11.2 McCabe-Thiele Method for Binary Distillation

11.2 McCabe-Thiele Method for Binary Distillation

The McCabe-Thiele method (developed in 1925 by Warren L. McCabe and Ernest W. Thiele) is the foundational graphical and analytical technique for evaluating binary distillation columns. While modern industrial simulations utilize rigorous matrix-based MESH (Mass, Equilibrium, Summation, Heat) solvers, the McCabe-Thiele framework remains the primary testing vehicle on the NCEES PE Chemical Exam because it visually and quantitatively links operating lines, thermodynamic equilibrium, reflux ratios, feed thermal conditions, and column tray requirements.

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1. Constant Molar Overflow (CMO) Assumptions

The mathematical elegance of the McCabe-Thiele method hinges on the Constant Molar Overflow (CMO) assumption. In an operating distillation tower, the molar flow rate of liquid ($L$) and vapor ($V$) in the rectifying section, and corresponding liquid ($\bar{L}$) and vapor ($\bar{V}$) in the stripping section, remain constant from tray to tray:

$L_1 = L_2 = \dots = L_n = L = \text{constant}$ $V_1 = V_2 = \dots = V_{n+1} = V = \text{constant}$ $\bar{L}_1 = \bar{L}_2 = \dots = \bar{L}_m = \bar{L} = \text{constant}$ $\bar{V}_1 = \bar{V}_2 = \dots = \bar{V}_{m+1} = \bar{V} = \text{constant}$

For CMO to hold rigorously, the system must satisfy four chemical engineering criteria:

1. Equal Molar Latent Heats: $\Delta H_{vap,1} \approx \Delta H_{vap,2}$. When one mole of heavy vapor condenses on a tray, the latent heat released is exactly sufficient to vaporize one mole of light liquid.
2. Negligible Heat of Mixing: $\Delta H_{mix} \approx 0$ across both liquid and vapor phases.
3. Negligible Sensible Heat Effects: Heat transferred to raise the temperature of incoming liquid from tray $n-1$ to tray $n$ is negligible compared to the latent heat of vaporization ($C_p \Delta T \ll \Delta H_{vap}$).
4. Adiabatic Operation: Heat losses through the column shell and insulation are zero ($Q_{loss} = 0$).

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2. Material Balances and Operating Line Equations

A continuous distillation column receives feed $F$ with light component mole fraction $z_F$, producing distillate $D$ at composition $x_D$ and bottoms $B$ at composition $x_B$.

                  [Condenser] <---- V_1 (y_1)
                       |
               +-------v-------+
               | Reflux Drum   |-----> Distillate D (x_D)
               +-------+-------+
                       |
                       v L_0 (Reflux, x_D)
             =====================
             |   Rectifying      |
             |   Section (L, V)  |
             =====================
             <--- Feed F (z_F, q)
             =====================
             |   Stripping       |
             | Section (L_bar,   |
             |          V_bar)   |
             =====================
                       |
               +-------v-------+
               | Reboiler      |-----> Bottoms B (x_B)
               +-------+-------+
                       |
                       +-----> V_bar (Boilup, y_B)

Overall Column Balances

• Total Mass Balance: $F = D + B$
• Component Light Balance: $F z_F = D x_D + B x_B$

Solving simultaneously yields product flow rates:

$D = F \left( \frac{z_F - x_B}{x_D - x_B} \right), \quad B = F \left( \frac{x_D - z_F}{x_D - x_B} \right)$

Rectifying Operating Line (ROL)

Taking a mass balance envelope around the condenser and tray $n$ in the rectifying (enriching) section:

$V y_{n+1} = L x_n + D x_D$ $y_{n+1} = \frac{L}{V} x_n + \frac{D}{V} x_D$

Define the Reflux Ratio ($R$) as the ratio of liquid returned to the column to distillate product withdrawn:

$R \equiv \frac{L}{D}$

From the condenser balance, $V = L + D = R D + D = (R + 1) D$. Substituting $L/V = R / (R + 1)$ and $D/V = 1 / (R + 1)$ yields the standard ROL Equation:

$y = \left( \frac{R}{R + 1} \right) x + \frac{x_D}{R + 1}$

• Slope: $\frac{L}{V} = \frac{R}{R + 1} < 1.0$
• y-Intercept: $\frac{x_D}{R + 1}$
• Intersection with $y = x$ line: Setting $y = x$ gives $x = \frac{R}{R+1} x + \frac{x_D}{R+1} \implies x = x_D$. The ROL always intersects the 45-degree diagonal at $(x_D, x_D)$.

