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23-Chem-A3 Heat and Mass Transfer · May 2015

Question 5 of 7: Packed-tower height for amine CO₂ absorption

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams — May 2015 — 04-Chem-A3 Mass Transfer Operations. Three-hour, open-book exam; any non-communicating calculator permitted. Format: seven questions in three parts — answer one of Q1–Q2 (Part A), one of Q3–Q4 (Part B) and two of Q5–Q7 (Part C); four questions of equal value constitute a complete paper. All seven are solved below for completeness. Every property datum (diffusivities, vapour pressures, solubilities, humid-air heat capacities) is stated in the question, so each answer is self-contained.

Reference texts: Geankoplis, Transport Processes and Separation Process Principles (4th ed., Prentice Hall) — molecular diffusion, convective mass transfer, gas absorption and drying; Welty, Wicks, Wilson & Rorrer, Fundamentals of Momentum, Heat and Mass Transfer (6th ed., Wiley) — boundary-layer and transient diffusion; Incropera & DeWitt, Fundamentals of Heat and Mass Transfer — transient conduction/diffusion charts and the sphere in a stagnant medium; Treybal, Mass-Transfer Operations (3rd ed.) — packed-tower design; Crank, The Mathematics of Diffusion (2nd ed.) — series solutions for the sphere; supporting data from Perry's Chemical Engineers' Handbook (9th ed.).

Question 5: Packed-tower height for amine CO₂ absorption (Part C — equal value)

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. A dilute countercurrent packed absorber with a linear equilibrium line through the origin. The tight outlet spec (0.04 mol%) drives a large number of transfer units. All quantities are in cgs/SI mole units.

QuantityValue
Gas in / out, $y_1,y_2$0.0126 / 0.0004
Gas flow $G$ / liquid flow $L$2.3 / 4.8 mol/s
Tower diameter40 cm ($S=1257$ cm²)
Overall coefficient $K_ya$$5\times10^{-5}$ mol/cm³·s
Equilibrium slope $m$1.58 ($y=1.58x^*$)

Find. The required packed height $z$.

packing, dia = 40 cm, z = ? gas in, y₁ = 0.0126 gas out, y₂ = 0.0004 amine in, x₂ = 0 liquid out, x₁ = 0.00585
Figure 5 — Countercurrent dilute CO₂ absorber. Rich gas enters the bottom; pure amine enters the top. The liquid exit composition $x_1$ closes the solute balance and sets the bottom driving force; the tight outlet spec makes $N_{OG}$ large.

Approach. Close the solute balance for $x_1$, evaluate the log-mean gas-phase driving force to get $N_{OG}$, form the height of a transfer unit $H_{OG}=(G/S)/K_ya$, then $z=H_{OG}N_{OG}$.

  1. Cross-section and liquid exit. $S=\tfrac{\pi}{4}(40)^2=1257\ \text{cm}^2$. Solute balance with $x_2=0$: $$x_1=\frac{G(y_1-y_2)}{L}=\frac{2.3(0.0126-0.0004)}{4.8}=0.00585.$$
  2. End driving forces. Using $y^*=1.58x$: bottom $\Delta y_1=y_1-1.58x_1=0.0126-0.00924=0.00336$; top $\Delta y_2=y_2-0=0.0004$.
  3. Log-mean driving force. $$\Delta y_\text{lm}=\frac{\Delta y_1-\Delta y_2}{\ln(\Delta y_1/\Delta y_2)}=\frac{0.00336-0.0004}{\ln(0.00336/0.0004)}=0.00139.$$
  4. Number of transfer units. $$N_{OG}=\frac{y_1-y_2}{\Delta y_\text{lm}}=\frac{0.0122}{0.00139}=8.77.$$
  5. Height of a transfer unit. Using the molar flux $G/S$: $$H_{OG}=\frac{G/S}{K_ya}=\frac{2.3/1257}{5\times10^{-5}}=36.6\ \text{cm}.$$
  6. Packed height. $$z=H_{OG}\,N_{OG}=(36.6)(8.77)=321\ \text{cm}=\boxed{3.2\ \text{m}}.$$
QuantityResult
Liquid exit $x_1$0.00585
Log-mean driving force0.00139
$N_{OG}$ / $H_{OG}$8.77 / 36.6 cm
Packed height $z$321 cm ≈ 3.2 m
Check
The 96.8% removal (1.26 → 0.04 mol%) demands a small top driving force ($\Delta y_2=0.0004$), which is what inflates $N_{OG}$ to nearly 9. The system is dilute (≤1.3 mol% CO₂), so treating $G$ and $L$ as constant and using the log-mean driving force is valid.