Unformatted text preview: a straight
operating line that passes through the operating point (Y0 = 0 , X1 = 0.01) at the bottom of the
column, as shown in Fig. 6.11(b) and is drawn tangent to the equilibrium curve, as shown. From
Eq. (6-5), the slope of the operating line = L'/V' = 3.0 mol B/mol of steam on a B-free basis.
(V’')min/L' = 0.33. This is compared to the given operating value of L'/V' = 2.0 or V'/L' = 0.5. Analysis: (continued)
(a) Plot of results: (b) Plot of results: Exercise 6.9 (continued) Exercise 6.10
Subject: Stripping of benzene (B) from straw oil by steam at 1 atm (101.3 kPa). Given: Oil enters stripper with 8 mol% benzene. 75% of B is stripped. Pure steam enters.
Steam exits with 3 mol% B. Sieve-plate column. Henry's law holds with a B partial pressure of
5.07 kPa when benzene mole fraction in the oil is 10 mol%.
Assumptions: Straw oil is not volatile and steam does not condense. Dalton's law.
Find: (a) Number of equilibrium stages required.
(b) Moles of steam required per 100 moles of benzene-oil feed to stripper.
(c) Number of equilibrium stages needed if 85% of B is stripped with same amount of
steam as in Part (b).
Analysis: With reference to the stripper in Fig. 6.11(b),
XN+1 = 8/92 = 0.087 mol B/mol B-free oil
Y0 = 0.0 mol B/mol B-free steam
YN = 3/97 = 0.0309 mol B/mol B-free steam
(b) For 100 moles of benzene-oil feed mixture, have 8 moles of benzene. Amount of benzene
stripped = 0.75(8) = 6 moles. Because the exit gas contains 0.0309 mol B/mol B-free steam, the
steam rate = 6/0.0309 = 194 moles.
Therefore, moles of steam/100 moles of benzene-oil feed = 194
Also, X1 = (8 - 6)/92 = 0.0217 mol B/mol B-free oil
(a) Assume Henry's law is given by Eq. (4-32), pB = HxB.. Because pB = 5.07 kPa when xB =
0.10, H = 5.07/0.1 = 50.7 kPa. Convert this to an a Y-X equilibrium equation. For a pressure of
101.3 kPa, assuming Dalton's law, yB = pB/P = 50.7xB/101.3 = 0.5xB . From Eq. (6-1), dropping
the benzene subscript, B, and noting that KB = 0.5,
KB = 0.5 = Y / (1 + Y )
X / (1 + X ) Solving, Y = 0.5 X
1 + 0.5 X (1) The Y-X plot includes the equilibrium curve given by Eq. (1), and a straight operating line that
connects the column terminal points (Y0 = 0.0, X1 = 0.0217) and (YN = 0.0309, XN+1 = 0.087).
The corresponding slope of the operating line is (0.0309 - 0.0)/(0.087 - 0.0217) = L'/G' = 0.474.
From the plot below, stepping off stages as in Fig. 6.11(b), required N = approximately 3 stages.
(c) For 85% stripping of B, amount of B stripped = 0.85(8) = 6.8 moles B. Not stripped is1.2
moles B. Therefore, for the same B-oil and steam feeds, X1 = 1.2/92 = 0.01304 mol B/mol oil
and YN =6.8/194 = 0.03505 mol B/mol steam. The Y-X plot for this case is also given below,
where the equilibrium curve is the same as in Part (a), and a straight operating line connects the
column terminal points (Y0 = 0.0, X1 = 0.01304) and (YN = 0.03505, XN+1 = 0.087). The
corresponding slope of the operating line is (0.03505 - 0.0)/(0.087 - 0.01304) = L'/G' = 0.474
(same as in Part (a)). From the plot below, stepping off stages as in Fig. 6.11(b), required N =
between 5 and 6 stages....
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