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关于大气干绝热递减率推导的一些典型争论

已有 1067 次阅读 2024-11-3 09:07 |个人分类:大气物理相关:干绝热递减率与大气稳定度|系统分类:论文交流

大气的干绝热递减率是用来判断大气垂向稳定性的标准。但是,笔者认为现有教科书中对这个参数的推导存在一些错误。下文汇总了最近一年多时间里,笔者在投稿多个外文期刊的过程中与多名学者的交流辩论的主要内容,同时,我把相关论文发布在了ESS Open Archive(地球与空间科学开放档案Earth and Space Science Open Archive详见:https://essopenarchive.org/users/643291/articles/656727-physical-mechanism-analysis-of-atmospheric-dry-adiabatic-lapse-rate

关于绝热递减率的争论集中在以下三个问题上。对这三个问题的分析决定了本文提出的推导方法是否有意义和价值。

1.有学者认为,大气的干绝热递减率的推导应遵循“绝热递减率”的定义。

我的回答:推导大气绝热递减率的目的,是分析和计算运动特性在垂直方向上处于中性稳定的大气的垂直温度递减率,而不是因为名称为“绝热递减率”而推导大气的绝热递减率。

2.绝热递减率的推导中忽略热交换合适吗?

我的回复:下任何结论都需要有理由,下特殊的结论更需要有特殊的理由。关于忽略热交换的问题,请支持者给出详细的分析计算和数据,以证明在气团升降的过程中,气团与周围大气之间的热交换速率,与他们之间的体积功交换相比,前者可以忽略不计。同时,请详细说明此次计算使用的气团大小,以及气团的大小是否可以小到微米或接近纳米尺度。

关于这个问题的一个简化思路是:物体的热交换率与其比表面积成正比,而对于一个气团来说,无论是球形还是立方体,其比表面积与其直径(或边长)正好成反比,这意味着当气团的尺寸从米减小到微米时,其比表面积将增加1000000倍,如果减小到接近纳米水平,这种增加就更大。因此,我们有什么理由得出以下结论:任意大小的气团与周围环境大气的体积功交换率一定远大于其热交换率,所以热交换可以忽略不计呢?这个问题解释不了的话,就已经足以推翻现有教材在干绝热递减率的推导方法。

3.有些学者坚持认为,在现有教材的推导中,dP=ρ*g*dz是指重力的做功,因此在现有教材的绝热递减率推导中,已经考虑了重力做功。

我的回复:dP=ρ*g*dz是在假设密度不随高度变化的情况下,对静力平衡公式P=ρ*g*z的近似计算,其中g*dz不是指重力因高度变化所做的功的意义,而只是指,在重力存在时的垂直压力分布结果的数值计算。更详细地说,P=ρ*g*z(微分分法为dP=ρ*g*dz,假设密度不变),是对静止流体中压力分布的数值计算,是对静态平衡的数值计算,而不是对流体中粒子运动过程中重力所做的功的计算。一个物理公式只能考虑一种物理机制,要么是对静力平衡结果的数值计算,要么是对物体运动过程中重力所做的功的计算,不能两者同时都是。重力势能引起的垂直压力分布的静力平衡计算,不能等同于重力做功的动态计算。静力平衡方程只计算由于引力场的存在(或引力势能的存在)而导致的压力的垂直静态分布,而不是计算物体在垂直运动过程中重力所做的动态功。

为了说明上述这个问题,我们还可以做一个假象实验:将一个直立的圆柱体装满水,圆柱体中的水压可以使用P=ρ*g*z计算(差分为dP=ρ*g*dz)。现在,假设在圆柱体的横截面(任意高度)上增加一个不透水、无重量、无厚度的圆形刚性膜,该膜完全切断了圆柱体内的水,且膜与圆筒内壁之间没有摩擦,则圆筒内水压的垂直分布应该完全不受影响,并且仍然可以使用P=ρ*g*z计算(微分法为dP=ρ*g*dz)。这个假设的例子即可证明,虽然圆柱体内的水压分布是由重力引起的,但它不等同于重力所做的动态功。

