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Workshop on Thermal Boundary Conditions

Workshop on Thermal Boundary Conditions. Y. Several thermal boundary conditions will be explored in this workshop modifying the flow on labyrinth case given on LE#1. This section gives a few definitions on thermal boundary conditions before to proceed to the workshop itself. . X.

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Workshop on Thermal Boundary Conditions

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  1. Workshop on Thermal Boundary Conditions Y Several thermal boundary conditions will be explored in this workshop modifying the flow on labyrinth case given on LE#1. This section gives a few definitions on thermal boundary conditions before to proceed to the workshop itself. X

  2. Blockages and Plates It is possible to specify at a blockage made of solid material (> 99) or at a plate : • A fixed temperature at the blockage or at the plate • A fixed heat flux W/m2 for plates and W/m3 for blockages • Adiabatic. For a blockage where a solid material was selected there is no heat source but the blockage may conduct heat! • Linear Heat Source specifies a heat flux due to the convective heat transfer coefficient; q = h(Tamb-Tcell)

  3. Special Case: Non-ParticipatingBlockages Non-participating blockages are defined by ‘other materials’ (not sólid, or líquid or gas) which, despite of not conducting heat they may have heat sources at their faces:

  4. Part I - IsothermalFlow • Upload exercise from LE#1 • Extend the domain accordingly to the figure • Insert: one blockage, one inlet and one outlet (they are reversed) • The fluid is air and the same inlet velocities are applied: 5m/s. • Adjust the grid for the new objects for 3 and 5 volumes along the y direction 0.005 0.003 0.015 Y X 0.025

  5. Part II(a) – HeatTransfer Case • Activate the energy equation • Upper channel: Tin = 200oC • Labyrinth: Tin = 20oC • Made the dividing wall and the pin of cooper, and set them adiabatic. • Keep B1 as material 198 but set the East face T = 200oC and the North face h = 100W/m2/oC with Tamb = 200oC. cooper Y B1 X

  6. Part II – Results (TemperatureProfile) • The hot stream transfer heat to the cold stream through convection-conduction-convection. • The dividing block and the pin transfer heat by conduction only. • Notice that dividing block and the pin were set as adiabatic, therefore there is no heat source on them. • The faces East and North of block B1 have energy sources. • The gray block is of material 198 without any heat source therefore it does not participate on the energy balance.

  7. Part III – HeatTransfer Case • Replace the upper channel fluid by water • Compare your results and comments the differences on the temperature field. • Case where heat convection is coupled with heat conduction are referred to conjugate heat transfer. Air-Air • Water-Air Upload this case q1 you need Upload this case q1 you need

  8. PossibleVariations • If you have a spare time try to: • Change the conducting properties of the dividing wall • Change the conducting properties of the pin (make it work as an heat insulator) • THE END

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