A feed mixture of thermal condition q = 1, consists of A (25 mol%), B (45 mol%), C (20 mol%) and D (10 mol%), respectively. The distillate contains 98 mol% B and 95 mol% C. The K-values are shown in Table 1.

Based on the K-values, the components are ordered by volatility: A > B > C > D.
If 98% of B and 95% of C go to the distillate, then B and C are primarily light components. A is even lighter. This means the separation occurs between C and D.

• Light Key (LK): C
• Heavy Key (HK): D

The heavy key is D, so $K_{HK} = K_D = 0.71$.
The relative volatility $\alpha_i = K_i / K_D$.

• $\alpha_A = K_A / K_D = 6.9 / 0.71 = 9.718$
• $\alpha_B = K_B / K_D = 3.2 / 0.71 = 4.507$
• $\alpha_C = K_C / K_D = 1.49 / 0.71 = 2.099$
• $\alpha_D = K_D / K_D = 0.71 / 0.71 = 1.000$

Basis: 100 mol of feed (F).
Given recoveries:

• Moles of B in distillate: $D x_{D,B} = 0.98 \times F z_B = 0.98 \times 100 \times 0.45 = 44.1$ mol
• Moles of C in distillate: $D x_{D,C} = 0.95 \times F z_C = 0.95 \times 100 \times 0.20 = 19.0$ mol

For non-key components:

• Component A is much lighter than LK (C). Assume all A goes to distillate. Moles of A in distillate: $D x_{D,A} = F z_A = 100 \times 0.25 = 25.0$ mol
• Component D is the HK. Its recovery in distillate is not specified. We need to determine it.

Total moles in distillate (D): $D = D x_{D,A} + D x_{D,B} + D x_{D,C} + D x_{D,D}$
We need to find $D x_{D,D}$.
Let's use the Fenske equation for the split of the heavy key (D) and light key (C) to determine the recovery of D in the distillate.
This is a circular dependency, as Fenske requires $N_{min}$, which requires the split.
A common approach for material balance in shortcut methods is to assume a small amount of HK in distillate and LK in bottom.
Let's assume a typical separation for the heavy key D. For example, 99% of D goes to the bottom.
So, recovery of D in bottom: $B x_{B,D} = 0.99 \times F z_D = 0.99 \times 100 \times 0.10 = 9.9$ mol.
Then moles of D in distillate: $D x_{D,D} = F z_D - B x_{B,D} = 10 - 9.9 = 0.1$ mol.

Now, sum the moles in the distillate:
$D = D x_{D,A} + D x_{D,B} + D x_{D,C} + D x_{D,D} = 25.0 + 44.1 + 19.0 + 0.1 = 88.2$ mol
Total moles in bottom: $B = F - D = 100 - 88.2 = 11.8$ mol

Now calculate mole fractions in distillate ($x_{D,i}$) and bottom ($x_{B,i}$):
Distillate (D = 88.2 mol):

• $x_{D,A} = 25.0 / 88.2 = 0.283$
• $x_{D,B} = 44.1 / 88.2 = 0.500$
• $x_{D,C} = 19.0 / 88.2 = 0.215$
• $x_{D,D} = 0.1 / 88.2 = 0.001$
• Sum $x_{D,i} = 0.283 + 0.500 + 0.215 + 0.001 = 0.999 \approx 1.000$

Bottom (B = 11.8 mol):

• Moles of A in bottom: $B x_{B,A} = F z_A - D x_{D,A} = 25.0 - 25.0 = 0.0$ mol $\implies x_{B,A} = 0.000$
• Moles of B in bottom: $B x_{B,B} = F z_B - D x_{D,B} = 45.0 - 44.1 = 0.9$ mol $\implies x_{B,B} = 0.9 / 11.8 = 0.076$
• Moles of C in bottom: $B x_{B,C} = F z_C - D x_{D,C} = 20.0 - 19.0 = 1.0$ mol $\implies x_{B,C} = 1.0 / 11.8 = 0.085$
• Moles of D in bottom: $B x_{B,D} = F z_D - D x_{D,D} = 10.0 - 0.1 = 9.9$ mol $\implies x_{B,D} = 9.9 / 11.8 = 0.839$
• Sum $x_{B,i} = 0.000 + 0.076 + 0.085 + 0.839 = 1.000$

Summary of Mole Fractions:

