On Sun, Jul 13, 2003 at 04:40:53PM -0400, Erik Reuter wrote: > But is it a phenomenological formula? I would think measurements were > made of the lapse rate, and then a curve was fit to the data.
I did some reading and it seems things are both simpler and more complex than my question implied. Simpler, in that with the right assumptions a simple formula CAN be derived for the earth. More complex because this formula doesn't always hold true (although under "normal" atmospheric conditions in the troposphere, with very dry air, it is a fairly good approximation). Anyway, the basic assumption is that the atmosphere may be modeled by a rising, dry parcel of air adiabatically expanding, reducing in pressure, and cooling. As I mentioned, this is often valid in the troposphere. It is NOT valid in the stratosphere, where ozone absorbs solar radiation and actually causes RISING temperature with increasing altitude. The applicable thermodynamic formula for this assumption is the adiabatic law, p ^ ( 1 - gamma ) = c1 T ^ (-gamma) where gamma is the ratio of the specific heat of the atmosphere at constant pressure to the specific heat at constant volume, gamma = cp / cv. This value is experimentally determined to be about 1.4 for N2 and O2 (it is 1.7 for He and 1.3 for CO2) Differentiating the adiabatic law gives ( 1 - gamma ) p ^ ( 1 - gamma ) dp / p = - gamma c1 T ^ (-gamma) dT / T and substituting c1 into this equation from the adiabatic law and solving for dp / p gives dp / p = ( 1 / ( 1 - 1 / gamma ) ) dT / T For earth, I previously gave an approximate formula for pressure, p/p0 = exp[ -0.115 h ] which may be written p/p0 = exp[ - h / hc ] where hc is 8.7km by my calculation ( or 8.5km by curve fit to experimental data). Differentiating this gives dp / p0 = - ( dh / hc ) exp [ - h / hc ] = - ( dh / hc ) p/p0 and solving for dp / p results in dp / p = - ( dh / hc ) This may be combined with the previous equation for dp / p to obtain - ( dh / hc ) = ( 1 / ( 1 - 1 / gamma ) ) dT / T Solving this for dT / dh gives dT / dh = - ( 1 - 1 / gamma ) T / hc If we use gamma = 1.4 , hc = 8.5km, and T=300K, then dT / dh = 10.1 deg/km which approximately agrees with experimental data (the number is often quoted as 9.8 deg/km), when the conditions fit the assumptions (one of which is that the air must be dry, another is that the atmosphere doesn't absorb any heat from outside sources). For the 5km habitat, the math gets more complicated since the "non-linear potential correction factor" cannot be neglected above about h=2km. Also, are the assumptions used in the deriviation valid for the habitat? If the heat for the habitat all comes from the surface (i.e., the endcaps are well insulated and unheated), and the air is maintained with very low moisture, then this could be a reasonable assumption. -- "Erik Reuter" <[EMAIL PROTECTED]> http://www.erikreuter.net/ _______________________________________________ http://www.mccmedia.com/mailman/listinfo/brin-l
