Understanding the Biological Reactor Problem

This problem asks us to calculate the effluent BOD5 concentration from a completely mixed biological reactor. We are given the reactor's tank volume, the flow rate, the incoming substrate concentration (BOD5), and kinetic parameters like half-saturation constant ($K_s$), decay coefficient ($k_d$), and maximum specific growth rate ($\mu_{max}$).

Calculating Key Reactor Parameters

First, we need to determine the Hydraulic Retention Time ($\theta$) for the reactor. This represents the average time water spends in the reactor.

The formula for hydraulic retention time is:

$$ \theta = \frac{V}{Q} $$

Where:

• $V$ = Tank Volume = 200 m³
• $Q$ = Flow Rate = 50 m³/day

Plugging in the values:

$$ \theta = \frac{200 \text{ m}^3}{50 \text{ m}^3/\text{day}} = 4 \text{ days} $$

The dilution rate ($D$) is the reciprocal of the hydraulic retention time:

$$ D = \frac{1}{\theta} = \frac{1}{4 \text{ days}} = 0.25 \text{ day}^{-1} $$

Applying Monod Kinetics for Effluent Calculation

For a completely mixed biological reactor operating at steady state, the effluent substrate concentration ($S$) can be related to the kinetic parameters using a modified Monod equation that includes cell decay ($k_d$). The relationship derived from the mass balance and kinetics is:

$$ k_d + D = \frac{\mu_{max} S}{K_s + S} $$

Where:

• $k_d$ = Decay coefficient = 0.06 day⁻¹
• $D$ = Dilution rate = 0.25 day⁻¹
• $\mu_{max}$ = Maximum specific growth rate = 3 day⁻¹
• $K_s$ = Half-saturation constant = 60 mg BOD5/L
• $S$ = Effluent substrate concentration (BOD5) in mg/L (what we need to find)

Step-by-Step Calculation of Effluent BOD5

Now, we substitute the known values into the equation:

$$ 0.06 \text{ day}^{-1} + 0.25 \text{ day}^{-1} = \frac{3 \text{ day}^{-1} \times S}{60 \text{ mg/L} + S} $$

Combine the terms on the left side:

$$ 0.31 \text{ day}^{-1} = \frac{3 S}{60 + S} $$

To solve for $S$, we rearrange the equation:

$$ 0.31 (60 + S) = 3 S $$

Distribute 0.31:

$$ 18.6 + 0.31 S = 3 S $$

Isolate the $S$ terms:

$$ 18.6 = 3 S - 0.31 S $$

$$ 18.6 = 2.69 S $$

Solve for $S$:

$$ S = \frac{18.6}{2.69} $$

$$ S \approx 6.9145 \text{ mg/L} $$

Final Result Comparison

The calculated effluent BOD5 concentration is approximately 6.92 mg/L. Comparing this value to the given options, it matches the first option.