Y = 1/3 bar retention = f (%0M, %F, Pb) 



OM , 

 15 



1 - I 



1 - I 



(1) Y = LEDG + (REDG - LEDG) - 



1 - e 



1 - I 



) 0.984 



55 



1 - RI 



10 



,10 



1 - RI 



(2) REDG = 15 + (RR - 15) 



1 



10 



1 - e 



RR = 26 + 9 .9812 (Pb - 0.9) 



II - RI 

 2.1 



RI = 0.05 + 0.295 e 



(1.8 - Pb) 

 0.9 



6.8 



0.63 





LN 



1 





1 - LI 



1 - LI 







LN 



(3) LEDG = LR 



- e 



,LN 



1 - e 



1 - LI 



LR = 2.9 + 5.278 (pb - 0.9) 



1 . 1 



LI = 0.4017 e 



Pb 

 0.9 



. 59 



0.0017 



LN = 4 - e 



Pb 

 1.34 

 . 1865 



4.6 



18 



