tag:blogger.com,1999:blog-937701726196626615.post2077978659556482960..comments2022-06-22T08:39:10.478-07:00Comments on A Skeptic's Guide to the Greenhouse Effect: The heat equation revisitedAndershttp://www.blogger.com/profile/15294862989593516422noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-937701726196626615.post-27220894972996868472011-08-02T08:26:24.890-07:002011-08-02T08:26:24.890-07:00Hi AT,
I shall confess that my experience in ther...Hi AT,<br /><br />I shall confess that my experience in thermodynamic modelling is limited, to say the least, so I'm quite open to the possibility that I will make some major and minor mistakes here and there. Nevertheless,<br /><br />I do not assume a priori that the process is diffussive, I merely try to construct a model, including backradiation as Roy and others advice us to do, and I end up with something that looks quite familiar to a diffussive heat equation.<br /><br />You are right that the constants divide out, but that changes quickly as one adds some further complications and inhomogeneties, as I do in the slightly more complicated model with variable absorption.<br /><br />I do not agree that "stationary" is the same as "equilibrium". A house can be heated by radiators and reach a stationary state without it beeing in equilibrium. The general solution to the equation d^2U/dx^2 = 0 is U = Bx + C which is not necessarily constant but can be linear.<br /><br />I am not so sure that diffusive heat transport is so insignificant as you state, especially in the thin atmosphere. But perhaps it is more a matter of definition, I argue that the raditive heat transfer processes that you think are so dominant in the atmosphere might actually be (formally) diffusive radiative processes.Andershttps://www.blogger.com/profile/15294862989593516422noreply@blogger.comtag:blogger.com,1999:blog-937701726196626615.post-28235251780968542242011-08-02T03:27:47.003-07:002011-08-02T03:27:47.003-07:00And that should be 'formulation', singular...And that should be 'formulation', singular, of course...Alien Technicianhttps://www.blogger.com/profile/06067794376743683491noreply@blogger.comtag:blogger.com,1999:blog-937701726196626615.post-5904607223401488542011-08-02T03:26:42.811-07:002011-08-02T03:26:42.811-07:00Hi Anders,
It's pleasing to see a quasi-famil...Hi Anders,<br /><br />It's pleasing to see a quasi-familiar formulations at the top of your post. However, I'm not sure that the way you use them illuminates the problem you have set yourself. <br /><br />As an aside, of course heat diffusion plays almost no role in atmospheric thermodynamics, so your approach will not in any case come near to representing the real processes at work. But you know this, I believe. So we won't worry about that, and just look at your maths :-)<br /><br />Perhaps you do not realise it, but you have implicitly assumed your conclusion in your starting conditions. In other words, you have discovered nothing but the situation you began with. I will add some quick notes that might help you see where your confusion comes from. <br /><br />You give a form of the stationary heat equation in one dimension as D*d^2U/dx^2 = 0. On reflection, you must realise that, by simple algebra, the factor D divides out.<br /><br />So you see that your factor A, which you use for emissivity by analogy to D, likewise vanishes in the stationary case.<br /><br />Remember that by using the homogeneous stationary heat equation, you have assumed that the system is in equilibrium, or in other words that its temperature is constant throughout.<br /><br />So, from the beginning you have assumed an imaginary static, homogeneous system at constant temperature. There cannot possibly be a 'greenhouse effect', or any other effect, in this system.<br /><br />Still you manage to 'discover' that<br /><br />AU(n-1) + AU(n+1) - 2AU(n) = 0<br /><br />But you fail to notice that, because the system is stationary, we can divide both sides by A and hey presto, A vanishes from the equation. <br /><br />So what you have 'discovered' is that U(n)=(U(n-1)+U(n+1))/2, or in fact that, in a stationary system in thermal equilibrium, the temperature at one point is the same as the average temperature of the surrounding point. <br /><br />Which is the assumption you began with.Alien Technicianhttps://www.blogger.com/profile/06067794376743683491noreply@blogger.com