Page 639 - Fundamentals of Water Treatment Unit Processes : Physical, Chemical, and Biological

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

detention time for the reactor is u ¼ 6 h and thus, J(O 2 transferred, ST) is the oxygen transfer rate for stand-
3
3
V(reactor) ¼ (333 m =h) 6h ¼ 2000 m . Assume further ard conditions (kg O 2 transferred=h)
that a single 2.13 m (7.0 ft) impeller is used and that J(O 2 delivered, ST) is the mass rate of oxygen delivered for
placed in the basin described results in a characteristic standard conditions (kg O 2 delivered=h)
curve as given by Figure 18.15. Let the basin be at sea
level and let T ¼ 208C. Assume that a direct current motor The mass fluxes of oxygen are calculated:
is used (for control of rpm) and the power expended is
proportional to the rotational velocity with 60 kW
expended at 60 rpm. J(O 2 transferred, ST)
Required ¼ K L a(20) C*(p ¼ 1:00 atm,20 C) V(reactor)

(a) Determine the K L a required (18:74)
Solution J(O 2 delivered, ST) ¼ Q r(O 2 air at 1:00 atm, T ¼ 20 C)
0

(a) The K L a(required) is given by the relation
(18:75)
J(O 2 , required) ¼ aK L a(bC s C)V(reactor)
¼ Q r(O 2 , ST) (18:76)
0
which with numerical data is
where
67 kg O 2 =h ¼ 1:0 K L a(1:0 0:008 0:002) kg O 2 =m 3 K L a(208C) is the mass transfer coefficient obtained in
2000 m 3 aeration test for actual system, usually in clean water,
1
converted to 208C(h )
which gives C*(p ¼ 1.00 atm, 208C) is the concentration of dissolved
oxygen in aeration basin for a hypothetical air pressure,
1
K L a(required) ¼ 5:6h : p(atm) ¼ 1.00 atm, T ¼ 208C, W ¼ clean water
3
(kg O 2 =m )
3
Discussion V(reactor) is the volume of aeration basin (m )
An aerator test, as outlined in Example 18.3, is necessary Q (ABasin) is the flow of air into diffuser system in
0
to verify whether K L a(actual) K L a(required) for a given aeration basin (m =s)
3
system. For comparison, a surface aerator test value of r(O 2 air at 1.00 atm, T ¼ 208C) is the density of oxygen
K L a ¼ 6.5 h 1 was obtained by Conway and Kumke based upon its partial pressure in air at p(air) ¼ 1.00 atm,
3
(1966) for a 1703 m (450,000 gal) basin and a 56 kW 3
(75 hp) turbine aerator. T ¼ 208C (kg O 2 =m )
r(O 2 , ST) is the same as previous definition, that is, density
of oxygen gas for standard conditions, only shorter
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18.3.2.3 Diffused Aeration notation (kg O 2 =m )
Diffused aeration is the generation of gas bubbles emanating
from pores or orifice openings submerged in a basin of water. The density of oxygen, r(O 2 ), for any pressure and
Since the advent of activated sludge in 1914, a variety of temperature may be calculated from the ideal gas law (see
diffusers have been used in practice to provide oxygen to Table B.7),
aeration basins, for example, porous plates, porous tubes,
saran-wrap tubes, pipes with orifices, etc. Clogging of the p(O 2 ) MW(O 2 )
(18:77)
porous diffusers prompted the use of orifice-type materials r(O 2 ) ¼ RT 1000
in the 1920s. Later, air filters were developed to protect the
fine-pore media. Surface aerators came on the scene primarily where
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to circumvent the clogging problem. Foregoing is from Boyle r(O 2 ) is the mass density of oxygen (kg O 2 =m )
et al. (1989, pp. 1, 2). p(O 2 ) is the partial pressure of oxygen in atmosphere
Evaluation of a diffused air system is commonly in terms (N=m )
2
of its oxygen transfer efficiency, defined as R is the gas constant (8.314 N m=mol K)
T is the absolute temperature (K)
J(O 2 transferred, ST) MW(O 2 ) is the molecular weight of oxygen (31.9988
(18:73)
J(O 2 delivered, ST)
E(O 2 , ST) g=mol)
where Example 18.5 Calculation of Gas Density
E(O 2 , ST) is the standard oxygen transfer efficiency at
T ¼ 208C, p ¼ 1.00 atm, W ¼ clean water (dimension- Given
less); the term, E(O 2 , ST) is the same as SOTE, the latter Air at standard temperature, T ¼ 208C, and standard pres-
being used by Boyle et al. (1989) sure, p(sea-level) ¼ 101.325 kPa.