Translation examples
Jos Q ˙ {\displaystyle {\dot {Q}}} on negatiivinen, lämpöä siirtyy nettomääräisesti pinnalta 2 pinnalle 1. Kun kaksi harmaapintaista kappaletta sulkee väliinsä tilan, niiden välinen nettomääräinen lämmön siirtyminen on: Q ˙ = σ ( T 1 4 − T 2 4 ) 1 − ϵ 1 A 1 ϵ 1 + 1 A 1 F 1 → 2 + 1 − ϵ 2 A 2 ϵ 2 {\displaystyle {\dot {Q}}={\dfrac {\sigma (T_{1}^{4}-T_{2}^{4})}{{\dfrac {1-\epsilon _{1}}{A_{1}\epsilon _{1}}}+{\dfrac {1}{A_{1}F_{1\rightarrow 2}}}+{\dfrac {1-\epsilon _{2}}{A_{2}\epsilon _{2}}}}}} missä ϵ 1 {\displaystyle \epsilon _{1}} ja ϵ 2 {\displaystyle \epsilon _{2}} ovat pintojen emissiivisyydet.
For two grey-body surfaces forming an enclosure, the heat transfer rate is: Q ˙ = σ ( T 1 4 − T 2 4 ) 1 − ϵ 1 A 1 ϵ 1 + 1 A 1 F 1 → 2 + 1 − ϵ 2 A 2 ϵ 2 {\displaystyle {\dot {Q}}={\dfrac {\sigma (T_{1}^{4}-T_{2}^{4})}{{\dfrac {1-\epsilon _{1}}{A_{1}\epsilon _{1}}}+{\dfrac {1}{A_{1}F_{1\rightarrow 2}}}+{\dfrac {1-\epsilon _{2}}{A_{2}\epsilon _{2}}}}}} where ϵ 1 {\displaystyle \epsilon _{1}} and ϵ 2 {\displaystyle \epsilon _{2}} are the emissivities of the surfaces.
Näin ollen Maan keskilämpötila riippuu aurinkovakiosta, Maan albedosta ja emissiivisyydestä pitkäaaltoiselle lähtösäteilylle. ( 1 − a ) S π r 2 = 4 π r 2 ϵ σ T 4 {\displaystyle \ (1-a)S\pi r^{2}=4\pi r^{2}\epsilon \sigma T^{4}} S on aurinkovakio, noin 1368 W·m-2 a {\displaystyle a} on Maan keskimääräinen albedo, mitattu 0.3 r on Maan säde, 6.371×106m π on piin arvo, noin 3.14159 σ {\displaystyle \sigma } on Stefanin-Boltzmannin vakio — joka on noin 5.67×10-8 J·K-4·m-2·s-1 ϵ {\displaystyle \epsilon } on Maan efektiivinen emissiivisyys 0.612, joka ottaa huomioon kasvihuone-ilmiön Vakio πr2 voidaan supistaa pois ( 1 − a ) S = 4 ϵ σ T 4 {\displaystyle \ (1-a)S=4\epsilon \sigma T^{4}} Ja lämpötila saadaan tästä yhtälöstä T = ( S ( 1 − a ) 4 ϵ σ ) 1 4 {\displaystyle \ T=\left({\frac {S(1-a)}{4\epsilon \sigma }}\right)^{\frac {1}{4}}} Tiedetään, että Maan aurinkovakio vaihtelee 1.4%, albedo 3.3% ja efektiivinen emissiivisyys 1.4%.
A very simple model of the radiative equilibrium of the Earth is ( 1 − a ) S π r 2 = 4 π r 2 ϵ σ T 4 {\displaystyle (1-a)S\pi r^{2}=4\pi r^{2}\epsilon \sigma T^{4}} where the left hand side represents the incoming energy from the Sun the right hand side represents the outgoing energy from the Earth, calculated from the Stefan-Boltzmann law assuming a model-fictive temperature, T, sometimes called the 'equilibrium temperature of the Earth', that is to be found, and S is the solar constant – the incoming solar radiation per unit area—about 1367 W·m−2 a {\displaystyle a} is the Earth's average albedo, measured to be 0.3. r is Earth's radius—approximately 6.371×106m π is the mathematical constant (3.141...) σ {\displaystyle \sigma } is the Stefan-Boltzmann constant—approximately 5.67×10−8 J·K−4·m−2·s−1 ϵ {\displaystyle \epsilon } is the effective emissivity of earth, about 0.612 The constant πr2 can be factored out, giving ( 1 − a ) S = 4 ϵ σ T 4 {\displaystyle (1-a)S=4\epsilon \sigma T^{4}} Solving for the temperature, T = ( 1 − a ) S 4 ϵ σ 4 {\displaystyle T={\sqrt{\frac {(1-a)S}{4\epsilon \sigma }}}} This yields an apparent effective average earth temperature of 288 K (15 °C; 59 °F).
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