Dear Leen, Jean, Marion, Benoit and others,
Many thanks for your reply!
I checked my Wall's thickness and thermal property etc. again. Why needs it 9 hours as the timebase? I think maybe it is due to my Ground floor (it is the basement's floor of the building) which includes about 0.99 m's Air Layer.
When I changed it into 0.1 m, the min. timebase 2.5 hours can work without the ERROR:Error creating the wall transfer function coeffients.
Therefore, I'd like to ask you,
1) Is it normal for TRNBuild setting how large the timebase is maximally? 5 hours or others?
2) How to solve this problem of the ground floor with about 0.99 m's Air Layer? But our building project indeed has
the ground floor with about 0.99 m's Air Layer. I can't reduce it into 0.1m? Is the only way to try the active layer approach of The group of Michael Kummert in Montreal?
Please see the Attached INF file.
Thanks in advance!
Br.
wang2012/11/17 Benoit Delcroix <benoit.delcroix@polymtl.ca>
Hello,
I'm one of the students of Michaël Kummert working on a new method to generate the tranfer functions. I thing I can help you for your problem. As Leen said, TRNSYS is unable to generate the transfer functions coefficients for a heavy and insulated wall with a low timebase. In the TRNBuild project, you can modify that value and increase it to allow TRNSYS to generate the transfer functions. Actually, TRNSYS is unable to generate the transfer function as soon as the Fourier number of the wall (thermal diffusivity * timebase / thhickness²) is lower than 0.0005. To check that, TRNBuild make the sum of one of the coefficients series (a, b or c). The 3 sums are equal and means, according to TRNBuild (I don't personally think so...but it's like this), the Fourier number.
As Leen said, it exists a trick. You can use an active layer to split the wall in 2 and then forcing TRNSYS to generate 2 series of transfer functions coefficients for the same wall. As the 2 parts are thinner, you could use a better timebase value. When you insert the active layer in the wall, try to split the layer with the lowest Fourier number. And then to insert the active layer, use the expert mode that allows you to insert an equivalent conductivity. Insert the biggest value you can. Like that, the active layer is negligible at the thermal point of view. Actually, the active layer is acting as a mini-zone to split the wall. It is a little tricky but it works...before having better solution.
Good luck for your project and have a nice week-end,
Benoit
Le 2012-11-17 14:43, trnsys-users-request@cae.wisc.edu a écrit :
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Today's Topics: 1. Re: Fwd: Error creating the wall transfer function coeffients (leen peeters)
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