Question

A sewage treatment plant employing activated sludge process is operated under the following conditions:
Wastewater flow rate into the aeration tank = 0.150 m³/s
Soluble BOD₅ in the influent = 84 mg/L
Soluble BOD₅ in the effluent = 11 mg/L
Mean cell residence time = 5 days
Total yield co-efficient = 0.5 kg of MLVSS/kg BOD₅ removed
Decay rate of microorganism = 0.050/day
The net waste activated sludge produced in the plant is ______ kg of VSS/day (rounded off to two decimal places).

Hint:

Use the net biomass production formula Px = YQ(S0-S)/(1+kd*theta_c), converting the flow rate to m3/day and the BOD difference to kg/m3 before multiplying.

Answer 1

Correct Answer: 378.43

Step 1: Write the standard formula for the net waste activated sludge production rate.
For an activated sludge system, the observed rate of biomass (VSS) production, accounting for endogenous decay over the mean cell residence time \(\theta_c\), is \[ P_x = \frac{Y \cdot Q (S_0 - S)}{1 + k_d \theta_c} \] where \(Y\) is the yield coefficient, \(Q\) is the flow rate, \(S_0\) and \(S\) are the influent and effluent soluble BOD5, and \(k_d\) is the endogenous decay rate.

Step 2: Convert the flow rate to a daily basis and find the BOD removed.
\[ Q = 0.150\ \text{m}^3/\text{s} \times 86400\ \text{s/day} = 12960\ \text{m}^3/\text{day} \] \[ S_0 - S = 84 - 11 = 73\ \text{mg/L} = 0.073\ \text{kg/m}^3 \]

Step 3: Compute the gross biomass production (before decay adjustment).
\[ Y \cdot Q (S_0-S) = 0.5 \times 12960 \times 0.073 = 473.04\ \text{kg/day} \]

Step 4: Divide by the decay factor \((1 + k_d \theta_c)\).
\[ 1 + k_d\theta_c = 1 + (0.050 \times 5) = 1 + 0.25 = 1.25 \] \[ P_x = \frac{473.04}{1.25} = 378.432\ \text{kg VSS/day} \]

Step 5: State the final answer.
Rounding to two decimal places, the net waste activated sludge produced is \(378.43\) kg of VSS/day.