Solved Example Problems on Design an oxidation ditch

PROBLEMS

11. Design an oxidation ditch for a community with following data:

(i) Population of community - 6000 persons.

(ii) Organic load of sewage - 40 g BOD per capita per day.

(iii) Sewage flow - 160 l/capita/day.

(iv) Permissible BOD of effluent - 20 mg/l.

Solution:

(i) Inflow Rate and Influent BOD:

Total Sewage Flow = Population x Sewage Contribution

= 6000 × 160 = 960 × 10³ l/d = 960 m³/d

Total BOD applied/day = Population x Organic load of Sewage

= 6000 x 40 g/c/d = 240 kg/d.

(ii) Volume of Ditch:

(iii) Volumetric Loading Rate:

(iv) Hydraulic Retention Time:

(v) Return Sludge Ratio:

(vi) Oxygen required and no. of rotors:

O₂ required = 1.2 kg/kg of BOD removed.

= 1.2 (250-20) x 0.96 = 265 kg/d.

= 11.04 kg/hr.

Oxygenation capacity of rotor, 70 cm dia @ 75 r.p.m.

= 2.8 kg of O2/hr/m/length.

Length of rotor = O₂ required / Oxygenation capacity of rotor 2.8 / 11.04

= 3.94 m (say 4 m)

150 m³ of ditch volume is required per metre length of rotor for circulation

800/150 = 5.33 m.

150

003-

Adopt 2 rotors of 2.7 m length each, provide clearance of 0.25 m on either side.

Width of ditch = 2.7 + 0.25 + 0.25 m

= 3.2 m

Assume depth = 1.5 m

Surface area required = Volume / Depth = 800 /1.5

= 533.3 m2.

Adopt 2 ditches, surface area of each ditch = 533.3/2 = 266.7 ^m2.

Length of ditch = 266.7/3.2  = 83.3 m. or 85 m.

(vii) Design of Settling Tank:

Assume SLR/SOR = 20 m³/m²/day and Detention Time = 2 hours

Surface area, A = Q /SOR =960/20 = 48 m²

Volume, V = 960/24 x 2 = 80 m².

Depth of tank = 80/48  =1.67 m.

Diameter of tank = ppppppppppp

Provide 8 m o tank of 1.7 m depth with hopper bottom.

12. Design a continuous flow type oxidation ditch to treat a domestic sewage flow of 2.5 MLD. Given the following data.

BOD, of raw sewage - 160 mg/l

Desired BOD₅ of effluent -  25 m

SS in raw sewage - 240 mg/l

Desired SS in effluent -30 mg/l

Mixed liquor suspended solids - 4600 mg/l

Growth yield co-efficient - 0.55 →Y

Micro organisms decay co-efficient  - 0.03 → k_d

Mean cell residence time  - 16days→ θ_c

Solution:

(i) BOD of influent and effluent.

(ii) Volume of Reactor (ditch)

(iii) Microbial Mass in Excess Sludge

(iv) Rate of Wasting of Excess Sludge

(v) Return Sludge Concentration.

(vi) Oxygen required (kg/d)

= 1.2 kg of O₂ per kg BOD₅ biloz babn

= 1.2 (160-25) x 2.5= 405 kg/d.

(vii) Check for HRT:

(viii) Rotor length:

Oxygenation capacity of 70 cm dia. rotor @ 75 rpm.

= 2.8 kg of O2/hr/m length.

Length of rotor = 405/24×2.8 = 6.03 m

(or)

150 m³ of ditch volume per m length of rotor for circulation 476.6 /150 = 3.18 m

Adopt 6.03 m.

Provide 2 rotors of 3 m length.

(ix) Ditch Dimensions:

Length of rotor = 3 m.

Clearance on both side = 3 + 2 x 0.25 = 3.5 m, keep depth = 1.2 m.

Surface area required = Volume / Depth 476.6 /1.2 = 397 ^m2.

Provide 2 ditches, surface area of each ditch

= 397 /2 = 200 m².

Length of each ditch = 200/3.5 = 58 m.