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--- basics:building_physics_-_basics:thermal_bridges:tbcalculation:examples:heatedb [2019/01/17 09:35] – cblagojevic
+++ basics:building_physics_-_basics:thermal_bridges:tbcalculation:examples:heatedb [2022/02/15 18:35] (current) – admin
@@ Line 23: / Line 23: @@
 <WRAP centeralign>
 **Calculating the conductance**
-<latex>
 $L_{2d}$
-</latex>
 </WRAP>
@@ Line 32: / Line 30: @@
 {{ :picopen:beheizter_keller_abb_4b.png?600 |}}
 <WRAP centeralign>
-<latex>
+$$
-$$\.q = 38{,}123 \, \frac{\text{W}}{\text{m}}$$ \\
+\large{\dot{q} = 38{,}123 \, \dfrac{\text{W}}{\text{m}}}\\
-$$L_{2d} = \frac{\.q}{T_i-T_e} = \frac{38{,}123}{30} = 1{,}2708 \, \frac{\text{W}}{\text{m} \cdot \text{K}}}$$
+$$
-</latex>
+</WRAP>
+<WRAP centeralign>
+$$
+\large{L_{2d} = \dfrac{\dot{q}}{T_i-T_e} = \dfrac{38{,}123}{30} = 1{,}2708 \, \dfrac{\text{W}}{\text{m} \cdot \text{K}}}
+$$
 </WRAP>
@@ Line 43: / Line 46: @@
 <WRAP centeralign>
 **Conductance of the basement floor slab and the basement wall**
-<latex>
 $L_{BP,KW}$
-</latex>
 </WRAP>
@@ Line 53: / Line 54: @@
 <WRAP centeralign>
-<latex>
+$$
-$$\.q = 22{,}190 \, \frac{\text{W}}{\text{m}}$$ \\
+\large{\dot{q} = 22{,}190 \, \dfrac{\text{W}}{\text{m}}}\\
-$$L_{BP,KW} = \frac{\.q}{T_i-T_e} = \frac{22{,}190}{30} = 0{,}7397 \, \frac{\text{W}}{\text{m} \cdot \text{K}}}$$
+$$
-</latex>
+</WRAP>
+<WRAP centeralign>
+$$
+\large{L_{BP,KW} = \dfrac{\dot{q}}{T_i-T_e} = \dfrac{22{,}190}{30} = 0{,}7397 \, \dfrac{\text{W}}{\text{m} \cdot \text{K}}}
+$$
 </WRAP>
@@ Line 63: / Line 69: @@
 <WRAP centeralign>
 **Calculating the 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}}$$
+\large{U_{AW} = 0{,}1205 \, \dfrac{\text{W}}{\text{m}^2 \cdot \text{K}}}
-$$L_{AW} = l_{AW} \cdot U_{AW} = 1{,}83 \cdot 0{,}1205 = 0{,}2205 \, \frac{\text{W}}{\text{m} \cdot \text{K}}}$$
+$$
-</latex>
+</WRAP>
+<WRAP centeralign>
+$$
+\large{L_{AW} = l_{AW} \cdot U_{AW} = 1{,}83 \cdot 0{,}1205 = 0{,}2205 \, \dfrac{\text{W}}{\text{m} \cdot \text{K}}}
+$$
 </WRAP>
@@ Line 81: / Line 90: @@
 </WRAP>
 <WRAP centeralign>
-<latex>
+$$
-$$\Psi_{gesamt} = L_{2d}-L_{AW}-L_{BP,KW}=1{,}2708-0{,}2205-0{,}7397=0{,}311 \, \frac{\text{W}}{\text{m} \cdot \text{K}}}$$
+\Psi_{overall} = L_{2d}-L_{AW}-L_{BP,KW}=1{,}2708-0{,}2205-0{,}7397=0{,}311 \, \dfrac{\text{W}}{\text{m} \cdot \text{K}}
-</latex>
+$$
 </WRAP>
@@ Line 97: / Line 106: @@
 <WRAP centeralign>
