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Posted 4 years ago by Catalin Daniel Iosif Technical University of Cluj-Napoca Student
Hello! How do you usually consider longitudinal/tranverse dispersivity values for heat transfer? Default, 0 or different?

When I use a simple model to see the influence of this parameter I observe huge differences in temperature field. The material I used is concrete, therefore I firstly set dispersivity to 0 in the first case (full homogenius) followed by setting default values (5m/0.5m) in the second case. The results are in the attached image.

There is any table showing the most correct values to consider based on different criteria?

Thank you!



Posted 4 years ago by Peter Schätzl Grundwassermodellierer
In a homogeneous material such as concrete, I'd expect dispersivity for heat transport to be small. If you think about heat plumes from open-loop geothermal installations, however, that move in a heterogeneous aquifer, I would put higher values. Note that there is a dependency between the length scale of the transport phenomenon and dispersivity, outlined for example in https://www3.epa.gov/ceampubl/learn2model/part-two/onsite/longdisp.html.
Posted 4 years ago by Catalin Daniel Iosif Technical University of Cluj-Napoca Student
Thank you!

I don't find anything clear about this parameter. Also, I don't clearly understand how heat conductivity along with volumetric heat capacity is calculated based on porosity and fluid/solid properties. I will do some simple models to better understand the last two terms. Please, help me if you have an answer at hand. Thank you in advance!
Posted 4 years ago by Catalin Daniel Iosif Technical University of Cluj-Napoca Student
I understand, I hope, how the software calculates the conductivity after modeling two simple models. My conclusions are: Solid thermal conductivity = Bulk thermal conductivity when water is not present in the layer and Solid thermal conductivity together with fluid thermal conductivity = Bulk thermal conductivity when layer is saturated. The formula used is the empirical one: n*lambdafluid + (1-n)*lambdasolid = lambda bulk. Am I right? Thank you!
Posted 4 years ago by Peter Schätzl Grundwassermodellierer
Yes, you are right. FEFLOW assumes air to be perfectly non-conductive and without capacity, thus the terms "solid" and "fluid" for the two phases with thermal conductivity/capacity.

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