SteadyStateThermalResults
Steady-state thermal solution and derived quantities
Description
A SteadyStateThermalResults object contains
the temperature and temperature gradient values in a form convenient for plotting and
postprocessing.
The temperature and its gradients are calculated at the nodes of the triangular or
tetrahedral mesh generated by generateMesh. Temperature values at the nodes
appear in the Temperature property. The three components of the
temperature gradient at the nodes appear in the XGradients,
YGradients, and ZGradients properties.
To interpolate the temperature or its gradients to a custom grid (for example,
specified by meshgrid), use interpolateTemperature or evaluateTemperatureGradient.
To evaluate heat flux of a thermal solution at nodal or arbitrary spatial locations,
use evaluateHeatFlux. To evaluate integrated heat
flow rate normal to a specified boundary, use evaluateHeatRate.
Creation
Solve a steady-state thermal problem using the solve function. This function returns a steady-state thermal solution as
a SteadyStateThermalResults object.
Properties
Object Functions
evaluateHeatFlux | Evaluate heat flux of thermal solution at nodal or arbitrary spatial locations |
evaluateHeatRate | Evaluate integrated heat flow rate normal to specified boundary |
evaluateTemperatureGradient | Evaluate temperature gradient of thermal solution at arbitrary spatial locations |
interpolateTemperature | Interpolate temperature in thermal result at arbitrary spatial locations |
Examples
Solve a 3-D steady-state thermal problem.
Create an femodel object for a steady-state thermal problem and include a geometry representing a block.
model = femodel(AnalysisType="thermalSteady", ... Geometry="Block.stl");
Plot the block geometry.
pdegplot(model.Geometry, ... FaceLabels="on", ... FaceAlpha=0.5) axis equal
Assign material properties.
model.MaterialProperties = ...
materialProperties(ThermalConductivity=80);Apply a constant temperature of 100 °C to the left side of the block (face 1) and a constant temperature of 300 °C to the right side of the block (face 3). All other faces are insulated by default.
model.FaceBC(1) = faceBC(Temperature=100); model.FaceBC(3) = faceBC(Temperature=300);
Mesh the geometry and solve the problem.
model = generateMesh(model); thermalresults = solve(model)
thermalresults =
SteadyStateThermalResults with properties:
Temperature: [12822×1 double]
XGradients: [12822×1 double]
YGradients: [12822×1 double]
ZGradients: [12822×1 double]
Mesh: [1×1 FEMesh]
The solver finds the temperatures and temperature gradients at the nodal locations. To access these values, use thermalresults.Temperature, thermalresults.XGradients, and so on. For example, plot temperatures at the nodal locations.
pdeplot3D(thermalresults.Mesh,ColorMapData=thermalresults.Temperature)
Analyze heat transfer in a rod with a circular cross-section and internal heat generation by simplifying a 3-D axisymmetric model to a 2-D model.
The 2-D model is a rectangular strip whose x-dimension extends from the axis of symmetry to the outer surface and whose y-dimension extends over the actual length of the rod (from -1.5 m to 1.5 m). Create the geometry by specifying the coordinates of its four corners. For axisymmetric models, the toolbox assumes that the axis of rotation is the vertical axis passing through r = 0.
g = decsg([3 4 0 0 .2 .2 -1.5 1.5 1.5 -1.5]');
Plot the geometry with the edge labels.
pdegplot(g,EdgeLabels="on") axis equal
Create an femodel object for steady-state thermal analysis and include the geometry.
model = femodel(AnalysisType="thermalSteady", ... Geometry=g);
Switch the type of the problem to axisymmetric.
model.PlanarType = "axisymmetric";The rod is composed of a material with these thermal properties.
k = 40; % thermal conductivity, W/(m*C) q = 20000; % heat source, W/m^3
For a steady-state analysis, specify the thermal conductivity of the material.
model.MaterialProperties = ...
materialProperties(ThermalConductivity=k);Specify the internal heat source.
model.FaceLoad = faceLoad(Heat=q);
Define the boundary conditions. There is no heat transferred in the direction normal to the axis of symmetry (edge 1). You do not need to change the default boundary condition for this edge. Edge 2 is kept at a constant temperature T = 100 °C.
model.EdgeBC(2) = edgeBC(Temperature=100);
Specify the convection boundary load on the outer boundary (edge 3). The surrounding temperature at the outer boundary is 100 °C, and the heat transfer coefficient is .
model.EdgeLoad(3) = edgeLoad(ConvectionCoefficient=50,...
AmbientTemperature=100);The heat flux at the bottom of the rod (edge 4) is .
model.EdgeLoad(4) = edgeLoad(Heat=5000);
Generate the mesh.
model = generateMesh(model);
figure
pdemesh(model)
axis equal
Solve the problem.
thermalresults = solve(model);
The solver finds the temperatures and temperature gradients at the nodal locations. To access these values, use thermalresults.Temperature, thermalresults.RGradients, and thermalresults.ZGradients. For example, plot temperatures at the nodal locations.
T = thermalresults.Temperature; figure pdeplot(thermalresults.Mesh,XYData=T,Contour="on") axis equal title("Steady-State Temperature")
