11.4 Multicomponent Distillation Fundamentals and Key Components

11.4 Multicomponent Distillation Fundamentals and Key Components

In chemical refineries, gas processing plants, and petrochemical complexes, distillation columns rarely process simple binary mixtures. Instead, columns fractionate multicomponent mixtures containing dozens of chemical species (e.g., crude atmospheric towers, deethanizers, depropanizers, and naphtha splitters). On the NCEES PE Chemical Exam, engineers are tested on shortcut design methods—specifically the Fenske-Underwood-Gilliland (FUG) sequence and the Kirkbride equation—to determine stage counts, reflux requirements, and feed tray locations without performing iterative plate-to-plate MESH calculations.

---

1. Key and Non-Key Component Classification

In multicomponent distillation, the chemical engineer cannot specify the complete recovery of every component independently; degrees of freedom permit specifying the separation split of only two components, designated as the key components:

   Volatility Order (alpha_i relative to Heavy Key):

   alpha_1 > alpha_2  >  alpha_LK  >  alpha_dist  >  alpha_HK  >  alpha_N
   [--------------]     [-------]     [--------]     [-------]     [-------]
      Light Non-          Light       Distributed      Heavy         Heavy
      Keys (LNK)        Key (LK)       Component      Key (HK)     Non-Keys (HNK)
          |                 |              |             |              |
          v                 v              v             v              v
   All to Distillate     Specified     Distributes   Specified    All to Bottoms
     (d_i = f_i)           Split        Top/Bottom     Split       (b_i = f_i)

Definitions

1. Light Key (LK): The more volatile of the two key components. A small, specified fraction is allowed in the bottoms product ($b_{LK}$ is specified and small; most enters distillate $d_{LK}$).
2. Heavy Key (HK): The less volatile of the two key components. A small, specified fraction is allowed in the distillate product ($d_{HK}$ is specified and small; most enters bottoms $b_{HK}$).
3. Light Non-Key (LNK): Components more volatile than the light key ($\alpha_i > \alpha_{LK}$). Because they are lighter than the LK, they distill overhead almost completely ($d_i \approx f_i, b_i \approx 0$).
4. Heavy Non-Key (HNK): Components less volatile than the heavy key ($\alpha_i < \alpha_{HK}$). Because they are heavier than the HK, they discharge in the bottoms almost completely ($b_i \approx f_i, d_i \approx 0$).
5. Distributed Components: Intermediate components with volatilities between the light key and heavy key ($\alpha_{HK} < \alpha_i < \alpha_{LK}$). These components split significantly between both distillate and bottoms products.

Material Balance Accounting

For a feed containing molar component flow rates $f_i = F z_{F,i}$:

• Distillate molar flows: $d_i = D x_{D,i}$
• Bottoms molar flows: $b_i = B x_{B,i}$
• Component conservation: $f_i = d_i + b_i$

---

2. Minimum Theoretical Stages: The Multicomponent Fenske Equation

At total reflux ($R \to \infty$), the minimum number of theoretical stages ($N_{min}$) required to separate the light key from the heavy key is calculated from the Fenske Equation:

$N_{min} = \frac{\ln\left[ \left( \frac{d_{LK}}{b_{LK}} \right) \left( \frac{b_{HK}}{d_{HK}} \right) \right]}{\ln(\alpha_{LK,HK,avg})}$

Where:

• $d_{LK}, b_{LK}$ = molar flow rates of light key in distillate and bottoms.
• $d_{HK}, b_{HK}$ = molar flow rates of heavy key in distillate and bottoms.
• $\alpha_{LK,HK,avg}$ = geometric mean relative volatility of LK relative to HK across column terminal conditions:

$\alpha_{LK,HK,avg} = \sqrt{\alpha_{top} \cdot \alpha_{bottom}}$

Component Distribution at Total Reflux

Once $N_{min}$ is determined, the Fenske equation allows exact calculation of the split for every non-key or distributed component $i$ at total reflux:

$\frac{d_i}{b_i} = \left( \frac{d_{HK}}{b_{HK}} \right) \left( \alpha_{i,HK} \right)^{N_{min}}$

Since $f_i = d_i + b_i = b_i (d_i / b_i) + b_i = b_i [1 + (d_i / b_i)]$, the individual flow rates are:

$b_i = \frac{f_i}{1 + (d_i / b_i)}, \quad d_i = f_i - b_i$

---

3. Minimum Reflux Ratio: The Underwood Equations

Unlike binary systems where $R_{min}$ is found graphically from a single pinch point, multicomponent systems experience two pinches (one above and one below the feed stage). In 1948, A. J. V. Underwood derived an analytical solution assuming constant molar overflow and constant relative volatilities.

