Please answer all questions and round final answer to 3 decimal places. Show calculations only and steps

Total Feed (F) = 500 kmol/hr
$F_{hexane} = 0.48 \times 500 = 240 \text{ kmol/hr}$
$F_{heptane} = 0.36 \times 500 = 180 \text{ kmol/hr}$
$F_{octane} = 0.16 \times 500 = 80 \text{ kmol/hr}$

Light Key (LK) = n-hexane, Heavy Key (HK) = n-octane.
Recovery of LK in distillate = 97%
$D_{hexane} = 0.97 \times F_{hexane} = 0.97 \times 240 = 232.8 \text{ kmol/hr}$
Recovery of HK in bottoms = 92%
$B_{octane} = 0.92 \times F_{octane} = 0.92 \times 80 = 73.6 \text{ kmol/hr}$

$B_{hexane} = F_{hexane} - D_{hexane} = 240 - 232.8 = 7.2 \text{ kmol/hr}$
$D_{octane} = F_{octane} - B_{octane} = 80 - 73.6 = 6.4 \text{ kmol/hr}$

The problem does not specify the recovery of n-heptane. For the purpose of completing the material balance and proceeding with Underwood's and Fenske's equations, we must make an assumption. A common approach is to assume that the intermediate component is split such that the total distillate and bottoms flow rates are consistent with the overall mass balance. However, this does not uniquely determine the split.
Given the lack of specific information, we will assume that n-heptane is split such that $D_{heptane} = 90 \text{ kmol/hr}$ and $B_{heptane} = 90 \text{ kmol/hr}$ (a 50/50 split), as this is a common simplifying assumption when no other data is provided for the intermediate component.

$D = D_{hexane} + D_{heptane} + D_{octane} = 232.8 + 90 + 6.4 = 329.2 \text{ kmol/hr}$
$B = B_{hexane} + B_{heptane} + B_{octane} = 7.2 + 90 + 73.6 = 170.8 \text{ kmol/hr}$
Check: $D + B = 329.2 + 170.8 = 500 \text{ kmol/hr}$ (Matches F)

Table 3.1: Material Balance and Mole Fractions

The equation is $\sum_{i=1}^{N_c} \frac{\alpha_i z_i}{\alpha_i - \theta} = 1 - q$.
Since $q=1$, $1-q=0$.
$\frac{\alpha_{hexane} z_{hexane}}{\alpha_{hexane} - \theta} + \frac{\alpha_{heptane} z_{heptane}}{\alpha_{heptane} - \theta} + \frac{\alpha_{octane} z_{octane}}{\alpha_{octane} - \theta} = 0$
$\frac{2.208 \times 0.48}{2.208 - \theta} + \frac{1 \times 0.36}{1 - \theta} + \frac{0.457 \times 0.16}{0.457 - \theta} = 0$
$\frac{1.05984}{2.208 - \theta} + \frac{0.36}{1 - \theta} + \frac{0.07312}{0.457 - \theta} = 0$

For $\theta = 1.3$: $1.16722 - 1.2 - 0.08674 = -0.11952$
For $\theta = 1.4$: $1.31168 - 0.9 - 0.07754 = 0.33414$
The root $\theta$ is between 1.3 and 1.4. Using linear interpolation:
$\theta = 1.3 + (1.4 - 1.3) \times \frac{0 - (-0.11952)}{0.33414 - (-0.11952)} = 1.3 + 0.1 \times \frac{0.11952}{0.45366} = 1.3 + 0.026345 = 1.326345$
We use $\theta = 1.326$.

$R_{min} + 1 = \sum_{i=1}^{N_c} \frac{\alpha_i x_{i,D}}{\alpha_i - \theta}$
$R_{min} + 1 = \frac{2.208 \times 0.707}{2.208 - 1.326} + \frac{1 \times 0.273}{1 - 1.326} + \frac{0.457 \times 0.019}{0.457 - 1.326}$
$R_{min} + 1 = \frac{1.560}{0.882} + \frac{0.273}{-0.326} + \frac{0.008683}{-0.869}$
$R_{min} + 1 = 1.7687 - 0.8374 - 0.00999$
$R_{min} + 1 = 0.92131$
$R_{min} = 0.92131 - 1 = -0.07869$

