Chemical Reactor Thermal Runaway Safeguards (Chemical Engineering) Solution

1. Initial Reactor Temperature ($y$-Intercept)

Evaluate $T(t)$ at $t = 0$:

2. Horizontal Asymptote & Adiabatic Ceiling

1. Name the Governing Formula: The Exponential Decay Limit Theorem:
   $$\lim_{t \to \infty} e^{-kt} = 0 \quad (\text{for } k > 0)$$
   (Formalized in Horizontal Asymptotes.)
2. Evaluate the Limit as $t \to \infty$:
   $$\lim_{t \to \infty} T(t) = \lim_{t \to \infty} \left[50 + \frac{120}{1 + 15e^{-2t}}\right] = 50 + \frac{120}{1 + 15(0)} = 50 + 120 = 170^\circ\text{C}$$
3. Physical Engineering Meaning: The horizontal asymptote is $T = 170^\circ\text{C}$. This represents the theoretical adiabatic ceiling temperature of the batch reaction. Once all chemical monomer reagents are completely consumed, exothermic enthalpy generation ceases, and the temperature stabilizes at $170^\circ\text{C}$.

3. Heating Rate via First Derivative

1. Name the Governing Formula: The Chain Rule and Reciprocal Differentiation Rule:
   $$\frac{d}{dt}[(u(t))^{-1}] = -[u(t)]^{-2} \cdot u'(t)$$
2. Differentiate $T(t)$: Rewrite $T(t) = 50 + 120(1 + 15e^{-2t})^{-1}$.
   $$\frac{dT}{dt} = 0 + 120(-1)(1 + 15e^{-2t})^{-2} \cdot \frac{d}{dt}[1 + 15e^{-2t}]$$
   $$\frac{d}{dt}[1 + 15e^{-2t}] = 15(-2)e^{-2t} = -30e^{-2t}$$
   Substitute:
   $$\frac{dT}{dt} = -120(1 + 15e^{-2t})^{-2} (-30e^{-2t}) = \frac{3600e^{-2t}}{(1 + 15e^{-2t})^2}$$
3. Sign Verification: Because $e^{-2t} > 0$ and $(1 + 15e^{-2t})^2 > 0$ for all real $t$, $\frac{dT}{dt} > 0$ strictly holds on $[0, \infty)$. The temperature is strictly monotonic increasing.

4. Inflection Point via Second Derivative

1. Name the Governing Formula: The Second Derivative Zero-Curvature Criterion:
   $$\frac{d^2T}{dt^2} = 0 \quad \text{with concavity sign flip}$$
   (Formalized in Concavity and the Second Derivative Test.)
2. Compute $\frac{d^2T}{dt^2}$: Let $u(t) = 3600e^{-2t} \implies u'(t) = -7200e^{-2t}$.
   Let $v(t) = (1 + 15e^{-2t})^2 \implies v'(t) = 2(1 + 15e^{-2t})(-30e^{-2t}) = -60e^{-2t}(1 + 15e^{-2t})$. Using the Quotient Rule:
   $$\frac{d^2T}{dt^2} = \frac{(-7200e^{-2t})(1 + 15e^{-2t})^2 - (3600e^{-2t})[-60e^{-2t}(1 + 15e^{-2t})]}{(1 + 15e^{-2t})^4}$$
   Factor out the common term $3600e^{-2t}(1 + 15e^{-2t})$ from the numerator:
   $$\text{Num} = 3600e^{-2t}(1 + 15e^{-2t}) \Big[ -2(1 + 15e^{-2t}) + 60e^{-2t} \Big]$$
   Simplify inside the brackets:
   $$-2 - 30e^{-2t} + 60e^{-2t} = 30e^{-2t} - 2$$
   Thus:
   $$\frac{d^2T}{dt^2} = \frac{3600e^{-2t}(30e^{-2t} - 2)}{(1 + 15e^{-2t})^3}$$
3. Solve for Zero Acceleration: Set $\frac{d^2T}{dt^2} = 0 \implies 30e^{-2t} - 2 = 0$:
   $$30e^{-2t} = 2 \implies e^{-2t} = \frac{2}{30} = \frac{1}{15}$$
   Take the natural logarithm:
   $$-2t = \ln\left(\frac{1}{15}\right) = -\ln(15) \implies t_{\text{crit}} = \frac{\ln(15)}{2} \approx \frac{2.708}{2} \approx 1.354\text{ hours}$$
4. Concavity Sign Transition:
   • For $t < t_{\text{crit}}$: $e^{-2t} > \frac{1}{15} \implies 30e^{-2t} - 2 > 0 \implies \frac{d^2T}{dt^2} > 0$ (Concave Upward, accelerating heat growth).
   • For $t > t_{\text{crit}}$: $e^{-2t} < \frac{1}{15} \implies 30e^{-2t} - 2 < 0 \implies \frac{d^2T}{dt^2} < 0$ (Concave Downward, decelerating heat growth). Because the concavity flips from positive to negative, an inflection point occurs at $t_{\text{crit}} \approx 1.354\text{ hours}$.
   • Temperature at inflection:
     $$T(t_{\text{crit}}) = 50 + \frac{120}{1 + 15(1/15)} = 50 + \frac{120}{2} = 110^\circ\text{C}$$

5. Chemical Safety & Coolant Capacity Sizing

The second derivative represents the thermal acceleration ($\frac{d}{dt}\left[\frac{dT}{dt}\right]$). Because the concavity changes from concave up to concave down at $t_{\text{crit}} = 1.354\text{ h}$, the first derivative $\frac{dT}{dt}$ reaches its absolute maximum value at this exact instant: