[Nurzhan]

Hello!

I am trying to simulate Newton's cooling process.

Background: I made an experiment where the temperature of a board's surface was monitored over time. Using the experimental data and applying Newton's equation of cooling I determined the heat transfer coefficient of the board. To do so, I fitted Newton's equation to the experimental data.

Currently, I would like to simulate the cooling process of the board using the obtained heat transfer coefficient.

Problem: Although I fitted the Newton's cooling equation into the experimental results quite accurately, my simulation does not match the experimental results in the same way as the fitted equation?

Newton's cooling equations was solved as: T=Ta+C exp(-h A t / V / rho / Cp)

where Ta - outside temperature, C - the coefficient of integration, h - heat transfer coefficient, t - the time, A - the are of the boards surface, V - the board's volume, rho - density, and Cp is heat capacity.

Thank you in advance.
Nurzhan

[ghorrocks]

Please post an image of what you are modelling, and include an image showing the boundary conditions. Also post an image of your mesh and your CCL.

[Nurzhan]

Hi Glenn.

Thanks for your reply.

Here all supporting images and CCL code

https://www.dropbox.com/s/qdpxhkzzhh...Block.msh?dl=0
https://www.dropbox.com/s/zfo794iihb...ation.ccl?dl=0
https://www.dropbox.com/s/6brz5hfeob...0view.png?dl=0
https://www.dropbox.com/s/nqrs7cyrc7...sides.png?dl=0
https://www.dropbox.com/s/53wx6i8bsw...0loss.png?dl=0


LIBRARY:
MATERIAL: Wood
Material Description = Wood
Material Group = User
Option = Pure Substance
Thermodynamic State = Solid
PROPERTIES:
Option = General Material
EQUATION OF STATE:
Density = 1040.79 [kg m^-3]
Molar Mass = 1.0 [kg kmol^-1]
Option = Value
END
SPECIFIC HEAT CAPACITY:
Option = Value
Specific Heat Capacity = 3527.76 [J kg^-1 K^-1]
END
THERMAL CONDUCTIVITY:
Option = Value
Thermal Conductivity = 0.504 [W m^-1 K^-1]
END
END
END
END
FLOW: Flow Analysis 1
SOLUTION UNITS:
Angle Units = [rad]
Length Units = [m]
Mass Units = [kg]
Solid Angle Units = [sr]
Temperature Units = [K]
Time Units = [s]
END
ANALYSIS TYPE:
Option = Transient
EXTERNAL SOLVER COUPLING:
Option = None
END
INITIAL TIME:
Option = Automatic with Value
Time = 0 [s]
END
TIME DURATION:
Option = Total Time
Total Time = 330 [s]
END
TIME STEPS:
Option = Timesteps
Timesteps = 0.5 [s]
END
END
DOMAIN: Default Domain
Coord Frame = Coord 0
Domain Type = Solid
Location = B16
BOUNDARY: Heat loss
Boundary Type = WALL
Location = F20.16
BOUNDARY CONDITIONS:
HEAT TRANSFER:
Heat Transfer Coefficient = 74.65 [W m^-2 K^-1]
Option = Heat Transfer Coefficient
Outside Temperature = 18 [C]
END
END
END
BOUNDARY: Insulated
Boundary Type = WALL
Location = Back,Bottom,Left,Right,Top
BOUNDARY CONDITIONS:
HEAT TRANSFER:
Option = Adiabatic
END
END
END
DOMAIN MODELS:
DOMAIN MOTION:
Option = Stationary
END
MESH DEFORMATION:
Option = None
END
END
SOLID DEFINITION: Solid 1
Material = Wood
Option = Material Library
MORPHOLOGY:
Option = Continuous Solid
END
END
SOLID MODELS:
HEAT TRANSFER MODEL:
Option = Thermal Energy
END
THERMAL RADIATION MODEL:
Option = None
END
END
END
INITIALISATION:
Option = Automatic
INITIAL CONDITIONS:
TEMPERATURE:
Option = Automatic with Value
Temperature = 60.2 [C]
END
END
END
OUTPUT CONTROL:
RESULTS:
File Compression Level = Default
Option = Standard
END
TRANSIENT RESULTS: Transient Results 1
File Compression Level = Default
Option = Standard
OUTPUT FREQUENCY:
Option = Timestep Interval
Timestep Interval = 30
END
END
END
SOLVER CONTROL:
ADVECTION SCHEME:
Option = High Resolution
END
CONVERGENCE CONTROL:
Maximum Number of Coefficient Loops = 10
Minimum Number of Coefficient Loops = 1
Timescale Control = Coefficient Loops
END
CONVERGENCE CRITERIA:
Residual Target = 1.E-4
Residual Type = RMS
END
TRANSIENT SCHEME:
Option = Second Order Backward Euler
TIMESTEP INITIALISATION:
Option = Automatic
END
END
END
END
COMMAND FILE:
Version = 14.5
Results Version = 14.5
END
SIMULATION CONTROL:
EXECUTION CONTROL:
EXECUTABLE SELECTION:
Double Precision = Off
END
INTERPOLATOR STEP CONTROL:
Runtime Priority = Standard
MEMORY CONTROL:
Memory Allocation Factor = 1.0
END
END
PARALLEL HOST LIBRARY:
HOST DEFINITION: chen005
Host Architecture String = winnt-amd64
Installation Root = C:\Apps\ANSYS Inc\v%v\CFX
END
END
PARTITIONER STEP CONTROL:
Multidomain Option = Independent Partitioning
Runtime Priority = Standard
EXECUTABLE SELECTION:
Use Large Problem Partitioner = Off
END
MEMORY CONTROL:
Memory Allocation Factor = 1.0
END
PARTITIONING TYPE:
MeTiS Type = k-way
Option = MeTiS
Partition Size Rule = Automatic
END
END
RUN DEFINITION:
Run Mode = Full
Solver Input File = Fluid Flow CFX_001.res
END
SOLVER STEP CONTROL:
Runtime Priority = Standard
MEMORY CONTROL:
Memory Allocation Factor = 1.0
END
PARALLEL ENVIRONMENT:
Number of Processes = 1
Start Method = Serial
END
END
END
END[IMG]