A negative $R_{min}$ indicates an issue with the chosen $\theta$ or the problem setup. The root $\theta$ must be between the relative volatilities of the key components. Here, $\alpha_{LK} = 2.208$ and $\alpha_{HK} = 0.457$. The intermediate component $\alpha_{heptane} = 1$.
For a multi-component system, there are $N_c-1$ roots for $\theta$. One root should be between $\alpha_{LK}$ and $\alpha_{IK}$ (n-hexane and n-heptane), and another between $\alpha_{IK}$ and $\alpha_{HK}$ (n-heptane and n-octane).
The given $\theta$ values (1.3, 1.4, 1.5) are between $\alpha_{hexane}$ (2.208) and $\alpha_{heptane}$ (1). This $\theta$ is used for the separation between the light key (n-hexane) and the intermediate component (n-heptane).
However, for $R_{min}$, we typically use the $\theta$ that lies between the relative volatilities of the designated light key and heavy key.
In this case, the designated LK is n-hexane ($\alpha=2.208$) and HK is n-octane ($\alpha=0.457$). The intermediate component is n-heptane ($\alpha=1$).
The root $\theta$ should be between $\alpha_{LK}$ and $\alpha_{HK}$.
The roots of the Underwood equation are between the $\alpha_i$ values.
The roots are $\theta_1 \in (1, 2.208)$ and $\theta_2 \in (0.457, 1)$.
The $\theta$ value we found (1.326) is between 1 and 2.208. This is the correct $\theta$ for the separation between n-hexane and n-heptane.
For $R_{min}$, we use the $\theta$ that lies between the relative volatilities of the key components.
The problem states n-hexane is LK and n-octane is HK.
The $\theta$ value used in the $R_{min}$ equation should be the one that lies between $\alpha_{LK}$ and $\alpha_{HK}$.
However, the given values for $\theta$ (1.3, 1.4, 1.5) are all greater than $\alpha_{heptane}=1$. This means we are looking for the root between $\alpha_{hexane}$ and $\alpha_{heptane}$.
If we use this $\theta$ for $R_{min}$, it implies that n-heptane is the heavy key for the top section.
Let's re-evaluate the $R_{min}$ calculation.
$R_{min} + 1 = \frac{2.208 \times 0.707}{2.208 - 1.326} + \frac{1 \times 0.273}{1 - 1.326} + \frac{0.457 \times 0.019}{0.457 - 1.326}$
$R_{min} + 1 = \frac{1.560}{0.882} + \frac{0.273}{-0.326} + \frac{0.008683}{-0.869}$
$R_{min} + 1 = 1.768707 - 0.837423 - 0.009992 = 0.921292$
$R_{min} = 0.921292 - 1 = -0.078708$

A negative $R_{min}$ is physically impossible. This indicates that the chosen $\theta$ is not the correct one for the $R_{min}$ calculation, or there's an issue with the problem statement or the assumed split of n-heptane.
In Underwood's method, the $\theta$ used for $R_{min}$ must be between the relative volatilities of the key components.
If n-hexane is LK and n-octane is HK, then the $\theta$ should be between 2.208 and 0.457.
The roots of the equation $\sum_{i=1}^{N_c} \frac{\alpha_i z_i}{\alpha_i - \theta} = 0$ are $\theta_1 \in (1, 2.208)$ and $\theta_2 \in (0.457, 1)$.
The $\theta$ we found (1.326) is $\theta_1$. This $\theta$ is used when the split is between n-hexane and n-heptane.
If the split is between n-heptane and n-octane, we would need $\theta_2$.
However, the problem defines LK as n-hexane and HK as n-octane.
Let's assume the problem expects us to use the $\theta$ that lies between the relative volatilities of the designated key components.
If the problem implies that the intermediate component (n-heptane) is not a key component, then the $\theta$ for $R_{min}$ should be between $\alpha_{hexane}$ and $\alpha_{octane}$.
This means we need to find the root of the Underwood equation that lies between 2.208 and 0.457.
The $\theta = 1.326$ is between 1 and 2.208.
Let's re-evaluate the function $f(\theta)$ for $\theta$ values between 0.457 and 1.
$f(0.5) = \frac{1.05984}{2.208 - 0.5} + \frac{0.36}{1 - 0.5} + \frac{0.07312}{0.457 - 0.5} = \frac{1.05984}{1.708} + \frac{0.36}{0.5} + \frac{0.07312}{-0.043} = 0.61934 + 0.72 - 1.6999 = -0.36056$
$f(0.6) = \frac{1.05984}{2.208 - 0.6} + \frac{0.36}{1 - 0.6} + \frac{0.07312}{0.457 - 0.6} = \frac{1.05984}{1.608} + \frac{0.36}{0.4} + \frac{0.07312}{-0.143} = 0.65910 + 0.9 - 0.51133 = 1.04777$
The second root is between 0.5 and 0.6.
Let's use linear interpolation for the second root:
$\theta_2 = 0.5 + (0.6 - 0.5) \times \frac{0 - (-0.36056)}{1.04777 - (-0.36056)} = 0.5 + 0.1 \times \frac{0.36056}{1.40833} = 0.5 + 0.1 \times 0.2560 = 0.5256$
Let's use $\theta = 0.526$. This $\theta$ is between $\alpha_{heptane}$ and $\alpha_{octane}$.

The problem states "estimated values of $\theta$ are 1.3, 1.4 and 1.5". This strongly suggests that the problem expects us to use the $\theta$ value that falls within this range, which is $\theta_1 = 1.326$.
If we use $\theta = 1.326$, and we get a negative $R_{min}$, it implies that the separation is not possible with this $\theta$ or that the problem is ill-posed for this specific $\theta$.
However, $R_{min}$ must be positive.
Let's re-check the formula for $R_{min}$.
$R_{min} = \frac{1}{\alpha_{LK} - \theta} \sum_{i=1}^{N_c} \frac{\alpha_i z_i}{\alpha_i - \theta} - 1$ is not correct.
The correct form for $R_{min}$ is:
$R_{min} = \frac{1}{x_{LK,D}} \sum_{i=1}^{N_c} \frac{\alpha_i x_{i,D}}{\alpha_i - \theta} - 1$ (This is also not correct)

Let's use the standard form:
$R_{min} = \frac{1}{D} \sum_{i=1}^{N_c} \frac{\alpha_i F z_i}{\alpha_i - \theta} - 1$ (This is for $q=0$)
For $q=1$, the equation for $\theta$ is $\sum_{i=1}^{N_c} \frac{\alpha_i z_i}{\alpha_i - \theta} = 0$.
Then, $R_{min} = \frac{1}{x_{