Page 180 - Separation process principles 2

4.8 Gas-Liquid Systems 145

mole fraction of the gas (solute) in the liquid The corresponding ratio of dissolved Con to water is
when the pressure of the gas is 1 atm. The curves of
5.7 10-3
Figure 4.27 can be used to estimate the solubility in water at = 5.73 x mol CO2/mol H20
1 - 5.7 x
other pressures and for mixtures of gases by applying Henry's
- -
law with the partial pressure of the solute, that The total number of moles of C02 to be absorbed is
mole-fraction solubilities are low and no chemical reactions
1,000 ft3 1,000
occur among the gas species or with water. Henry's law, dis- - = 2.79 lbmol
-
cussed briefly in Chapter 2 and given in Table 2.3, is rewrit- 359 ft3/lbmol (at STP) 359
ten for use with Figure 4.27 as
or (2.79)(44)(0.454) = 55.73 kg.
Assuming all the absorbed C02 is vented at the mountain top, the
(4-32)
number of moles of water required is 2.79/(5.73 x =
458 lbmol = 8,730 lb = 3,963 kg.
where Hi = Henry's law constant, atm.
~
~
~
'
~
For gases with a high solubility, such as H ~ If one corrects for the fact that the pressure on top of the mountain
is 101 kPa, so that not all of the C02 is vented, 4,446 kg (9,810 lb)
law may not be applicable, even at low partial pressures. In
of water are required.
that case, experimental data for the actual conditions of pres-
sure and temperature are necessary as in Example 4.14. In
either case, calculations of equilibrium conditions are made
EXAMPLE 4.14
in the manner illustrated in previous sections of this chapter
by combining material balances with equilibrium relation- The partial pressure of ammonia (A) in air-ammonia mixtures
ships or data. The following two examples illustrate single- in equilibrium with their aqueous solutions at 20°C is given in
stage, gas-liquid equilibria calculation methods. Table 4.7. Using these data, and neglecting the vapor pressure of
water and the solubility of air in water, construct an equilibrium
diagram at 101 kPa using mole ratios YA = rnol NHs/mol air, and
EXAMPLE 4.13 XA = rnol NH3/mol H20 as coordinates. Henceforth, the subscript
A is dropped. If 10 rnol of gas, of composition Y = 0.3, are con-
An ammonia plant, located at the base of a 300-ft (91.44-m)-high tacted with 10 rnol of a solution of composition X = 0.1, what
mountain, employs a unique absorption system for disposing of are the compositions of the resulting phases at equilibrium? The
by-product COz. The COz is absorbed in water at a C02 partial
process is assumed to be isothermal and at atmospheric pressure.
pressure of 10 psi (68.8 kPa) above that required to lift water to the
top of the mountain. The C02 is then vented to the atmosphere at
SOLUTION
the top of the mountain, the water being recirculated as shown in
Figure 4.28. At 25"C, calculate the amount of water required to dis-
The equilibrium data given in Table 4.7 are recalculated in terms of
pose of 1,000 ft3 (28.31 m3)(S~P) of C02.
mole ratios in Table 4.8 and plotted in Figure 4.29.
Mol NH3 in entering gas = 10[Y/(1 + Y)] = 10(0.3/1.3) = 2.3
SOLUTION
Mol NH3 in entering liquid = 10[X/(1 + X)] = 10(0.1/1.1) = 0.91
Basis: 1,000 ft3 (28.31 m3) of C02 at 0°C and 1 atrn (STP). From
Figure 4.27, the reciprocal of the Henry's law constant for CO2 at
25°C is 6 x mole fractionlatm. The C02 pressure in the ab- Table 4.7 Partial Pressure of Ammonia over
Ammonia-Water Solutions at 20°C
sorber (at the foot of the mountain) is
10
Pcoz = - + 300ftH20 = 9.50 atm = 960 kPa NH3 Partial Pressure, kPa g NH3/g H20
14.7 34 ft H20/atm 4.23
0.05
At this partial pressure, the equilibrium concentration of COz in the
water is
xco2 = 9.50(6 x = 5.7 x mole fraction C02 in water

COP vent
Table 4.8 Y-X Data for Ammonia-Water, 20°C

Y, mol NH31mol Air X, mol NH3/mol H20
0.044 0.053
0.101 0.106
0.176 0.159
0.279 0.212
Plant
0.426 0.265
Figure 4.28 Flowsheet for Example 4.13.