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Benutzer:Stephan Schwarzbold

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de Dieser Benutzer spricht Deutsch als Muttersprache.
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Ich: Baujahr 1977; Tischlermeister, Student der Architektur in Weimar; viel unterwegs real und hier.



Vereinfachung Wärmeträgheitswert für mehrschichtige Bauteile:

D = \sum{D_i} = \sum{R_i \cdot S_i} Formel.Waermetraegheitswert.einfach.png

R = \sum R_i Formel.R_Summe.png

S_{24} = \frac{D}{R} = \frac{\sum{R_i \cdot S_i}}{\sum R_i}Formel.S24.png

\Theta = 0{,}45 \bigg( 1 + \frac{S}{\alpha_{a}} + \frac{\alpha_{i}}{S} + \frac{\alpha_{i}}{\alpha_{a}} \bigg) \cdot \exp{\frac{D}{\sqrt{2}}}Formel.TAV.einfach.png

HC = \sum {\rho_i c_i s_i}Formel.HC.png

S = max(S24 & [sqrt((0,0085² HC) / R]) S = \max\left(S_{24}, \frac{\sqrt{0{,}0085^2 \cdot HC}}{R}\right)Formel.Differenzierung.S_max.png


näherungsweise Vereinfachung Phasenverschiebung:

\nu \approx 0{,}113t_0 \cdot D - 0{,}017t_0 \approx 2{,}7 D - 0,4 bei t0 = 24h

Phasenverschiebung einschalig: \nu = \frac{1}{15} \cdot \left( 40{,}5 \cdot \sum D - \arctan \frac{R_{si}}{R_{si} + S \cdot \sqrt{2}} + \arctan \frac{S}{S + R_{se} \cdot \sqrt{2}} \right)

Phasenverschiebung zweischalig: \nu = \frac{1}{15} \cdot \left( 40{,}5 \cdot \sum D - \arctan \frac{R_{si}}{R_{si} + U_i \cdot \sqrt{2}} + \arctan \frac{U_a}{U_a + R_{se} \cdot \sqrt{2}} \right)

das TAV ist der Kehrwert der TAD: \nu = \frac{1}{\Theta}

bei d-größergleich-1: U_i = \frac{R_i \cdot S^2_i + U_{i-1}}{1 + R_i \cdot U_{i-1}}

TAD-mehrschichtige Wand: \Theta = 0{,}9 \cdot \left( \frac{S_{1} + R_{si}}{S_{1} + U_{1}} \cdot \left( \prod \frac {S_{n} + U_{n-1}}{S_{n} + U_{n}} \right) \cdot \frac{R_{se} + U_n}{R_{se}} \right) \cdot \exp \left( \sum \frac{D_i}{\sqrt{2}} \right)

Wärmespeicherkennwert: für D-größergleich-1 U = S

für D<1 U = \frac{R \cdot S^2 + R_{si}}{1 + R \cdot R_{se}}

Wärmespeicherkennwert S = 0,0085\cdot b

Wärmedurchlasswiederstand D = S \cdot R_i

TAD-einschichtig: \Theta = 0{,}9\cdot \frac{(S+R_{si})\cdot(R_{se}+U)}{(S+U)\cdot R_{se}}\cdot \exp \left(\frac{D}{\sqrt{2}}\right)

U = \frac{1}{R_{se} + \sum \frac{s_i}{\lambda_i} + R_{si}}

U = \frac{1}{R_{se} + \sum R_i + R_{si}}

R_{nom} = \frac{d_{nom}}{\lambda_{nom}}\qquad \swarrow

R_{nom} = \frac{d_{nom}}{\lambda_{nom}}

\Delta R = \frac{\partial R}{\partial d} \Delta d - \frac{\partial R}{\partial \lambda} \Delta \lambda \qquad \swarrow

\Delta R = \frac{\partial R}{\partial d} \Delta d - \frac{\partial R}{\partial \lambda} \Delta \lambda

\Delta R = \frac{1}{\lambda} \Delta d - \frac{d}{\lambda^2} \Delta \lambda \qquad \swarrow