Step 1: The Underwood Feed Equation (Finding the Active Root $\theta$)

The feed equation determines the characteristic root $\theta$:

$\sum_{i=1}^C \frac{\alpha_i z_{F,i}}{\alpha_i - \theta} = 1 - q$

Where:

• $\alpha_i$ = relative volatility of component $i$ relative to the heavy key (so $\alpha_{HK} = 1.0$).
• $z_{F,i}$ = mole fraction of component $i$ in the total feed.
• $q$ = feed thermal quality factor ($q = 1.0$ for saturated liquid, so $1 - q = 0$).

[!IMPORTANT] Selection of the Active Underwood Root ($\theta$):
The polynomial equation has $C$ roots. To calculate $R_{min}$ for key components that are adjacent in volatility, the active root $\theta$ must be selected strictly between the relative volatilities of the heavy key and light key: $\alpha_{HK} < \theta < \alpha_{LK} \iff 1.0 < \theta < \alpha_{LK}$

Step 2: The Underwood Distillate Equation (Finding $V_{min}$ and $R_{min}$)

Once the active root $\theta$ is calculated, it is substituted into the Underwood distillate equation to determine the minimum overhead vapor rate ($V_{min}$):

$V_{min} = \sum_{i=1}^C \frac{\alpha_i d_i}{\alpha_i - \theta}$

Where $d_i$ are the molar flow rates of each component in the distillate ($d_i \approx 0$ for heavy non-keys). The minimum reflux ratio ($R_{min}$) is then computed from:

$R_{min} = \frac{L_{min}}{D} = \frac{V_{min} - D}{D} = \frac{V_{min}}{D} - 1$

---

4. Stage-Reflux Trade-Off: The Gilliland Correlation

With $N_{min}$ and $R_{min}$ established, the chemical engineer specifies an operating reflux ratio ($R = 1.15\text{--}1.40 R_{min}$). The required number of theoretical stages ($N$) is determined from the empirical Gilliland Correlation.

   Y = (N - N_min) / (N + 1) ^
                         1.0 |*
                             |  *  Gilliland Correlation
                             |    * 
                         0.5 |      *.
                             |        *.
                             |           *--..
                         0.0 +-----------------*----+---> X = (R - R_min) / (R + 1)
                             0.0              0.5  1.0

Dimensionless Gilliland Parameters

• Reflux parameter: $X = \frac{R - R_{min}}{R + 1}$
• Stage parameter: $Y = \frac{N - N_{min}}{N + 1}$

Analytical Formulations (Molokanov and Eduljee Equations)

Because reading graphs on the computer-based PE exam introduces interpolation error, use the standard analytical approximation developed by H. E. Eduljee (1975):

$Y = \frac{N - N_{min}}{N + 1} = 0.75 \cdot \left[ 1 - X^{0.5668} \right]$

Rearranging to solve explicitly for actual theoretical stages $N$:

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

---

5. Optimal Feed Stage Location: The Kirkbride Equation

Introducing feed at the wrong tray forces operating lines away from equilibrium, wasting stages. The empirical Kirkbride Equation determines the optimal ratio of theoretical stages in the rectifying section ($N_R$, above feed) to stripping section ($N_S$, below feed):

$\log_{10}\left( \frac{N_R}{N_S} \right) = 0.206 \cdot \log_{10}\left[ \left( \frac{z_{HK}}{z_{LK}} \right) \left( \frac{x_{B,LK}}{x_{D,HK}} \right)^2 \left( \frac{B}{D} \right) \right]$

Where:

• $N_R$ = number of theoretical stages above the feed stage.
• $N_S$ = number of theoretical stages below the feed stage (including reboiler).
• $z_{LK}, z_{HK}$ = feed mole fractions of light key and heavy key.
• $x_{B,LK}$ = mole fraction of light key in bottoms.
• $x_{D,HK}$ = mole fraction of heavy key in distillate.
• $B, D$ = total bottoms and distillate molar flow rates.

Since total stages equal $N = N_R + N_S$, the individual section stage counts are resolved immediately:

$N_S = \frac{N}{1 + (N_R / N_S)}, \quad N_R = N - N_S$

The feed enters at stage $N_R + 1$ counting down from the top.

---

6. Summary Comparison: Shortcut Methods in Multicomponent Distillation

---

7. Comprehensive Worked Numerical Example: Deethanizer Column Design

Problem Statement

A petroleum refinery designs a deethanizer fractionator to separate $F = 100.0\text{ kmol/h}$ of a hydrocarbon feed entering as a saturated liquid ($q = 1.00$).

