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--- basics:building_physics_-_basics:thermal_bridges:tbcalculation:examples:unheatedb [2025/01/23 11:48] – [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 32: / Line 32: @@
 ==== Calculation of the conductance ====
-{{  :picopen:unbeheizter_keller_abb_4.png?200}}For the unheated basement, the overall value $L_{L_{2d}}$ must also be calculated as it is for the floor slab and the heated basement. The required heat flow from the heated space consists of the conductances $L_{ie}$ and $L_{iu}$:
+{{  :picopen:unbeheizter_keller_abb_4.png?200}}For the unheated basement, the overall value $L_{2d}$ must also be calculated as it is for the floor slab and the heated basement. The required heat flow from the heated space consists of the conductances $L_{ie}$ and $L_{iu}$:
 <WRAP centeralign> $$ \large{\Phi = L_{iu} \cdot (\theta_i - \theta_u) + L_{ie} \cdot (\theta_i - \theta_e)} $$ </WRAP>
@@ 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: