152 Dr. G. Bakker on the 



The capillary energy per unit of mass, or the volume 

 integral of the departure p N — p? from the law of Pascal is : 



HS= f V fa-pr) dh = & t J 2 (l+ ~) (p*-pr) dh 



or: H = f(^ N -p T )-^{(p N - / > T )-(p N '- i V)}. (44) 

 From (43) and (44) we easily find by elimination of p N ': 

 ^/=(R + ir)p 2 -(R-^)px-|H. . (45) 



The expression for the work done by the pressures p u p 2i 

 and p T becomes therefore : 



Now we have : ^ T f= (R + Jf )^ 2 - (R-J?) ^. . (37) 

 By substitution in (46) : 



9^dl+?d8)-iBdS-l-RdS(p 1 -p s )-^ i (p 1 -p 1 ){28dS 4 ?rfS 



.... (47). 

 Further we have : 



V =7r(R 2 2 -R 1 2 > = 7r?(R 1 H-R 2 >=27rfR^ = Sf: 



hence: <Jt? = Sd?+?dS. 



The expression for the work done by the capillary layer 

 becomes therefore : 



,*-iH(l + |)*-iS=l\^ (?„).. 



The equation of energy of the cylindrical capillary layer 

 per unit of mass becomes consequently : 



Td v =d e + V dv-iB.(l+^\dS~£i^d(tv). (48) 



