Hi Loïc,
If you look to your inf file you find :
THERMAL CONDUCTANCE, U= 37.80000 kJ/h m2K; U-Wert= 3.77020 W/m2K
(incl. alpha_i=7.7 W/m^2 K and alpha_o=25 W/m^2 K)
I think that’s means in this case that the thermal conductance refers only
to your wall and U-wert, in addition to thermal conductance, includes the inside and outside heat transfer coefficients:
1/U-wert=(1/7.7)+(25)+(1/thermal conductance)
Kind regards
Hassan
Dr. Hassan Mahach
wissenschaftlicher Mitarbeiter
Arbeitsfeld Technische Innovationen
+49 681 844 972 23
mahach@izes.de
www.izes.de
IZES gGmbH | Altenkesseler Straße 17, Geb. A1 | 66115 Saarbrücken
Büro Berlin | Albrechtstraße 22 | 10117 Berlin
Telefon: +49 681 844 972 0 | Fax: +49 681 761 799 9
Rechtsform: gGmbH | Registergericht Saarbrücken HRB 8499
Steuernummer: 040 140 241 53 | USt.-IdNr.: DE 138 113 796
Aufsichtsratsvorsitz: Minister Jürgen Barke
Geschäftsführung: Prof. Frank Baur | Dr. Lesya Matiyuk
Von: Loic Tachon [mailto:dr.loic.tachon@gmail.com]
Gesendet: Dienstag, 5. Juli 2022 15:09
An: Mahach, Hassan <mahach@izes.de>
Cc: TRNSYS users mailing list at OneBuilding.org <trnsys-users@lists.onebuilding.org>
Betreff: Re: [TRNSYS-users] TRNSYS ground floor
Hi Hassan,
Thanks for your reply.
What is the difference between THERMAL CONDUCTANCE, U= 37.80000 kJ/h m2K; U-Wert= 3.77020 W/m2K ? (we find theses wall value in the inf file) This is not just unit conversion because 37.8 / 3.6 is different that 3.77.
Thanks for the help.
Best regards,
Le mar. 5 juil. 2022 à 10:36, Mahach, Hassan <mahach@izes.de> a écrit :
Hi Dr. Tachon,
The U-Wert means how your wall transfer the heat and it’s normal that you get this change because you changed the thickness of your materials as well as his type (U-wert= Thermal conductivity/thickness). You may also consider your ground-on-slab as boundary wall by setting to the inside (front) and outside (back) convective heat transfer coefficients. You can refer to volume, page 143 of the documentation
“WALLS WITH KNOWN BOUNDARY CONDITION
A wall having a known boundary condition might be a concrete floor resting on ground of a known temperature or a wall adjacent to a zone whose temperature is known. The front of the wall is considered to be at the inside of the zone. Normally, the boundary condition is the temperature of a node connected to the back surface of the wall through a pure resistance. It is possible to specify the surface temperature of the outside by setting the back side heat transfer coefficient of the WALL Type (HBACK) to a very small value (less than 0.001 kJ/h/m²). This WALL Type is then only appropriate for use as a wall with a known boundary. The use of a very small value may be confusing but was kept for backwards compatibility reasons”
I Hope that’s helps
Hassan
Dr. Hassan Mahach
wissenschaftlicher Mitarbeiter
Arbeitsfeld Technische Innovationen
+49 681 844 972 23
IZES gGmbH | Altenkesseler Straße 17, Geb. A1 | 66115 Saarbrücken
Büro Berlin | Albrechtstraße 22 | 10117 Berlin
Telefon: +49 681 844 972 0 | Fax: +49 681 761 799 9
Rechtsform: gGmbH | Registergericht Saarbrücken HRB 8499
Steuernummer: 040 140 241 53 | USt.-IdNr.: DE 138 113 796
Aufsichtsratsvorsitz: Minister Jürgen Barke
Geschäftsführung: Prof. Frank Baur | Dr. Lesya Matiyuk
Von: TRNSYS-users [mailto:trnsys-users-bounces@lists.onebuilding.org] Im Auftrag von Loic Tachon via TRNSYS-users
Gesendet: Dienstag, 5. Juli 2022 09:38
An: TRNSYS users mailing list at OneBuilding.org <trnsys-users@lists.onebuilding.org>
Cc: Loic Tachon <dr.loic.tachon@gmail.com>
Betreff: [TRNSYS-users] TRNSYS ground floor
Dear users,
I want to simulate a huge building (5500 m2) with a 20 cm concrete ground without isolation.
I don't really understand how the direct contact boundary condition works.
I made a ground wall composed of 20 cm of concret and 80 cm of calcar.
I realize that the simulation result (Tzone temperature) is very dependent of the ground temperature boundary condition.
I use type 77 with the depth of ground temperature of 1m, the amplitude of 12 and the mean temperature of 12.
If I change the parameters of type 77, the zone temperature results change a lots.
In the inf file, i get u value result of my floor : THERMAL CONDUCTANCE, U= 4.21086 kJ/h m2K; U-Wert= 0.97567 W/m2K
If I don't put the 80 cm of calcar, I get the following u value :
THERMAL CONDUCTANCE, U= 37.80000 kJ/h m2K; U-Wert= 3.77020 W/m2K
Is this normal ? What means U-Wert ?
--
Dr. Loïc Tachon
(+33) 6 7440-0536 (France)
--
Dr. Loïc Tachon
(+33) 6 7440-0536 (France)