Cooling problem

Numerical methods and mathematical models of Elmer
tibich72
Posts: 63
Joined: 07 Dec 2009, 05:16

Re: Cooling problem

Post by tibich72 »

> 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.
raback
Site Admin
Posts: 4832
Joined: 22 Aug 2009, 11:57
Antispam: Yes
Location: Espoo, Finland
Contact:

Re: Cooling problem

Post by raback »

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
tibich72
Posts: 63
Joined: 07 Dec 2009, 05:16

Re: Cooling problem

Post by tibich72 »

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
tibich72
Posts: 63
Joined: 07 Dec 2009, 05:16

Re: Cooling problem

Post by tibich72 »

I've re-modeled the heatsink to have only 5 fins in the direction of the airflow, with plenty of space between the fins.
heatsink3d.jpg
Heatsink Model
(215.37 KiB) Not downloaded yet
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.
velocity_slice_25mm.jpg
Slice through middle of heatsink
(55.9 KiB) Not downloaded yet
velocity_slice_44mm.jpg
Slice 1mm below the top of heatsink
(52.41 KiB) Not downloaded yet
Thanks.
tibich72
Posts: 63
Joined: 07 Dec 2009, 05:16

Re: Cooling problem

Post by tibich72 »

The last slice picture:
velocity_slice_46mm.jpg
Slice 1mm above the top of the heatsink
(53.3 KiB) Not downloaded yet
Thanks,
Tibi
raback
Site Admin
Posts: 4832
Joined: 22 Aug 2009, 11:57
Antispam: Yes
Location: Espoo, Finland
Contact:

Re: Cooling problem

Post by raback »

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
tibich72
Posts: 63
Joined: 07 Dec 2009, 05:16

Re: Cooling problem

Post by tibich72 »

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.

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

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).
velocity_slice_25mm vectors.jpg
Slice middle heatsink w/ vectors
(249.29 KiB) Not downloaded yet
Maybe I should change the type of flow for the air?

Thanks,
Tibi
tibich72
Posts: 63
Joined: 07 Dec 2009, 05:16

Re: Cooling problem

Post by tibich72 »

Forgot to mention: the distance between the fins in the direction of the air flow is 3.25cm.
Post Reply