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--- basics:building_physics_-_basics:thermal_bridges:tbcalculation:examples:unheatedb [2019/01/24 09:40] – [See also] cblagojevic
+++ basics:building_physics_-_basics:thermal_bridges:tbcalculation:examples:unheatedb [2022/02/15 18:58] (current) – admin
@@ Line 10: / Line 10: @@
 <WRAP centeralign>
-<latex>
+$$
 \large{\dfrac{1}{U} = \dfrac{1}{U_f} + \dfrac{A}{(A \cdot U_{bf}) + (z \cdot P \cdot U_{bw}) + (h \cdot P \cdot U_W) + (0{,}33 \cdot n \cdot V)}}
-</latex>
+$$
 \\
 \\
@@ Line 36: / Line 36: @@
 <WRAP centeralign>
-<latex>
+$$
 \large{\Phi = L_{iu} \cdot (\theta_i - \theta_u) + L_{ie} \cdot (\theta_i - \theta_e)}
-</latex>
+$$
 </WRAP>
@@ Line 44: / Line 44: @@
 <WRAP centeralign>
-<latex>
+$$
 \large{\Phi = \left(\frac{L_{iu} \cdot L_{ue}}{L_{iu} + L_{ue}}\right) \cdot (\theta_i - \theta_e) \quad \Rightarrow \quad L_{2d} =  \left(\frac{L_{iu} \cdot L_{ue}}{L_{iu} + L_{ue}} + L_{ie}\right)}
-</latex>
+$$
 </WRAP>
 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>
-<latex>
+$$
-$$ \bordermatrix{
+\begin{matrix}
-    & L_{iu} & L_{ie} & L_{ue} \cr
+& \begin{matrix}L_{iu}&L_{ie}&L_{ue}\end{matrix} \\\\
-L_1 & 1      & 1      & 0 \cr
+\begin{matrix}L_1\\\\L_2\\\\L_3\end{matrix}
-L_2 & 0      & 1      & 1 \cr
+& \begin{pmatrix}1\quad&1\quad&0\quad\\\\0\quad&1\quad&1\quad\\\\1\quad&0\quad&1\quad\end{pmatrix}\\\\
-L_3 & 1      & 0      & 1 \cr
+\end{matrix}
-}
 \quad \Rightarrow \quad
 \begin{matrix}
-L_{iu} = 0{.}5 \cdot (L_1-L_2+L_3) \\
+L_{iu} = 0{,}5 \cdot (L_1-L_2+L_3) \\
-L_{is} = 0{.}5 \cdot (L_1+L_2-L_3) \\
+L_{ie} = 0{,}5 \cdot (L_1+L_2-L_3) \\
-L_{us} = 0{.}5 \cdot (L_2+L_3-L_1)
+L_{ue} = 0{,}5 \cdot (-L_1+L_2+L_3)
 \end{matrix}
 $$
-</latex>
 </WRAP>
@@ Line 73: / Line 71: @@
 <WRAP centeralign>
 **Determining the conductance**
-<latex>
+$$
 \Large{_{2d}}
-</latex>
+$$
 </WRAP>
@@ Line 82: / Line 80: @@
 <WRAP centeralign>
-<latex>
+$$
 \large{
-$$L_{iu} = 0{.}5 \cdot (L_1-L_2+L_3) = 0{.}5540 \, \frac{\text{W}}{\text{m} \cdot \text{K}}} $$\\
+L_{iu} = 0{.}5 \cdot (L_1-L_2+L_3) = 0{.}5540 \, \frac{\text{W}}{\text{m} \cdot \text{K}} \\
-$$L_{is} = 0{.}5 \cdot (L_1+L_2-L_3) = 0{.}2314 \, \frac{\text{W}}{\text{m} \cdot \text{K}}} $$\\
+L_{is} = 0{.}5 \cdot (L_1+L_2-L_3) = 0{.}2314 \, \frac{\text{W}}{\text{m} \cdot \text{K}} \\
-$$L_{us} = 0{.}5 \cdot (-L_1+L_2+L_3) = 2{.}6177 \, \frac{\text{W}}{\text{m} \cdot \text{K}}}$$}
+L_{us} = 0{.}5 \cdot (-L_1+L_2+L_3) = 2{.}6177 \, \frac{\text{W}}{\text{m} \cdot \text{K}}}
-</latex>
+$$
 </WRAP>
 <WRAP centeralign>
-<latex>
+$$
-\large{L_{2d} =  \left(\frac{L_{iu} \cdot L_{ue}}{L_{iu} + L_{ue}} + L_{ie}\right) = 0{.}6886 \, \frac{\text{W}}{\text{m} \cdot \text{K}}}}
+\large{L_{2d} =  \left(\frac{L_{iu} \cdot L_{ue}}{L_{iu} + L_{ue}} + L_{ie}\right) = 0{.}6886 \, \frac{\text{W}}{\text{m} \cdot \text{K}}}
-</latex>
+$$
