最近30从www.hoelymoley.com 2023 - 04 - 01 - t02:18:09z //www.hoelymoley.com/feeds/question/21213 https://creativecommons.org/licenses/by-sa/4.0/rdf //www.hoelymoley.com/q/21213 1 Letitbe //www.hoelymoley.com/users/22424 2021 - 05 - 04 - t13:35:28z 2021 - 08年- 08 - t21:41:16z

< p >我有一个问题从去年在竞争激烈的考试:< / p > < p >的变化率计算地球的对流层温度和压力(<跨类= " math-container " > $ \压裂{dT} {dP} $ < / span >),鉴于< / p > < ul > <李>标高,< span class = " math-container " > H < / span >,美元在< span class = " math-container " > $ p(\文本{H}) =(\压裂{1}{e} \ cdot P_0) = 8 \文本{公里}$ < / span >(的位势高度压力等于美元<跨类= " math-container " > \压裂{1}{e} < /跨度>最初的一美元)< /李> <李>递减率<跨类=“math-container”> = \γ= 6.5美元\压裂{°C}{公里}$ < / span >(温度随高度的变化)< /李> < / ul > < p >我试着使用潜在的温度方程和压力曲线方程,但是我不能连接它。< / p > < p >请帮我或者只是给一个建议。谢谢! < / p >

//www.hoelymoley.com/questions/21213/-/21315 # 21315 1 User123 //www.hoelymoley.com/users/19122 2021 - 05 - 27 - t17:51:40z 2021 - 08年- 08 - t21:41:16z

< p >哇,@John附带了一个简单的答案。< / p > < p >我们的目标是找到< / p > < p > <跨类= " math-container " > $ $ \压裂{dT} {dP} = \压裂{dT} {dz} \ cdot \压裂{dz} {dP} $ $ < / span > < / p > < p >压力方程的失误:< span class = " math-container " > $ $ p = P_0 \ cdot e ^{- \压裂{z} {H}} $ $ < / span >, < span class = " math-container " > H = 8公里= 8000美元< / span >(标高)和<跨类= " math-container " > z < / span >是美元表面的高度。< / p > < p >我们区分这一次,得到:< span class = " math-container " > $ $ \压裂{dP} {dz} = \压裂{P_0}} {H e ^{- \压裂{z} {H}} $ $ < / span > < / p > < p >逆值<跨类= " math-container " > $ $ \压裂{dz} {dP} = \压裂{H} {P_0 \ cdot e ^{- \压裂{z} {H}}} $ $ < / span > < / p > < p >和<跨类= " math-container " > $ \压裂{dT} {dz} $ < / span > ?它只是递减率<跨类= " math-container " > \γ< / span >美元。< / p > < p >使用上面,我们重申:< / p > < p > <跨类= " math-container " > $ $ \压裂{dT} {dP} = \压裂{dT} {dz} \ cdot \压裂{dz} {dP} = \压裂{\伽马H} {P_0 \ cdot e ^{- \压裂{z} {H}}} $ $ < / span >但不是我们有神秘类< span = " math-container " > z < / span >美元我们不想要的变量。所以,< span class = " math-container " > ^ $ $ P = P_0 e{- \压裂{z} {H}} $ $ < / span >和<跨类= " math-container " > $ $ z = - H \ ln \压裂{P} {P_0} $ $ < / span >我们插入< span class = " math-container " > $ \压裂{dT} {dP} $ < / span >方程得到类< span = " math-container " > $ $ \压裂{dT} {dP} = \压裂{\伽马H} {P} $ $ < / span >, < span class = " math-container " > H = 8000美元\ rm \, m $ < / span >, < span class = " math-container " > $ \γ= 6.5 \ rm \ \压裂{K}{公里}$ < / span >和<跨类= " math-container " > \ rm \ P = 100美元,美元kPa < / span >。因此,< span class = " math-container " > $ $ \压裂{dT} {dP} = 0.052 \ rm \ \压裂{K} {hPa} $ $ < / span >(我相信)足够近。< / p >

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