Page 211 - Fundamentals of Water Treatment Unit Processes : Physical, Chemical, and Biological

166 Fundamentals of Water Treatment Unit Processes: Physical, Chemical, and Biological

Example 8.3 Determine an Equivalent As to dealing with a range of temperatures for ‘‘H(air),’’
‘‘H(air, 208C)’’ Table CD8.3, Part (a) gives that capability by substituting
different values of temperature in ‘‘cell B3’’ to generate a
Problem range of results for H(air), plotting H(air) vs. T, and then
Determine a Henry’s constant for air applying the coefficients, e.g., ‘‘a’’ and ‘‘b’’ in a best-fit
equation.
Given
3
H(O 2 ) ¼ 0.0004383 (kg O 2 dissolved=m water=kPa O 2 )—
Table H.5, 208C 8.3.1.2 Application of Henry’s Law to Saturator
X(O 2 ) ¼ 0.209 mol O 2 =molesall gasesinair—Table B.7
3
H(N 2 ) ¼ 0.0001876 (kg N 2 dissolved=m water=kPa N 2 )— The ‘‘saturator’’ is a tank where oxygen and nitrogen (and
minor gases) in air under high pressure are transferred to
Table H.5, 208C
X(N 2 ) ¼ 0.781 mol N 2 =moles all gases in air—Table B.7 water in accordance with Henry’s law. The gas-transfer rate,
i.e., from gas to aqueous solution, is governed by the pressure
Solution in the saturator and the water surface area that has contact
1. C(O 2 þ N 2 ) ¼ H(O 2 ) X(O 2 ) P(total) þ H(N 2 ) X(N 2 )
with the gases (i.e., air, which is a mixture). The dissolved gas
P(total) concentration leaving the saturator, C(saturator), is always
¼ [0:0004383 0:209 þ 0:0001876 less than the equilibrium level as stated by Henry’s law. The
0:781] 101:325 kPa ratio of these two values is the ‘‘efficiency-factor,’’ f, which is
5
¼ [9:16 10 5 þ 14:65 10 Š defined (Edzwald, 1995, p. 7) by
101:325 kPa
3
¼ 0:0241 kg (O 2 þ N 2 )=m water C(saturator, gas A)
(8:4)
f ¼
(Ex8:3:1) H(gas A) X(gas A) P(n gases)
2. Substitute in Equation 8.3, in which
f is the ratio of gas concentration leaving saturator to gas
C(air) ¼ H(air) X(air) P(total) concentration by Henry’s law
3
0:0241 kg (O 2 þ N 2 )=m water C(saturator, gas A) is the concentration of dissolved gas
‘‘A’’ in water leaving the saturator, which is the same as
¼ H(air) X(air) P(total) 3
C(saturator), the collective term (kg dissolved gas A=m
3
0:0241 kg (O 2 þ N 2 )=m water
water)
¼ H(air) 1:00 101:325 kPa
H(O 2 þ N 2 ) 0:000238 (kg O Equation 8.3 thus has a modified form for a saturator, i.e.,
2
3
þ N 2 dissolved=m water=kPa air) (Ex8:3:2)
C(saturator, gas A) ¼ f H(gas A)
Discussion X(gas A) P(n gases) (8:5)
The result is for 208C. Such an equivalent Henry’s constant
for air is an ‘‘artifice’’ that some may prefer as opposed For packed-bed saturators, f 0.9, and for unpacked satura-
to dealing with the individual gases. In dissolving, however, tors, f 0.7 (Edzwald, 1995). The value for f depends upon
the gases act individually, i.e., oxygen and nitrogen mostly
(which comprise 0.99032 mol fraction of air). The reason the packing, the hydraulic loading rate, and the saturator
for considering ‘‘air’’ as a dissolved gas, i.e., in lieu of depth. The gas A may be any gas, e.g., O 2 ,N 2 ,Ar, CO 2 ,or
oxygen and nitrogen (and other minor gases) independ- in terms of practice, ‘‘air,’’ which is the aggregate of the
ently, is that the literature refers frequently to ‘‘dissolved component gases. For the usual case of ‘‘air’’ as the gas,
air.’’ The minor gases include CO 2 , Ar, Ne, He, Kr, Xe, CH 4 , X(gas A) ¼ 1.0 (which is the approximate sum of the major
and H 2 . As a matter of interest, the sum of the mole fractions component gases), i.e., X(N 2 ) ¼ 0.78084, X(O 2 ) ¼ 0.209476,
of O 2 ,N 2 ,CO 2 , Ar equals, 0.9999700, i.e., X(Ar) ¼ 0.00934, X(CO 2 ) ¼ 0.000314; the sum of the mole
fractions for these four gases out of the 12 listed in Table
[X(O 2 ) ¼ 0:2094760 þ X(N 2 ) ¼ 0:7808400 þ X(CO 2 )
H.1 is, X(N 2 ,O 2 ,CO 2 ,Ar) ¼ 0.99997.
¼ 0:0003140 þ X(Ar)] ¼ 0:0093400]
The saturator capital cost is about 12% of the cost of a
¼ 0:9999700: DAF plant and about 50% of the operating cost (Haarhoff
and Rykaart, 1995). Therefore, a higher f will provide a cost
The equivalent Henry’s coefficient, if CO 2 and Ar are
saving in operation that may justify the added capital cost.
included, is (from Table CD8.3)
8.3.1.3 Saturator
H(O 2 þ N 2 þ CO 2 þ Ar) ¼ 0:0002561 (kg O 2 þ N 2
3
þ CO 2 þ Ar) dissolved=m water=kPa air) Henry’s law, modified for saturator application, is the basis for
calculating C(saturator), i.e.,
The other gases, i.e., Ne, He, Kr, Xe, CH 4 , and H 2 , com-
prise only 0.0000300 mol fraction. C(saturator) ¼ f H(air) P(saturator) (8:6)