更进一步地,在干绝热递减率的推导中,引入公式P=ρ*g*z(差分法为dP=ρ*g*dz,假设密度不变)只是人为地规定了不同高度的气团压力(即在任意高度,气团压力等于同一高度周围大气的压力),而不是计算气团在垂直运动时的能量转换过程。

雅各布的书中推导过程也能证明我的上述观点:

在雅各布的书中(Jacob D.J.1999Introduction to Atmospheric Chemistry Princeton University Press 53-55),构建了一个热力学循环来推导干绝热递减率,该书中的那个热力学循环过程不涉及任何对重力做功的考虑,但获得了与其他教科书相同的结果。在雅各布的书中,推导干绝热递减率的物理过程完全是围绕书中所提出的热力学循环的。虽然在其后来的推导过程中,引入了公式dP=ρ*g*dz,但之所以引入这个公式,只是用于人为地规定气团的压力变化等于周围相同高度的大气压力变化(即在升降过程中,气团的压力与同等高处的环境大气压力处处相等),而不是为了考虑重力功。更为具体地:雅各布书中的热力学循环与气团的运动方向无关,并不特指垂向运动,气团的水平运动也可以建立这样的热力学循环。因此,在这个推导过程中,运动方向的确定仅由dP=ρ*g*dz引入,但事实上,该公式计算的是静止大气中的静力平衡,而不是重力所做的动态功。该书的推导进一步说明,为了获得0.98/100米的计算结果,并不需要考虑重力所做的功。此外,从物理模型建模的角度来说,对于垂向运动的气团的能量变化过程的计算,应将与能量变化相关的所有项目都放在一起进行能量转换和能量守恒的计算,而不可以是单独地计算每个项目,但不计算不同项目之间的能量转换关系”,也不能是简单地指定一个能量转换过程(这里指的是热交换)可以忽略,但却没有足够的依据

上述分析表明,现有教科书在推导大气垂直运动特性的关键参数——绝热递减率时,虽然考虑了气团与周围环境的体积功交换等能量转换过程,但没有考虑重力所做的功,因此这是一种错误的做法。

4.对书上的理论推导的反驳,只需要从理论的角度进行反驳即可,但是,对于书上的理论推导的实验支持和论证不能只是用文字描述,必须要有详实的真实的数据。

中高层大气的常见平均温度分布是0.65/100米左右,不仅小于0.98,更大大小于1.353。因此,不论是以0.98还是1.353作中性大气的临界值,都说明在大多数的常见天气下,垂直温度递减率等于0.65°C/100m的大气是稳定大气,而不是中性大气。只是,如果是以1.353为临界值的话,0.65°C/100m的大气的稳定度显得更加大。在近地表的大气中(主要指高度在1000米以下的大气中),在日出或日落的前后,应该可以比较容易地观测到1.353°C/100m乃至更大的垂向温度分布现象的出现,而且应该可以在这个时间段更加容易地观测到近地表大气在稳定和不稳定之间切换的现象。

同样重要的是,处于中性稳定度的或是处于轻度稳定的大气,并非不会产生垂向的物质(包括示踪剂、污染物等)混合现象,因为由于各种原因所导致的弥散作用(包括浓度梯度、微风等)在大气流体中会一直存在。因此判断大气垂向运动特征是稳定的、中性的还是不稳定的,往往需要用比较宏观的和粗略的尺度来进行判断。

5.理解干绝热递减率的物理意义的最好方法是:以物理图像清晰的和合乎逻辑的方式证明干绝热递减率与大气稳定性之间的关系。这是因为,在推导和证明为什么不同的垂直温度分布导致不同程度的大气稳定性的过程中,会涉及到如何理解大气干绝热递减率的内在物理机制的问题。

基础物理教材是传播物理学知识体系的基础,所有的基础物理教材都经过了大量聪明头脑的检验,因此出错的概率极低,但我们并不能由此断定基础物理教材就一定不会有缺陷。欢迎理性的且有理有据的公开辩论。

 

以下是上文的英文版,内容基本相同:

In the autumn of 2011, in the process of preparing to teach a graduate course, I realized for the first time that the existing textbooks had serious shortcomings in the derivation of the adiabatic lapse rate of atmosphere and the proof of atmospheric stability, and made what I thought was the correct derivation.

