A waste stabilization pond is used to treat a dilute municipal wastewater before the liquid is discharged into a river. The inflow to the pond has a flow rate of Q=4,000 m3 / day and a BOD concentration of Cin = 25 mg/L. The volume of the pond is 20,000 m3. The purpose of the pond is to allow time for the decay of BOD to occur before discharge into the environment. BOD decays in the pond with a first-order rate constant equal to 0.25/day. What is the BOD concentration at the outflow, Cout of the pond, in units of mg/L?

Question

Engineering

Chanel Walton

10 October, 20:11

A waste stabilization pond is used to treat a dilute municipal wastewater before the liquid is discharged into a river. The inflow to the pond has a flow rate of Q=4,000 m3 / day and a BOD concentration of Cin = 25 mg/L. The volume of the pond is 20,000 m3. The purpose of the pond is to allow time for the decay of BOD to occur before discharge into the environment. BOD decays in the pond with a first-order rate constant equal to 0.25/day. What is the BOD concentration at the outflow, Cout of the pond, in units of mg/L?

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Answers (1)

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Dennis Hart · Answer 1

Cout = 11mg/L

Explanation:

Assume Steady State and CSTR, this means Qin = Qout

Given: Qin = 4 000m3/day

Cin = 25mg/L

V = 20 000m3

k = 0.25/day

Find: Cout = ?

outStarting from the mass balance equation for steady state 0 = QCin - QCout - kCoutV

manipulate the equation to get:

/ (Cout=Cin (Q/Q+kV) / )

Then simple plug in your givens:

Cout = (25mg/L) ((4000m3/day) / (4000m3/day + 0.25/day (20000m3))

Cout = 11mg/L