Page 210 - Fundamentals of Water Treatment Unit Processes : Physical, Chemical, and Biological

Flotation 165

a gas, ‘‘A,’’ in the aqueous phase is proportional to the partial Given 3
pressure of gas ‘‘A’’ at the gas–water interface, i.e., H(O 2 ) ¼ 0.0004383 (kg dissolved oxygen=m water=kPa
oxygen) at 208C
C(gas A) ¼ H(gas A) P(gas A) (8:1) X(O 2 ) ¼ 0.209 mol O 2 =mole air (Table B.7)
P(atmosphere) ¼ 101.325 kPa (stated)
in which Required
C(gas A) is the concentration of dissolved gas A in aque- Calculate the concentration of dissolved oxygen at equi-
3
ous solution (kg gas A=m water) librium by Henry’s law.
H(gas A) is the Henry’s constant for a given gas, e.g., ‘‘A’’
3 Solution
(kg gas A=m water=kPa gas A)
Substitute given data, for oxygen, in Equation 8.3,
P(gas A) is the partial pressure of gas A above gas–water
interface (kPa gas A)
C(O 2 ) ¼ H(O 2 ) X(O 2 ) P(total)
A is the designation for a particular species of a gas, e.g.,
3
¼ 0:0004383 (kg dissolved O 2 =m water=kPa O 2 )
O 2 or N 2
(Ex8:1:1)
Table H.5 gives Henry’s constants for different gases with
. 0.209 mol O 2 =moles all gases in air
temperature as a variable. In addition, coefficients, i.e., ‘‘A’’
. 101.325 kPa total pressure
and ‘‘B,’’ are given at the top of Table H.5 for ‘‘best-fit’’ 3
¼ 0:00928 kg dissolved O 2 =m water (9:3mg=L)
empirical equations for H(gas A) vs. temperature, i.e.,
H(gas A) ¼ A exp(B T8C). As a historical note, Henry’s
Discussion
law was mentioned by Masterson and Pratt (1958, p. 233) As stated, the foregoing simple calculation demonstrates
and by Eckenfelder et al. (1958, p. 251). In other words, the application of Henry’s law. Two ideas are involved.
application of Henry’s law to flotation has been established First, in accordance with Henry’s law, each gas dissolves
since the 1950s. in water in proportion to its partial pressure. Second,
The partial pressure of ‘‘gas A’’ is implied in Dalton’slaw, Dalton’s law states that each gas in a mixture of gases exerts
i.e., it states that the total pressure in a mixture of gases is the a partial pressure in accordance with its respective mole
sum of the respective partial pressures, expressed for a single fraction, e.g., X(O 2 ). In addition, note that the units are
delineated fully, which helps in applying Henry’s law.
gas as
P(gas A) ¼ X(gas A) P(n gases) (8:2)
Example 8.2 Calculation of Dissolved Nitrogen
in which
Concentration at Sea Level by Henry’s Law at 208C
X(gas A) is the mole fraction of gas A in gas phase (mol
3
3
gas A=m =sum of mol n gases=m )
The problem refers to water at equilibrium with air at sea
P(n gases) is the total pressure of all gases above gas–water
level at 208C based on 101.325 kPa total atmospheric
interface (kPa n gases) pressure.
n is the number of species of gases in a given volume
Given
3
H(N 2 ) ¼ 0.0001875 (kg dissolved nitrogen=m water=kPa
Then combining Equations 8.1 and 8.2, nitrogen) at 208C
a
X(N 2 ) ¼ 0.78084 moles N 2 =mole air (Table B.7)
C(gas A) ¼ H(gas A) X(gas A) P(n gases) (8:3)
P(atmosphere) ¼ 101.325 kPa (stated)
Equation 8.3 is the operational equation. If it happens that Required
Calculate the concentration of dissolved nitrogen at equi-
P(n ¼ 1), i.e., a ‘‘pure’’ gas A is involved, then it follows
librium by Henry’s law.
that X(gas A) ¼ 1.0. Examples 8.1 and 8.2 illustrate the appli-
cation of Henry’s law, which is straightforward in accordance Solution
with Equation 8.3. Example 8.3 shows the calculation of Substitute given data, for nitrogen, in Equation 8.3,
a ‘‘pseudo’’ Henry’s constant for air, H(air), which again is
straightforward, albeit involving mole fraction weighted aver- C(N 2 ) ¼ H(N 2 ) X(N 2 ) P(total)
age of the respective Henry’s constant. 3
¼ 0:0001875 (kg dissolved N 2 =m
water=kPa air)
Example 8.1 Calculation of Dissolved Oxygen
Concentration at Sea Level by Henry’s Law at 208C . 0.78084 moles N 2 =moles all gases in air
. 101.325 kPa
3
The problem refers to water at equilibrium with oxygen ¼ 0:01483 kg dissolved N 2 =m water (14:8mg=L)
at sea level at 208C based on 101.325 kPa total atmos-
pheric pressure and demonstrates the application of Discussion
Henry’s law. The relevant points are stated in Example 8.1.