Feed Composition & Relative Volatilities (Referenced to n-Butane, $\alpha_3 = 1.00$):

1. Ethane (LNK): $z_1 = 0.100$ ($f_1 = 10.0\text{ kmol/h}$), $\alpha_1 = 5.00$
2. Propane (LK): $z_2 = 0.400$ ($f_2 = 40.0\text{ kmol/h}$), $\alpha_2 = 2.20$
3. n-Butane (HK): $z_3 = 0.350$ ($f_3 = 35.0\text{ kmol/h}$), $\alpha_3 = 1.00$
4. n-Pentane (HNK): $z_4 = 0.150$ ($f_4 = 15.0\text{ kmol/h}$), $\alpha_4 = 0.40$

Separation Targets:

• Recover $98.0%$ of Propane (LK) in the distillate ($d_2 = 0.980 \times 40.0 = 39.20\text{ kmol/h}$, $b_2 = 0.80\text{ kmol/h}$).
• Recover $98.0%$ of n-Butane (HK) in the bottoms ($b_3 = 0.980 \times 35.0 = 34.30\text{ kmol/h}$, $d_3 = 0.70\text{ kmol/h}$).
• Light Non-Key (Ethane) distributes entirely overhead: $d_1 = 10.00\text{ kmol/h}$, $b_1 = 0.00\text{ kmol/h}$.
• Heavy Non-Key (n-Pentane) discharges entirely in bottoms: $b_4 = 15.00\text{ kmol/h}$, $d_4 = 0.00\text{ kmol/h}$.

Calculate:

1. Total distillate ($D$) and bottoms ($B$) flow rates.
2. Minimum theoretical stages ($N_{min}$) via the Fenske equation.
3. The exact active root $\theta$ from the Underwood feed equation.
4. The minimum vapor rate ($V_{min}$) and minimum reflux ratio ($R_{min}$).
5. For an operating reflux ratio $R = 1.30 , R_{min}$, find the actual theoretical stages ($N$) using the Eduljee-Gilliland correlation.
6. The optimal feed stage location using the Kirkbride equation.

---

Step 1: Total Product Flow Rates

$D = d_1 + d_2 + d_3 + d_4 = 10.00 + 39.20 + 0.70 + 0.00 = \mathbf{49.90\text{ kmol/h}}$ $B = b_1 + b_2 + b_3 + b_4 = 0.00 + 0.80 + 34.30 + 15.00 = \mathbf{50.10\text{ kmol/h}}$ Check: $D + B = 49.90 + 50.10 = 100.00\text{ kmol/h} = F$.

Distillate mole fractions: $x_{D,1} = 10.00 / 49.90 = 0.2004, \quad x_{D,2} = 39.20 / 49.90 = 0.7856$ $x_{D,3} = 0.70 / 49.90 = 0.01403, \quad x_{D,4} = 0.00$

Bottoms mole fractions: $x_{B,1} = 0.00, \quad x_{B,2} = 0.80 / 50.10 = 0.01597$ $x_{B,3} = 34.30 / 50.10 = 0.6846, \quad x_{B,4} = 15.00 / 50.10 = 0.2994$

---

Step 2: Minimum Stages ($N_{min}$) via Fenske Equation

$\alpha_{LK,HK} = \frac{\alpha_2}{\alpha_3} = \frac{2.20}{1.00} = 2.20$ $\frac{d_{LK}}{b_{LK}} = \frac{39.20}{0.80} = 49.00, \quad \frac{b_{HK}}{d_{HK}} = \frac{34.30}{0.70} = 49.00$

$N_{min} = \frac{\ln(49.00 \times 49.00)}{\ln(2.20)} = \frac{\ln(2,401.0)}{\ln(2.20)} = \frac{7.7836}{0.78846} = \mathbf{9.872\text{ stages}}$

---

Step 3: Underwood Active Root ($\theta$)

Since feed is saturated liquid ($q = 1.00$), $1 - q = 0.0$. The feed equation is: $\sum_{i=1}^4 \frac{\alpha_i z_{F,i}}{\alpha_i - \theta} = \frac{5.00(0.100)}{5.00 - \theta} + \frac{2.20(0.400)}{2.20 - \theta} + \frac{1.00(0.350)}{1.00 - \theta} + \frac{0.40(0.150)}{0.40 - \theta} = 0$ $\frac{0.500}{5.00 - \theta} + \frac{0.880}{2.20 - \theta} + \frac{0.350}{1.00 - \theta} + \frac{0.060}{0.40 - \theta} = 0$

The root must lie between $\alpha_{HK} = 1.00$ and $\alpha_{LK} = 2.20$. Test $\theta = 1.325$:

• Term 1: $0.500 / (5.00 - 1.325) = 0.500 / 3.675 = +0.13605$
• Term 2: $0.880 / (2.20 - 1.325) = 0.880 / 0.875 = +1.00571$
• Term 3: $0.350 / (1.00 - 1.325) = 0.350 / (-0.325) = -1.07692$
• Term 4: $0.060 / (0.40 - 1.325) = 0.060 / (-0.925) = -0.06486$

$\text{Sum} = 0.13605 + 1.00571 - 1.07692 - 0.06486 = \mathbf{0.0000}$ The exact active root is $\mathbf{\theta = 1.3250}$.

