Smithsonian miscellaneous collections . amber system. To find the final stage of two masses of air under constant pressurewith initial linear vertical diminution of temperature. If tlie vertical gradient of temperature within the masses 1 and2 (fig. 4) is smaller than that for neutral equilibrium then the entropy 574 SMITHSONIAN MISCELLANEOUS COLLECTIONS VOL. 51 increases with the altitude and it can happen that the entropy atthe altitude cx in the cooler mass 1 is as large as that at the base of the mass 2, i. e., at the alti-tude (h — c2); the upper layersof 1 may one after the other ser-ial
Smithsonian miscellaneous collections . amber system. To find the final stage of two masses of air under constant pressurewith initial linear vertical diminution of temperature. If tlie vertical gradient of temperature within the masses 1 and2 (fig. 4) is smaller than that for neutral equilibrium then the entropy 574 SMITHSONIAN MISCELLANEOUS COLLECTIONS VOL. 51 increases with the altitude and it can happen that the entropy atthe altitude cx in the cooler mass 1 is as large as that at the base of the mass 2, i. e., at the alti-tude (h — c2); the upper layersof 1 may one after the other ser-ially have respectively the sameentropy as the layers in 2 up tothe altitude (h — c2). In thefinal stage the lower part of 1will become spread out at thebase; over it will lie strata con-sisting of 1 and 2 mixed; above this will rest that portion of 2that initially lay between (h — c2) and h. At the boundaries be-tween these three layers the temperature changes are the temperatures diminish linearly as in the equations. fig. 4. T, - Thl + nxC. (*-«)? T2 = Th2+-gc {h-z) then we have to seek the altitude at which the entropies are equalor S1=S2 for the same value of ph in the equations S, = K + Cp K log Thx - K - 1) log rj52 = K + Cp [n2 log Th2 - («2 - 1) log T2] Let us assume nx = n2 = n; let 8t be the temperature of the mass 1at the altitude cx and 02 the temperature of the mass 2 at the altitude h — c, then we have log-. = log—= Cp 0x) * , Th2 ^1log tZ c, C2 J (^ - Th2) Hence for n = 2 and a vertical temperature gradient of per 100 meters and for h = 3000 meters Thl = 2630, Th2= 2730,we find cx = 2154 meters and c2 = 2090 meters. In this case the greater part of the masses 1 and 2 remain unmixed,the available kinetic energy will not be much smaller than if in thefinal stage the whole of mass 1 lies below and the whole of 2 above. Again, for h = 6000 meters Thl = 2480, Th2 = 2580, we find cx =2390 meters and c2 = 2096 meters. ON THE ENERGY OF S
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