Example 31.3-1

1.3-1. Constant Underflow in Leaching Oil from Meal. Use the same conditions as given in Example 12.10-1, but assume constant underflow of N=1.85kg solid /kg solution. Calculate the exit flows and compositions and the number of stages required. Compare with Example 31.3-1. Ans. yAN=0.111,xA1=0.623,4.3 stages EXAMPLE 31.3-1. Countercurrent Leaching of Oil from Meal A continuous countercurrent multistage system is to be used to leach oil from meal by benzene solvent (B3). The process is to treat 2000 kg/h of inert solid meal (B) containing 800kg oil (A) and also 50kg benzene (C). The inlet flow per hour of fresh solvent mixture contains 1310kg benzene and 20kg oil. The leached solids are to contain 120kg oil. Settling experiments similar to those in the actual extractor show that the solution retained depends upon the concentration of oil in the solution. The data (B3) are tabulated below as Nkg inert solid B/kg solution and yAkg oil A/kg solution: Calculate the amounts and concentrations of the stream leaving the process and the number of stages required. Solution: The underflow data from the table are plotted in Fig. 31.3-3 as N versus yA. For the inlet solution with the untreated solid, L0= 800+50=850kg/h,yA0=800/(800+50)=0.941,B=2000kg/h, and N0=2000/(800+50)=2.36. For the inlet leaching solvent, VN+1 =1310+20=1330kg/h and xAN+1=20/1330=0.015. The points VN+1 and L0 are plotted. The point LN lies on the N-versus- yA line in Fig. 31.3-3. Also for this point, the ratio NNyAN=(kg solid /kg solution )(kg oil /kg solution =kg solid /kg oil =2000/120=16.67. Hence, a dashed line through the origin at yA=0 and N=0 is plotted with a slope of 16.67 and it intersects the N-versus- yA line at LN. The coordinates of LN at this intersection are NN=1.95kg solid /kg solution and yAN=0.118 kgoil/kgsolution. Making an overall balance by substituting into Eq. (31.3-4) to determine point M, L0+VN+1=850+1330=2180kg/h=M Figure 31.3-3. Grophical construction for number of stages for Example 31.3-1. Substituting into Eq. (31.3-5) and solving. 14.yai+VN1xN+1xAs=850(0.941)+1330(0.015)=2180xsw=0.376 Substituting into Eq. (31.3-6) and solving. B=2000=NkM=Ns(2180)Nk=0.918 The point M is plotted with the coordinates xAM=0.36 and NM= 0.918 in Fig. 31.3-3. The line VN+1ML0 is drawn, as is line LNM which intersects the abscissa at point V1 where xA1=0.600. The amounts of streams V1 and LN are calculated by substituting into Eqs. (31,34) and (31,35), and solving simultaneotsly: Ls+Vi=M=2180LsVAs+VixA1=Lsw(0.118)+V1(0.600)=21.00(0.376)1477 Hence, LN=1016kg solution in the outlet underflow stream and V1=1164kg solution/h in the exit overflow stream. Alternatively, the amounts could have been calculated using the lever-arm rule. The operating point is obtained as the intersection of lines L0VI and LNVN+1 in Fig. 31.3-3. Its coordinates can also be calculated from Eqs. (31.3-11) and (31.3-12). The stages are stepped off as shown. The fourth stage for L4 is slightly past the desired LN. Hence, about 3.9 stages are required