---

Step 4: Minimum Reflux Ratio ($R_{min}$)

Substitute $\theta = 1.3250$ into the distillate equation ($d_4 = 0$): $V_{min} = \frac{\alpha_1 d_1}{\alpha_1 - \theta} + \frac{\alpha_2 d_2}{\alpha_2 - \theta} + \frac{\alpha_3 d_3}{\alpha_3 - \theta}$ $V_{min} = \frac{5.00(10.00)}{5.00 - 1.325} + \frac{2.20(39.20)}{2.20 - 1.325} + \frac{1.00(0.70)}{1.00 - 1.325}$ $V_{min} = \frac{50.00}{3.675} + \frac{86.24}{0.875} - \frac{0.70}{0.325} = 13.605 + 98.560 - 2.154 = \mathbf{110.011\text{ kmol/h}}$

Minimum reflux ratio: $R_{min} = \frac{V_{min} - D}{D} = \frac{110.011 - 49.90}{49.90} = \frac{60.111}{49.90} = \mathbf{1.2046}$

---

Step 5: Actual Theoretical Stages via Gilliland Correlation

Operating reflux ratio: $R = 1.30 \times R_{min} = 1.30 \times 1.2046 = \mathbf{1.5660}$

Dimensionless reflux parameter: $X = \frac{R - R_{min}}{R + 1} = \frac{1.5660 - 1.2046}{1.5660 + 1.00} = \frac{0.3614}{2.5660} = \mathbf{0.1408}$

Using Eduljee's equation: $Y = 0.75 \left[ 1 - X^{0.5668} \right] = 0.75 \left[ 1 - (0.1408)^{0.5668} \right]$ $(0.1408)^{0.5668} = 0.3299 \implies Y = 0.75 \times (1 - 0.3299) = 0.75 \times 0.6701 = \mathbf{0.5026}$

Calculate actual stages $N$: $Y = \frac{N - N_{min}}{N + 1} \implies N = \frac{N_{min} + Y}{1 - Y} = \frac{9.872 + 0.5026}{1 - 0.5026} = \frac{10.3746}{0.4974} = \mathbf{20.86 \approx 21\text{ theoretical stages}}$

---

Step 6: Feed Stage Location via Kirkbride Equation

$\log_{10}\left( \frac{N_R}{N_S} \right) = 0.206 \cdot \log_{10}\left[ \left( \frac{z_{HK}}{z_{LK}} \right) \left( \frac{x_{B,LK}}{x_{D,HK}} \right)^2 \left( \frac{B}{D} \right) \right]$

Input values:

• $z_{HK} / z_{LK} = 0.350 / 0.400 = 0.875$
• $x_{B,LK} / x_{D,HK} = 0.01597 / 0.01403 = 1.1383$
• $(x_{B,LK} / x_{D,HK})^2 = (1.1383)^2 = 1.2957$
• $B / D = 50.10 / 49.90 = 1.0040$

Term inside brackets: $\text{Bracket} = 0.875 \times 1.2957 \times 1.0040 = 1.1383$ $\log_{10}(1.1383) = 0.05626$ $\log_{10}\left( \frac{N_R}{N_S} \right) = 0.206 \times 0.05626 = 0.01159$ $\frac{N_R}{N_S} = 10^{0.01159} = \mathbf{1.0271}$

With $N_R + N_S = 21$: $1.0271 N_S + N_S = 21 \implies 2.0271 N_S = 21 \implies N_S = 10.36 \approx \mathbf{10\text{ stages (stripping)}}$ $N_R = 21 - 10 = \mathbf{11\text{ stages (rectifying)}}$

The feed should enter at Stage 11 or 12 from the top.

---

7. Critical PE Exam Traps & Pitfalls

Trap 1: Selecting the Wrong Underwood Root ($\theta$)
The Underwood feed equation has multiple roots. A critical exam error is picking a root outside the interval between the heavy key and light key ($\alpha_{HK} < \theta < \alpha_{LK}$). Selecting $\theta < \alpha_{HK}$ or $\theta > \alpha_{LK}$ produces nonsensical or negative minimum reflux ratios.

Trap 2: Misidentifying Distributed Components
Do not assume all non-keys go $100%$ to top or bottom! If a feed component has a relative volatility intermediate between the light key and heavy key, it is a distributed key and splits between both products. Its distribution must be checked using Fenske's equation before evaluating Underwood's distillate equation.

Trap 3: Inverting the Ratios in the Kirkbride Equation
Notice carefully: Kirkbride's equation uses $(z_{HK} / z_{LK})$—heavy key over light key—multiplied by $(x_{B,LK} / x_{D,HK})^2$—bottoms light key over distillate heavy key. Inverting either ratio flips $N_R / N_S$, placing the feed tray in the completely wrong half of the column!