Over the years, I have comprehensively reflected on this issue many times. I think my views on the adiabatic lapse rate of atmosphere are correct and logical. For details, please see my article published online: https://www.authorea.com/656727

https://essopenarchive.org/users/643291/articles/656727-physical-mechanism-analysis-of-atmospheric-dry-adiabatic-lapse-rate

The debate on the adiabatic lapse rate focuses on the following three issues. The analysis of these three issues determines whether the derivation method proposed in this article is meaningful and valuable.

1. Some scholars argue that the derivation of the atmospheric adiabatic lapse rate should follow the definition of the "adiabatic lapse rate".

My reply: The purpose of the derivation of the adiabatic lapse rate of atmosphere is to analyze and calculate the vertical temperature lapse rate of air whose motion characteristics are at neutral stability in the vertical direction, instead of deriving the adiabatic lapse rate of the atmosphere because of the name is 'adiabatic lapse rate'.

2. Some scholars insist that, it is appropriate to ignore heat exchange in the derivation of adiabatic lapse rate.

My reply: Conclusions need to have reasons, and special conclusions need to have special reasons. Regarding the problem of ignoring heat exchange, please give detailed analysis calculations and data to prove that the rate of heat exchange is negligible compared to the volume work exchange during the vertical motion of the air parcel. At the same time, please elaborate on the size of the air parcel used for this calculation and whether the size of the air parcel can be in the micron or even nanometer scale.

The rate of heat exchange of an object is directly proportional to its specific surface area. For an air parcel, whether spherical or cubic, its specific surface area is exactly inversely proportional to its diameter (or side length), which means that when the size of an air parcel decreases from meters to microns, its specific surface area will increase by a factor of 1000000, and if it is reduced to close to the nanometer level, this increase is even greater. Therefore, what reason do we have to conclude that the volume work exchange rate between an air parcel of any size and its surroundings must be much greater than its heat exchange rate, so the heat exchange can be ignored?

3. Some scholars argue that, in the existing derivation of the textbook, dP= ρ*g*dz means the consideration of gravity work.

My reply: dP = ρ*g*dz is an approximate calculation of the static equilibrium formula P= ρ*g*z when it is assumed that the density does not change with height, where g*dz does not refer to the meaning of the work done by gravity due to the change in height, but only to the numerical calculation of the result of the vertical pressure distribution due to the presence of gravity.

In more detail, the P= ρ*g*z (the differential method is dP= ρ*g*dz, assuming that the density does not change), is a numerical calculation of the pressure distribution in a stationary fluid, a numerical calculation of the static equilibrium, and not a calculation of the work done by gravity during the movement of particles in the fluid. A physical formula can only consider one physical mechanism, either the numerical calculation of the results of static equilibrium, or the work done by gravity during object motion, but not both.

The calculation of the static equilibrium of the vertical pressure distribution caused by the gravitational potential energy is not equivalent to the dynamic calculation of the work done by gravity. The static equilibrium equation only calculates the vertical static distribution of pressure due to the presence of a gravitational field (or the existence of gravitational potential energy), rather than the dynamic work done by gravity during the vertical motion of an object.

To illustrate this problem, we can also do a hypothetical experiment: fill an upright cylinder with water, and the pressure of water in the cylinder can be calculated using P= ρ*g*z (the differential is dP = ρ*g*dz). Now, let's imagine that a circular rigid membrane with impermeability, no weight and no thickness is added to the cross-section of the cylinder (at any height), the membrane completely cuts off the water in the cylinder, and the membrane is completely lubricated with no friction between the membrane and the inner wall of the cylinder, and the vertical distribution of water pressure in the cylinder should be completely unaffected, and can still be calculated using P= ρ*g*z (the differential method is dP = ρ*g*dz). This hypothetical example can prove that although the water pressure distribution in the cylinder is caused by gravity, it is completely independent of the work done by gravity.

