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2,427 The location of P3 is obtained by measurement, using a ratio, similar to (86), L3 P3 L3 +P3 S3 5,185
L S 2.37 − 1.00
=
=
=
= 2.37 or 3 =
= 0.578
L3
L3 2,188
2.37
SP3
L3 P3
Using this ratio, with the measured length of L3 S , the point P3 is located as shown on the
triangular plot below. In Fig. 8.44, the point V3 is in equilibrium with L3 and, therefore, is
located at the end of a tie line drawn from L3. V3 operates with L2 , where L2 is the intersection
on the equilibrium curve of an extended line through V3 and, again, P3.
From the triangular diagram, the composition of V2 is obtained from a tie line with L2, and the
composition of L1 is obtained from a tie line with V1. The results are as follows, including the
results of material balances around stage 1, with flows in kg/h
Component L1 , wt% T
B
W
Total 32
59
9
100 Flow in L1 ,
kg/h
1,199
2,210
337
3,746 V2 , wt%
16.7
1.3
82.0
100.0 Flow in V2 ,
kg/h
1,133
88
5,562
6,783 Flow in S1 ,
kg/h
0
0
1,983
1,983 Since S1 = 1,983 kg/h and, from above, S1 + S2 = 2,023, S2 = 2,023 – 1,983 = 40 kg/h. Exercise 8.19
Subject: Analysis of a multiplefeed countercurrent extraction cascade shown in Fig. 8.45.
Given:
solute. Feed F' contains solvent and solute. Feed F " contains unextracted raffinate and Find: Equations to establish the three reference (operating) points on a righttriangle graph.
Analysis: Referring to Fig. 8.45, let:
P' be the operating point from the left side of stage 1 to between stages m1 and m.
P" be the operating point from between stages m and m+1 to between stages p1 and p.
P''' be the operating point from between stages p and p+1 to the right side of stage n.
The equations for locating these three operating points are obtained by combining total material
balance equations as follows, using the nomenclature of Fig. 8.45:
Overall material balance:
Mixing point = M = L0 + F '+ F "+Vn +1 = V1 + Ln
(1)
(2)
Material balance around stage m:
F '+Vm+ 1 + Lm− 1 = Vm + Lm
Material balance around stage p:
(3)
F "+Vp + 1 + Lp −1 = Vp + Lp
Operating point P' : P' = V1  L0 = ……. = Vm  Lm1
(4)
(5)
Operating point P'' : P'' = Vm+1  Lm = ……. = Vp  Lp1
Operating point P''' : P''' = Vp+1  Lp = ……. = Vn+1  Ln
(6)
(7)
From Eqs. (1) and (4): V1  L0 = M  (L0 + Ln) = M  J = P'
where J = (L0 + Ln)
JLn L0
Locate J by the inverse lever arm rule,
=
(8)
JL0 Ln
Then operating point P' is the intersection of lines through V1 and L0 , and through M and J
Operating point P'' is obtained by combining Eqs. (2) and (5) to give:
(9)
P " = Vm+1 − Lm = Vm − Lm−1 − F ' = P ' − F '
Operating point P''' is obtained by combining Eqs. (3) and (6) to give:
P ''' = V p − L p −1 − F '' = P '' − F '' = P ' − ( F ' + F '' ) = P ' − K
(10)
where K = (F' + F") F ''
(11)
'
KF '' F
Then operating point P''' is the intersection of lines through Vn+1 and Ln , and through
P' and K
Operating point P''' is the intersection of lines through P' and F' , and through P''' and F'' Locate K by the inverse lever arm rule, KF ' = These constructions are illustrated in the righttriangle diagram on the next page. Exercise 8.19 (continued)
Analysis: (continued) Exercise 8.20
Subject: Extra...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
 Spring '11
 Levicky
 The Land

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