• Feed: $z_A = 0.250$, $z_B = 0.450$, $z_C = 0.200$, $z_D = 0.100$
• Distillate: $x_{D,A} = 0.283$, $x_{D,B} = 0.500$, $x_{D,C} = 0.215$, $x_{D,D} = 0.001$
• Bottom: $x_{B,A} = 0.000$, $x_{B,B} = 0.076$, $x_{B,C} = 0.085$, $x_{B,D} = 0.839$

For a saturated liquid feed ($q=1$), the Underwood equation for $\theta$ is:

$$\sum_{i=1}^{N_c} \frac{\alpha_i z_i}{\alpha_i - \theta} = 1-q = 0$$

Using the given $\theta = 1.5$:

$$\frac{\alpha_A z_A}{\alpha_A - \theta} + \frac{\alpha_B z_B}{\alpha_B - \theta} + \frac{\alpha_C z_C}{\alpha_C - \theta} + \frac{\alpha_D z_D}{\alpha_D - \theta}$$ $$= \frac{9.718 \times 0.25}{9.718 - 1.5} + \frac{4.507 \times 0.45}{4.507 - 1.5} + \frac{2.099 \times 0.20}{2.099 - 1.5} + \frac{1.000 \times 0.10}{1.000 - 1.5}$$ $$= \frac{2.4295}{8.218} + \frac{2.02815}{3.007} + \frac{0.4198}{0.599} + \frac{0.10}{-0.5}$$ $$= 0.2956 + 0.6744 + 0.7008 - 0.2000 = 1.4708$$

This sum is not zero, which means $\theta = 1.5$ is not the correct root for $q=1$. The problem states "trial and error value of $\theta$ is 1.5", which implies we should use this value. This is a contradiction. If $q=1$, the root $\theta$ must be between $\alpha_{HK}$ and $\alpha_{LK}$ (i.e., between $\alpha_D=1$ and $\alpha_C=2.099$). Since $\theta=1.5$ is within this range, we will proceed with it, assuming the problem implies this value should be used despite the sum not being zero.

The Underwood equation for minimum reflux ratio ($R_{min}$) is:

$$R_{min} + 1 = \sum_{i=1}^{N_c} \frac{\alpha_i x_{D,i}}{\alpha_i - \theta}$$ $$R_{min} + 1 = \frac{\alpha_A x_{D,A}}{\alpha_A - \theta} + \frac{\alpha_B x_{D,B}}{\alpha_B - \theta} + \frac{\alpha_C x_{D,C}}{\alpha_C - \theta} + \frac{\alpha_D x_{D,D}}{\alpha_D - \theta}$$ $$= \frac{9.718 \times 0.283}{9.718 - 1.5} + \frac{4.507 \times 0.500}{4.507 - 1.5} + \frac{2.099 \times 0.215}{2.099 - 1.5} + \frac{1.000 \times 0.001}{1.000 - 1.5}$$ $$= \frac{2.749}{8.218} + \frac{2.2535}{3.007} + \frac{0.4513}{0.599} + \frac{0.001}{-0.5}$$ $$= 0.3345 + 0.7494 + 0.7534 - 0.0020 = 1.8353$$ $$R_{min} = 1.8353 - 1 = 0.8353$$

Minimum reflux ratio ($R_{min}$): $\boxed{0.835}$

The Fenske equation for minimum stages ($N_{min}$) is:

$$N_{min} = \frac{\ln \left[ \left( \frac{x_{D,LK}}{x_{B,LK}} \right) \left( \frac{x_{B,HK}}{x_{D,HK}} \right) \right]}{\ln \alpha_{LK,HK}}$$

Here, LK = C, HK = D. So $\alpha_{LK,HK} = \alpha_C = 2.099$.

$$N_{min} = \frac{\ln \left[ \left( \frac{x_{D,C}}{x_{B,C}} \right) \left( \frac{x_{B,D}}{x_{D,D}} \right) \right]}{\ln \alpha_C}$$ $$N_{min} = \frac{\ln \left[ \left( \frac{0.215}{0.085} \right) \left( \frac{0.839}{0.001} \right) \right]}{\ln 2.099}$$ $$N_{min} = \frac{\ln \left[ (2.5294) \times (839) \right]}{\ln 2.099} = \frac{\ln (2122.2366)}{\ln 2.099}$$ $$N_{min} = \frac{7.660}{0.741} = 10.337$$

