# Mixed BC - heat transfer - laplacianFoam

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May 12, 2011, 05:39
What is valueExpression?
#21
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Grim
Join Date: Mar 2011
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Can someone explain what the line valueExpression "Tinf"; does?
I see it used repeatedly, such as in the example below, but don't understand what it's for.

Quote:
 Originally Posted by herbert Code: ```rightWall { type groovyBC; variables "htot=1000.0;Tinf=20.0;rho=800.0;cp=385.0;k=DT*rho*cp;"; valueExpression "Tinf"; fractionExpression "1.0/(1.0 + k/(mag(delta())*htot))"; }```

 May 12, 2011, 05:51 #22 New Member   Raimonds Vilums Join Date: Oct 2010 Posts: 17 Rep Power: 7 Tinf stand for T_infinity. This notation is quite often used for room or ambient temperature

 May 12, 2011, 06:10 #23 New Member   Grim Join Date: Mar 2011 Posts: 6 Rep Power: 6 Why are both valueExpression and fractionExpression used? Could valueExpression be left out? Also, there seems to be a need for a Neumann condition, why is there no gradientExpression? I'm confused as to how fractionExpression works in combination with valueExpression.

 May 12, 2011, 10:17 #24 New Member   Brent Craven Join Date: Mar 2009 Location: University Park, PA, USA Posts: 21 Rep Power: 8 It is somewhat explained here: Mixed BC - heat transfer - laplacianFoam

 May 12, 2011, 11:37 #25 Senior Member   Dr. Alexander Vakhrushev Join Date: Mar 2009 Posts: 213 Rep Power: 10 To implement 2nd order on the boundary you need bigger stencil to be used. To get this you need a fictitious cell from "outside", while for the unstructured grid it is not possible to understand which neighbor is correct from the side of the internal field. Mixed BC can not treat this equation at the boundary, because you cannot include variable delta vectors in your settings (as well as "k" and "h" can be variable). __________________ Best regards, Dr. Alexander VAKHRUSHEV Christian Doppler Laboratory for "Advanced Process Simulation of Solidification and Melting" Simulation and Modelling of Metallurgical Processes Department of Metallurgy University of Leoben Franz-Josef-Str. 18 A - 8700 Leoben Österreich / Austria Tel.: +43 3842 - 402 - 3125 http://smmp.unileoben.ac.at Last edited by makaveli_lcf; May 13, 2011 at 02:30.

 May 29, 2011, 04:32 #26 Senior Member   Fabian Braennstroem Join Date: Mar 2009 Posts: 407 Rep Power: 10 Hello to all, I have some questions as well... The explanation of Raimonds was quite interesting, but what I do not understand is where the outer heat transfer coefficient and wall heat conductivity and thickness come into the formula? Or better, how can I remove the calculated inner heat transfer coefficient from the total heat transfer coefficient used in the above equation? I would rather use: type groovyBC; value uniform 1; gradientExpression "gradT"; fractionExpression "0"; variables "Cp=-1006;alphaa=10.0;d=0.1;lambda=1.7;kValue=(1.0/((1.0/(alphaEff*Cp*mag(delta())))+d/lambda+1/alphaa));Tamb=270;heatFlux=kValue*(T-Tamb);gradT=heatFlux/(alphaEff * Cp);" Would be nice, if you can comment on this! Fabian

 May 29, 2011, 04:36 #27 Senior Member   Dr. Alexander Vakhrushev Join Date: Mar 2009 Posts: 213 Rep Power: 10 Here this case is for zero-thickness walls. You just balance your heat fluxes from fluid side and wall side. __________________ Best regards, Dr. Alexander VAKHRUSHEV Christian Doppler Laboratory for "Advanced Process Simulation of Solidification and Melting" Simulation and Modelling of Metallurgical Processes Department of Metallurgy University of Leoben Franz-Josef-Str. 18 A - 8700 Leoben Österreich / Austria Tel.: +43 3842 - 402 - 3125 http://smmp.unileoben.ac.at

