Number of Transfer Units for a Packed Distillation Tower: 23-Chem-A2 Unit Operations and Separation Processes, Undated paper, Q4

Question B1: Number of Transfer Units for a Packed Distillation Tower (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Value
Feed $F$ / composition $z$	100 mol / 0.40 acetone (saturated liquid, $q=1$)
Distillate $x_D$ / bottoms $x_W$	0.95 / 0.05
Reflux ratio $R=L/D$	2.5 (total condenser)
$k_ya$ / $k_xa$	0.2 / 1.8 kmol·s⁻¹m⁻³(mol frac)⁻¹
Equilibrium data	acetone–ethanol $x$–$y$ table (11 points)

Find. The number of (overall gas-phase) transfer units $N_{OG}$ for the separation.

Operating lines from the material balance; $N_{OG}$ is the integral of $dy/(y^*-y)$ up each line, where $y^*$ is read off the equilibrium curve. Driving forces pinch near the feed and near pure acetone, which is where most transfer units accumulate.

Approach. Close the overall and component balances to get $D$, $W$ and the internal flows; build the rectifying and stripping operating lines; then integrate $N_{OG}=\int dy/(y^*-y)$ up both lines using the equilibrium table.

Product split. Overall and acetone balances: $$D=F\frac{z-x_W}{x_D-x_W}=100\frac{0.40-0.05}{0.95-0.05}=38.9\ \mathrm{mol},\qquad W=61.1\ \mathrm{mol}.$$
Internal flows (equimolar overflow, $q=1$). $$L=RD=97.2,\quad V=L+D=136.1\ \mathrm{mol};\qquad \bar L=L+F=197.2,\quad \bar V=V=136.1\ \mathrm{mol}.$$
Operating lines. Rectifying $y=\dfrac{L}{V}x+\dfrac{Dx_D}{V}=0.714x+0.271$; stripping $y=\dfrac{\bar L}{\bar V}x-\dfrac{Wx_W}{\bar V}=1.449x-0.0224$. With the vertical $q$-line at $x=0.40$ they meet at the feed point $(0.40,\,0.557)$.
Overall gas-phase transfer units. Stepping $y$ up each line, reading $y^*$ from the equilibrium table at the operating $x$, and integrating numerically (Simpson): $$N_{OG}=\int_{x_W}^{x_D}\frac{dy}{y^*-y}=\underbrace{5.15}_{\text{stripping}}+\underbrace{13.4}_{\text{rectifying}}.$$ $N_{OG}\approx 18.6$ overall transfer units. Most units pile up in the rectifying section, where the equilibrium curve crowds the operating line near pure acetone (small $y^*-y$).
Coefficient context (height, if required). The individual film coefficients combine, using a representative equilibrium slope $m\approx1.3$, into an overall coefficient $$K_ya=\left(\frac1{k_ya}+\frac{m}{k_xa}\right)^{-1}=\left(\frac1{0.2}+\frac{1.3}{1.8}\right)^{-1}=0.175\ \mathrm{kmol\,s^{-1}m^{-3}}.$$ The packed height would then be $Z=H_{OG}N_{OG}$ with $H_{OG}=\bar V/(K_ya\,S)$ — but since the feed is stated as a fixed quantity (100 mol) rather than a rate, and no column cross-section $S$ is given, the height is indeterminate and the transfer-unit count is the deliverable.

Quantity	Value
Distillate / bottoms $D$ / $W$	38.9 / 61.1 mol
Rectifying / stripping vapour $V$ / $\bar V$	136.1 / 136.1 mol
$N_{OG}$ stripping / rectifying	5.15 / 13.4
Total $N_{OG}$	≈ 18.6
Overall coefficient $K_ya$ (for $H_{OG}$)	0.175 kmol·s⁻¹m⁻³

Check. The feed prose names the second component “water,” but the equilibrium table supplied is explicitly labelled acetone–ethanol; this solution uses the table as given (the only self-consistent choice). The uppermost three equilibrium rows ($x\ge0.85$) are noted in the source as very faint; because the rectifying pinch sits in exactly that region, $N_{OG}$ is sensitive to those points — the reported 18.6 should be read as “of order 18–20 transfer units.” If the intended system were truly acetone–water (a much wider-boiling, near-ideal-then-pinching pair), the numeric $N_{OG}$ would differ although the method is identical.

Concept & Theory

Topic. Packed-tower distillation by the transfer-unit (HTU–NTU) method.

Key relations. Operating lines from mass balances; $N_{OG}=\int dy/(y^*-y)$; $Z=H_{OG}N_{OG}$; $1/K_ya=1/k_ya+m/k_xa$; $H_{OG}=V/(K_ya\,S)$.

Why this works. A packed column exchanges mass continuously rather than in discrete stages, so the separation is measured by how many times the local driving force $(y^*-y)$ must be “used up” going from bottom to top — that integral is $N_{OG}$. The operating lines give the bulk vapour $y$ paired with liquid $x$; the equilibrium curve gives $y^*$. Where the two lines approach the equilibrium curve (feed pinch, near-pure-acetone pinch) the driving force collapses and transfer units accumulate rapidly, which is why $N_{OG}$ is large for this high-purity split.

Common pitfall. Using $(y^*-y)$ with $y^*$ evaluated at the wrong $x$ (it must be the liquid composition on the operating line at that height), or confusing $N_{OG}$ (overall gas) with the number of theoretical stages. A McCabe–Thiele stage step-off here gives $\sim18$ stages, consistent with $N_{OG}\approx18.6$ because the stripping/rectifying slopes are close to unity.

Source. Treybal, Mass-Transfer Operations 3rd ed., Ch. 9; Geankoplis §11.5–11.6.

Question 4 of 6: Number of Transfer Units for a Packed Distillation Tower

Question B1: Number of Transfer Units for a Packed Distillation Tower (25 marks)