Minimum number of stages ($N_{min}$): $\boxed{10.337}$

Use Figure 1 below to solve the number of stages.
Actual reflux ratio $R = 1.3 \times R_{min} = 1.3 \times 0.8353 = 1.08589$.
Calculate the Gilliland parameter:

$$\frac{R-R_{min}}{R+1} = \frac{1.08589 - 0.8353}{1.08589 + 1} = \frac{0.25059}{2.08589} = 0.1201$$

From Figure 1 (Gilliland correlation), locate 0.1201 on the x-axis. Read the corresponding value on the y-axis, $\frac{N-N_{min}}{N+1}$.
Looking at the graph, for $x = 0.12$, the y-value is approximately 0.58.

$$\frac{N-N_{min}}{N+1} = 0.58$$

Substitute $N_{min} = 10.337$:

$$\frac{N - 10.337}{N+1} = 0.58$$ $$N - 10.337 = 0.58(N+1)$$ $$N - 10.337 = 0.58N + 0.58$$ $$N - 0.58N = 0.58 + 10.337$$ $$0.42N = 10.917$$ $$N = 10.917 / 0.42 = 25.993$$

Number of ideal stages (N): $\boxed{25.993}$

First, calculate the actual number of stages:
$N_{actual} = N / \text{efficiency} = 25.993 / 0.50 = 51.986$ stages.
Kirkbride's equation:

$$\log \left( \frac{N_p}{N_s} \right) = 0.206 \log \left[ \left( \frac{B}{D} \right) \left( \frac{x_{B,HK}}{x_{D,LK}} \right) \left( \frac{F_s}{F_p} \right)^2 \right]$$

Where:

• $N_p$: number of stages above the feed (rectifying section)
• $N_s$: number of stages below the feed (stripping section)
• $B = 11.8$ mol, $D = 88.2$

$$\log \left( \frac{N_p}{N_s} \right) = 0.206 \log \left[ \left( \frac{B}{D} \right) \left( \frac{x_{B,HK}}{x_{D,LK}} \right) \left( \frac{F_s}{F_p} \right)^2 \right]$$

Where:

• $B = 11.8$ mol (total moles in bottom)
• $D = 88.2$ mol (total moles in distillate)
• $x_{B,HK} = x_{B,D} = 0.839$ (mole fraction of heavy key D in bottom)
• $x_{D,LK} = x_{D,C} = 0.215$ (mole fraction of light key C in distillate)
• $F_s = F z_D = 100 \times 0.10 = 10$ mol (moles of heavy key D in feed)
• $F_p = F z_C = 100 \times 0.20 = 20$ mol (moles of light key C in feed)

Substitute the values into the equation:

$$\log \left( \frac{N_p}{N_s} \right) = 0.206 \log \left[ \left( \frac{11.8}{88.2} \right) \left( \frac{0.839}{0.215} \right) \left( \frac{10}{20} \right)^2 \right]$$ $$\log \left( \frac{N_p}{N_s} \right) = 0.206 \log \left[ (0.133787) \times (3.902326) \times (0.25) \right]$$ $$\log \left( \frac{N_p}{N_s} \right) = 0.206 \log \left[ 0.13059 \right]$$ $$\log \left( \frac{N_p}{N_s} \right) = 0.206 \times (-0.8841)$$ $$\log \left( \frac{N_p}{N_s} \right) = -0.1821$$ $$\frac{N_p}{N_s} = 10^{-0.1821} = 0.6575$$

The total number of ideal stages is $N = 25.993$. The number of ideal stages in the rectifying section ($N_p$) and stripping section ($N_s$) are related by $N_p + N_s = N - 1$ (excluding the feed stage itself).

$$N_p + N_s = 25.993 - 1 = 24.993$$

We have a system of two equations:

1. $N_p = 0.6575 N_s$
2. $N_p + N_s = 24.993$ Substitute (1) into (2):

$$0.6575 N_s + N_s = 24.993$$ $$1.6575 N_s = 24.993$$ $$N_s = \frac{24.993}{1.6575} = 15.079$$ $$N_p = 0.6575 \times 15.079 = 9.914$$

The feed tray is the $(N_p+1)^{th}$ ideal stage from the top. Number of ideal stages from top to feed tray (inclusive) = $N_p + 1 = 9.914 + 1 = 10.914$. Given a tray efficiency of 50%, the number of actual trays from the top to the feed tray is:

$$N_{feed,actual} = \frac{N_p + 1}{\text{efficiency}} = \frac{10.914}{0.50} = 21.828$$

Rounding to the nearest whole number, the feed tray is the 22nd actual tray from the top.

The location of the feed tray (actual trays): $\boxed{\text{22nd tray from the top}}$