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--- basics:building_physics_-_basics:thermal_bridges:tbcalculation:examples:ewfs [2019/02/13 09:17] – cblagojevic
+++ basics:building_physics_-_basics:thermal_bridges:tbcalculation:examples:ewfs [2022/08/08 12:14] (current) – yaling.hsiao@passiv.de
@@ Line 19: / Line 19: @@
 ===== Calculation model =====
-First,  an overall model is prepared for calculating the  conductance $L_{2d}$. If the actual size of the building is not known, B‘ can be selected for the option B, but it should not be too small ($B' > 5 \, \text{m}$). $B' = 7{,}992 \, \text{m}$ was selected here (including the exterior wall) //[Note: With some heat flow software programs the dimensions in the models may have inconvenient decimal places; however this is not of any significance]//. The floor slab is only modelled up to the axis of symmetry and thus has the length $\frac{B'}{2}=4{,}996 \, \text{m}$. The height of the exterior wall should be three times as high as the area to be examined. It is essential that the length is measured based on the exterior references, that is, up to the lower edge of the perimeter insulation. The same dimensional reference should be used in the respective [[planning:calculating_energy_efficiency:phpp_-_the_passive_house_planning_package |PHPP]] calculation. In the example shown here, the length of the exterior wall is 2.03 m. The block of earth should be modelled with a size of at least $2{,}5 \cdot B'$.
+First,  an overall model is prepared for calculating the  conductance $L_{2d}$. If the actual size of the building is not known, B‘ can be selected for the option B, but it should not be too small ($B' > 5 \, \text{m}$). $B' = 7{,}992 \, \text{m}$ was selected here (including the exterior wall) //[Note: With some heat flow software programs the dimensions in the models may have inconvenient decimal places; however this is not of any significance]//. The floor slab is only modelled up to the axis of symmetry and thus has the length $\frac{B'}{2}=3{,}996 \, \text{m}$. The height of the exterior wall should be three times as high as the area to be examined. It is essential that the length is measured based on the exterior references, that is, up to the lower edge of the perimeter insulation. The same dimensional reference should be used in the respective [[planning:calculating_energy_efficiency:phpp_-_the_passive_house_planning_package |PHPP]] calculation. In the example shown here, the length of the exterior wall is 2.03 m. The block of earth should be modelled with a size of at least $2{,}5 \cdot B'$.
 In the following illustration all relevant reference dimensions and boundary conditions are included. Ultimately, it is irrelevant which boundary conditions for the temperature are used for determining the conductances; however, it is recommended that those should be used which will be needed later on for determining the surface temperatures. The lower part of the illustration shows the calculation of the conductance $L_{2d}$, this results from the heat flow calculated using the heat flow software program.
 <WRAP centeralign>
 **Calculating the conductance**
-<latex>
 $L_{2d}$
-</latex>
 </WRAP>
-{{ :picopen:aussenwand_auf_bodenplatte_abb_8a.png?600 |}}
+{{ :picopen:01_updated_-_aussenwand_auf_bodenplatte_abb_8a.png?600 |}}
 {{ :picopen:aussenwand_auf_bodenplatte_abb_8b.png?600 |}}
 <WRAP centeralign>
-<latex>
+$$
-$$\.q = 19{,}488 \, \frac{\text{W}}{\text{m}}$$ \\
+\dot{q} = 19{,}488 \, \frac{\text{W}}{\text{m}}
-$$L_{2d} = \frac{\.q}{T_i-T_e} = \frac{19{,}488}{30} = 0{,}6496 \, \frac{\text{W}}{\text{m} \cdot \text{K}}}$$
+$$
-</latex>
+</WRAP>
+<WRAP centeralign>
+$$
+\large{L_{2d} = \frac{\dot{q}}{T_i-T_e} = \frac{19{,}488}{30} = 0{,}6496 \, \frac{\text{W}}{\text{m} \cdot \text{K}}}
+$$
 </WRAP>
@@ Line 44: / Line 49: @@
 <WRAP centeralign>
 **Conductance of the floor slab**
-<latex>
 $L_{BP}$
-</latex>
 </WRAP>
-{{ :picopen:aussenwand_auf_bodenplatte_abb_9a.png?600 |}}
+{{ :picopen:02_updated_-_aussenwand_auf_bodenplatte_abb_9a.png?600 |}}
 {{ :picopen:aussenwand_auf_bodenplatte_abb_9b.png?600 |}}
 <WRAP centeralign>
-<latex>
+$$
-$$\.q = 14{,}023 \, \frac{\text{W}}{\text{m}}$$ \\
+\dot{q} = 14{,}023 \, \frac{\text{W}}{\text{m}}
-$$L_{BP} = \frac{\.q}{T_i-T_e} = \frac{14{,}023}{30} = 0{,}4674 \, \frac{\text{W}}{\text{m} \cdot \text{K}}}$$
+$$
-</latex>
+</WRAP>
+<WRAP centeralign>
+$$
+\large{L_{BP} = \frac{\dot{q}}{T_i-T_e} = \frac{14{,}023}{30} = 0{,}4674 \, \frac{\text{W}}{\text{m} \cdot \text{K}}}
+$$
 </WRAP>
@@ Line 65: / Line 73: @@
 <WRAP centeralign>
 **Conductance of the exterior wall**
-<latex>
 $L_{AW}$
-</latex>
 </WRAP>
 <WRAP centeralign>
-<latex>
+$$
-$$U_{AW} = 0{,}1205 \, \frac{\text{W}}{\text{m}^2 \cdot \text{K}}$$
+U_{AW} = 0{,}1205 \, \frac{\text{W}}{\text{m}^2 \cdot \text{K}}
-$$L_{AW} = l_{AW} \cdot U_{AW} = 2{,}03 \cdot 0{,}1205 = 0{,}2446 \, \frac{\text{W}}{\text{m} \cdot \text{K}}}$$
+$$
-</latex>
+</WRAP>
+<WRAP centeralign>
+$$
+\large{L_{AW} = l_{AW} \cdot U_{AW} = 2{,}03 \cdot 0{,}1205 = 0{,}2446 \, \frac{\text{W}}{\text{m} \cdot \text{K}}}
+$$
 </WRAP>
@@ Line 83: / Line 94: @@
 <WRAP centeralign>
-<latex>
+$$
-$$\Psi = L_{2d}-L_{AW}-L_{BP}=0{,}6496-0{,}2446-0{,}4674=-0{,}062 \, \frac{\text{W}}{\text{m} \cdot \text{K}}}$$
+\large{\Psi = L_{2d}-L_{AW}-L_{BP}=0{,}6496-0{,}2446-0{,}4674=-0{,}062 \, \frac{\text{W}}{\text{m} \cdot \text{K}}}
-</latex>
+$$
 </WRAP>
@@ Line 103: / Line 114: @@
 <WRAP centeralign>
 ** Determining the minimum surface temperature and**
-<latex>
 $f_{Rsi}$
-</latex>
 </WRAP>
-{{ :picopen:aussenwand_auf_bodenplatte_abb_10a.png?600 |}}
+{{ :picopen:updated_03-_aussenwand_auf_bodenplatte_abb_10a.png?600 |}}
 {{ :picopen:aussenwand_auf_bodenplatte_abb_10b.png?600 |}}
 <WRAP centeralign>
-<latex>
+$$
-$$f_{Rsi \, = \, 0{,}25} = \frac{17{,}4-(-10)}{20-(-10)} = 0{,}91$$
+f_{Rsi \, = \, 0{,}25} = \frac{17{,}4-(-10)}{20-(-10)} = 0{,}91
-</latex>
+$$
 </WRAP>