Stripping Operating Line (SOL)

Taking a mass balance envelope around the bottom of the column and tray $m$ in the stripping section:

$\bar{L} x_m = \bar{V} y_{m+1} + B x_B$ $y_{m+1} = \frac{\bar{L}}{\bar{V}} x_m - \frac{B}{\bar{V}} x_B$

Define the Boilup Ratio ($V_B$) as the ratio of vapor returned to bottoms withdrawn: $V_B \equiv \bar{V} / B$. Since $\bar{L} = \bar{V} + B$, the slope is $\bar{L} / \bar{V} = (V_B + 1) / V_B > 1.0$. The standard SOL Equation is:

$y = \left( \frac{\bar{L}}{\bar{V}} \right) x - \frac{B}{\bar{V}} x_B$

• Slope: $\frac{\bar{L}}{\bar{V}} > 1.0$
• Intersection with $y = x$ line: Setting $y = x$ confirms that the SOL always intersects the 45-degree diagonal at $(x_B, x_B)$.

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3. Feed Thermal Quality ($q$) and the $q$-Line Equation

The transition between the rectifying section and stripping section occurs at the feed stage. The shift in liquid and vapor molar flow rates depends entirely on the thermal condition (quality $q$) of the feed.

Thermodynamic Definition of $q$

The quality factor $q$ is defined as the heat required to convert $1\text{ mole}$ of feed into saturated vapor, divided by the molar latent heat of vaporization ($\lambda$):

$q \equiv \frac{H_V - H_F}{H_V - H_L}$

Where:

• $H_V$ = molar enthalpy of feed at its dew point (saturated vapor).
• $H_L$ = molar enthalpy of feed at its bubble point (saturated liquid).
• $H_F$ = actual molar enthalpy of the feed.
• $\lambda = H_V - H_L$ = molar latent heat of vaporization.

Flow Rate Changes Across the Feed Stage

An enthalpy and mass balance over the feed plate demonstrates that:

$\bar{L} = L + q F$ $\bar{V} = V - (1 - q) F \iff V = \bar{V} + (1 - q) F$

Derivation of the $q$-Line Equation

Subtracting the SOL component balance from the ROL component balance and substituting the flow rate jump relations yields the locus of intersections between the ROL and SOL, termed the $q$-line:

$y = \left( \frac{q}{q - 1} \right) x - \frac{z_F}{q - 1}$

• Intersection with $y = x$: Setting $y = x$ yields $x(q - 1) = q x - z_F \implies x = z_F$. The $q$-line always originates at $(z_F, z_F)$ on the 45-degree diagonal.
• Slope of the $q$-line: $\text{Slope} = \frac{q}{q - 1}$.

The Five Feed Thermal Conditions

               y ^                     (q > 1) Subcooled Liquid
                 |                    ^  (Slope > 1)
             1.0 |                   /   
                 |                  /|   (q = 1) Saturated Liquid
                 |                 / |   (Slope = infinity)
                 |   (0 < q < 1)  /  |   
                 |   Part. Vapor /   |  
                 |   (Slope < 0)<    |   
                 |        |          |  
                 |        |          |   (q = 0) Saturated Vapor
                 |        v          |   (Slope = 0)
             z_F |-----------+-------+----------------->
                 |          / (z_F, z_F)
                 |         /         
                 |        / (q < 0) Superheated Vapor (0 < Slope < 1)
                 +-------+-----------+--------------------------> x
                 0      z_F         1.0

For subcooled liquid with specific heat $C_{p,L}$: $q = 1 + \frac{C_{p,L} (T_{bubble} - T_F)}{\lambda}$

For superheated vapor with specific heat $C_{p,V}$: $q = -\frac{C_{p,V} (T_F - T_{dew})}{\lambda}$

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4. Minimum Reflux ($R_{min}$), Total Reflux, and Optimum Reflux Ratio

Minimum Reflux Ratio ($R_{min}$)

As the reflux ratio $R$ is decreased, the ROL slope $R/(R+1)$ decreases, rotating the ROL downward toward the equilibrium curve. Eventually, the ROL, SOL, and $q$-line intersect at a single point $(x_p, y_p)$ lying directly on the VLE curve. This point is the pinch point.

