DESIGN OF CONSTRUCTED WETLAND FOR THE CITY OF TLAXIACO,

OAXACA, MEXICO

DESIGN NOTES, PROCEDURE, CRITERIA

Version 1.5:   June 11, 2004

  • Spatial design: The spatial design of the wetland should be such that it consists of N long, narrow channels, each of width w, in parallel. This is to ensure plug-flow conditions. The actual wetland can be of irregular shape, but rectangular or quasi-rectangular shapes are preferred. The width W of the wetland is the sum of the widths of the N narrow channels.

  • Inflow: Inflow is distributed across the width W of the wetland by means of a perforated pipe.

  • Hydraulic gradient: The system is built with a hydraulic gradient between inflow and outflow (longitudinal gradient). The hydraulic gradient should be at least 1 percent, and not more than 3 percent, depending on the topography.

  • Outflow: The effluent is collected at an outlet channel, which is filled with coarse gravel and spans the width of the wetland. The effluent is discharged directly into the river or conveyed for downstream use in irrigation.

  • Soil permeability: Soil permeability should be tested to ensure 0.0001 to 0.00001 cm/sec, i.e., sandy clays or silty clays.

  • Sewage load:Assume 0.15 m3/day/person of water supply in Tlaxiaco; factor 0.6 of water-to-sewage conversion; 0.09 m3/day/person of sewage load; assume 2,500 families draining to municipal sewer, 4 persons per family, 10,000 persons total; 900 m3/day sewage load; or 0.0104 m3/sec. Assume a sewage load of 0.01 m3/sec for design purposes.

  • Water balance -- Theory: Influent Qi minus effluent Qo plus precipitation P minus evapotranspiration ET should largely cancel out. The equation is: Qo = Qi + P - ET.   Precipitation P in Tlaxiaco is 900 mm/yr. Evapotranspiration ET is estimated at ET= 1350 mm/yr. Qi = 0.01 m3/sec = 864 m3/day.

  • Water balance -- Practice: Assume an area of 10,000 m2. Therefore: Qo = 0.00986 m3/sec. This means that the outflow is not very sensitive to the difference between P and ET. Since Qo does not differ much from Qi, Qi will be the design Q.

  • Performance: A constructed wetland can reduce BOD, SS, N, metals (cadmium, copper, mercury, zinc, chromium, lead, and selenium), trace organics, and pathogens (bacteria, viruses, protozoa). Phosphorous removal is not expected to be substantial.

  • Design (actual) water depth d: Assume d= 0.6 m.

  • BOD removal: Described by a first-order model BODe/BODi = e-KT t

  • Hydraulic residence time t: t should be at least 5-6 days to reduce BOD to acceptable levels. t = (LWdn)/Q; L= length; W= width; d= depth; n= porosity, defined as voids volume over total volume; w= cell width; N= number of cells; W= wN.

  • Porosity: Asssume porosity n= 0.75.

  • Wetland size calculations: Assume w= 5 m; N= 20; W= wN= 100 m; L= 100 m; LW= 10,000 m2; d= 0.6 m; n= 0.75. Then: dn= 0.45 m; LWdn= 4,500 m3; t= (LWdn)/Q = 4500/864 = 5.2 days.

  • Performance formula: BODe/BODi = r = A e(-0.7 KT Av1.75 t)
    A= fraction of BOD not removed as settleable solids; assume A= 0.52.
    KT = rate constant, a function of temperature.
    Av = specific area for microbial activity; assume 15.7 m2/m3.

  • Calculation of rate constant:
    KT = K20 (1.1)(T - 20)
    K20 = 0.0057 day-1. Assume T = 13 oC.
    Then: KT = 0.0029 day-1.

  • Performance calculations:
    BODe/BODi = r = A e(-0.7 KTAv1.75 t) = 0.52 e[(-0.7) (0.0029) (15.7)1.75 (5.2)] = 0.14

  • Sensitivity: The equation is not sensitive to Av, but it is sensitive to temperature T.

  • Design of constructed wetland for Tlaxiaco:

    BOD ratio r = A e(-0.7 Av1.75 KT t)

    t = (LWdn)/(86400*Q) days

    Total width W= wN

    KT = K20 (1.1)(T - 20) day-1

    ln(r)= ln(A) - 0.7 Av1.75 KT t

    ln(r)= ln(A) - 0.7 Av1.75 K20 (1.1)(T-20) t

    ln(r)= ln(A) - 0.7 Av1.75 K20 (1.1)(T-20) (LWdn)/(86400*Q)

    0.7 Av1.75 K20 (1.1)(T-20) (LWdn)/(86400*Q) = [ln(A) - ln(r)]

    L = 86400 Q [ln(A) - ln(r)] /[0.7 Av1.75 K20 (1.1)(T-20) wNdn]

    Assume: Q= 0.01; A = 0.52; r = 0.14; Av= 15.7; K20= 0.0057; T= 13; w= 5; N = 20; d= 0.6; n=0.75

    L = 86400 (0.01) [ln(0.52) - ln(0.14)] / [0.7 (15.7)1.75 (0.0057) (1.1)(-7) (5)(20)(0.6)(0.75)]

    L = 864 [ln(0.52) - ln(0.14)] / [0.7 (123.83) (0.0057) (0.513) (5)(20)(0.6)(0.75)]

    L = 864 [ln(0.52) - ln(0.14)] / [(0.253) (5)(20)(0.6)(0.75)]

    L = 864 ln(3.71) / [(0.253) (5)(20)(0.6)(0.75)]

    L = 864 (1.331) / [(0.253) (5)(20)(0.6)(0.75)]

    L = 1150 / 11.38 = 101 m. Assume L = 100 m.

    t= (LWdn)/(86400*Q) = (100)(100)(0.6)(0.75) / [(86400) (0.01)] = 5.2 days.

    Check performance: r = A e(-0.7 Av1.75 KT t) = 0.52 e[(-0.7) (123.83) (0.0057) (0.513) (5.2)] = 0.14    OK.

    Effluent BOD= 0.14 X influent BOD= 0.14 X 80 ppm = 11 ppm. Efficiency of pond/wetland treatment system = [1 - (11/350)] X 100 = 96.8% This is slightly better than secondary treatment (90-95% BOD removal).

    Conclusion: With Q= 0.01 m3/sec and T = 13 oC, the wetland will have dimensions of 100 m x 100 m = 1 ha, t = 5.2 days, BOD ratio r = 0.14, and effluent BOD from wetland = 11 ppm. Total efficiency of oxidation pond/constructed wetland system is 96.8% (secondary).

  • Calculation of volume of dike: Assume depth of dike = 0.9 m; top width= 2 m; upstream slope 2:1; downstream slope 2:1; total area 3.42 m3/m. Length of dike = 400 m. Total volume of dike= 1368 m3. Unit cost 40 Mexican pesos per cubic meter. Therefore, cost is $ 54,720 (Mexican pesos).

    Depth of excavation = 1368 m3 / 10,000 m2 = 0.14 m.

    Estimated project cost = 4 x 54,720 = $ 218,880 (U.S. $ 24,320).