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basics:building_physics_-_basics:thermal_bridges:tbcalculation:examples:unheatedb

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basics:building_physics_-_basics:thermal_bridges:tbcalculation:examples:unheatedb [2019/01/24 08:47] – [Calculation of the conductance] cblagojevicbasics:building_physics_-_basics:thermal_bridges:tbcalculation:examples:unheatedb [2019/01/24 09:30] – cblagojevic
Line 11: Line 11:
 <WRAP centeralign>  <WRAP centeralign> 
 <latex> <latex>
-$$\frac{1}{U} = \frac{1}{U_f} + \frac{A}{(A \cdot U_{bf}) + (z \cdot P \cdot U_{bw}) + (h \cdot P \cdot U_w) + (0{.}33 \cdot n \cdot V)}+\large{\dfrac{1}{U} = \dfrac{1}{U_f} + \dfrac{A}{(A \cdot U_{bf}) + (z \cdot P \cdot U_{bw}) + (h \cdot P \cdot U_W) + (0{,}33 \cdot n \cdot V)}}
 </latex> </latex>
 \\ \\
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 <WRAP centeralign>  <WRAP centeralign> 
 <latex> <latex>
-$$\Phi = L_{iu} \cdot (\theta_i - \theta_u) + L_{ie} \cdot (\theta_i - \theta_e)$$+\large{\Phi = L_{iu} \cdot (\theta_i - \theta_u) + L_{ie} \cdot (\theta_i - \theta_e)}
 </latex> </latex>
 </WRAP> </WRAP>
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 <WRAP centeralign>  <WRAP centeralign> 
 <latex> <latex>
-$$\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)$$+\large{\Phi = \left(\frac{L_{iu} \cdot L_{ue}}{L_{iu} + L_{ue}}\right) \cdot (\theta_i - \theta_e) \quad \Rightarrow \quad L_{2d} =  \left(\frac{L_{iu} \cdot L_{ue}}{L_{iu} + L_{ue}} + L_{ie}\right)}
 </latex> </latex>
 </WRAP> </WRAP>
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 **Determining the conductance**  **Determining the conductance** 
 <latex> <latex>
-$L_{2d}$ +\Large{_{2d}}
 </latex> </latex>
 </WRAP> </WRAP>
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 <WRAP centeralign>  <WRAP centeralign> 
 <latex> <latex>
 +\large{
 $$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_2+L_3-L_1) = 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> </latex>
 </WRAP> </WRAP>
 +
  
 <WRAP centeralign>  <WRAP centeralign> 
 <latex> <latex>
-$$L_{2d} =  \left(\frac{L_{iu} \cdot L_{ue}}{L_{iu} + L_{ue}} + L_{ie}\right) = 0{.}6886 \, \frac{\text{W}}{\text{m} \cdot \text{K}}}$$+\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> </latex>
 </WRAP> </WRAP>
Line 101: Line 102:
  
