Chemical Reactor Hot-Spot Optimization (Chemical Engineering) Solution

Continuous temperature monitoring in exothermic reactors is critical for safety. The temperature profile across reaction zones is modeled by piecewise functions, which must be continuous and differentiable to avoid physical discontinuities in heat transfer models.

An exothermic chemical reactor has a temperature profile along its central axis modeled by the piecewise function:

where $x$ is the distance in meters from the inlet, and $T(x)$ is the temperature in degrees Celsius.

Determine the following:

1. Find the values of $a$ and $b$ such that the temperature profile is continuous and differentiable at the boundary $x = 3$ meters.
2. Optimize the reactor design by finding the maximum temperature along the axis in the interval $0 \le x \le 5$.

Step-by-Step Solution:

1. Find $a$ and $b$ for Continuity and Differentiability:

   • Continuity Condition: The left-hand limit must equal the right-hand limit at $x = 3$:

     $$\lim_{x \to 3^-} (-x^2 + 6x + 80) = \lim_{x \to 3^+} (ax + b)$$
     $$-3^2 + 6(3) + 80 = 3a + b$$
     $$-9 + 18 + 80 = 3a + b \implies 89 = 3a + b \quad \text{--- (Equation 1)}$$

   • Differentiability Condition: The derivative from the left must equal the derivative from the right at $x = 3$:

     $$T'(x) = \begin{cases} -2x + 6 & : 0 \le x < 3 \\ a & : x > 3 \end{cases}$$
     $$\lim_{x \to 3^-} (-2x + 6) = \lim_{x \to 3^+} (a)$$
     $$-2(3) + 6 = a \implies a = 0$$

   • Solve for $b$: Substitute $a = 0$ into Equation 1:

     $$89 = 3(0) + b \implies b = 89$$

     So, the parameters are $a = 0$ and $b = 89$.

2. Find the Maximum Temperature on $0 \le x \le 5$: With $a = 0$ and $b = 89$:

   $$T(x) = \begin{cases} -x^2 + 6x + 80 & : 0 \le x < 3 \\ 89 & : x \ge 3 \end{cases}$$

   To find the maximum temperature:

   • For $0 \le x < 3$, find critical points of $T(x) = -x^2 + 6x + 80$:
     $$T'(x) = -2x + 6 = 0 \implies x = 3$$
     Since $x=3$ is the boundary, the function increases on $[0, 3)$ because $T'(x) > 0$ for $x < 3$.
   • At $x = 0$, $T(0) = 80^\circ\text{C}$.
   • For $x \ge 3$, the temperature is constant at $T(x) = 89^\circ\text{C}$.

   Therefore, the maximum temperature along the axis is $89^\circ\text{C}$, reached at $x \ge 3$.

Meaning of the Answer & Real-Life Application:

Continuity and differentiability of thermal profiles prevent heat flux discontinuities in modeling. A smooth temperature transition ensures that the physical catalyst beds do not experience localized thermal shock, which could lead to reactor runaway.