Mech-HT 120 WS03 Thermal Contact

Workshop 3

Thermal Contact

WS 3-1ANSYS, Inc. Proprietary© 2009 ANSYS, Inc. All rights reserved.

August 2009Inventory #002667

ANSYS Mechanical Heat Transfer

Workshop 3 - Thermal Contact

Training ManualProblem Description

• A set of electrical contacts is to be modeled in both an open an dclosed configuration.

– We’ll assume both parts are structural steel.

• While open, the heat load flows only through the “Arm” of theassembly.

• When closed, the contacts allow heat flow through the “Arm” a nd“Base”.

• The contacts are modeled with a gap, so we will use the contact



• The contacts are modeled with a gap, so we will use the contactpinball control to simulate a closed contact configuration .

Base

ArmGap


Training ManualProblem Description

• Boundary Conditions:– Heat flux on end face of arm = 0.5 W/mm 2

– Temperature fixed on 2 bottom surfaces of base = 30 ºC– Convection on 5 faces on arm: h = 2e- 4 W/(mm 2* ºC)

• No convection on heat flux surface or circular cont act region.

– Convection on 8 faces on base: h = 4e -4 W/(mm 2* ºC)• No convection on 2 temperature surfaces or circular contact region.




Training ManualUnits Setup

• Open Workbench and specify the unit system (Metric, kg, mm, s, ºC, mA, N, mV).

• Choose to “Display Values in Project Units”.




Training ManualModel Setup

1. From the Workbench project page toolbox, select a Steady State Thermal analysis system.



2. Right click the Geometry cell and import geometry “Contacts_WS3.x_t”.

3. Double click the Model cell to open the Mechanical application.



4. When the import is complete you will notice that no contact has been defined. This is because the gap is larger than the default search tolerance. We begin by adding manual contact.

5. Highlight the “Connections” branch RMB > Insert > Manual



branch RMB > Insert > Manual Contact Region.

6. Scope one circular face to the contact region and the other to the target (note, the order is unimportant in this example).



7. In the Advanced contact details change the “Pinball Region” setting to “Radius”.– Recall from chapter 3 that, with bonded

and no separation contact, the pinball controls heat flow.

– As long as the contact and target surfaces are within the pinball radius, heat transfer can occur.



heat transfer can occur.

8. Enter a pinball radius = 0.5 mm.• Note a sample pinball is displayed for

visual reference once a radius is entered.



9. Highlight the “Steady State Thermal” branch and add boundary conditions as described:

10.Highlight the end face of the arm, RMB > Insert > Heat Flux.– Magnitude = 0.5 W/mm 2



11.Highlight the 2 bottom surfaces of the base, RMB > Insert > Temperature.– Magnitude = 30 ºC



12.Select 5 faces on the arm (top, bottom, both sides and the opposite end from where the heat flux was applied).

13.RMB > Insert > Convection– Film coefficient = 2e -4 W/(mm 2 * ºC)– Ambient Temperature = 30 ºC

14.Select the 8 faces on the rectangular



14.Select the 8 faces on the rectangular section of the base (not the 2 ends where the temperature load is applied).

15.RMB > Insert > Convection– Film coefficient = 4e -4 W/(mm 2 * ºC)– Ambient Temperature = 30 ºC


Training ManualSolution

• Note, a default mesh is used of this example. In actual practice, always check the mesh quality and add mesh controls as needed.



• Since we assume a steady state linear analysis, no analysis settings will be modified.

16.Solve


Training ManualPostprocessing

17.When the solution completes, drag and drop the temperature and both convection loads onto the Solution branch to setup reaction probes.

18.Evaluate All Results.• Note the reactions from both boundary

conditions on the base are essentially zero.



• Note that the applied heat flux is 0.5 W/mm2 and the area of the face is 40 mm 2, which yields a total heat load of 20 W.

• Reaction on the arm convection = -20 W (steady state verification).



19.Add a temperature result and evaluate the result.– Note the temperature distribution within the

assembly. – As expected, no heat is flowing from the

arm to the base since the contact pinball is smaller than the geometric gap.



• Next we will model a condition where the contacts are closed. While we could modify the geometry to close the gap, instead we will increase the pinball size to simulate the gap is closed. Continued on the next page.



20.Return to the schematic and highlight the top cell in the system, RMB > Rename.

21.Rename the system “Contact Open”.



22.Again highlight the top cell in the system, RMB > Duplicate.

23.Repeat the steps above to rename the new system “Contacts Closed”.


Training ManualModel 2 Setup

24.Double click the Model cell in the “Contact Closed” system to open the new analysis.




Training ManualModel 2 Setup

• Since the model is a duplicate of our original, only the contact needs to be changed.

25.Highlight the contact region and change the pinball radius to 1mm.

• The bonded contact type assumes closed contact whenever contact and target



whenever contact and target faces are within the pinball. we can use this to simulate a closed contact condition here.

26.Solve.


Training Manual

• Again checking the reactions at the boundary conditions we can verify a steady state heat balance.

• RT = 2.3222 W, RC1 = 10.62 W, RC2 = 7.0582 W• RT + RC1 + RC2 = 20 W

Model 2 Postprocessing



– Note the temperature distribution now shows heat transfer from the contact arm to the base as expected.

Mech-HT 120 WS03 Thermal Contact

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