 Therefore: Therefore:
- 
 <WRAP centeralign>  <WRAP centeralign> 
 <latex> <latex>
-$$\frac{1}{U} = \frac{1}{U_f} + \frac{A}{(A \cdot U_{bf}) + (z \cdot P \cdot U_{bw}) + (h \cdot P \cdot U_w) + (0{.}33 \cdot n \cdot V)} +$$\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 \quad \Rightarrow \quad
-U = 0{.}1273 \, \frac{\text{W}}{\text{m}^2 \cdot \text{K}}} $$+U = 0{.}1273 \, \dfrac{\text{W}}{\text{m}^2 \cdot \text{K}}}$$
 </latex> </latex>
 </WRAP> </WRAP>
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 <WRAP centeralign>  <WRAP centeralign> 
 <latex> <latex>
-$$\Psi_g = L_{2d}-l_{EW} \cdot U_{EW}-0{.}5 \cdot B' \cdot U$$+$$\Psi_g = L_{2d}-l_{AW} \cdot U_{AW}-0{.}5 \cdot B' \cdot U$$
 </latex> </latex>
 </WRAP> </WRAP>
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 <WRAP centeralign>  <WRAP centeralign> 
 <latex> <latex>
-$$\Psi_g = 0{.}687 \, \frac{\text{W}}{\text{m} \cdot \text{K}}} \, - \, 1{.}830 \, \text{m} \, \cdot \, 0{.}120 \, \frac{\text{W}}{\text{m}^2 \cdot \text{K}}} \, - \, 0{.}5 \, \cdot \, 8 \, \text{m} \, \cdot \, 0{.}1273 \, \frac{\text{W}}{\text{m}^2 \cdot \text{K}}} = -0{.}042 \, \frac{\text{W}}{\text{m} \cdot \text{K}}}$$+$$\Psi_g = 0{.}687 \, \dfrac{\text{W}}{\text{m} \cdot \text{K}}} \, - \, 1{.}830 \, \text{m} \, \cdot \, 0{.}120 \, \dfrac{\text{W}}{\text{m}^2 \cdot \text{K}}} \, - \, 0{.}5 \, \cdot \, 8 \, \text{m} \, \cdot \, 0{.}1273 \, \dfrac{\text{W}}{\text{m}^2 \cdot \text{K}}} = -0{.}042 \, \dfrac{\text{W}}{\text{m} \cdot \text{K}}}$$
 </latex> </latex>
 </WRAP> </WRAP>
Line 139: Line 139:
 **Determining the conductance**  **Determining the conductance** 
 <latex> <latex>
-$L_{2d}$ +\Large{L_{2d}}
 </latex> </latex>
 </WRAP> </WRAP>
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 **Ψ-value (proportion of exterior wall)**  **Ψ-value (proportion of exterior wall)** 
 </WRAP> </WRAP>
- 
 <WRAP centeralign>  <WRAP centeralign> 
 <latex> <latex>
-$$\Psi_g = L_{ie}-l_{EW} \cdot U_{EW}$$ +\large{\Psi_g = L_{ie}-l_{AW} \cdot U_{AW}$} 
-$$\Psi_{exterior wall} = 0{.}231-1{.}830 \cdot 0{.}120 = 0{.}0114$$+</latex> 
 +</WRAP> 
 + 
 +<WRAP centeralign> 
 +<latex> 
 +\large{\Psi_{exterior wall} = 0{.}231-1{.}830 \cdot 0{.}120 = 0{.}0114}
 </latex> </latex>
 </WRAP> </WRAP>
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 <WRAP centeralign>  <WRAP centeralign> 
 <latex> <latex>
-$$\Psi_g = L_{iu}-0{.}5 \cdot B' \cdot U_{basement ceiling}$$ +\large{\Psi_g = L_{iu}-0{.}5 \cdot B' \cdot U_{basement ceiling}} 
-$$\Psi_{ basement ceiling } = 0{.}5543-0{.}5 \cdot 8 \cdot 0{.}148 = -0{.}0377$$+</latex> 
 +</WRAP> 
 + 
 +<WRAP centeralign> 
 +<latex> 
 +\large{\Psi_{basement ceiling} = 0{.}5543-0{.}5 \cdot 8 \cdot 0{.}148 = -0{.}0377}
 </latex> </latex>
 </WRAP> </WRAP>
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 <WRAP centeralign>  <WRAP centeralign> 
 <latex> <latex>
-$$U_{f,corrected} = U_f + \frac{\Psi_{basement ceiling} \cdot P}{A}$$+\large{U_{f,corrected} = U_f + \dfrac{\Psi_{basement ceiling} \cdot P}{A}}
 </latex> </latex>
 </WRAP> </WRAP>
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 <WRAP centeralign>  <WRAP centeralign> 
 <latex> <latex>
-$$\theta_{Keller} = \theta_i-f_x \cdot (\theta_i - \theta_e) = 20 \, ^\circ C - 0{.}6 \cdot (20 \, ^\circ C - (-10 \, ^\circ C)) = 2 \, ^\circ C$$+\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> </latex>
 </WRAP> </WRAP>
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 **Determining the minimum surface temperature and**  **Determining the minimum surface temperature and** 
 <latex> <latex>
-$f_{Rsi}$ +\Large{f_{Rsi}}
 </latex> </latex>
 </WRAP> </WRAP>
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 <WRAP centeralign>  <WRAP centeralign> 
 <latex> <latex>
-$$f_{Rsi} = \frac{17{.}6 - (-10)}{20-(-10)}=0{.}92$$+\large{f_{Rsi} = \dfrac{17{.}6 - (-10)}{20-(-10)}=0{.}92}
 </latex> </latex>
 </WRAP> </WRAP>
basics/building_physics_-_basics/thermal_bridges/tbcalculation/examples/unheatedb.txt · Last modified: 2022/02/15 18:58 by admin