 **Determining the conductance**
-<latex>
 $L_{2d}$
-</latex>
 </WRAP>
@@ Line 106: / Line 113: @@
 {{ :picopen:beheizter_keller_abb_7b.png?600 |}}
 <WRAP centeralign>
-<latex>
+$$
-$$\.q = 29{,}588 \, \frac{\text{W}}{\text{m}}$$ \\
+\Large{\dot{q} = 29{,}588 \, \dfrac{\text{W}}{\text{m}}} \\
-$$L_{2d} = \frac{\.q}{T_i-T_e} = \frac{29{,}588}{30} = 0{,}9863 \, \frac{\text{W}}{\text{m} \cdot \text{K}}}$$
+$$
-</latex>
+</WRAP>
+<WRAP centeralign>
+$$
+\large{L_{2d} = \dfrac{\dot{q}}{T_i-T_e} = \dfrac{29{,}588}{30} = 0{,}9863 \, \dfrac{\text{W}}{\text{m} \cdot \text{K}}}
+$$
 </WRAP>
@@ Line 119: / Line 131: @@
 </WRAP>
 <WRAP centeralign>
-<latex>
+$$
-$$\Psi_{unten} = L_{2d}-L_{BP,KW}=0{,}9863-0{,}7397 = 0{,}247 \, \frac{\text{W}}{\text{m} \cdot \text{K}}}$$
+\large{\Psi_{lower} = L_{2d}-L_{BP,KW}=0{,}9863-0{,}7397 = 0{,}247 \, \dfrac{\text{W}}{\text{m} \cdot \text{K}}}
-</latex>
+$$
 </WRAP>
@@ Line 132: / Line 144: @@
 <WRAP centeralign>
-<latex>
+$$
-$$\Psi_{upper} = \Psi_{overall}-\Psi_{lower} = 0{,}311-0{,}247 = 0{,}064 \, \frac{\text{W}}{\text{m} \cdot \text{K}}}$$
+\large{\Psi_{upper} = \Psi_{overall}-\Psi_{lower} = 0{,}311-0{,}247 = 0{,}064 \, \frac{\text{W}}{\text{m} \cdot \text{K}}}
-</latex>
+$$
 </WRAP>
@@ Line 146: / Line 158: @@
 <WRAP centeralign>
 **Determining the minimum surface temperature **
-<latex>
 $f_{Rsi}$
-</latex>
 </WRAP>
@@ Line 157: / Line 167: @@
 {{ :picopen:beheizter_keller_abb_8b.png?600 |}}
 <WRAP centeralign>
-<latex>
+$$
-$$f_{Rsi,A} = \frac{18{,}5-(-10)}{20-(-10)} = 0{,}94$$
+\large{f_{Rsi,A} = \dfrac{18{,}5-(-10)}{20-(-10)} = 0{,}94}
-$$f_{Rsi,B} = \frac{16{,}9-(-10)}{20-(-10)} = 0{,}89$$
+$$
-$$f_{Rsi,C} = \frac{11{,}6-(-10)}{20-(-10)} = 0{,}72$$
-</latex>
 </WRAP>
+<WRAP centeralign>
+$$
+\large{f_{Rsi,B} = \dfrac{16{,}9-(-10)}{20-(-10)} = 0{,}89}
+$$
+</WRAP>
+<WRAP centeralign>
+$$
+\large{f_{Rsi,C} = \dfrac{11{,}6-(-10)}{20-(-10)} = 0{,}72}
+$$
+</WRAP>
 **Note!** //areas which are near the ground surface are subject to bigger temperature fluctuations than those which are further away from the surface. The steady-state calculated surface temperatures of areas further away are therefore less meaningful, but are usually on the safe side because in reality the boundary conditions necessary for this would have had to prevail for months. If more exact surface temperatures are required, then the solution may lie in two or three-dimensional transient calculations.//