Imagine a shallow pond that is well mixed due to wind and a steady flow of a cre
ID: 513410 • Letter: I
Question
Imagine a shallow pond that is well mixed due to wind and a steady flow of a creek that flows through the pond. Given the parameters below and if the microbes in the pond consume the inflowing biodegradable organic matter according to typical kinetics, determine the following:
The BOD5 leaving the pond.
The biodegradable organic matter removal efficiency of the pond.
The concentration of volatile suspended solids leaving the pond.
Parameters:
Q = 75.0 m3/d
X0 = 0.0 mg VSS/L
S0 = 115 mg BOD5/L
Xf = ? mg VSS/L
Sf = 115 mg BOD5/L
V = 200 m3
Explanation / Answer
Flow rate of influent and effluent water 75.0 m3/d
S0 115 mg BOD5/L Sf = Need to calculate
Tank Volume = 200 m3
Biomass = 0.0 mg VSS/L Biomass leaving the pond = Need to calculate
You have written in the question that Sf = 115 mg BOD5 / L which might be wrong since that would completely mess up the efficiency (If Sf is indeed 115, the efficiency would be 0%.) I think we need to calculate the Sf using hydraulic retention time.
Hydraulic Retention time (theta) = Volume of aeration tank in m3 / Sewage inflow in m3/d
Thus, Hydraulic Retention time = 200 m3 / 75.0 m3/d
Hydraulic Retention time (HRT) = 2.67 days
1 / HRT = [(umSf) / (Ks + Sf)] - Kd
They must have given standard values of Ks, um and Kd which are standard parameter values for conventional activated sludge systems using a mixed flow reactor. Even if they are not mentioned in the question, they may have been mentioned somewhere in your textbook. You can easily find these values if you look up the correct table. Hence, using the values of Ks = 60 mg BOD5/L, um = 3 / day, K = 5 and Kd = 0.06 / day, solve for Sf.
Sf = [Ks (1 + theta . Kd)] / [(theta . um - 1 - theta . Kd)]
Sf = [60 ( 1 + 0.1602)] / [(8.01 - 1 - 0.1602)]
Sf = (60 + 9.612) / 6.85
Sf = 69.612 / 6.85
Sf = 10.162 mg BOD5/L
Thus biodegradable organic matter removal efficiency = 1 - (Sf / S0) X 100
biodegradable organic matter removal efficiency = 1 - (10.162 / 115) X 100
biodegradable organic matter removal efficiency = 91.163 %
Rate of substrate utilization, Rs is calculated as: - Q (S0 - Sf) / V
Rs is also calculated as - KXfSf / [Ks + Sf]
Equate the two equations:
- Q (S0 - Sf) / V = - KXfSf / [Ks + Sf]
Xf = [(Ks + Sf) (S0 - Sf) (Q/V)] / (KSf)
But Q/V = 1 / theta
Xf = [70.162 X 104.838 X 0.375] / [5 X 10.162]
Xf = 2758.366 / 50.81
Xf = 54.288 mg VSS/L