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--- basics:building_physics_-_basics:thermal_bridges:tbcalculation:examples:unheatedb [2024/10/09 09:58] – 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
@@ Line 41: / Line 41: @@
 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> $$ \begin{matrix} & \begin{matrix}L_{iu}&L_{ie}&L_{ue}\end{matrix} \\\ begin{matrix}L_1\\\\L_2\\\\L_3\end{matrix} & \begin{pmatrix}1\quad&1\quad&0\quad\\\\0\quad&1\quad&1\quad\\\\1\quad&0\quad&1\quad\end{pmatrix}\\\ end{matrix} \quad \Rightarrow \quad \begin{matrix} L_{iu} = 0{,}5 \cdot (L_1-L_2+L_3) \\ L_{ie} = 0{,}5 \cdot (L_1+L_2-L_3) \\ L_{ue} = 0{,}5 \cdot (-L_1+L_2+L_3) \end{matrix} $$ </WRAP>
 <WRAP centeralign>
@@ Line 128: / Line 126: @@
 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>
-**Ψ-value (proportion of exterior wall)**
-</WRAP>
-<WRAP centeralign>
+<WRAP centeralign> $ \large{\Psi_g = L_{ie}-l_{AW} \cdot U_{AW}} $ </WRAP>
-$$
-\large{\Psi_g = L_{ie}-l_{AW} \cdot U_{AW}}
-$$
-</WRAP>
-<WRAP centeralign>
+<WRAP centeralign> $ \large{\Psi_{exterior wall} = 0{.}231-1{.}830 \cdot 0{.}120 = 0{.}0114} $ </WRAP>
-$$
-\large{\Psi_{exterior wall} = 0{.}231-1{.}830 \cdot 0{.}120 = 0{.}0114}
-$$
-</WRAP>
 For the floor slab:
-<WRAP centeralign>
+<WRAP centeralign> **Ψ-value (proportion of floor slab)** </WRAP>
-**Ψ-value (proportion of floor slab)**
-</WRAP>
-<WRAP centeralign>
+<WRAP centeralign> $ \large{\Psi_g = L_{iu}-0{.}5 \cdot B' \cdot U_{basement ceiling}} $ </WRAP>
-$$
-\large{\Psi_g = L_{iu}-0{.}5 \cdot B' \cdot U_{basement ceiling}}
-$$
-</WRAP>
-<WRAP centeralign>
+<WRAP centeralign> $ \large{\Psi_{basement ceiling} = 0{.}5543-0{.}5 \cdot 8 \cdot 0{.}148 = -0{.}0377} $ </WRAP>
-$$
-\large{\Psi_{basement ceiling} = 0{.}5543-0{.}5 \cdot 8 \cdot 0{.}148 = -0{.}0377}
-$$
-</WRAP>
-In the [[planning:calculating_energy_efficiency:phpp_-_the_passive_house_planning_package |PHPP]] $\Psi_{exterior wall}$  is entered as a normal thermal bridge. With $\Psi_{basement ceiling}$, $U_f$ for the basement ceiling is corrected in the Ground worksheet so that the overall resistance of the basement level can be calculated afterwards with the aid of the approximation functions.
+In the [[:planning:calculating_energy_efficiency:phpp_-_the_passive_house_planning_package|PHPP]] $\Psi_{exterior wall}$ is entered as a normal thermal bridge. With $\Psi_{basement ceiling}$, $U_f$ for the basement ceiling is corrected in the Ground worksheet so that the overall resistance of the basement level can be calculated afterwards with the aid of the approximation functions.
+<WRAP centeralign> $ \large{U_{f,corrected} = U_f + \dfrac{\Psi_{basement ceiling} \cdot P}{A}} $ </WRAP>
-<WRAP centeralign>
-$$
-\large{U_{f,corrected} = U_f + \dfrac{\Psi_{basement ceiling} \cdot P}{A}}
-$$
-</WRAP>
 ===== Determining the minimum surface temperature and the f_Rsi factor =====