[TRNSYS-users] Stability criteria not fullfilled

[David BRADLEY]

Stéphan,
  The energy transfer through walls in Type56 is modeled using Conduction Transfer Functions. These basically identify the main factors that influence the energy transfer (temperature difference across the wall, the energy that is currently stored in the wall, etc.) and then assign a weighting factor to each one to tell how important its influence is. Because the wall has a time history of energy transferring through it, weighting factors are not applied to the current time but they are applied to the same factors back through the history of the wall, stepping backwards in increments called a timebase. If your timebase is one hour then weighting factors would be determined for the current time, for one hour ago, for two hours ago, etc. If you have a thermally massive wall, you have to step further back in history in order to characterize the wall's response. If you have a thin wall, you don't have to go far back at all and in fact, if the wall's history can be completely summed up within one timebase (like an uninsulated steel roof, for example) then you can't use Conduction Transfer Functions at all and the wall should be treated as purely resistive (ie no mass). By default, Type56 looks back in history for 20 timebases and by default the timebase is one hour.
   
   In your case, you have a wall whose response cannot be adequately represented by looking 20 hours back in history so you need to increase your timebase, probably to 2 hours. To do so, click on the "outputs" button in the main Project window in TRNBuild.

Best,
  David


On 6/27/2012 02:31, struong at student.ulg.ac.be wrote:
> Dear TRNSYS-users,
>
> I'm currently encoding my building in TRNBuild and in a wall I have a layer of cellulose of 40 cm thickness. When i try to save my project, I get the error message "Stability criteria not fullfilled" (see picture in attached file "error.JPG). If i reduce this thickness to 20 cm, the problem is solved so it seems that it is a problem due to the thickness of the layer. How can I solve the problem? I'm thinking about encoding a thickness of 20 cm and divide the conductivity of the material by 2 in order to have an identical u-value for my wall but I don't know if it is correct! Indeed if I do so, the inertia of my wall (hence my building) will not be the same! Do you have some advice?
>
> Best regards,
>
> Truong Van Quynh Stéphan
> 2 ème Master STE
> Gembloux Agro Bio-Tech
> Belgium
>
>
>
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> TRNSYS-users at cae.wisc.edu
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-- 
***************************
David BRADLEY
Principal
Thermal Energy Systems Specialists, LLC
22 North Carroll Street - suite 370
Madison, WI  53703 USA

P:+1.608.274.2577
F:+1.608.278.1475
d.bradley at tess-inc.com

http://www.tess-inc.com
http://www.trnsys.com



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