 May 29, 2011, 06:17 #28 Senior Member   Fabian Braennstroem Join Date: Mar 2009 Posts: 407 Rep Power: 10 Hi Alexander, thanks for the quick response! But how would I set the value for htot? Is there the fluid-side (inner) wall heat transfer coefficient included? Do you think my mentioned groovyBC approach is good for non-zero-thickness walls? Or is it maybe not needed for room walls in HVAC related problems and I just can use the zero-thickness approach!? Best Regards Fabian

 May 29, 2011, 10:45 #29 Senior Member   Ben K Join Date: Feb 2010 Location: Ottawa, Canada Posts: 140 Rep Power: 10 Fabian: Your method is like setting a fixedGradient boundary condition which updates the gradient value after each internal iteration, alternatively you could do the same with a fixedValue boundary condition. This should work, but I believe it will be slower to converge than the other method mentioned in this post which is more implicit. Read this post: Mixed BC - heat transfer - laplacianFoam

 May 29, 2011, 10:48 #30 Senior Member   Dr. Alexander Vakhrushev Join Date: Mar 2009 Posts: 213 Rep Power: 10 Hi Fabian! Let's clarify the BC which are discussed here: you have your heat transfer coefficient (HTC) defined for the heat exchange between solid and liquid regions. It is already includes wall thickness, the only thing which is also defined is external temperature. You can suppose that it is you room temperature or cooling system liquid temperature. It is a kind of the simplified approach: you have HTC parameter coming e.g. from experimental measurements and you neglect (do not model) heat conduction processes through your wall. The liquid flux is also estimated here assuming "laminar" model for turbulent flow: q = lambda_eff * sGrad(T). If you include wall functionsm it becomes more complicate when discretized. Considering your question: what model and equations do intend to simulate for the wall. It can be: 1) Full 3D conjugated heat transfer task (see some solvers and tutorials in OF). 2) To simplify the calculations you can assume 2D or even 1D wall heat conduction. __________________ Best regards, Dr. Alexander VAKHRUSHEV Christian Doppler Laboratory for "Advanced Process Simulation of Solidification and Melting" Simulation and Modelling of Metallurgical Processes Department of Metallurgy University of Leoben Franz-Josef-Str. 18 A - 8700 Leoben Österreich / Austria Tel.: +43 3842 - 402 - 3125 http://smmp.unileoben.ac.at

 May 29, 2011, 14:49 #31 Senior Member   Fabian Braennstroem Join Date: Mar 2009 Posts: 407 Rep Power: 10 Hello to you both, thanks for the explanation! Though I still have some troubles to really understand :-( Let's say I would like to model a wall with: alphaOuter=10W/m2K a assumed (not modelled) thickness of d=0.1m with a heat conduction coefficient lambda=2.3W/mK and an ambient temperature of 273.15K... What would I need to set for 'htot'? Is it just htot= 1/(d/lambda+1/alphaOuter)? rightWall { type groovyBC; variables "htot=1000.0;Tinf=273.15;cp=1006.0;k=alphaEff*cp;" ; valueExpression "Tinf"; fractionExpression "1.0/(1.0 + k/(mag(delta())*htot))"; }Thanks for your help! Fabian

 May 30, 2011, 04:57 #32 New Member   Raimonds Vilums Join Date: Oct 2010 Posts: 17 Rep Power: 7 Let's get clear with mathematics. The BC, which I implemented, is k, h, Tinf are known What is your formula to be implemented? P.S. I found this useful link to generate image url which displays latex formula

May 31, 2011, 03:12
#33
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Fabian Braennstroem
Join Date: Mar 2009
Posts: 407
Rep Power: 10
Hello,

Quote:
 Originally Posted by argonaut Let's get clear with mathematics. The BC, which I implemented, is k, h, Tinf are known What is your formula to be implemented? P.S. I found this useful link to generate image url which displays latex formula
I would say, that I have almost the same, but for h I used:
h=1/(1/alphaInner+d/lambda+1/alphaOuter)

With alphaInner=alphaEff*Cp*mag(delta()))
which is probably not correct to include in h (k-value), as I use the wall temperature (T) and not the fluid temperature.