 </WRAP>
@@ Line 103: / Line 101: @@
 Therefore:
 <WRAP centeralign>
-<latex>
+$$
-$$\dfrac{1}{U} = \dfrac{1}{U_f} + \dfrac{A}{(A \cdot U_{bf}) + (z \cdot P \cdot U_{bw}) + (h \cdot P \cdot U_W) + (0{.}33 \cdot n \cdot V)}
+\dfrac{1}{U} = \dfrac{1}{U_f} + \dfrac{A}{(A \cdot U_{bf}) + (z \cdot P \cdot U_{bw}) + (h \cdot P \cdot U_W) + (0{.}33 \cdot n \cdot V)}
 \quad \Rightarrow \quad
-U = 0{.}1273 \, \dfrac{\text{W}}{\text{m}^2 \cdot \text{K}}}$$
+U = 0{.}1273 \, \dfrac{\text{W}}{\text{m}^2 \cdot \text{K}}
-</latex>
+$$
 </WRAP>
@@ Line 113: / Line 111: @@
 <WRAP centeralign>
-<latex>
 $$\Psi_g = L_{2d}-l_{AW} \cdot U_{AW}-0{.}5 \cdot B' \cdot U$$
-</latex>
 </WRAP>
@@ Line 124: / Line 120: @@
 <WRAP centeralign>
-<latex>
+$$\Psi_g = 0{.}689 \, \dfrac{\text{W}}{\text{m} \cdot \text{K}} \, - \, 1{.}830 \, \text{m} \, \cdot \, 0{.}120 \, \dfrac{\text{W}}{\text{m}^2 \cdot \text{K}} \, - \, 0{.}5 \, \cdot \, 8 \, \text{m} \, \cdot \, 0{.}1273 \, \dfrac{\text{W}}{\text{m}^2 \cdot \text{K}} = -0{.}042 \, \dfrac{\text{W}}{\text{m} \cdot \text{K}}$$
-$$\Psi_g = 0{.}687 \, \dfrac{\text{W}}{\text{m} \cdot \text{K}}} \, - \, 1{.}830 \, \text{m} \, \cdot \, 0{.}120 \, \dfrac{\text{W}}{\text{m}^2 \cdot \text{K}}} \, - \, 0{.}5 \, \cdot \, 8 \, \text{m} \, \cdot \, 0{.}1273 \, \dfrac{\text{W}}{\text{m}^2 \cdot \text{K}}} = -0{.}042 \, \dfrac{\text{W}}{\text{m} \cdot \text{K}}}$$
-</latex>
 </WRAP>
@@ Line 138: / Line 132: @@
 <WRAP centeralign>
 **Determining the conductance**
-<latex>
+$
 \Large{L_{2d}}
-</latex>
+$
 </WRAP>
@@ Line 160: / Line 154: @@
 **Ψ-value (proportion of exterior wall)**
 </WRAP>
 <WRAP centeralign>
-<latex>
+$$
-\large{\Psi_g = L_{ie}-l_{AW} \cdot U_{AW}$}
+\large{\Psi_g = L_{ie}-l_{AW} \cdot U_{AW}}
-</latex>
+$$
 </WRAP>
 <WRAP centeralign>
-<latex>
+$$
 \large{\Psi_{exterior wall} = 0{.}231-1{.}830 \cdot 0{.}120 = 0{.}0114}
-</latex>
+$$
 </WRAP>
@@ Line 179: / Line 174: @@
 <WRAP centeralign>
-<latex>
+$$
 \large{\Psi_g = L_{iu}-0{.}5 \cdot B' \cdot U_{basement ceiling}}
-</latex>
+$$
 </WRAP>
 <WRAP centeralign>
-<latex>
+$$
 \large{\Psi_{basement ceiling} = 0{.}5543-0{.}5 \cdot 8 \cdot 0{.}148 = -0{.}0377}
-</latex>
+$$
 </WRAP>
@@ Line 193: / Line 188: @@
 <WRAP centeralign>
-<latex>
+$$
 \large{U_{f,corrected} = U_f + \dfrac{\Psi_{basement ceiling} \cdot P}{A}}
-</latex>
+$$
 </WRAP>
@@ Line 205: / Line 200: @@
 <WRAP centeralign>
-<latex>
+$$
 \large{\theta_{basement} = \theta_i-f_x \cdot (\theta_i - \theta_e) = 20 \, ^\circ C - 0{.}6 \cdot (20 \, ^\circ C - (-10 \, ^\circ C)) = 2 \, ^\circ C}
-</latex>
+$$
 </WRAP>
@@ Line 214: / Line 209: @@
 <WRAP centeralign>
 **Determining the minimum surface temperature and**
-<latex>
+$$
 \Large{f_{Rsi}}
-</latex>
+$$
 </WRAP>
@@ Line 224: / Line 219: @@
 <WRAP centeralign>
-<latex>
+$$
 \large{f_{Rsi} = \dfrac{17{.}6 - (-10)}{20-(-10)}=0{.}92}
-</latex>
+$$
 </WRAP>