> The normal velocity should be set to zero by some means, there are more than one way to do that though.
I've tried all combinations of setting/not setting the velocity and "normal-tangential velocity", but I'm stumped. Basically when I set "normal-tangential velocity", the entire area of the heatsink acts as if it's solid and the air has a speed of zero between the fins. I could not find another way of setting the speed at the boundary to zero -- if you could point me to some alternate way, it'd be appreciated.
Cooling problem
-
- Site Admin
- Posts: 4832
- Joined: 22 Aug 2009, 11:57
- Antispam: Yes
- Location: Espoo, Finland
- Contact:
Re: Cooling problem
Hi
How many elements do you have between the fins? If you have linear basis having just one element cannot give any flow. I would try to start with fever fins to get an estimate how many elements you need to describe the flow between the fins.
-Peter
How many elements do you have between the fins? If you have linear basis having just one element cannot give any flow. I would try to start with fever fins to get an estimate how many elements you need to describe the flow between the fins.
-Peter
Re: Cooling problem
In the slice through the middle of the fins I can see at least 7 triangles on a few spot checks -- so I assume 7 tetrahedras. That's a good suggestion (starting with fewer fins), but it's still strange there's no flow.
Unfortunately, for now, I cannot go with a much finer mesh than 308K elements because my machine cannot handle more than that (Elmer fails in allocating the data structures).
Thanks,
Tibi
Unfortunately, for now, I cannot go with a much finer mesh than 308K elements because my machine cannot handle more than that (Elmer fails in allocating the data structures).
Thanks,
Tibi
Re: Cooling problem
I've re-modeled the heatsink to have only 5 fins in the direction of the airflow, with plenty of space between the fins.
This time I've done the simulation on a simplified system model, which had only the airbox and this new heatsink (no other components -- the model discussed above had about 200 extra bodies). However, the results are pretty much the same -- there seems to be something that just slows down the flow through the fins. I'm attaching three images, maybe it helps with visualizing what's happening. The first image is through the middle of the heatsink. The next two (44mm and 46mm) are slices through the system showing velocity at 1mm below the top of the heatsink and 1mm above the top of the heatsink. As you can see, even there at the top the velocity is very low, which I don't think it should be the case.
Thanks.Re: Cooling problem
The last slice picture:
Tibi
Thanks,Tibi
-
- Site Admin
- Posts: 4832
- Joined: 22 Aug 2009, 11:57
- Antispam: Yes
- Location: Espoo, Finland
- Contact:
Re: Cooling problem
Hi Tibi
The results look reasonable to me. Assume you have laminar flow in a channel. The max. velocity should scale with the square of the channel width. Looking at the widths between the channels and outside the cooler a ratio 1/4 seems rather close which would give speed ratio 1/16. Of course the geometry is more complex than this but I really don't see a problem myself.
Now of course the flow need not be laminar. What's the Reynolds number? Turbulence will give more plug-flow kind of average velocity profiles which will reduce the difference in maximum velocities in the system.
-Peter
The results look reasonable to me. Assume you have laminar flow in a channel. The max. velocity should scale with the square of the channel width. Looking at the widths between the channels and outside the cooler a ratio 1/4 seems rather close which would give speed ratio 1/16. Of course the geometry is more complex than this but I really don't see a problem myself.
Now of course the flow need not be laminar. What's the Reynolds number? Turbulence will give more plug-flow kind of average velocity profiles which will reduce the difference in maximum velocities in the system.
-Peter
Re: Cooling problem
Hi Peter,
thanks for the feedback. As I've mentioned, I'm not an expert in thermal analysis, so can't answer your questions regarding laminar flow and Reynolds number. I'm including the .sif file, maybe it has the answers to your questions.
To me it still seems a bit strange that the air simply cannot flow between the fins -- the inlet velocity is pretty high, and there should be something going in between the fins. It just seems to hit a wall in front of the heatsink, as is shown in this image (again a slice through the middle of the heatsink, but with vectors based on velocity).
Maybe I should change the type of flow for the air?
Thanks,
Tibi
thanks for the feedback. As I've mentioned, I'm not an expert in thermal analysis, so can't answer your questions regarding laminar flow and Reynolds number. I'm including the .sif file, maybe it has the answers to your questions.
