Unformatted text preview: cause of the
dilute condition,
x
V′ y
V′
y
X=
≈x=
− Yout = 0.01896
− 0.000095 (3)
L′ 1 − y
L′
1− x
1− y
which is very close to the given operating line.
(b) From Eq. (6138), N OG = Y = 0.25 dY
Y −Y *
Y = 0.00503 (4) 1
(Y − Y *)
as a function of Y, where for a given value of Y, the value of X is obtained from Eq. (3). Then, x
is computed from X. Then, y = y* is computed from the given equilibrium equation, and Y* is
1
then obtained from the equilibrium curve. Below is a plot of
versus Y. The area under
(Y − Y *)
the curve is the value of the integral for NOG . In the table below, the trapezoidal method is used
Eq. (4) can be solved graphically or numerically. Both require a table of values of Exercise 6.39 (continued) Analysis: (b) (continued)
to break the curve into segments of ∆Y. each, varying in width from 0.0103 to 0.0413. For each
segment (width), the average value of 1/(YY*) is used to compute the product = Area = ∆Y times
1/(YY*), which is given in the last column. The sum of these areas is NOG = 5.16. Exercise 7.1
Subject: Comparison of absorption, distillation, and stripping Find: Differences between absorption and distillation, and stripping and distillation
Analysis:
Absorption Stripping Distillation Feed is gas Feed is liquid Feed is liquid, vapor or a
mixture of the two Second phase (absorbent) is
added Second phase (stripping agent) Second phase is created by
is added
heat transfer Operation can be adiabatic Operation can be adiabatic Must have heat transfer at the
top and bottom stages Singlesection cascade Singlesection cascade Twosection cascade Can be almost isothermal Can be almost isothermal Can have a large temperature
range Can not separate a closeboiling mixture Can not separate a closeboiling mixture Can separate a closeboiling
mixture Tray efficiency can be low Tray efficiency can be
moderate Tray efficiency can be high Exercise 7.2
Subject: Emergence of packing to replace trays. Find: Reasons why some existing trayed towers are being retofitted with packing, and some
largediameter columns are being designed for packing.
Analysis: New random packings and structured packings have been introduced with higher
capacity, lower pressure drop, and higher efficiency than trays. Also for packed columns, liquid
distributors have been greatly improved. Channeling of liquid in packed columns is now much
less of a problem. Exercise 7.3
Subject: Condenser coolant for distillation of a methaneethane mixture. Find: Appropriate coolant.
Analysis: Assume that the distillate is nearly pure methane, the most volatile of the two
components. To use cooling water, the distillate temperature would be approximately 120oF.
But the critical temperature of methane is 115oF. Therefore could not condense it. The critical
pressure of methane is 673 psia. Using Fig. 7.16, could consider operation at 415 psia because
that is safely below the critical pressure (P/Pc = 0.62 and could operate as high as a P/Pc = 0.8).
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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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