature of all nodes to ambient, 300 K u0(1:numNodes) = 300; % Find all nodes along the bottom edge and set their initial temperature % to the value of the constant BC, 1000 K nodesY0 = abs(p(2,  ) < 1.0e-5;u0(nodesY0) = 1000; ic = @(~) u0; setInitialConditions(pdem,ic); % Set solver options pdem.SolverOptions.RelativeTolerance = 1.0e-3; pdem.SolverOptions.AbsoluteTolerance = 1.0e-4; % |solvepde| automatically picks the parabolic solver to obtain the solution. R = solvepde(pdem,tlist); u = R.NodalSolution; figure; plot(tlist,u(3, );grid on title 'Temperature Along the Top Edge of the Plate as a Function of Time' xlabel 'Time, seconds' ylabel 'Temperature, degrees-Kelvin' % figure; pdeplot(pdem,'xydata',u(:,end),'contour','on','colormap','jet'); title(sprintf('Temperature In The Plate, Transient Solution( %d seconds)\n', ... tlist(1,end))); xlabel 'X-coordinate, meters' ylabel 'Y-coordinate, meters' axis equal; % fprintf('\nTemperature at the top edge of the plate(t = %5.1f secs) = %5.1f degrees-K\n', ... tlist(1,end), u(4,end)); |
3楼2016-06-22 06:30:10
FMStation
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4楼2016-06-22 06:42:36
FMStation
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5楼2016-06-22 06:43:32
