Question

8.4. A free water surface constructed wetland is to be designed to reduce the TP concentration from 1 mg/L in the influent to 0.1 mg/L in the effluent. The influent flow rate is 2700 L/min, ET is 900 mm/ yr, the transpiration fraction is 0.6, infiltration is negligible, and rainfall is 1500 mm/yr. It is antici­pated that the TP background concentration in the wetland is 0.04 mg/L, the temperature of the water is 20"C, the areal rate coefficient is 18 m/yr, and that the wetland can be designed as a 1-tank system.

Determine the area (ha)of the wetland that is required.

Answer 1

Given that:

the influent flow rate=2700 L/min

ET=900 mm/year

the transpiration fraction=0.6

rainfall=1500 mm/year

TP background concentration in the wetland=0.04 mg/L

temparature of the water=20"c

areal rate coefficient=18 m/year

Answer: A free water surface constructed wetland is to be designed to reduce the TP concentration from longle in the influent to oil mg/L. Here the infiltration is negligible So, we have to determine the area (ha) of the wetland that is required: 46: - 0.0365 de Aw, bo 305.98 xs les l Ci-CP — - KA l cerct , there, : A is in hectanes background concentration 0.04 mgle Ci inflow concentration Ce effluent concentration KA = Areal coefficient 8 = total flow which comes from water balance Qe = din +(P-E) AW

Qin 2700 L/min = 3888 m3/day in the same way: KA = 18 m/ meer 30.049/ day Pe - 3888 +0.004 -0.0001) Aw -(2) PE'1500 mm / year 26.oou my day ET = 500 ml year - G-6.6Jx500 = 360 mm Iyear = 0.0009 ml day de = 3888 +0.004 -0.0009) Aw All these values are substitute in equation (1) 0.0365 AWS 0.049 V (1-0.04 y (3868 +0.6031 AW AN 1 20.1-6.04

0.96 Aw = 0.145 x en L 0.06 1 (3888 +0.0031 Aw Aw = 0.745 x In [16] x (3886 +0.0031 Aw) Aw = 0.745x 2.773 X (3868 +0.0031 Aw) Aw = 2.065 x 3888 + 6,402 x103 Aw Aw = 8080 ha The area of the weltand is required is 8060 ha. no