basics:building_physics_-_basics:thermal_bridges:tbcalculation:examples:unheatedb
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basics:building_physics_-_basics:thermal_bridges:tbcalculation:examples:unheatedb [2024/10/07 11:18] – [Calculation of the conductance] yaling.hsiao@passiv.de | basics:building_physics_-_basics:thermal_bridges:tbcalculation:examples:unheatedb [2025/01/24 13:55] (current) – old revision restored (2025/01/23 11:52) mpatyna | ||
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Thus $L_{ie}$ , $L_{iu}$ and $L_{ue}$ need to be determined. For this purpose, first the conductances of the separate partial areas are ascertained and the following system of equations is solved: | Thus $L_{ie}$ , $L_{iu}$ and $L_{ue}$ need to be determined. For this purpose, first the conductances of the separate partial areas are ascertained and the following system of equations is solved: | ||
- | <WRAP centeralign> | + | <WRAP centeralign> |
+ | $$ | ||
+ | \begin{matrix} | ||
+ | & \begin{matrix}L_{iu}& | ||
+ | \begin{matrix}L_1\\\\L_2\\\\L_3\end{matrix} | ||
+ | & \begin{pmatrix}1\quad& | ||
+ | \end{matrix} | ||
+ | \quad \Rightarrow \quad | ||
+ | \begin{matrix} | ||
+ | L_{iu} = 0{,}5 \cdot (L_1-L_2+L_3) \\ | ||
+ | L_{ie} = 0{,}5 \cdot (L_1+L_2-L_3) \\ | ||
+ | L_{ue} = 0{,}5 \cdot (-L_1+L_2+L_3) | ||
+ | \end{matrix} | ||
+ | $$ | ||
+ | </ | ||
In order to be able to read the conductances of the individual areas directly from the heat flow program, it makes sense to choose the temperatures so that the temperature difference to the respective neighbouring rooms amounts to one Kelvin. These calculations are demonstrated using an example: | In order to be able to read the conductances of the individual areas directly from the heat flow program, it makes sense to choose the temperatures so that the temperature difference to the respective neighbouring rooms amounts to one Kelvin. These calculations are demonstrated using an example: | ||
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The conductance $L_{iu}$ is now assigned to the floor slab, $L_{ie}$ is assigned to the exterior wall. The effect of this thermal bridge is therefore expressed by two Ψ-values. The following applies for the exterior wall: | The conductance $L_{iu}$ is now assigned to the floor slab, $L_{ie}$ is assigned to the exterior wall. The effect of this thermal bridge is therefore expressed by two Ψ-values. The following applies for the exterior wall: | ||
- | <WRAP centeralign> | + | <WRAP centeralign> |
- | **Ψ-value (proportion of exterior wall)** | + | |
- | </ | + | |
- | <WRAP centeralign> | + | <WRAP centeralign> |
- | $$ | + | |
- | \large{\Psi_g = L_{ie}-l_{AW} \cdot U_{AW}} | + | |
- | $$ | + | |
- | </ | + | |
- | <WRAP centeralign> | + | <WRAP centeralign> |
- | $$ | + | |
- | \large{\Psi_{exterior wall} = 0{.}231-1{.}830 \cdot 0{.}120 = 0{.}0114} | + | |
- | $$ | + | |
- | </ | + | |
For the floor slab: | For the floor slab: | ||
- | <WRAP centeralign> | + | <WRAP centeralign> |
- | **Ψ-value (proportion of floor slab)** | + | |
- | </ | + | |
- | <WRAP centeralign> | + | <WRAP centeralign> |
- | $$ | + | |
- | \large{\Psi_g = L_{iu}-0{.}5 \cdot B' \cdot U_{basement ceiling}} | + | |
- | $$ | + | |
- | </ | + | |
- | <WRAP centeralign> | + | <WRAP centeralign> |
- | $$ | + | |
- | \large{\Psi_{basement ceiling} = 0{.}5543-0{.}5 \cdot 8 \cdot 0{.}148 = -0{.}0377} | + | |
- | $$ | + | |
- | </ | + | |
- | In the [[planning: | + | In the [[:planning: |
+ | |||
+ | <WRAP centeralign> | ||
- | <WRAP centeralign> | ||
- | $$ | ||
- | \large{U_{f, | ||
- | $$ | ||
- | </ | ||
===== Determining the minimum surface temperature and the f_Rsi factor ===== | ===== Determining the minimum surface temperature and the f_Rsi factor ===== |
basics/building_physics_-_basics/thermal_bridges/tbcalculation/examples/unheatedb.1728292689.txt.gz · Last modified: by yaling.hsiao@passiv.de