ON THE ENERGY OF STORMS MARGULES 561 



If h 2 = h l and T i = \ {T ix + T i2 ) this last equation becomes 



T — T 



If this approximate formula be applied to the preceding example 

 as computed in §21, we have 



h i = 2000 meters *] 



T i2 — T ix = 3 J- whence V = 14.82 meters per second. 



^ = 26 1. 6 Z ° J 



Chapter III 



METHODS OF COMPUTING THE AVAILABLE KINETIC ENERGY FOR A 

 MASS OF AIR THAT PASSES ADIAB ATICALLY FROM ANY GIVEN 

 INITIAL STAGE INTO THAT OF EQUILIBRIUM 



§(23) Let there be given a mass of dry air, bounded by the 

 horizontal ground plane and vertical walls, that is, initially at rest 

 but not in equilibrium. We assume an initial condition of rest in 

 order to be able to make use of the equation (a) of the preceding 

 chapter, in our computation of the potential energy. We assume 

 furthermore that that portion of the mass that is above the level 

 surface h is already initially in equilibrium, and that during the 

 changes that occur in the lower portion of the mass, this upper 

 part acts like a piston of constant weight ^.B^ where* B is the area 

 of the base or of any horizontal section. 



Using a notation analogous to that of §13 we now have as before 

 the following expression for the potential energy corresponding to 

 the position of the column of air above the surface element dB 



dP = dB { £pdz + (Z - h) p h } 



Whence follows the P for the whole mass from the ground up to 

 the level surface at the altitude h 



P = j* B j pd B dz + B (Z - h) p h = J pdk + Constant 

 = R j Tdm + Constant. 



Similarly for the initial stage we obtain 



(P + I) = C p j Tdm + Constant 