In the derivation of the dry adiabatic lapse rate, the formula P=ρ*g*z (the differential method is dP =ρ*g*dz) is introduced to artificially specify the pressure of the air parcel at different altitude (i.e., at any altitude, the pressure of the air parcel is equal to the pressure of the surrounding atmosphere at the same altitude), rather than to calculate the energy conversion process of the air parcel during vertical motion. The derivation made in Jacob's book is a proof of my above view:

In Jacob's book (Jacob D.J., 1999: Atmospheric transport. Introduction to Atmospheric Chemistry, Princeton University Press, 53-55), using a thermodynamic cycle to derive the dry adiabatic lapse rate, in which the book does not involve any consideration of gravitational work, but obtains the same results as other textbooks. In Jacob's book, the considering the physical process of deducing the dry adiabatic lapse rate is around the thermodynamic cycle proposed in the book. In the later derivation process, although the formula dP= ρ*g*dz was introduced, it was introduced because it was artificially specified that the change of air parcel pressure was equal to the pressure of the atmosphere at the same height around it, it was not for the consideration for gravity work. The thermodynamic cycle in Jacob's book has nothing to do with the direction of motion of the air parcel, not specifically the vertical direction, and the horizontal movement of the air parcel can also establish such a thermodynamic cycle. Therefore, in this derivation process, the determination of the direction of motion is only introduced by dP= ρ*g*dz, and, this formula calculates the static equilibrium in the resting atmosphere, not the work done by gravity. The derivation of the book further illustrates that in order to obtain the calculation result of 0.98 degrees/100 meters, it is not necessary to consider the work done by gravity.

In addition, for the calculation of the energy change process of the vertically moving air parcel, all the items related to the energy change should be put together for the calculation of energy conversion and energy conservation, instead of ‘calculating each item separately, but do not comprehensively calculating the energy conversion relations between different items’, let alone ‘simply specifying that one of the energy conversion processes (here refers to heat exchange) can be ignored, but without sufficient basis’.

4.The refutation of 'the theoretical deduction in textbooks' only needs to be refuted from the perspective of theory, but the experimental support and demonstration of 'the theoretical deduction in textbooks' cannot only be described in words, but must have detailed and real data.

The common average temperature distribution in the middle and upper atmosphere is about 0.65 °C/100 m, which is not only less than 0.98, but also much less than 1.353. Therefore, whether the cut-off value of 0.98 or 1.353 for a neutral atmosphere indicates that an atmosphere with a vertical temperature decline rate equal to 0.65°C/100m is stable rather than a neutral atmosphere under common weather conditions. However, if the cut-off value of 1.353 is used, the stability of the atmosphere at 0.65°C/100m is even greater. In the near-surface atmosphere (mainly in the atmosphere below 1000 m), the occurrence of vertical temperature distributions of 1.353°C/100m or more should be observed relatively easily around sunrise or sunset, and the phenomenon of switching between stable and unstable in the near-surface atmosphere should be more easily observed during these two periods.

It is also important to note that a mildly stable atmosphere is not immune to vertical mixing (including tracers, pollutants, etc.), since dispersion due to various causes is always present in the fluid (including concentration gradients, breezes, etc.). Therefore, to judge whether the vertical motion characteristics of the atmosphere are stable, neutral or unstable, it is often necessary to judge with relatively macroscopic and rough scales.

5.The best way to illustrate the physical meaning of the dry adiabatic lapse rate is to prove the relationship between the dry adiabatic lapse rate and the atmospheric stability in a clear and logical way. This is because in the process of deriving and proving why different vertical temperature distributions lead to different degrees of atmospheric stability, it involves the question of how to understand the underlying physical mechanism of the dry adiabatic lapse rate of the atmosphere.

Basic physics textbooks are the basis for disseminating the knowledge system of physics, and all basic physics textbooks have been tested by a large number of intelligent people, so the probability of error is extremely low, but we cannot conclude that basic physics textbooks will not have defects. We welcome a rational and well-founded public debate. 

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