DESIGN OF OXIDATION POND FOR THE CITY OF TLAXIACO, OAXACA, MEXICO |
DESIGN NOTES, PROCEDURE, CRITERIA Version 2.0: September 10, 2004 |
Depth: Aerobic ponds should not be too shallow or too deep. If they are too shallow, they may be destroyed by emergent vegetation. If they are too deep, light penetration may be impaired and the lower layer may become anaerobic. A depth of 1 m is not too shallow nor too deep. Assume a depth = 1.2 m. Mixing: Mixing of pond contents contributes to the distribution of dissolved oxygen throughout the profile. Mixing is primarily by wind action and secondarily by temperature gradients. The maximum effect of wind action can be obtained when there is an unobstructed wind path of 100 to 200 m. Ponds should be designed with a longitudinal orientation parallel to the prevailing winds. The direction of the prevailing winds in Tlaxiaco will have to be determined experimentally, since there is not wind direction data available. Trees provide a barrier to wind action, so care should be taken to minimize the presence of large trees upwind of the flow path. Design procedure: Use the Wilhelm-Werner equation for chemical reactor design (Reed, S. C., R. W. Crites, and E. J. Middlebrooks. 1995. Natural Systems for Waste Management and Treatment, McGraw-Hill, Inc., New York).
Sewage load count: To determine the sewage load, the APASZU 96 map of the city of Tlaxiaco (December 1996) was provided by the Office of Water and Sewage of Tlaxiaco. The city was divided into five regions:
Load calculation: Assume 0.15 m3/day/person of water consumption in Tlaxiaco; assume factor 0.6 of water-to-sewage conversion; then, 0.09 m3/day/person of sewage load. The pond will be designed for 2,500 families draining to the municipal sewer system. With an estimate of 4 persons per family, this results in 10,000 persons total, or 900 m3/day sewage load; or 0.0104 m3/sec. Assume a sewage load of 0.01 m3/sec for design purposes. Sewage concentration calculation: Assume a load of 30 grams/person/day of dry fecal matter, applicable to developing countries such as Mexico. Then, the sewage concentration will be: 30 gr/person/day X 10 000 persons X 1000 mg/gr / (900 m³/day X 1000 liter/m³) = 333 mg/liter (ppm). Assume 350 ppm for design purposes. Design example: The pond has length L, width W, and depth d. Assume three cells in series and three cells in parallel, for a total of nine cells. The mean temperature of the coldest month (January) is T= 13 oC. Assume that the design dispersion number D= 0.1 (plug flow), subject to verification. Assume a hydraulic retention time t= 20 days. Therefore, the BOD removal efficiency is 76.57%. With design Q= 0.01 m3/sec, the volume of the pond is V= LWd= 0.01 m3/sec X 20 days X 86400 sec/day = 17,280 m3. With a design depth d= 1.2 m, then, the area of the oxidation pond is LW= 14,400 m2. Assume a width W= 90 m. Then, L= 160 m. Assume n= 3 cells in parallel. Therefore each cell has a width w= W/n= 30 m.
Calculation of D: D= 0.184 [tv(w + 2d)]0.489w1.511 (Ld)-1.489 v= kinematic viscosity (m2/day)
At 13 oC, v = 1.21 centistokes = 0.0121 stoke = 0.0121 cm2/sec = 0.0121 cm2/sec
X 0.0001 m2/cm2 X 86400 sec/day= D = 0.184 [20 X 0.1045 (30 + 2.4)]0.489 (30)1.511 (160 X 1.2)-1.489 = D = 0.184 (67.716)0.489 (30)1.511 (192)-1.489 = D = (0.184) X (7.856) X (170.58) X (0.0003982) = 0.0982 Since calculated D= 0.0982 is sufficiently close to design D= 0.1, the effective BOD removal for t= 20 days is 76.6% = 77% For influent BOD= 350 ppm, the effluent BOD from the oxidation pond is = 350 X (1 - 0.77) = 80 ppm Remarks: The design pond dimensions are L= 162 m, d= 1.2 m, and W = 90 m, with n= 3 parallel cells of w= 30 m each, and three cells in series of l= 54 m each. The actual BOD removal is 77% and the effluent BOD is 80 ppm. Calculation of volume of dike: Assume depth of dike = 1.8 m; top width= 3 m; upstream slope 2:1; downstream slope 3:1; total area 13.5 m3/m. Length of dike = 520 m. Total volume of dike= 7014 m3. Unit cost 40 Mexican pesos per cubic meter. Therefore, cost is $ 280,560 (Mexican pesos).
Depth of excavation = 7014 m3 / 14,580 m2 = 0.48 m.
Estimated project cost = 2 x 280,560 = $ 561,120 (U.S. $ 62,347). |