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--- basics:building_physics_-_basics:thermal_bridges:tbcalculation:examples:unheatedb [2025/01/23 11:50] – [Calculation of the conductance] mpatyna
+++ 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 42: / Line 42: @@
 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>
+$$
+\begin{matrix}
+& \begin{matrix}L_{iu}&L_{ie}&L_{ue}\end{matrix} \\\\
+\begin{matrix}L_1\\\\L_2\\\\L_3\end{matrix}
+& \begin{pmatrix}1\quad&1\quad&0\quad\\\\0\quad&1\quad&1\quad\\\\1\quad&0\quad&1\quad\end{pmatrix}\\\\
+\end{matrix}
+\quad \Rightarrow \quad
+\begin{matrix}
+L_{iu} = 0{,}5 \cdot (L_1-L_2+L_3) \\
+L_{ie} = 0{,}5 \cdot (L_1+L_2-L_3) \\
+L_{ue} = 0{,}5 \cdot (-L_1+L_2+L_3)
+\end{matrix}
+$$
+</WRAP>
 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:
-<WRAP centeralign> **Determining the conductance** $ \Large{L_{2d}} $ </WRAP>
+<WRAP centeralign> **Determining the conductance** $$ \Large{_{2d}} $$ </WRAP>
 {{  :picopen:unbeheizter_keller_abb_5a.png?600  }}{{  :picopen:unbeheizter_keller_abb_5b.png?600  }}