ON THE ENERGY OF STORMS MARGULES 575 



In this case the descent of the mass i, and the ascent of the other 

 mass 2, are much smaller than for the same values of h in the first 

 and second analyses §(24) and §(25); hence also the available 

 kinetic energy cannot be evaluated by using the equations there 

 deduced. 



§(30) Two masses of air having constant temperatures. In order 

 to be able to express more accurately the influence of the formation 

 of three strata on — d(P + I) I have computed an example for 

 the case of constant 7\ and T 2 . In this case we have 



S, = K + C p \ogT 1 



S 2 = K + C p \ogT 2 - 



h — z 



h — z 



T 



1 2 



gh ^ g(h - c t ) _ gc 2 Jj_ 



f y t ^p 6 n- 



1 2 1 1 1 2 1 1 



In the final stage the temperature of the median layer isconstant,viz., 



(TO 



o 



\Jx + T 2 ) 



For ^ = 3000 meters, T t = 258 , T 2 = 268°, we find c x = 1099.5 nieters 

 c 2 = 1025.9 meters, (T') = 263. 12 . 



The available kinetic energy is to be found as in the first computa- 

 tion but by a much longer route. The velocity for the average 

 kinetic energy and for two chambers of equal volume is V = 9.5 

 meters per second or only half as much as for the more complete 

 overflow from 2 and underflow from 1 as computed by equation 

 (I) of section 25. 



FOURTH ANALYSIS. EQUALIZATION OF THE PRESSURES IN A HORI- 

 ZONTAL LAYER THAT RETAINS THE CONSTANT VOLUME 



§(31) Computation of the available kinetic energy. 



Initial stage . Assume a thin horizontal layer, bounded by rigid 

 walls, divided by a screen into two chambers having volumes k t 

 and k 2 , whose masses are under the pressures p x and p 2 respectively. 

 After removal of the dividing screen the pressure throughout the 

 whole volume k = k t + k 2 becomes p' ; if the expansion of one mass 



