Rayleigh Equation for Batch Distillation

The Rayleigh equation is used to calculate the composition and amount of liquid remaining in a batch distillation process. It is expressed as:

$ \ln \left( \frac{W}{F} \right) = \int_{x_F}^{x_W} \frac{dx_A}{y_A^* - x_A} $

Where:

• $F$ is the initial moles of the feed.
• $W$ is the final moles of the residue.
• $x_F$ is the mole fraction of component A in the feed.
• $x_W$ is the mole fraction of component A in the residue.
• $y_A^*$ is the mole fraction of component A in the vapor in equilibrium with the liquid of composition $x_A$.

Batch Distillation Parameters

The problem provides the following information:

• Initial feed moles, $F = 100 \text{ moles}$.
• Feed composition, $x_{A,F} = 0.28$.
• Final residue composition, $x_{A,W} = 0.2$.
• Linear vapor-liquid equilibrium (VLE) approximation: $y_A^* = 1.325x_A + 0.121$.

First, we determine the term $(y_A^* - x_A)$ from the given VLE equation:

$ y_A^* - x_A = (1.325x_A + 0.121) - x_A = 0.325x_A + 0.121 $

Integral Calculation

Substitute this expression into the Rayleigh equation integral:

$ \ln \left( \frac{W}{F} \right) = \int_{0.28}^{0.2} \frac{dx_A}{0.325x_A + 0.121} $

To solve this integral, we use the substitution $u = 0.325x_A + 0.121$. Differentiating gives $du = 0.325 dx_A$, which implies $dx_A = \frac{du}{0.325}$.

The integral transforms to:

$ \int \frac{1}{u} \frac{du}{0.325} = \frac{1}{0.325} \int \frac{du}{u} = \frac{1}{0.325} \ln|u| $

Substituting back for $u$, the antiderivative is $\frac{1}{0.325} \ln(0.325x_A + 0.121)$. Since $0.325x_A + 0.121$ is positive for the given range of $x_A$, the absolute value is not needed.

Now, evaluate the definite integral using the limits $x_F = 0.28$ and $x_W = 0.2$:

$ \left[ \frac{1}{0.325} \ln(0.325x_A + 0.121) \right]_{0.28}^{0.2} $

$ = \frac{1}{0.325} \left[ \ln(0.325 \times 0.2 + 0.121) - \ln(0.325 \times 0.28 + 0.121) \right] $

$ = \frac{1}{0.325} \left[ \ln(0.065 + 0.121) - \ln(0.091 + 0.121) \right] $

$ = \frac{1}{0.325} \left[ \ln(0.186) - \ln(0.212) \right] $

$ = \frac{1}{0.325} \ln \left( \frac{0.186}{0.212} \right) $

$ \approx \frac{1}{0.325} \ln(0.877358) \approx \frac{1}{0.325} (-0.13083) \approx -0.40255 $

Calculate Final Residue Moles

The result of the integral gives $\ln \left( \frac{W}{F} \right) \approx -0.40255$.

To find the ratio $W/F$, we take the exponential of this value:

$ \frac{W}{F} = e^{-0.40255} \approx 0.66853 $

Given the initial feed $F = 100$ moles, we can now calculate the final residue moles $W$:

$ W = F \times 0.66853 = 100 \times 0.66853 \approx 66.853 \text{ moles} $

Rounding the result to two decimal places, the number of moles of residue left behind is $66.85$.