Chemical Reactor Continuous Yield Optimization (Chemical & Environmental Engineering) Solution

An environmental chemical process engineer at the UNLV Howard R. Hughes College of Engineering is designing a continuous fixed-bed catalytic reactor column to remove toxic volatile organic compounds (VOCs) from industrial flue gas emissions.

The active catalyst bed is contained within a 3D spatial prism $E$ with dimensions in meters ($\text{m}$):

with volumetric reaction rate density $R(x, y, z) = \frac{e^{x^2}}{z}\text{ mol}/(\text{m}^3\cdot\text{hr})$.

Step-by-Step Solution:

1. Analytical Obstacle in the Given Order:
   In the given order:

   $$Y = \int_{1}^{4} \int_{0}^{1} \int_{2y}^{2} \frac{e^{x^2}}{z} \, dx \, dy \, dz$$
   the innermost integral requires evaluating $\int e^{x^2} \, dx$. As established in calculus, the function $e^{x^2}$ does not possess an elementary antiderivative (its antiderivative is expressible only via the non-elementary error function $\text{erfi}(x)$). Closed-form analytical evaluation in the order $dx \, dy \, dz$ is therefore impossible.

2. Re-slice the Planar Shadow Domain:
   State the Re-slicing Protocol in the Projection Plane (defined in Changing the Order of Integration in Triple Integrals):

   • The outer variable $z$ is independent of $x$ and $y$, remaining bounded on the interval:
     $$1 \le z \le 4$$
   • In the $xy$-plane, the region is bounded by the line $x = 2y$, the vertical line $x = 2$, and the lower boundary $y = 0$.
   • Invert the boundary function by solving for $y$ in terms of $x$:
     $$x = 2y \implies y = \frac{x}{2}$$
   • When $y = 0$, $x = 0$; when $y = 1$, $x = 2$.
   • Thus, for a vertical slice in the new order ($dy \, dx$), $y$ enters at $y = 0$ and exits at $y = \frac{x}{2}$, while $x$ spans the constant interval:
     $$0 \le y \le \frac{x}{2}, \quad 0 \le x \le 2$$

3. Formulate the Reordered Triple Integral:
   Reorder the integration differentials to $dy \, dx \, dz$:

   $$Y = \int_{1}^{4} \int_{0}^{2} \int_{0}^{x/2} \frac{e^{x^2}}{z} \, dy \, dx \, dz$$

4. Evaluate the Reordered Triple Integral:

   Evaluate inner $y$-integral:
   Treat $x$ and $z$ as constants:

   $$\int_{0}^{x/2} \frac{e^{x^2}}{z} \, dy = \left[ \frac{e^{x^2}}{z} y \right]_{0}^{x/2} = \frac{e^{x^2}}{z} \left(\frac{x}{2}\right) = \frac{x e^{x^2}}{2z}$$
   Notice that integrating with respect to $y$ has produced the algebraic factor $x$, which immediately unlocks $u$-substitution for the next layer.

   Evaluate middle $x$-integral:
   

   $$\int_{0}^{2} \frac{x e^{x^2}}{2z} \, dx = \frac{1}{2z} \int_{0}^{2} x e^{x^2} \, dx$$
   Apply $u$-substitution: let $u = x^2 \implies du = 2x \, dx \implies x \, dx = \frac{du}{2}$.
   When $x = 0, u = 0$; when $x = 2, u = 2^2 = 4$.
   
   $$\frac{1}{2z} \int_{0}^{4} \frac{1}{2} e^u \, du = \frac{1}{4z} [e^u]_{0}^{4} = \frac{e^4 - 1}{4z}$$

   Evaluate outer $z$-integral:
   

   $$Y = \int_{1}^{4} \frac{e^4 - 1}{4z} \, dz = \frac{e^4 - 1}{4} \int_{1}^{4} \frac{1}{z} \, dz = \frac{e^4 - 1}{4} \Big[ \ln |z| \Big]_{1}^{4}$$
   $$Y = \frac{e^4 - 1}{4} (\ln 4 - \ln 1) = \frac{(e^4 - 1)\ln 4}{4} = \frac{(e^4 - 1)\ln 2}{2}\text{ mol/hr}$$

   Using numerical approximations ($e^4 \approx 54.598, \ln 4 \approx 1.3863$):

   $$Y \approx \frac{(53.598)(1.3863)}{4} \approx 18.57\text{ mol/hr}$$

Meaning of the Answer & Real-Life Application:

• Quantitative Interpretation: The total chemical conversion capacity of the fixed-bed catalytic reactor is exactly $\frac{(e^4 - 1)\ln 4}{4}\text{ mol/hr} \approx 18.57\text{ mol/hr}$.
• Engineering Application: In chemical plant and environmental process design, analytical solutions allow engineers to perform rapid parametric sensitivity analysis—such as varying bed length $L$, width $W$, and column height $H$—to determine the optimal reactor size required to achieve $99.9\%$ pollutant destruction efficiency without solving expensive 3D Navier-Stokes numerical grids for every geometric iteration.