\Delta R = \frac{1}{\lambda} \Delta d - \frac{d}{\lambda^2} \Delta \lambda

ΔReff = R + ΔR

R_{eff} \ge R_{erf}

Z_{90%} = \frac{90%\cdot1}{80%}Z = 1{,}125\cdot Z

\lambda_{10,tr}= \frac{\lambda_{10,g}}{1+6\cdot u_{v}} \qquad \swarrow

\lambda_{10,tr}= \frac{\lambda_{10,g}}{1+6\cdot u_{v}}

\lambda_{10,tr}= \frac{\lambda_{10,g}}{1+u_{m}} \qquad \swarrow

\lambda_{10,tr}= \frac{\lambda_{10,g}}{1+u_{m}}

\lambda_{10,tr}= \frac{\lambda_{10,g}}{1+\frac{u_{m}}{2}} \qquad \swarrow

\lambda_{10,tr}= \frac{\lambda_{10,g}}{1+\frac{u_{m}}{2}}

\lambda_{10,tr}= \frac{\lambda_{10,g}}{1+6\cdot u_{v}} \qquad \swarrow

\lambda_{10,tr}= \frac{\lambda_{10,g}}{1+6\cdot u_{v}}

\lambda_{Z90%} = \lambda_{10,tr}\cdot (1+Z_{90%}) \qquad \swarrow

\lambda_{Z90%} = \lambda_{10,tr}\cdot (1+Z_{90%})

\lambda_{R} = 1{,}05\lambda_{10,tr}\cdot (1+Z_{90%}) \qquad \swarrow

\lambda_{R} = 1{,}05\lambda_{10,tr}\cdot (1+Z_{90%})

\lambda_{10,tr} = \frac{\lambda_{4108}}{1+Z} \qquad \swarrow

\lambda_{10,tr} = \frac{\lambda_{4108}}{1+Z}

\Delta\lambda = \lambda_{R}-\lambda_{4108} \qquad \swarrow

Δλ = λR − λ4108

\Delta\lambda = 1{,}05\frac{\lambda_{4108}}{1+Z}\cdot (1+Z_{90%})-\lambda_{4108} \qquad \swarrow

\Delta\lambda = 1{,}05\frac{\lambda_{4108}}{1+Z}\cdot (1+Z_{90%})-\lambda_{4108}

\Delta\lambda = \left( \frac{1{,}05\lambda_{4108}}{1+Z}\cdot (1+Z_{90%})\right)-\lambda_{4108} \qquad \swarrow

\Delta\lambda = \left( \frac{1{,}05\lambda_{4108}}{1+Z}\cdot (1+Z_{90%})\right)-\lambda_{4108}

\Delta\lambda = \left( \frac{1{,}05\lambda_{4108}}{1+Z}\cdot (1+1{,}125Z)\right)-\lambda_{4108} \qquad \swarrow

\Delta\lambda = \left( \frac{1{,}05\lambda_{4108}}{1+Z}\cdot (1+1{,}125Z)\right)-\lambda_{4108}

Z = \frac{\lambda_{4108}}{\lambda_{10,tr}}-1

\Delta R_{1} = \frac{1}{1{,}00}\cdot (-0{,}005) - \frac{0{,}02}{1{,}00^2}\cdot 0{,}078 = -0{,}00656

\Delta R_{2} = \frac{1}{0{,}23}\cdot (-0{,}01) - \frac{0{,}365}{0{,}23^2}\cdot 0{,}015 = -0{,}14698

\Delta R_{3} = \frac{1}{0{,}70}\cdot (-0{,}005) - \frac{0{,}015}{0{,}70^2}\cdot 0{,}053 = -0{,}00877

R_{1} = \frac{0{,}02}{1{,}00} = 0{,}02 \qquad R_{eff,1} = 0{,}02 + (-0{,}00656) = 0{,}01344

R_{2} = \frac{0{,}365}{0{,}23} = 1{,}587 \qquad R_{eff,2} = 1{,}587 + (-0{,}14698) = 1{,}43998

R_{3} = \frac{0{,}015}{0{,}70} = 0{,}021 \qquad R_{eff,3} = 0{,}021 + (-0{,}00877) = 0{,}01266

U_{eff} = \frac{1}{0{,}13 + 1{,}46608 + 0{,}04} = 0{,}61

Wärmeeindringungskoeffizient: b = \sqrt{c \cdot \lambda \cdot \rho}

porosität: c_p = 1 - \frac{\rho_{n}}{\rho_{g}}

Temperaturamplitudendämpfung \Theta = \frac{1}{\nu} = \frac{\Delta \hat \theta_{e}}{\Delta \hat \theta_{i}} \ge 1{,}0

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