Page 640 - Fundamentals of Water Treatment Unit Processes : Physical, Chemical, and Biological

Gas Transfer 595

Required 45 A=aerator; annual operating days were 191 days
Calculate the density of air, r(air, ST) and of oxygen in air, with two aerators in operation and 174 days with
r(O 2 , ST) for standard conditions three aerators; power cost was $0.05=kWh.
(b) Retrofit—Diffused aeration: In 1985, the aeration
Solution basins were retrofitted with a fine-pore diffused aer-
Substituting numerical data in Equation 18.75 yields ation system consisting of 1350 230 mm (9 in.)
r(air, ST) and r(O 2 , ST), respectively. For reference, the
diameter diffusers with ratio diffuser surface area to
equivalent MW(air) ¼ 29.964 g=mole (Table B.7) and
tank floor area ¼ 0.116. Each aeration basin had
the mole fraction of oxygen in air, X(O 2 =air) ¼ 0.209476
installed 450 ceramic disk diffusers, equally spaced
(also Table B.7)
at 660 mm cc (26 in.). Equipment was installed for
in situ gas cleaning and for pressure monitoring.
N
101,325 28:964g air kg (c) Retrofit data: Three positive displacement blowers
m 2
0
Nm mol air 1000g
r(air,ST) ¼ were installed, each 37 kW (50 hp) with Q (air per
3
8:314 (273:15 þ 20)K blower) ¼ 0.400 m =s=blower (850 scfm=blower).
mol K
0
¼ 1:204kg air=m 3 Each blower was expandable to Q (air per
3
blower) ¼ 0.802 m =s=blower (1700 scfm=blower)
N by retrofitting with a higher wattage electric motor.
0:209476 101,325 3
m 2 31:9988g O 2 kg The design criteria were: Q ¼ 0.072 m =s (1.65 mgd);
Nm 1000g
r(O 2 ,ST) ¼
8:314 (273:15þ20)K mol O 2 J(BOD) ¼ 614 kg BOD=day (1,356 lb=day); J(NH 3 -
mol K N)¼ 136 kg NH 3 -N=day (300 lb=day); J(O 2
¼ 0:2787kg O =m 3
2 required) ¼ 1.15 kg O 2 =kg BOD appliedþ 4.6 kg
O 2 =kg NH 3 -N applied. Also, Q (air)=diffuser¼
0
3
Discussion 0.063 m air=s=diffuser (1.33 scfm=diffuser).
Application of the ideal gas law, as applied above is
straightforward. The key points relate to (1) using the Required
partial pressure of the gas, that is, oxygen in air; (2) apply- Estimate K L a for an aeration basin based upon the given
ing the molecular weight of the gas to convert from molar data. Assume that E(O 2 , ST) ¼ 0.15 for the purpose of the
density to mass density. Such calculations, as shown, are calculation.
useful in determining the ‘‘mass flow of oxygen deliv-
ered,’’ that is, J(O 2 , delivered, ST), as in Equation (18.76), Solution
and in doing calculations that may involve the flow of air Apply Equation 18.76. First determine the input data:
under any conditions, for example, as related to compres-
sors. As with all calculations, any final numbers, for (1) From Table H.5, C*(O 2 =air, ST) ¼ 0.209476 43.39
example, for mass flow, J, or power required, should be mg O 2 =L ¼ 9.07 mg O 2 =L ¼ 0.00907 kg O 2 =m 3
rounded to the number of significant figures applicable (2) V(reactor) ¼ 12.2 m 12.2 m 4.6 m ¼ 684.66 m 3
3
(commonly three for engineering purposes). (3) Q (air)=blower ¼ 0.400 m =s=blower
0
3
Combining Equations 18.71 through 18.73 gives the (4) r(O 2 , ST) ¼ 0.2787 kg O 2 =m (Example 18.5)
oxygen transfer efficiency yields
Now apply Equation (18.78):
K L a(ST) C*(O 2 in air, ST) V(reactor)
Q (air) r(O 2 , ST) *
E(O 2 , ST)
K L a(ST) C (O 2 in air, ST) V(reactor)
0
(Ex18:5:1) E(O 2 , ST) Q (air) r(O 2 , ST)
0
Equation 18.76 provides a relation for conversion between kg O 2 3
the oxygen transfer efficiency, E(O 2 , ST), and K L a, the K L a(ST) 0:00907 m 3 684:66 m
mass-transfer coefficient for oxygen. Once K L a is deter- 0:15 ¼ m 3 kg O 2
mined, the oxygen transfer efficiency may be determined 0:400 s 0:2787 m 3
for any other condition, which includes calculating Q (air) 1
0
and compressor power required, etc. K L a(ST) ¼ 0:027 s
¼ 9:7h 1
Example 18.6 K L a from Plant Data (Data Taken
from Boyle et al., 1989) Discussion
Once K L a is determined for the conditions described, a
Given variety of what-if questions may be addressed (also called
(a) Background: The Plymouth, WI activated sludge ‘‘scenarios’’). For example, suppose pure oxygen is used
plant operated between 1978 and 1985 using three instead of air. What happens if the airflow is increased (for
aeration basins, each 12.2 m 12.2 m 4.6 m air, not pure oxygen)? What if the elevation of the aeration
(40 ft 40 ft 15 ft), and each with one 30 kW basin is 1610 m (5280 ft)? At high elevations is there merit
(40 hp) surface aerator. Annual energy costs for in using a deeper basin? Note that the round off was
aeration were estimated at $37,500 based on applied at the last step for K L a.