Page 781 - Fundamentals of Water Treatment Unit Processes : Physical, Chemical, and Biological

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

Y(O 2 -stoichiometric) [O 2 -demand=ammonia]
¼ 0.66 g O 2 used=g substrate degraded Y(O 2 -stoichiometric= ammonia) C(NH 3 -as-N) Q
Endogenous respiration: In addition, the oxygen demand ¼ [4.3 kg O 2 used=kg ammonia oxidized] [0.022 kg
3
3
for cell oxidation may also be calculated from stoichiom- NH 3 -as-N=m ] (18,925 m =day)
etry (Section 22.4.2.2; Example 22.2=3; Table 22.8) based ¼ 1790 kg O 2 used=day
on the equation for the conversion of cells to carbon
dioxide and water. As seen, the conversion is Total Oxygen Demand (Substrate, Endogenous Respir-
ation, Ammonia)
Y(O 2 -stoichiometric=cell oxidation) The total oxygen required per day is
¼ 1.42 kg O 2 used=kg cells oxidized [Total O 2 demand rate] ¼ [Substrate O 2 demand]
þ [O 2 -demand=decay] þ [O 2 -demand=ammonia]
Ammonia oxidation: If ammonia is oxidized to nitrite=
nitrate, an additional oxygen demand is exerted. The oxygen ¼ 4731 kg O 2 substrate=day þ 336 kg O 2 decay=day
demand for ammonia oxidation may be calculated from þ 1790 kg O 2 NH 3 =day
stoichiometry (Section 22.3.3.3=Nitrification, Equation ¼ 6857 kg O 2 =day
22.9) based on the equation for the conversion of ammonia
to cells, nitrate, water. As seen, the conversion gives Discussion
In estimating oxygen demand based on BOD, the BOD
Y(O 2 -stoichiometric=ammonia) model, i.e., measuring dissolved oxygen reduction in
a 300 mL bottle (seeded and incubated at 208C over a
¼ 4:3g O 2 used=g ammonia oxidized
5-day period) is not necessarily accurate; but it may be a
reasonable estimate. Another approach in estimating oxy-
Example 23.2 Oxygen Required gen demand would be to measure the TOC of the incom-
ing substrate and apply the general substrate formula,
C 10 H 19 O 3 N, and use the carbon in the formula as the
Given
S o 280 mg BOD=L or 0.280 kg BOD=m 3 basis for an oxygen demand calculation as in Section
22.4.5, Table 22.8, for a stoichiometric equation for the
S 30 mg BOD
3
Q ¼ 18,925 m =day (5.0 mgd) oxidation of domestic wastewater.
b 0.10 day 1 x The rate of oxygen delivered, J, is calculated by Equation
18.33, J ¼ (dC=dt) V ¼ K L a(C s C) V, Sections 18.2.2.5
C(NH 3 -as-N) 22 mg=L (Section 22.1.5.3=Ammonia)
and 18.3.2.3; the value of K L a depends on whether a
surface aerator or a diffused aeration is used. For diffused
Required
Oxygen utilization rate (OUR) due to substrate, endogen- aeration, Equation 18.45 may be used to calculate
0
ous respiration of cells, and ammonia oxidation K L a, which is, K L a ¼ (k L =V) [zh o Q =(v o þv w )d]. Another
approach is to assume an oxygen transfer efficiency, e.g.,
Solution E(oxygen to water)¼ J(oxygen transferred to water=J(oxygen
Substrate oxidation: See Section 22.4.5, Table 22.8, for a delivered). With diffused air, the oxygen transfer efficiencies
stoichiometric equation for the oxidation of domestic was- are about 0.06 for coarse-bubble aeration and 0.10–0.12 for
tewater, using the formula, C 10 H 19 O 3 N for domestic waste fine-bubble aeration (from Morgan and Bewtra, 1960); see
and C 5 H 7 O 2 N for cells. But since the BOD equivalents are also Table 18.7. If we assume E 0.06, then J(demand)=
not given, the BOD given is used as an index of substrate J(delivered) ¼ (6857 kg O 2 =day)=J(delivered) 0.06, then
BOD demand (for this problem). J(O 2 delivered) 114,283 kg O 2 =day. From Table B.7,
3
Since the substrate mass is given in terms of BOD, we r(O 2 , STP) ¼ 0.278 kg O 2 =m and, r(air, STP) ¼ 1.204 kg
3
can evaluate oxygen demand, O 2 =m , which gives J(air delivered) ¼ 494,952 kg air=d
500,000 kg air=d. The airflow, Q(air) may be calculated by
[Substrate O 2 demand] Q(S o S) the ideal gas law, PV¼ nRT, Appendix H.
3
3
¼ (18,925 m =day) [(0.280–0.030) kg BOD=m ]
Sludge volume index
¼ 4731 kg O 2 =day
A routine measure of final clarifier underflow sludge con-
Endogenous respiration: See Section 22.4.5, Table 22.8, centration is the sludge volume index, SVI. This is the
for stoichiometric equation for endogenous respiration volume of 1 g of sludge after settling 30 min in a 1000 mL
(oxidation of cell matter). cylinder. It is measured by obtaining the oven-dry weight of
From Example 23.1, DX(decay rate) 237 kg cells the settled sludge and dividing this into the volume previ-
decay=day. ously noted for the corresponding mass of settled sludge.
Usually it is desired that
[O 2 -demand=decay] Y(O 2 -stoichiometric=cell oxidation)
DX(decay rate) SVI < 100:
¼ [1.42 kg O 2 used=kg cells oxidized]
237 kg cells decay=day Larger values imply bulking sludge. Also it should be
noted that
¼ 336 kg O 2 used=day
Ammonia oxidation: See Section 22.3.3.3 for stoichiomet- 10 6
ric equation for the oxidation of ammonia. X r ¼ SVI