At the pinch point, the vertical distance between the operating line and equilibrium line becomes zero ($y^* - y = 0$), requiring an infinite number of equilibrium stages ($N \to \infty$) to cross the pinch. The slope of the ROL at minimum reflux is:

$\left( \frac{R_{min}}{R_{min} + 1} \right) = \frac{x_D - y_p}{x_D - x_p}$

Solving for $R_{min}$:

$R_{min} = \frac{x_D - y_p}{y_p - x_p}$

For a saturated liquid feed ($q = 1.00$), $x_p = z_F$, and $y_p = \frac{\alpha z_F}{1 + (\alpha - 1) z_F}$. Under this specific condition, $R_{min}$ evaluates algebraically to:

$R_{min} = \frac{1}{\alpha - 1} \left[ \frac{x_D}{z_F} - \alpha \left( \frac{1 - x_D}{1 - z_F} \right) \right]$

Total Reflux ($R \to \infty$) and Minimum Stages ($N_{min}$)

If the distillate withdrawal rate is set to zero ($D = 0$), all overhead vapor is condensed and returned as reflux ($L = V$). Then:

$\frac{L}{V} = \frac{R}{R + 1} \to 1.0, \quad \frac{x_D}{R + 1} \to 0$

Both the ROL and SOL collapse directly onto the $y = x$ diagonal line. Under total reflux, the driving force between equilibrium and operating lines is maximized, yielding the absolute minimum number of equilibrium stages ($N_{min}$). For constant relative volatility, $N_{min}$ is calculated analytically via the Fenske Equation:

$N_{min} = \frac{\ln\left[ \left( \frac{x_D}{1 - x_D} \right) \left( \frac{1 - x_B}{x_B} \right) \right]}{\ln(\alpha_{avg})}$

Where $\alpha_{avg} = \sqrt{\alpha_{top} \cdot \alpha_{bottom}}$.

Optimum Operating Reflux Ratio

In industrial column design, an economic trade-off governs the selection of reflux ratio:

• Lower $R$: Reduces utility costs (condenser cooling water and reboiler steam), but increases required stages ($N$), necessitating a taller tower shell (higher capital expenditure).
• Higher $R$: Reduces required stages ($N$), but increases column diameter (higher vapor flow) and drastically escalates reboiler steam consumption.

The economic optimum reflux ratio typically falls within the range:

$R_{opt} = (1.10 \text{ to } 1.50) \cdot R_{min}$

Most industrial and PE exam designs specify $R = 1.20 \text{ to } 1.30 , R_{min}$.

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5. Stage Stepping, Condenser Types, and Reboiler Counting

Stages are stepped off on the McCabe-Thiele diagram between the equilibrium curve and the active operating line, starting from $(x_D, x_D)$ and descending to $(x_B, x_B)$.

   y ^              x_D (ROL)
 1.0 |             /|
     |            /-+ Stage 1
     |           /  |
     |          /---+ Stage 2
     |         /    |
     |        /-----+ Stage 3 (Crosses Feed -> Switch to SOL)
     |       /      |
     |      /-------+ Stage 4
     |     /        |
     |    /---------+ Stage 5 (Reboiler)
 x_B |---+----------+------------------------>
     0  x_B        x_D                      1.0 x

Condenser Classifications

• Total Condenser: Completely condenses overhead vapor into liquid. Liquid distillate has the exact same composition as the overhead vapor ($x_D = y_1$). A total condenser provides zero theoretical stages ($N_{cond} = 0$).
• Partial Condenser: Condenses only enough liquid to provide reflux; distillate is withdrawn as vapor ($y_D$). The vapor product is in equilibrium with the reflux liquid ($y_D = y_1^*$). A partial condenser acts as one theoretical equilibrium stage ($N_{cond} = 1$).