Many thanks.
Nurzhan

[Nurzhan]

In addition, I would like to ask how to remove heat conduction term from the governing equation using GUI?

[ghorrocks]

Have you done all the normal sensitivity checks? In other words:

* Is you mesh fine enough?
* Are you converging tight enough?
* Is you time step small enough?

This is an FAQ: http://www.cfd-online.com/Wiki/Ansys..._inaccurate.3F

[Nurzhan]

Hi Glenn.

Please correct me, if I am wrong. Since, I used Newton's law of cooling, which basically describes the surface cooling, then to simulate this process, the heat equation at the last node must be desctized. In order to simulate it correctly, the governing equation at the surface of the block is described as follows:



Because I am only interested in the surface temperature and not in the conduction within the body, then using FVM - implicit scheme:

,

where , , (as mentioned earlier only the surface temperature is objective), , and

will be reduced to .

Based on this, the precision of my result will predominantly dependence on the time step rather on the mesh spacing, won't it?

I would appreciate if you could advise me how to add the source terms Sp and Su or just one of them?

Cheers,
Nurzhan

[ghorrocks]

Why not just use the convective boundary condition built into CFX then you don't need any source terms?

Because I am only interested in the surface temperature and not in the conduction within the body

I do not understand this statement. The convective boundary condition is heat transfer from the heat leaving the surface through convection against heat reaching the surface from conduction. So how can ignore conduction?

Your equations appear to assume that T is constant at all locations in the body. CFX does not work this way, if needs to model conduction and thermal mass in the body to get it right. But if the thermal conductivity is high then the temperature gradient due to conduction will be small.

[Nurzhan]

Hi Glenn,

The equation written previously shows the discretisation of the last finite volume within my domain. I do agree that there is conduction within the body as the heat leaving the system. However, the heat flux out of the body is controlled by the boundary condition, where the higher the heat transfer coefficient and the lower the ambient temperature, then the higher is the heat loss.

Nonetheless, this is what I want to do:

During my experiment, I measured the temperature decrease of the block's surface. The the obtained experimental temperature drop can be explained by Newton's law of cooling, which does not take into account any heat conduction inside the body



This equation was fit to the experimental data end h value was obtained.

Assume, I do not know the thermal conductivity of that material (in my case I know it roughly and used it in ANSYS as the software does not allow me to do transient measurement without specified heat conduction), how then I can simulate this cooling process using ANSYS? How would you do it then? Ideally, using the obtained heat transfer coefficient in ANSYS, then the result must be the same as the Newton's equation.

Best regards and many thanks for brainstorming my ideas.

Nurzhan

[ghorrocks]

I am glad to try to help straighten your brainstorm into a nice laminar flow

Newton's Law of cooling as you have written it assumes that the body can be taken as a control volume where a single temperature value represents the whole body. In other words there is no temperature variation in the body and the body can be represented as a single control volume.

CFX does not work this way, you have to discretise your body into a mesh. So if you want to compare your equation to CFX then you are going to have to account for this. I recommend you use a very high thermal conductivity as this will minimise the thermal gradients inside the body. But as long as the thermal conductivity is high enough then the results should match Newton's Law of cooling very accurately.

[Nurzhan]

Hi Glenn,

I was planning to tell you that I finally got what I want, when I saw your post.

As you said, I increased thermal conductivity 1000 W/m/K and it worked very well for me.

Thanks Glenn.

As always you helped me a lot.