Re: Psi-value

[archi...@gmail.com]

Hi Frank.

This excell file you are using is psiBase_052013.xls (as stated in your pdf file properties). It was developed as part of a diploma thesis at the Rosenheim University of Applied Sciences, was provided as freeware to be used for training, education or for personal use. No problem if you are the author of this software, but if you're not, then the footer "developed by Frank Gergaud for ......" sounds a little dishonest to me.

Fabrizio

Il giorno mercoledì 29 settembre 2021 alle 13:43:04 UTC+2 Gergaud Frank ha scritto:

Hi all,

I am using therm and an excel table to calculate the psi-value. I would be very thankful to have your point of view and advices in this calculation. You can find in attached files, the summary of my calculation for this specific case and the therm file. Below my questions: 

The program therm only provide the heat flow (W) and heat flux (W/m2) so I need to calculate the psi-value or L2D value according to ISO 10211:

 

The psi-value (Ψ) is calculated according to the ISO 10211: Ψ = L2D – L1D

 

L2D

L2D (W/m.K) = Heat flow / ΔT

 

Heat flow (W/m) = Get it from therm

ΔT (K)= 30 (According to ISO 10211, the temperature inside the room is 20deg and outside is -10deg so the delta temperature will always be 30deg)

 

L1D

L1D =U1 x L1 + U2 x L2 +…

 

U1 = U-value of first construction element

L1 = Length of first construction element

 

U2 = U-value of second construction element

L2 = Length of first construction element

 

The problem comes to determine L2D and heat flow. See question below:

 

QUESTIONS:

 1. The first question I have is how do I set the boundary condition in this case?

The thermal bridge is in between external and internal wall and internal slab and roof:

 I assume the boundary condition like this:

In red, it’s outside and in green it’s inside.

2. The second question is which heat flow shall I use? The one to the inside or the one to the outside, how do I know which heat flow to use?

3. The heat flow is in W so I need to divide by the length. I think I need to divide by the total length used for L1D calculation so L1 + L2 (In this case, L1= 1,1 and L2= 0,8). The total length is 1,9m. Am I correct? Shall I use the total length given by therm?

 4. The last question is about the boundary condition for the internal wall. As the 2 rooms below are heated, can I assume that the internal wall is adiabatic?

 

 

In attached file you can find the summary of my calculation for this specific case, the therm file and the excel spreadsheets. To summarize, my problem is to determine the boundary condition in this case and the heat flow that I need to use.

 

Sorry for the long mail, it has been a long time I’m working on it and I couldn’t find the right answer to my problem. Thanks a lot in advance for your time and support.

 

Have a nice day,

Frank

[archi...@gmail.com]

By the way Frank, the simulation you have just removed had been incorrectly set up.

The detail you had chosen in the above mentioned excell file is not the appropriate one. Actually your detail is not included in psiBase_052013.xls

Fortunately I had already downloaded your therm file and I had started to work on it so I can answer the technical questions that might be of some interst to everyone.

About temperature

First fo all the standard EN ISO 10211 does not fix the external temperature to -10°C, the Passivhaus Institut does. To determine the Psi value would be sufficient a deltaT of only 1°C.

The external temperature is needed only if you have to calculate the minimal internal superficial temperature of the detail, or better its fRsi factor: in this case all the internal superficial resistances should be set 0.25 m2K/W (in Therm Film Coefficient = 4 W/m2K).  External temperature should be representative of the coldest avarage monthlywinter temperature. Let assume that for your detail is actually -10°C.

Type of detail

In the attached figure I schematized the detail, with the according Temperatures and Superficial Resistances (according to EN ISO 6946); internal and external dimensions for Psi(i) and Psi(e) values calculation are also shown.

Cutting edges

According to EN ISO 10211 the detail adiabatic edges should be cut at least at 3 times their thickness from the detail core: this is to be sure that isothermal lines are perfectly parallel near the cut and that the edge is truly adiabatic.

In your detail: approximately 1400mm for the upward and downward walls, approximately 1100mm for the leftward internal floor and almost 3900 mm for the rightward flat roof (considering the green cover as part of the element's stratigraphy).

Consistently to these dimensions I modified the Therm model: so the external gross dimensions are x=3900mm y=1400mm and internal internale dimensions are x=4357mm y=2695mm

Element's U-values (1D)

According to EN ISO 6946 U values are:

U-wall = 1/Rtot = 1 / (0.13 + 0.013/0.21 + 0.265/0.044 + 0.009/0.21 + 0.02/0.026 + 0.15/0.894 + 0.04) = 0.1382 W/m2K

U-flat roof = 1/Rtot = 1 / (0.1 + 0.395/2.0 + 0.30/0.026 + 0.60/2.0 + 0.04) = 0.0821 W/m2K

Therm calculations

Heat flow with Therm Φ = 27.0397 W/m       deltaT = 30 K

L2D = Φ /deltaT = 27.0397/30 = 0.9013 W//mK

Psi(e) = L2D - Σ U x l_ext = 0.9013 -  0.1382x1.40 - 0.0821x3.90 = 0.3876 W/mK

Psi(i) = L2D - Σ U x l_int = 0.9013 - 0.1382x2.695 - 0.0821x4.357 = 0.1711 W/mK

That's all.

ciao Fabrizio