Determine the volume (in m3) required for an aerated lagoon to treat a domestic

Question

Determine the volume (in m3) required for an aerated lagoon to treat a domestic plus industrial waste flow of 0.30 Mgal/d with an average BOD of 600 mg/l. The temperature extremes anticipated for the lagoon contents range from 10C in winter to 35C in summer. Minimum BOD reduction through the lagoon should be 75%. During laboratory treatability studies, the wastewater exhibited the following characteristics: k20°C (reaction rate coefficient at 20°C) = 0.68 per day, (temperature coefficient) =1.047.

Explanation / Answer

S/S0 = 1 / (1 + (kTV/Q))

where,

S0 = BOD in domestic and industrial waste water

= 600 mg/l

S = Final BOD after minimum 75% reduction

= 150 mg/l

Q = Flow rate of waste water

= 0.30 Mgal/day

kT = rection rate coefficient for specific temperature (day-1)

Now, kT = k20 (T-20)

k20 = rection rate coefficient at 20 deg C (day-1)

= 0.68 day-1

= temperature coefficient

= 1.047

Hence, k30 = 0.68 (1.047)(30-20)

= 1.076 day-1

and, k10 = 0.68 (1.047)(10-20)

= 0.430 day-1

Volume of Lagoon during summer (30 deg C),

S/S0 = 1 / (1 + (k30V/Q))

(150/600) = 1 / (1 + (1.076 x V/0.3))

V = 0.84 Mgallon

Volume of Lagoon during Winter (10 deg C),

S/S0 = 1 / (1 + (k10V/Q))

(150/600) = 1 / (1 + (0.430 x V/0.3))

V = 2.09 Mgallon

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