So it should rather be:

variables "Cp=-1006;alphaa=10.0;d=0.1;lambda=1.7;kValue=(1.0/(d/lambda+1/alphaa));Tamb=270;heatFlux=kValue*(T-Tamb);gradT=heatFlux/(alphaEff * Cp);"

And for the existing wallHeatTransfer alphaWall would be the same as the above kValue. (the missing part for the wallHeatTransfer bc is the use of alphaEff)

Is this correct?
Thanks! Fabian

 May 31, 2011, 09:19 #34 Senior Member   Ben K Join Date: Feb 2010 Location: Ottawa, Canada Posts: 140 Rep Power: 10 Fabian: Just out of interest, are you trying to come up with a modified heat transfer coefficient (h) instead of directly modelling the temperature drop through the wall which I'm guessing is like a thin interface? If so, have you considered making a thermal conductivity field as a function of position where the thermal conductivity changes from one value to another at the inner wall interface? For this, you might also need to refine your mesh a bit around the wall, but then you can just use a normal definition of the heat transfer coefficient.

 May 31, 2011, 11:30 #35 Senior Member   Fabian Braennstroem Join Date: Mar 2009 Posts: 407 Rep Power: 10 Hello Ben, sorry.. I have trouble to follow :-( I try it in other words, my scenario looks like this: I have a really simple room with walls (like the hotRoom) tutorial. In addition I have a free cylinder placed in the middle of the room with an heatflux boundary condition. I am running buoyantSimpleFoam. Now I would like to set for the walls a boundary condition for temperature, which adjusts the temperature (or gradient) of these walls, according to a given heat transfer coefficient. In the end of the hotroom and heated cylinder (1000W) I would like set the boundary condition that I will get a kind of mean air temperature of 20 Celsius with an outer ambient temperature of 10 Celsius. This would work out with a total heat transfer coefficient of 1W/m2K when the walls have an total area of 100m2. Best Regards! Fabian

 January 14, 2012, 14:16 #36 Member   ehsan Join Date: Mar 2009 Posts: 98 Rep Power: 8 Hello I like to imply a simple adiabatic wall B.C for rhoCenteralFoam? Could you please give me a clue? Thanks

 February 9, 2012, 10:03 #37 New Member   Join Date: Feb 2012 Posts: 4 Rep Power: 5 Dear All, I am quite new to OpenFoam and I try to implement heat transfer coefficient in laplacian foam via groovybc. I downloaded and compiled Swak4Foam sucessfully and added following expressions in "0/T": __________________________________________________ _________ corner_wide_side { type groovyBC; value uniform 1800; variables ( "htot = 1200.0;" "Tinf = 323.15;" "rho = 7200.0;" "cp = 700.0;" "k = 30.0;" ); valueExpression "Tinf"; fractionExpression "1.0/(1.0 + k/(mag(delta())*htot))"; __________________________________________________ __________ I receive the following error message: Calculating temperature distribution Time = 0.04 word::stripInvalid() called for word htot For debug level (= 2) > 1 this is considered fatal Abgebrochen Any ideas? Best regards!

 February 9, 2012, 10:30 #38 New Member   Join Date: Feb 2012 Posts: 4 Rep Power: 5 Well, I think I was able to solve the problem by deleting the spaces - seems to work now. Best regards

 September 20, 2012, 17:59 changing refValue #39 Member   ,... Join Date: Apr 2011 Posts: 92 Rep Power: 5 Hi Do you guys have any idea how I can change the value of refValue to some other value than the default 0? When I am solving the heat transfer problem with convective bc following the discussions above, my final steady temperature is always 0, which seems to be wrong; the temperature should, ultimately, reach the ambient temperature.

 September 20, 2012, 18:05 #40 Senior Member   Dr. Alexander Vakhrushev Join Date: Mar 2009 Posts: 213 Rep Power: 10 For convective BC you have to write your own implementation based on mixed type __________________ Best regards, Dr. Alexander VAKHRUSHEV Christian Doppler Laboratory for "Advanced Process Simulation of Solidification and Melting" Simulation and Modelling of Metallurgical Processes Department of Metallurgy University of Leoben Franz-Josef-Str. 18 A - 8700 Leoben Österreich / Austria Tel.: +43 3842 - 402 - 3125 http://smmp.unileoben.ac.at

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