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 × influent BOD = 0.14 × 80 ppm = 11 ppm.
Efficiency of pond/wetland treatment system = [1 - (11/350)] × 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 × 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).