Reboiler Counting

Nearly all distillation columns utilize a partial reboiler (such as a kettle reboiler or thermosiphon reboiler), where liquid bottoms ($x_B$) is withdrawn in equilibrium with returning boilup vapor ($y_B$). Therefore, the reboiler always counts as one theoretical equilibrium stage ($N_{reb} = 1$).

To find the number of actual physical trays inside the tower:

$N_{trays} = \frac{N_{stages} - N_{reb} - N_{cond}}{E_o}$

For a column with a total condenser ($N_{cond} = 0$) and partial reboiler ($N_{reb} = 1$):

$N_{trays} = \frac{N_{stages} - 1}{E_o}$

Efficiencies: Murphree vs. Overall Tray Efficiency

• Murphree Vapor Tray Efficiency ($E_{MV}$): Measures the approach to equilibrium for a single tray $n$: $E_{MV} = \frac{y_n - y_{n+1}}{y_n^* - y_{n+1}}$ Where $y_{n+1}$ is vapor entering tray $n$, $y_n$ is actual vapor leaving tray $n$, and $y_n^*$ is theoretical vapor in equilibrium with liquid $x_n$.
• Overall Column Efficiency ($E_o$): The ratio of total theoretical stages to actual physical trays: $E_o = \frac{N_{ideal}}{N_{actual}}$ $E_o$ is frequently estimated on the PE exam using the O'Connell Correlation: $E_o \approx 0.492 \cdot (\mu_F \cdot \alpha_{LK,HK})^{-0.245}$ Where $\mu_F$ is feed liquid viscosity in centipoise ($\text{cP}$). Higher viscosity and higher relative volatility decrease tray efficiency due to reduced mass transfer rates.

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6. Comprehensive Worked Numerical Example: Binary Benzene-Toluene Column Sizing

Problem Statement

A continuous fractionating column separates $F = 100.0\text{ kmol/h}$ of a liquid mixture containing $40.0\text{ mol}%$ benzene (1) and $60.0\text{ mol}%$ toluene (2). Distillate product must contain $x_D = 0.950$ benzene, and bottoms must contain $x_B = 0.050$ benzene. The feed is introduced as a saturated liquid ($q = 1.00$). The average relative volatility is constant at $\alpha = 2.50$.

The column is equipped with a total condenser and a partial kettle reboiler. The operating reflux ratio is set to $R = 1.30 , R_{min}$. The overall column tray efficiency is $E_o = 0.650$.

Calculate:

1. Product flow rates $D$ and $B$ in $\text{kmol/h}$.
2. The minimum reflux ratio $R_{min}$.
3. The operating reflux ratio $R$, and the ROL and SOL equations.
4. The minimum number of theoretical stages $N_{min}$ at total reflux.
5. Step off stages manually to find total theoretical stages ($N_{stages}$).
6. The number of actual physical trays required in the column shell ($N_{trays}$).

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Step 1: Overall Material Balances

$F = D + B \implies 100.0 = D + B$ $F z_F = D x_D + B x_B \implies 100.0(0.400) = 0.950 D + 0.050 B$ $40.0 = 0.950 D + 0.050(100.0 - D) = 0.950 D + 5.0 - 0.050 D = 0.900 D + 5.0$ $0.900 D = 35.0 \implies D = \mathbf{38.89\text{ kmol/h}}$ $B = 100.0 - 38.89 = \mathbf{61.11\text{ kmol/h}}$

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Step 2: Minimum Reflux Ratio ($R_{min}$)

Because the feed is saturated liquid ($q = 1.00$), the $q$-line is vertical at $x = z_F = 0.400$. The pinch point occurs at $x_p = 0.400$. The equilibrium vapor composition at the pinch point is:

$y_p = \frac{\alpha x_p}{1 + (\alpha - 1) x_p} = \frac{2.50 \times 0.400}{1 + (1.50 \times 0.400)} = \frac{1.000}{1.600} = 0.625$

Pinch coordinates: $(x_p, y_p) = (0.400, 0.625)$.

