Unformatted text preview: The solvent in the final total raffinate = 724.9  657.5 = 67.4 lb/h.
The total flow rate of the exiting solvent, SD , can not be found yet because the flow rate, VN , can
not be determined until the extract reflux rate is known. However, for, YA = 0.83 in the extract
leaving Stage N, the wt% C for an extract saturated with solvent = 87.5. Similarly, the entering
fresh solvent rate, SB can not be determined yet.
From the shape of the equilibrium curve on the previous page, at minimum extract reflux, the
intersection of the operating lines for the extractor sections above and below the feed would
appear to be at the intersection with a vertical line for the wt% A (solventfree) in the feed, and
with the equilibrium curve. Assume this is so. Then at minimum extract reflux, the pinch point
will be at the feed stage. Accordingly, the composition of raffinate stream, LF+1, entering the
feed stage F (in Fig. 8.26b) is identical to the composition of the feed. From the Y  X plot on the
previous page, the solventfree composition of the extract, VF , is that in equilibrium with the X =
0.35 of the feed, or Y = 0.56. From the figures on the next page, the wt% C in this extract is
91.3. Now, the entering and leaving flow rates can be calculated by material balances for the
section of the extractor above the feed stage in Fig. 8.26b.
An overall solventfree material balance gives the following, where the prime superscript
indicates a flow rate on a solventfree basis:
VF' = L'F +1 + D = L'F +1 + 342.5
(3)
A similar balance on the solute, A, gives:
YVF VF' = 0.56VF' = X LF +1 L'F +1 + YD D = 0.35 L'F +1 + 0.83(342.5) = 0.35 L'F +1 + 284.3
(4)
Solving Eqs. (3) and (4), gives: VF' = 782.9 lb / h and L'F +1 = 440.4 lb / h
The corresponding total flow rates (including solvent) are:
VF = 782.9/(1  0.913) = 8,998.9 lb/h of which, 8,998.9  782.9 = 8,216.0 is solvent
LF+1 = 440.4/(1  0.125) = 503.3 lb/h of which 503.3  440.4 = 62.9 is solvent
Therefore, the exiting solvent rate, SD = 8,216.0  62.9 = 8,152.1 lb/h Analysis: (continued) Exercise 8.26 (continued) Exercise 8.26 (continued)
Analysis: (continued)
The extract, VN , contains 83 wt% A on a solventfree basis. Therefore, From the
equilibrium plots on a previous page, the corresponding wt% C in this extract is 87.5 wt%.
Therefore, the total flow rate of VN = 8,152.1/(0.875) = 9,316.7 lb/h
By total material balance around the solvent removal and stream divider above the top of the
cascade (as shown in Fig. 8.26b), LD = VN  SD  D = 9,316.7  8,152.1  342.5 = 822.1 lb/h.
Therefore, the minimum extract reflux rate = 822.1 lb/h,
and the minimum extract reflux ratio = 822.1/342.5 = 2.40
To compute the number of equilibrium stages, take twice the minimum extract reflux
flow rate or 2(822.1) = 1,644.2 lb/h with the same composition of 83 wt% A and 17 wt% B. The
solventfree extract is the same flow rate as before of D = 342.5 lb/h. Therefore, the solventfree
flow rate of V'N = 1,644.2 + 342.5 = 1,986.7 lb/h.
Since, this flow is 87.5 wt% solvent, total flow rate VN = 1,986.7/(1  0.875) = 15,893.6 lb/h.
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 Spring '11
 Levicky
 The Land

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