THE EXTRACTION OF THE JUICE BY MILLS 235 



As a numerical example, let /= oi, w = o-i, and m = 0.5. Then, 



>- = wm - = 0-500, and i (i n) = 0-500 = recovery. 



wm 4- / (i m) 



zewi + 2/(i - 



= 0.333, and i - (i - r 2 ) 2 =0-555 = recovery. 



wm 



= 0-250, and i (i rg) 3 = 0-578 = recovery. 

 + 3/(i -w) 



= 0-200, and i (i r 4 }* = o-58q = recovery. 



wm + 4 / (i m) 

 The expression for compound imbibition may be thus obtained.* 

 Let there be two mills in series, to the first one of which is delivered bagasse 

 containing unit quantity of sugar. To this mill is returned also the dilute 

 juice recovered through the addition of water at the mill second in series. 



Let the constant factor of recovery (i.e., - , - N ) be dt noted by r. 



wm 4~ j (i m) 



Let e l and e^ be the actual quantities of material obtained at the first and 

 second mills. Then e l is the recovery, and it is desired to express e l in terms 

 of r and n, where n is the number of mills, in this case. two. 



Now i -f 2 is the quantity presented to the first mill, when ^ is recovered. 



Therefore r = ^ , or ^ = r (i + e 2 ). 



I ~T" ^2 

 The quantity passed on to the mill second in series is 



Of this quantity r is obtained, so that 



o 



r =^ 



(i + e z ) (i - r) 

 whence e 2 = r (i + ^) (i r) = 



1 ~ r + r * 



But 1 = r (i 4- ^2) 



/ r r 2 \ r r 



wherefore ^ = r ( i 4- -; 9 ) = ; , = ; r,- 



V i r 4 r*J i r+r 2 r 4- (i r) 2 



In the case of three mills in series the solution appears as below, it being 

 understood that water is used at the last mill only, the dilute juice there 

 expressed being employed as the diluent at the mill second in series, the pro- 

 duct obtained here being returned to the first mill. 



In the first mill r = - - or e. A r (i 4- ^2) 



The second mill receives i -{- e 2 r (i + e 2 ) = (i + e*>) (i r) 



At the second mill r = - r e ' 2 



whence % = r (i + e 2 ) (i r) -4- e. 3 r. 



The third mill receives e. A 4- (i + 2) ( x r ) -- { r (* + ^2) (i r) 4- 



At the third mill r = . . ^ ^ 



(i + e 2 ) (i - r) 2 -f e s (i - r) 



whence e. A = r (i 4- e 2 ) (i r) 2 + e. A r (i r). 



* For this solution I am indebted to Mr. Lewis Wachenberg of the Reserve Refinery, Louisiana. 



