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May 24, 2017, 00:56 
Heat transfer

#1 
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BlackHeartInertia
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HI. How do you simulate heat transfer in fluent?
In my case Im using a wall made of 100 mm of steel and 200 mm of firebrick but In my 3d model I just made the fluid domain because Im using the option shell conduction. I thought about 1. Define a new material with an equivalent k 2. Modify the 3d model and made a wall for steel and a wall for firebrick Sent from my SMG570M using CFD Online Forum mobile app 

May 25, 2017, 21:40 

#2 
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Svetlana Tkachenko
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You can do a hand calculation to merge these two walls into one, and supply that to Fluent.
Particularly for one wall the fourier law says q = k * dT / L < Fourier law For two walls you have q1 = k1 * dT1 / L1 < Fourier law wall 1 q2 = k2 * dT2 / L2 < Fourier law wall 2 But in equilibrium these two amounts are the same so k1 * dT1 / L1 = k2 * dT2 / L2 < energy equilibrium at wall interface And we also know what dT2 + dT1 is, let's call it dTtotal (we know it because the solver will rely on 'T internal  T ambient' and this is the total T difference) Two equations now: 1) k1 * dT1 / L1 = k2 * dT2 / L2 2) dT1 + dT2 = dTtotal We have 2 equations and 2 unknowns here. These unknowns are dT1 and dT2. We need to find value of either of them and plug it back into the "Fourier law wall 1" equation. Therefore (from equation 2) dT2 = dTtotal  dT1 Therefore (plug this back into equation 1) k1 * dT1 / L1 = k2 * (dTtotal  dT1) / L2 Therefore (open the brackets) k1 * dT1 / L1 = k2 * dTtotal / L2  k2 * dT1 / L2 Therefore (move one term to the LHS) k1 * dT1 / L1 + k2 * dT1 / L2 = k2 * dTtotal / L2 Therefore (group) dT1 * (k1 / L1 + k2 / L2) = k2 * dTtotal / L2 Therefore now we know dT1 dT1 = k2 * dTtotal / (L2 * (k1 / L1 + k2 / L2)) Therefore we can now put this back into the heat flux formula as told before q = k1 * dT1 / L1 = k1 * k2 * dTtotal / (L2 * L1 * (k1 / L1 + k2 / L2)) = k1 * k2 * dTtotal / (k1 * L2 + k2 * L1) Now what is the meaning of this? We can write it in a more meaningful form: q = k1 * k2 * dTtotal / (k1 * L2 + k2 * L1) = dTtotal / (L2/k2 + L1/k1) It means that L/k is the material resistance to conduction and it is allowed to add them for multi layer walls. You can see the same thing explained (but not derived) at this page: http://scientificsentence.net/Thermo...Thermal_Energy Then you can tell Fluent any wall conduction and any wall thickness you like, for one walls only, as long as L/k for the new wall (its thickness divided by its material conductivity) equals to (L2/q2 + L1/q1). Please correct me if I am wrong here. I am not familiar with your problem so I may have missed something essential. 

May 26, 2017, 00:53 

#3 
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Lucky
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I would recommend to always start with 2 until you run into some serious technical issues like running out of RAM or something. Fluent is a 3D tool and you need to be using it as a 3D tool, otherwise you shouldn't be using Fluent.


May 28, 2017, 20:02 
mesh wall?

#4 
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Svetlana Tkachenko
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Do you mean that it would be better to include the wall into the computational domain, mesh it, and include it in the calculations?
If so, it could be perhaps interesting to check how much the results differ from those obtained with a 1d Fourier law assumption. In some cases this is very important but not in others. 

May 28, 2017, 20:29 

#5  
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Lucky
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Quote:
Again Fluent is a 3D tool (or 2D, but definitely more than 1D). 1D calculations can be done on pen and paper (or excel/mathcad/matlab). You don't walk into a casino, sit at a poker table, and try to play blackjack. 

May 28, 2017, 21:07 
it's a bc

#6 
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Svetlana Tkachenko
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I think this is a boundary condition. They have a multi dimensional computational domain and 1d heat transfer on one of its boundaries.


May 29, 2017, 20:00 

#7 
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BlackHeartInertia
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Ok, I have included the walls as a solid with design geometry but Im not sure if I just have to specify the material in the cell zone conditions.. or i need to specify wall boundary conditions too at thermal like shell conduction? Because my results are different now
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May 29, 2017, 23:52 
convection

#8 
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Svetlana Tkachenko
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Oh, sorry! For me shell conduction is confusing. It expects you to specify a heat generation rate as a constant. I don't know what to put there.
Instead, I personally select convection option in the thermal tab, and supply the value of 1/ (L2/k2 + L1/k1) as the heat transfer coefficient there. The free stream temperature is the ambient temperature on the outside of the wall. 

June 19, 2017, 09:30 

#9  
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BlackHeartInertia
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Quote:
Sent from my SMG570M using CFD Online Forum mobile app 

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