Slope of ROL at minimum reflux: $\frac{R_{min}}{R_{min} + 1} = \frac{x_D - y_p}{x_D - x_p} = \frac{0.950 - 0.625}{0.950 - 0.400} = \frac{0.325}{0.550} = 0.59091$

$R_{min} = \frac{0.59091}{1 - 0.59091} = \frac{0.325}{0.550 - 0.325} = \frac{0.325}{0.225} = \mathbf{1.444}$

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Step 3: Operating Lines Formulation

Operating reflux ratio: $R = 1.30 \times R_{min} = 1.30 \times 1.4444 = \mathbf{1.878}$

Rectifying Operating Line (ROL): $\text{Slope} = \frac{R}{R + 1} = \frac{1.878}{2.878} = 0.6525$ $\text{Intercept} = \frac{x_D}{R + 1} = \frac{0.950}{2.878} = 0.3301$ $\mathbf{y = 0.6525 x + 0.3301}$

Stripping Operating Line (SOL): Since $q = 1.00$: $\bar{L} = L + F = (R \cdot D) + F = (1.878 \times 38.89) + 100.0 = 73.04 + 100.0 = 173.04\text{ kmol/h}$ $\bar{V} = V = (R + 1) D = 2.878 \times 38.89 = 111.93\text{ kmol/h}$ $\text{Slope} = \frac{\bar{L}}{\bar{V}} = \frac{173.04}{111.93} = 1.5460$ $\text{Intercept} = -\frac{B x_B}{\bar{V}} = -\frac{61.11 \times 0.050}{111.93} = -0.0273$ $\mathbf{y = 1.5460 x - 0.0273}$

Check intersection at $x = 0.400$:

• ROL: $y = (0.6525 \times 0.400) + 0.3301 = 0.5911$
• SOL: $y = (1.5460 \times 0.400) - 0.0273 = 0.5911$ Identical values confirm mathematical alignment at the $q$-line.

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Step 4: Minimum Theoretical Stages ($N_{min}$) via Fenske Equation

$N_{min} = \frac{\ln\left[ \left( \frac{x_D}{1 - x_D} \right) \left( \frac{1 - x_B}{x_B} \right) \right]}{\ln(\alpha)} = \frac{\ln\left[ \left( \frac{0.950}{0.050} \right) \left( \frac{0.950}{0.050} \right) \right]}{\ln(2.50)} = \frac{\ln(19.0 \times 19.0)}{\ln(2.50)} = \frac{\ln(361.0)}{\ln(2.50)} = \frac{5.8889}{0.91629} = \mathbf{6.43\text{ stages}}$

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Step 5: Manual Step-by-Step Stage Calculation

For an ideal binary system, liquid in equilibrium with vapor $y$ is: $x = \frac{y}{\alpha - (\alpha - 1) y} = \frac{y}{2.50 - 1.50 y}$