Code: Select all
Header
CHECK KEYWORDS Warn
Mesh DB "." "demo"
Include Path ""
Results Directory ""
End
Simulation
Max Output Level = 40
Coordinate System = Cartesian
Coordinate Mapping(3) = 1 2 3
Simulation Type = Steady State
Steady State Max Iterations = 1
Output Intervals = 1
Timestepping Method = BDF
BDF Order = 1
Solver Input File = demo.sif
End
Constants
Gravity(4) = 0 -1 0 9.82
Stefan Boltzmann = 5.67e-08
Permittivity of Vacuum = 8.8542e-12
Boltzmann Constant = 1.3807e-23
Unit Charge = 1.602e-19
End
Solver 1
Equation = Navier-Stokes
Procedure = "FlowSolve" "FlowSolver"
Variable = Flow Solution[Velocity:3 Pressure:1]
Exec Solver = Always
Stabilize = True
Bubbles = False
Lumped Mass Matrix = False
Optimize Bandwidth = True
Steady State Convergence Tolerance = 5.0e-2
Nonlinear System Convergence Tolerance = 5.0e-2
Nonlinear System Max Iterations = 5
Nonlinear System Newton After Iterations = 10
Nonlinear System Newton After Tolerance = 1.0e-3
Nonlinear System Relaxation Factor = 1
Linear System Solver = Iterative
Linear System Iterative Method = BiCGStabL
BiCGStabL Polynomial Degree = Integer 4
Linear System Max Iterations = 500
Linear System Convergence Tolerance = 1.0e-6
Linear System Preconditioning = ILUT
Linear System ILUT Tolerance = 1.0e-1
Linear System Abort Not Converged = True
Linear System Residual Output = 1
Linear System Precondition Recompute = 1
End
Solver 4
Equation = KEpsilon
Procedure = "KESolver" "KESolver"
Variable = KE[Kinetic Energy:1 Kinetic Dissipation:1]
Exec Solver = Never
Bubbles = True
Lumped Mass Matrix = False
Optimize Bandwidth = True
Steady State Convergence Tolerance = 5.0e-2
Nonlinear System Convergence Tolerance = 5.0e-2
Nonlinear System Max Iterations = 20
Nonlinear System Newton After Iterations = 30
Nonlinear System Newton After Tolerance = 1.0e-3
Nonlinear System Relaxation Factor = .25
Linear System Solver = Iterative
Linear System Iterative Method = BiCGStabL
BiCGStabL Polynomial Degree = Integer 4
Linear System Max Iterations = 500
Linear System Convergence Tolerance = 1.0e-8
Linear System Preconditioning = ILUT
Linear System ILUT Tolerance = 1.0e-2
Linear System Abort Not Converged = True
Linear System Residual Output = 1
Linear System Precondition Recompute = 1
End
Solver 2
Equation = Heat Equation
Procedure = "HeatSolve" "HeatSolver"
Variable = -dofs 1 Temperature
Exec Solver = Always
Stabilize = True
Bubbles = False
Lumped Mass Matrix = False
Optimize Bandwidth = True
Steady State Convergence Tolerance = 5.0e-2
Nonlinear System Convergence Tolerance = 1.0e-2
Nonlinear System Max Iterations = 1
Nonlinear System Newton After Iterations = 10
Nonlinear System Newton After Tolerance = 1.0e-3
Nonlinear System Relaxation Factor = 1
Linear System Solver = Iterative
Linear System Iterative Method = BiCGStab
BiCGStabL Polynomial Degree = Integer 2
Linear System Max Iterations = 5000
Linear System Convergence Tolerance = 1.0e-8
Linear System Preconditioning = ILUT
Linear System ILUT Tolerance = 1.0e-2
Linear System Abort Not Converged = True
Linear System Residual Output = 1
Linear System Precondition Recompute = 1
End
Solver 3
Exec Solver = After Simulation
Procedure = "ResultOutputSolve" "ResultOutputSolver"
Output File Name = "..\output"
Output Format = "vtk"
End
Equation 1
Name = "Solids"
Active Solvers(1) = 2
Convection = None
End
Equation 2
Name = "Air"
Convection = Computed
NS Convect = False
Active Solvers(3) = 1 4 2
End
Material 1
Name = "Aluminum"
Heat Capacity = 921
Density = 2700
Heat Conductivity = 200
End
Material 2
Name = "Air"
Heat Capacity = 1005
Density = 1.1141
Heat Conductivity = 0.0271
Compressibility Model = Incompressible
Viscosity = 1.7014E-05
Heat Expansion Coefficient = 0.0
Specific Heat Ratio = 1.4
Reference Temperature = 25
KE Cmu = .09
KE C1 = 1.44
KE C2 = 1.92
KE SigmaK = 1
KE SigmaE = 1.3
KE Clip = 1.0e-6
Viscosity Model = KE
Heat Conductivity Model = KE
End
Initial Condition 1
Name = "Initial Condition"
Velocity 1 = -13.1
Temperature = 40
Kinetic Energy = 0.024025
Kinetic Dissipation = 0.0865271957591413
End
Body Force 1
Name = "Buoyancy"
Boussinesq = True
End
Body 2
Name = "HS_1"
Equation = 1
Material = 1
Target Bodies(1) = 2
End
Body 1
Name = "airbox_full"
Equation = 2
Body Force = 1
Initial Condition = 1
Material = 2
Target Bodies(1) = 1
End
Boundary Condition 1
Target Boundaries(1) = 5
Name = "Airbox Boundary 1"
Normal-Tangential Velocity = True
Velocity 1 = 0
Wall Law = True
Boundary Layer Thickness = 0.00353585422641078
Surface Roughness = 9
Temperature = 40
End
Boundary Condition 2
Target Boundaries(1) = 3
Name = "Airbox Boundary 2"
Normal-Tangential Velocity = True
Velocity 1 = 0
Wall Law = True
Boundary Layer Thickness = 0.00353585422641078
Surface Roughness = 9
Temperature = 40
End
Boundary Condition 3
Target Boundaries(1) = 2
Name = "Airbox Boundary 3"
Normal-Tangential Velocity = True
Velocity 1 = 0
Wall Law = True
Boundary Layer Thickness = 0.00353585422641078
Surface Roughness = 9
Temperature = 40
End
Boundary Condition 4
Target Boundaries(1) = 6
Name = "Airbox Boundary 4"
Normal-Tangential Velocity = True
Velocity 1 = -13.1
Temperature = 40
End
Boundary Condition 5
Target Boundaries(1) = 4
Name = "Airbox Boundary 5"
Normal-Tangential Velocity = True
Velocity 1 = 0
Wall Law = True
Boundary Layer Thickness = 0.00353585422641078
Surface Roughness = 9
Temperature = 40
End
Boundary Condition 6
Target Boundaries(1) = 1
Name = "Airbox Boundary 6"
End
Boundary Condition 7
Name = "Air2Solids 7"
Target Boundaries(93) = 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100
Normal-Tangential Velocity = True
Velocity 1 = 0
Wall Law = True
Boundary Layer Thickness = 0.00001
End
Thanks,
Tibi
Re: Cooling problem
Forgot to mention: the distance between the fins in the direction of the air flow is 3.25cm.