Solved Example Problems on Design an oxidation pond

PROBLEMS

13. Design an oxidation pond for treating sewage from a hot climatic residential colony with 5000 persons contributing sewage @ 120 lpcd. The five day BOD of sewage is 300 mg/l.

Solution:

(i) Quantity of sewage treated per day = Population x Sewage contribution. = 5000 x 120 lpcd = 6,00,000 litres. = 600 m3/d or 0.6 MLD.

(ii) BOD content per day.

Total BOD = Discharge x BOD concentration

(iii) Assume organic loading in pond (Hot climate) = 300 kg/ha/day.

Surface area required = Total BOD of sewage/ Organic loading rate

= 180 kg/d/ 300kg/ha/d = 6000 m2.

(iv) Assume L/B = 2.

L x B = 6000 m²

2B x B = 6000

B = 55 m

L= 2 x 55 110 m.

(v) Assume effective depth = 1.2 m.

(vi) Volume (capacity) of pond. = L x B x H

= 110 x 55 x 1.2 = 7260 m3.

(vii) Capacity = Sewage flow per day x Detention time (days).

D.T= Capacity/Flow =7260 m³ 600m³/d =12.1 days = 12 days.

LotfupzoM diwong legiA anobo bad

Result:

Use oxidation pond with

Length = 110 m; Width = 55 m.

Overall depth = (1.2 + 1) = 2.2 m.

Detention period = 12 days.