Chemical Bioreactor Dissolved Oxygen Depletion (Chemical & Environmental Engineering) Solution

Problem Overview

We evaluate the dynamic dissolved oxygen profile $C(t) = \frac{8t}{t^2 + 4}$ ($\text{mg/L}$) for $t \ge 0$. We compute $\frac{dC}{dt}$, construct a sign chart, apply the First Derivative Test and Single Critical Point Rule to find the global peak oxygen level, and assess aeration turbine controls.

Step-by-Step Mathematical Solution

Step 1: Differentiate via the Quotient Rule

The governing formulation is the Quotient Rule:

1. Define numerator and denominator:
   $$u(t) = 8t \implies u'(t) = 8$$
   $$v(t) = t^2 + 4 \implies v'(t) = 2t$$
2. Substitute into the Quotient Rule:
   $$C'(t) = \frac{8(t^2 + 4) - 8t(2t)}{(t^2 + 4)^2} = \frac{8t^2 + 32 - 16t^2}{(t^2 + 4)^2} = \frac{32 - 8t^2}{(t^2 + 4)^2}$$
3. Factor the numerator:
   $$C'(t) = \frac{8(4 - t^2)}{(t^2 + 4)^2} = \frac{8(2 - t)(2 + t)}{(t^2 + 4)^2}$$

Step 2: Locate and Filter Critical Numbers

The governing concept is Critical Numbers:

1. Stationary Points ($C'(t) = 0$): Set the factored numerator to zero:
   $$8(2 - t)(2 + t) = 0 \implies t = 2 \quad \text{or} \quad t = -2$$
2. Domain Filter: The physical operational domain is $t \in [0, \infty)$. Therefore, $t = -2$ is discarded.
3. Singular Points ($C'(t) \text{ DNE}$): The denominator is $(t^2 + 4)^2$. Because $t^2 + 4 \ge 4 > 0$ for all real $t$, the denominator is never zero. There are no singular points.
4. Valid Critical Number: The function has exactly one critical number in its domain:
   $$t = 2\text{ hours}$$

Step 3: Construct Sign Chart & Apply First Derivative Test

Evaluate the sign of $C'(t) = \frac{8(2 - t)(2 + t)}{(t^2 + 4)^2}$ on $[0, \infty)$ around $t = 2$:

• Note that the denominator $(t^2 + 4)^2 > 0$ and the factor $(2 + t) > 0$ for all $t > 0$. The sign is controlled strictly by $(2 - t)$.
• Interval $(0, 2)$: Test $t = 1 \implies C'(1) = \frac{8(1)(3)}{(1 + 4)^2} = \frac{24}{25} > 0 \implies (+)$. Concentration is increasing ($\nearrow$).
• Interval $(2, \infty)$: Test $t = 3 \implies C'(3) = \frac{8(-1)(5)}{(9 + 4)^2} = -\frac{40}{169} < 0 \implies (-)$. Concentration is decreasing ($\searrow$).

Extremum Classification:

• Because $C'(t)$ transitions from positive to negative ($+ \to -$) at $t = 2$, $C(2)$ is a local maximum.
• By the Single Critical Point Rule on $[0, \infty)$, because $t = 2$ is the only interior critical number and $C(0) = 0$, $C(2)$ is guaranteed to be the absolute maximum over the entire domain:
  $$C(2) = \frac{8(2)}{(2)^2 + 4} = \frac{16}{4 + 4} = \frac{16}{8} = 2.0\text{ mg/L}$$
• Absolute Maximum Value: $2.0\text{ mg/L}$ at location $t = 2\text{ hours}$.

Step 4: Asymptotic Behavior & Operational Control

1. Long-Term Limit: Evaluate the limit as $t \to \infty$:
   $$\lim_{t \to \infty} C(t) = \lim_{t \to \infty} \frac{8t}{t^2 + 4} = \lim_{t \to \infty} \frac{8/t}{1 + 4/t^2} = \frac{0}{1} = 0\text{ mg/L}$$
2. Process Engineering Control Decision: After the initial aeration surge reaches peak saturation ($2.0\text{ mg/L}$) at $t = 2\text{ hours}$, active biological consumption drives the dissolved oxygen steadily downward toward zero. Environmental protection regulations require a minimum DO concentration of $C_{\min} \ge 1.0\text{ mg/L}$ to prevent microbial die-off. Setting $C(t) = 1.0 \implies \frac{8t}{t^2 + 4} = 1 \implies t^2 - 8t + 4 = 0 \implies t = 4 \pm 2\sqrt{3} \approx 7.46\text{ hours}$. Thus, the supervisory SCADA control system must automatically restart auxiliary air compressors before $t \approx 7.46\text{ hours}$.