不会增加潜在的温度随高度与对流层内绝热过程的本质?- 江南体育网页版- - - - -地球科学堆江南电子竞技平台栈交换 最近30从www.hoelymoley.com 2023 - 07 - 07 - t22:12:08z //www.hoelymoley.com/feeds/question/20180 https://creativecommons.org/licenses/by-sa/4.0/rdf //www.hoelymoley.com/q/20180 0 不会增加潜在的温度随高度与对流层内绝热过程的本质? MichaelW //www.hoelymoley.com/users/21047 2020 - 09 - 08 - t08:53:46z 2020 - 09 - 08 - t15:51:44z < p >我提到这个问题,我问最近在物理委员会:< / p > < p > < a href = " https://physics.stackexchange.com/questions/578308/doesnt-increase-of-potential-temperature-with-height-contradict-adiabatic-natur " > https://physics.stackexchange.com/questions/578308/doesnt-increase-of-potential-temperature-with-height-contradict-adiabatic-natur < / > < / p > < p >类型再没有意义,所以我链接到它。也许这个问题可以在这里回答。< / p > //www.hoelymoley.com/questions/20180/-/20181 # 20181 4 答案由Joscha Fregin不会增加潜在的温度随高度矛盾在对流层绝热过程的本质? Joscha Fregin //www.hoelymoley.com/users/18833 2020 - 09 - 08 - t13:57:27z 2020 - 09 - 08 - t15:47:38z < p >如果我们生活在一个干燥的气氛确实你的推理是正确的。空气绝热地上升和空气会松约9.8°C /公里(干绝热温度梯度)。这意味着温度恒定的潜力。然而,地球大气不干燥。一旦上升,潮湿的空气包裹达到饱和时,它将上升潮湿绝热递减率(6 - 7°C /公里)。潜在的温度将上升(因为如果包裹再次摔倒时,这将是温暖的在同一水平(见下文)。这导致云,有时下雨,因此,减少了总体水含量的空气包裹。< / p > < p >如果这样一个包裹现在要再次下降将获得根据干绝热温度递减率(因为它失去了水在提升)。假设最初一个包裹开始上升与湿绝热递减率与温度<跨类= " math-container " > T (T = 0)美元< / span >,然后再次摔倒,达到相同的水平在<跨类= " math-container " > T = 1美元< / span >。然后,您将观察到< span class = " math-container " > $ T (T = 0) & lt; T(t=1)$. Thus, potential temperature must have increased as the parcel was rising.

//www.hoelymoley.com/questions/20180/-/20182 # 20182 4 回答由BarocliniCplusplus不会增加潜在的温度随高度与对流层内绝热过程的本质? BarocliniCplusplus //www.hoelymoley.com/users/704 2020 - 09 - 08 - t15:51:44z 2020 - 09 - 08 - t15:51:44z < p >你可能忘记压力也随高度(指数)。另外,因为< span class = " math-container " > P = \ρrt美元< / span >, < span class = " math-container " > $ \压裂{dP} {dT} = \ρ< / span >雷亚尔(也就是说,< span class = " math-container " > c_p < / span >美元不出现)。但在回答你的问题我跑题了。< / p > < p >让我们打破为什么潜在温度随高度增加。先方程:< span class = " math-container " > $ $ \θ= T \离开(\压裂{P_0} {P} \右)^{\压裂{R_d} {c_p}}{1} \标签$ $ < / span >现在,如果我们想问为什么< span class = " math-container " > \θ< / span >美元增加高度,让我们换种<跨类= " math-container " > (1) < / span >美元作为高度的函数:< span class = " math-container " > $ $ \θ(z)左= T (z) \[\压裂{P_0} {P (z)} \右]^{\压裂{R_d} {c_p}} \标记{2}$ $ < / span >现在如果你遵循< a href = " //www.hoelymoley.com/questions/19517/derivative-of-exner-function/19519 # 19519 " >我推导< / >寻找< span class = " math-container " > $ $ \压裂{\部分\θ}{\部分z} = \压裂{\θ}{T} \离开(\压裂{\部分T}{\部分z} + \压裂{g} {c_p} \右)$ $ < / span > < / P > < P >您可能会注意到,< span class = " math-container " > $ \压裂{\部分\θ}{\部分z} = 0 < / span >敌我识别美元。< span class = " math-container " > $ \压裂{\部分T}{\部分z} = - \压裂{g} {c_p} = -9.8 \ textrm {K公里}^ {1}$ < / span >(干绝热温度梯度)。If $\frac{\partial T}{\partial z}>-\frac{g}{c_p}$ then $\frac{\partial \theta}{\partial z}>0$.

This does not mean that the parcel is not adiabatic, though. For a parcel to be truly adiabatic, $\frac{\partial \theta}{\partial t}+u\frac{\partial \theta}{\partial x}+v\frac{\partial \theta}{\partial y}+w\frac{\partial \theta}{\partial z}=0$ Therefore, the only time a certain location can be considered adiabatic and have $\frac{\partial \theta}{\partial z}=0$ is when there is no heating ($\frac{\partial \theta}{\partial t}=0$), and no horizontal advection of $\theta$ ($u\frac{\partial \theta}{\partial x}+v\frac{\partial \theta}{\partial y}=0$). Note, I did not exclude vertical advection of $\theta$, because that is implied in the condition that $\frac{\partial \theta}{\partial z}=0$.

I'll also take pause here and list a few diabatic processes, and why there may be some justification to ignore them for your mental image.

  • Radiation: with the exception of greenhouse gases, the atmosphere is mostly transparent.
  • Surface heating/ turbulent transport: more important near the surface (note that $\theta$ is a minimum at the surface).
  • Clouds/microphysical processes: discrete and are never everywhere all the time and moisture is a completely separate variable that varies all the time. There is a way around that (to some extent- the equivalent potential temperature, for latent heat exchange).
  • Chemical processes: See below about how I remember $\theta$ increases with height.

If you want a good reminder of why $\theta$ increases with height, it might be easier to remember that $sgn \left( \frac{\partial \theta}{\partial z}\right)$ is a good indicator of static stability. And thunderstorms need to stop at some point, even if that is the tropopause (where $\frac{\partial \theta}{\partial z}>0$ due to ozone heating).

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