Page 229 - Separation process principles 2

194 Chapter 6 Absorption and Stripping of Dilute Mixtures

For a packed column, define the "height equivalent to a theoretical (equilibrium) stage (plate)," HETP, and
explain how it and the number of equilibrium stages differ from "height of a transfer unit," HTU, and "number
of transfer units," NTU, respectively.
Explain differences between "loading point" and "flooding point" in a packed column.
Estimate packed height, packed-column diameter, and pressure drop across the packing.
Estimate HTU from correlations of mass-transfer coefficients.
Explain how the number of theoretical stages is computed for concentrated solutions in which equilibrium and
operating lines are curved.

Industrial Example in this example is greater than the energy of condensation
liberated from the absorption of acetone.
A typical absorption operation is shown in Figure 6.1. The
As was shown in Figure 5.9, the fraction of a component
feed, which contains air (21% 02, 78% N2, and 1% Ar),
absorbed in a countercurrent cascade depends on the number
water vapor, and acetone vapor, is the gas leaving a dryer
of equilibrium stages and the absorption factor, A = L/(KV),
where solid cellulose acetate fibers, wet with water and
for that component. For the conditions of Figure 6.1, using
acetone, are dried. The purpose of the 30-tray (equivalent to
L = 1943 kmolh and V = 703 kmolh, estimated K-values
10 equilibrium stages) absorber is to remove the acetone by
and absorption factors, which range over many orders of
contacting the gas with a suitable absorbent, water. By using
magnitude, are
countercurrent flow of gas and liquid in a multiple-stage
device, the material balance, shown in Figure 6.1, indicates Component A = L/(KV) K-value
that 99.5% of the acetone is absorbed. The gas leaving Water 89.2 0.03 1
the absorber contains only 143 ppm (parts per million) by Acetone 1.38 2.0
weight of acetone vapor and can be recycled to the dryer or Oxygen 0.00006 45,000
Nitrogen 0.00003 90,000
exhausted to the atmosphere. Although the major component
Argon 0.00008 35,000
transferred between phases is acetone, the material balance
indicates that small amounts of oxygen and nitrogen are also For acetone, the K-value is based on Eq. (4) of Table 2.3, the 4
absorbed by the water solvent. Because water is present in modified Raoolt's law, K = y PSI P, with y = 6.7 for a 1
both the feed gas and the absorbent, it can be both absorbed dilute solution of acetone i11 water at 25OC and 101.3 kPa.
and stripped. As seen in Figure 6.1, the net effect is that For oxygen and nitrogen, K-values are based on the use of
water is stripped because more water appears in the exit gas Eq. (6) of Table 2.3, Henry's law, K = HIP, using constants
than in the feed gas. The exit gas is almost saturated with from Figure 4.27 at 25OC. For water, the K-value is obtained
water vapor and the exit liquid is almost saturated with air. from Eq. (3) of Table 2.3, Raoult's law, K = PS/P, which ap-
The temperature of the absorbent decreases by 3°C to supply plies because the mole fraction of water in the liquid phase is
the energy of vaporization needed to strip the water, which close to 1. For argon, the Henry's law constant at 25OC was
obtained from the International Critical Tables [I].
Figure 5.9 shows that if the value of A is greater than 1,
Exit gas
25 "C any degree of absorption can be achieved: the larger the
90 kPa
> value ofA, the fewer the number of stages required to absorb
Liquid absorbent kmollh
25 "C Araon 6.9 a desired fraction of the solute. However, very large values of
101.3 kPa 0, 144.291 A can correspond to absorbent flow rates that are larger than
N2 535.983
krnollh Water 22.0 necessary. From an economic standpoint, the value of A, for
Water 1943 Acetone 0.05
the main (key) species to be absorbed, should be in the range
of 1.25 to 2.0, with 1.4 being a frequently recommended
Feed gas value. Thus, the above value of 1.38 for acetone is favorable.
For a given feed-gas i3ow rate and choice of absorbent,
krnollh Exit liquid factors that influence the value of A are absorbent flow rate,
Argon 25 "C, temperature, and pressure. Because A = L/(KV), the larger
144.3 101.3 kPa
536.0 the absorbent flow rate is, the larger will be the value of A.
Water 5.0 krnollh The required absorbent flow rate can be reduced by reducing
Acetone 10.3 02 0.009
N 2 0.017 the K-value of the solute. Because the K-value for many
Water 1,926.0
Acetone 10.25 solutes varies exponentially with temperature and is inversely
Figure 6.1 Typical absorption process. proportional to pressure, this reduction can be achieved by