A lagoon (oxidation pond) system has three cells, each 115,000 m2 in area, a min
ID: 1713966 • Letter: A
Question
A lagoon (oxidation pond) system has three cells, each 115,000 m2 in area, a minimum operating depth of 0.6 m and a maximum operating depth of 1.5 m. It receives 1,900 m3 /day of wastewater having an average BOD5 of 122 mg/L. What is the BOD5 loading and what is the detention time in the lagoon? Consider the following guidelines:
a) BOD loading to small lagoons should not exceed 22 kg BOD5/hectare-d.
b) Detention time in lagoon, excluding the bottom 0.6 m in the volume calculation. It should be at least 6 months.
Explanation / Answer
Area of each cell in lagoon = 115000 sqm
Number of cells in each lagoon =3
Total area= 3*115000= 345000 sqm
Wastewater = 1900 m3/ day average BOD5 = 122 Mg/l
Total BOD 5 loading rate is = 1900*1000* 122/1000000 = 231.8 Kg/ d or 231.8/ 34.5 = 6.718 kg / hectare- day
Total area of lagoon = 115000*3= 345000 sqm
Considering an average operating depth of 1.05 m total volume of lagoon = 1.05* 345000= 362250 cubic metres
Total water per day = 1900 cubic metres
Thus detention time = 362250/ 1900 = 190.65 or 191 days > 6 months so OK
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