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0.95 − 0.192 = 195.3 lbmol Exercise 13.17
Subject: Calculation procedure for binary batch stripping with constant boilup ratio.
Given: Equipment arrangement of Fig. 13.8, where the initial charge is placed in the condenser
Assumptions: No holdup on the trays or in the reboiler or condenser. No pressure drop.
Equilibrium stages. Constant molar overflow.
Find: A calculation procedure like that developed in Section 13.2 for batch rectification with
constant reflux ratio, but with constant boilup ratio, VB = V/B or from Eq. (7-12),
L/V = (1 + VB)/VB.
Analysis: Refer to Fig. 13.4. Plot the y-x equilibrium curve for the more volatile component on
a McCabe-Thiele diagram. Add the 45o line. From the specified boilup ratio, compute the slope
of the operating line, L/V, from Eq. (7-12). Locate an arbitrary value of the instantaneous
bottoms composition, xB on the 45o line and draw an operating line from this point to an
intersection with the equilibrium curve. Starting at the point xB on the 45o line, step off a number
of stages = number of equilibrium stages in the column plus one for the reboiler, as shown in the
diagram on the next page. The point for the uppermost stage is the mole fraction for the vapor
leaving the top stage, which, with the total condenser, is the composition in feed tank, xC,
(condenser accumulator). Locate several other operating lines, all of the same slope. Each xC
must be equal to or greater than the initial feed tank composition, x0. The range of the
corresponding xB values depends on the operation step stop specification. With constant molar
overflow, the following form of Eq. (13-2) applies, where W0 is the initial moles of charge to the
feed tank and Wt is the moles left in the feed tank at the end of the operation step: ln W0
xC − x B The cumulative bottoms composition is given by material balance in a manner used to develop
Eq. (13-12): x Bavg = W0 x0 − Wt xCt
W0 − Wt This procedure is applied in Exercise 13.18. Analysis: (continued) Exercise 13.17 (continued) Exercise 13.18
Subject: Batch stripping of a mixture of normal hexane (H) and normal octane (O).
Given: Charge of 100 kmol of 10 mol% H and 90 mol% O. Boilup up rate, V = 30 kmol/h.
Boilup ratio, V/L = 0.5. Partial reboiler and column with 2 equilibrium stages (total of 3 stages).
Assumptions: No holdup except in feed tank. Perfect mixing in the feed tank. 1 atm pressure.
No pressure drop. Configuration is in Fig. 13.8.
Find: Instantaneous bottoms and cumulative bottoms compositions at 2 h of operation.
Analysis: By the end of 2 hours of operation, the total amount of boilup = 2(30) = 60 kmol.
The total liquid rate running down the column is 60/0.5 = 120 kmol. Therefore, the bottoms
produced = 120 - 60 = 60 kmol. The liquid left in the feed tank will be 100 - 60 = 40 kmol. If a
perfect separation took place, the bottoms would be pure O and the liquid left in the feed tank
would be 25 mol% H. Therefore the mole fraction of H in the...
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