Question

Consider the following figure of an activated sludge process (ASP), depicting the flow (Q), substrate (S), and microorganism concentration (X) at various points in the system, where subscripts “0”, “r”, “w”, “e”, and “i” indicate influent, recycle line, wastage, effluent, and flow from aeration tank to settling tank, respectively. Note that influent to the ASP has microbes too. V is volume of aeration tank. Choose the correct option for net rate of formation of microorganisms in the system at steady state, from the following

• A. $\frac{(Q-Q_w)X_e + Q_wX_r - QX_0}{V}$
• B. $\frac{(Q-Q_w)X_e + Q_wX_r}{V}$
• C. $\frac{(Q-Q_w)X_e + Q_wX_r}{VX_e}$
• D. $\frac{(Q+Q_r)X_i + Q_wX_r}{VX}$

Hint:

For mass balance problems at steady state: - Rate of accumulation = 0 - Rate in - Rate out + Rate of reaction = 0 In this case, the "Rate of reaction" corresponds to the net formation of microorganisms. Net rate of formation = Rate out - Rate in

Answer 1

The Correct Option is A

Step 1: Define the net rate of formation of microorganisms at steady state.
At steady state, the rate of accumulation of microorganisms in the aeration tank is zero. The net rate of formation is the difference between the rate at which microorganisms leave the system and the rate at which they enter the system.

Step 2: Perform a mass balance of microorganisms around the aeration tank.
The rate of microorganisms entering the aeration tank is the sum of the microorganisms in the influent flow: Rate in = \( QX_0 \) (Note: The question states that the influent to the ASP has microbes too, with concentration \( X_0 \))

The rate of microorganisms leaving the system is through the effluent and the wastage flow. Rate out (effluent) = \( (Q-Q_w)X_e \)
Rate out (wastage) = \( Q_wX_r \)
Total rate out = \( (Q-Q_w)X_e + Q_wX_r \)

Step 3: Calculate the net rate of formation.
The net rate of formation of microorganisms in the system is the difference between the mass flow rate of microorganisms leaving the system and the mass flow rate of microorganisms entering the system. Net rate of formation = Total rate out - Rate in
Net rate of formation = \( (Q-Q_w)X_e + Q_wX_r - QX_0 \)

Step 4: Express the net rate of formation per unit volume.
The question asks for the net rate of formation of microorganisms in the system, and the options are given per unit volume of the aeration tank (\( V \)). Therefore, we divide the net rate of formation by the volume \( V \): Net rate of formation per unit volume = \( \frac{(Q-Q_w)X_e + Q_wX_r - QX_0}{V} \)

This expression matches option (A).