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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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Answer:

Cout = 11mg/L

Step-by-step 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

User Pierpaolo Ercoli
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