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--- basics:building_physics_-_basics:thermal_bridges:tbcalculation:examples:unheatedb [2024/10/03 08:32] – old revision restored (2024/10/03 08:30) jschnieders
+++ 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|}}
+{{  :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 $L_{iu}$:
-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>
+<WRAP centeralign> $$ \large{\Phi = L_{iu} \cdot (\theta_i - \theta_u) + L_{ie} \cdot (\theta_i - \theta_e)} $$ </WRAP>
-$$
-\large{\Phi = L_{iu} \cdot (\theta_i - \theta_u) + L_{ie} \cdot (\theta_i - \theta_e)}
-$$
-</WRAP>
-However, the indoor temperature $\theta_u$ of the basement is unknown, therefore $L_{2d}$  cannot be calculated directly, but with the use of $L_{ue}$ it is possible to express the heat flow or the conductance using the following equation:
+However, the indoor temperature $\theta_u$ of the basement is unknown, therefore $L_{2d}$ cannot be calculated directly, but with the use of $L_{ue}$ it is possible to express the heat flow or the conductance using the following equation:
-<WRAP centeralign>
+<WRAP centeralign> $$ \large{\Phi = \left(\frac{L_{iu} \cdot L_{ue}}{L_{iu} + L_{ue}}\right) \cdot (\theta_i - \theta_e) \quad \Rightarrow \quad L_{2d} = \left(\frac{L_{iu} \cdot L_{ue}}{L_{iu} + L_{ue}} + L_{ie}\right)} $$ </WRAP>
-$$
-\large{\Phi = \left(\frac{L_{iu} \cdot L_{ue}}{L_{iu} + L_{ue}}\right) \cdot (\theta_i - \theta_e) \quad \Rightarrow \quad L_{2d} =  \left(\frac{L_{iu} \cdot L_{ue}}{L_{iu} + L_{ue}} + L_{ie}\right)}
-$$
-</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:
+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>
@@ Line 66: / Line 57: @@
 $$
 </WRAP>
 In order to be able to read the conductances of the individual areas directly from the heat flow program, it makes sense to choose the temperatures so that the temperature difference to the respective neighbouring rooms amounts to one Kelvin. These calculations are demonstrated using an example:
-<WRAP centeralign>
+<WRAP centeralign> **Determining the conductance** $$ \Large{_{2d}} $$ </WRAP>
-**Determining the conductance**
-$$
-\Large{_{2d}}
-$$
-</WRAP>
-{{ :picopen:unbeheizter_keller_abb_5a.png?600 |}}
+{{  :picopen:unbeheizter_keller_abb_5a.png?600  }}{{  :picopen:unbeheizter_keller_abb_5b.png?600  }}
-{{ :picopen:unbeheizter_keller_abb_5b.png?600 |}}
-<WRAP centeralign>
+<WRAP centeralign> $$ \large{ L_{iu} = 0{.}5 \cdot (L_1-L_2+L_3) = 0{.}5540 \, \frac{\text{W}}{\text{m} \cdot \text{K}} \\ L_{ie} = 0{.}5 \cdot (L_1+L_2-L_3) = 0{.}2314 \, \frac{\text{W}}{\text{m} \cdot \text{K}} \\ L_{ue} = 0{.}5 \cdot (-L_1+L_2+L_3) = 2{.}6177 \, \frac{\text{W}}{\text{m} \cdot \text{K}}} $$ </WRAP>
-$$
-\large{
-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_{us} = 0{.}5 \cdot (-L_1+L_2+L_3) = 2{.}6177 \, \frac{\text{W}}{\text{m} \cdot \text{K}}}
-$$
-</WRAP>
+<WRAP centeralign> $$ \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}}} $$ </WRAP>
-<WRAP centeralign>
-$$
-\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}}}
-$$
-</WRAP>
 ==== Calculating the Ψ-value ====
@@ Line 151: / Line 126: @@
 The conductance $L_{iu}$ is now assigned to the floor slab, $L_{ie}$ is assigned to the exterior wall. The effect of this thermal bridge is therefore expressed by two Ψ-values. The following applies for the exterior wall:
-<WRAP centeralign>
+<WRAP centeralign> **Ψ-value (proportion of exterior wall)** </WRAP>
-**Ψ-value (proportion of exterior wall)**
-</WRAP>
-<WRAP centeralign>
+<WRAP centeralign> $ \large{\Psi_g = L_{ie}-l_{AW} \cdot U_{AW}} $ </WRAP>
-$$
-\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 =====