• Stage 1 (Top Tray): Overhead vapor from Tray 1 is condensed in a total condenser: $y_1 = x_D = 0.9500$. $x_1 = \frac{0.9500}{2.50 - 1.50(0.9500)} = \frac{0.9500}{1.0750} = 0.8837$ From ROL: $y_2 = 0.6525 x_1 + 0.3301 = 0.6525(0.8837) + 0.3301 = 0.9067$
• Stage 2: $x_2 = \frac{0.9067}{2.50 - 1.50(0.9067)} = \frac{0.9067}{1.1400} = 0.7954$ From ROL: $y_3 = 0.6525(0.7954) + 0.3301 = 0.8491$
• Stage 3: $x_3 = \frac{0.8491}{2.50 - 1.50(0.8491)} = \frac{0.8491}{1.2264} = 0.6924$ From ROL: $y_4 = 0.6525(0.6924) + 0.3301 = 0.7819$
• Stage 4: $x_4 = \frac{0.7819}{2.50 - 1.50(0.7819)} = \frac{0.7819}{1.3272} = 0.5891$ From ROL: $y_5 = 0.6525(0.5891) + 0.3301 = 0.7145$
• Stage 5: $x_5 = \frac{0.7145}{2.50 - 1.50(0.7145)} = \frac{0.7145}{1.4283} = 0.5002$ From ROL: $y_6 = 0.6525(0.5002) + 0.3301 = 0.6565$
• Stage 6: $x_6 = \frac{0.6565}{2.50 - 1.50(0.6565)} = \frac{0.6565}{1.5153} = 0.4333$ From ROL: $y_7 = 0.6525(0.4333) + 0.3301 = 0.6128$
• Stage 7 (Feed Stage): $x_7 = \frac{0.6128}{2.50 - 1.50(0.6128)} = \frac{0.6128}{1.5808} = 0.3876$ Notice $x_7 = 0.3876 < z_F = 0.400$. We cross the feed stage and switch to the SOL! From SOL: $y_8 = 1.5460(0.3876) - 0.0273 = 0.5720$
• Stage 8: $x_8 = \frac{0.5720}{2.50 - 1.50(0.5720)} = \frac{0.5720}{1.6420} = 0.3484$ From SOL: $y_9 = 1.5460(0.3484) - 0.0273 = 0.5113$
• Stage 9: $x_9 = \frac{0.5113}{2.50 - 1.50(0.5113)} = \frac{0.5113}{1.7331} = 0.2950$ From SOL: $y_{10} = 1.5460(0.2950) - 0.0273 = 0.4288$
• Stage 10: $x_{10} = \frac{0.4288}{2.50 - 1.50(0.4288)} = \frac{0.4288}{1.8568} = 0.2309$ From SOL: $y_{11} = 1.5460(0.2309) - 0.0273 = 0.3297$
• Stage 11: $x_{11} = \frac{0.3297}{2.50 - 1.50(0.3297)} = \frac{0.3297}{2.0055} = 0.1644$ From SOL: $y_{12} = 1.5460(0.1644) - 0.0273 = 0.2269$
• Stage 12: $x_{12} = \frac{0.2269}{2.50 - 1.50(0.2269)} = \frac{0.2269}{2.1597} = 0.1051$ From SOL: $y_{13} = 1.5460(0.1051) - 0.0273 = 0.1352$
• Stage 13: $x_{13} = \frac{0.1352}{2.50 - 1.50(0.1352)} = \frac{0.1352}{2.2972} = 0.0588$ From SOL: $y_{14} = 1.5460(0.0588) - 0.0273 = 0.0636$
• Stage 14: $x_{14} = \frac{0.0636}{2.50 - 1.50(0.0636)} = \frac{0.0636}{2.4046} = 0.0264 < x_B = 0.0500$

Interpolating between Stage 13 ($x = 0.0588$) and Stage 14 ($x = 0.0264$): $\text{Fraction} = \frac{0.0588 - 0.0500}{0.0588 - 0.0264} = \frac{0.0088}{0.0324} = 0.27$ Total theoretical stages: $N_{stages} = 13 + 0.27 = \mathbf{13.3\text{ theoretical stages}}$.

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Step 6: Physical Trays Required

The partial kettle reboiler provides $1$ theoretical equilibrium stage. The total condenser provides $0$ stages. $N_{ideal,trays} = N_{stages} - 1.0 = 13.3 - 1.0 = 12.3\text{ stages}$ $N_{trays} = \frac{N_{ideal,trays}}{E_o} = \frac{12.3}{0.650} = 18.92 \to \mathbf{19\text{ actual physical trays}}$

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7. Critical PE Exam Traps & Pitfalls

Trap 1: Forgetting to Subtract the Partial Reboiler from Stage Count
When a problem asks "How many trays must be installed in the column shell?" and provides theoretical stages stepped off on a McCabe-Thiele diagram, you MUST subtract 1 stage for the partial reboiler before applying tray efficiency. The reboiler is an external vessel or bottom pool that generates an equilibrium stage independently of tray hydraulics.

Trap 2: Misinterpreting $q$-Line Slopes for Two-Phase Feeds
For a feed that is $40%$ vapor, the vapor fraction is $f = 0.40$, which means liquid fraction is $q = 1 - f = 0.60$. The slope of the $q$-line is $q / (q - 1) = 0.60 / (0.60 - 1.00) = 0.60 / (-0.40) = -1.50$. A frequent exam trap is confusing $q$ with the fraction vaporized $f$, resulting in inverted slopes and incorrect pinch points.

Trap 3: Counting a Total Condenser as a Theoretical Stage
A total condenser simply converts vapor $y_1$ into liquid of the exact same composition ($x_D = y_1$). No phase equilibrium separation occurs. It provides ZERO stages ($N = 0$). Only a partial condenser provides an equilibrium stage ($N = 1$).