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23-Chem-A3 Heat and Mass Transfer · May 2013

Question 5 of 7: Height of a Packed Ammonia Absorption Tower

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

Notes on this paper

National Exams — May 2013 — 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 constitute a complete paper (Parts A/B carry 20% each, Part C 30% each). All seven are solved below for completeness. Property data are stated in each Given block. Two chart appendices (SI humidity–temperature charts) accompany Q6.

Reference texts: Geankoplis, Transport Processes and Separation Process Principles (4th ed., Prentice Hall) — molecular diffusion, mass-transfer coefficients, absorption, humidification; Welty, Wicks, Wilson & Rorrer, Fundamentals of Momentum, Heat and Mass Transfer (5th/6th ed., Wiley) — transient diffusion, boundary-layer mass transfer; Treybal, Mass-Transfer Operations (3rd ed., McGraw-Hill) — packed-tower and cooling-tower design; supporting data from Perry's Chemical Engineers' Handbook (9th ed.) and Incropera & DeWitt, Fundamentals of Heat and Mass Transfer.

Question 5: Height of a Packed Ammonia Absorption Tower (Part C — 30%)

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. Dilute counter-current absorption of NH₃ into water at 293 K, 1 atm. Gas in $V_1=57.8$ kmol/h at $y_1=0.040$, gas out $y_2=0.005$; pure water in $L_2=68.0$ kmol/h at $x_2=0$. Film coefficients $k'_y a=0.0739$, $k'_x a=0.169\ \text{kmol}/(\text{m}^3\cdot\text{s}\cdot\text{mol frac})$. Equilibrium (20 °C column, since $T=293$ K):

$x_A$00.02080.02580.03090.04050.05030.07370.0960
$y_A^{*}$ (20 °C)00.01580.01970.02390.03280.04160.06570.0915

Find. The packed height $Z$ of the tower.

packedbedgas out y₂=0.005water in x₂=0 (68 kmol/h)gas in y₁=0.040 (57.8 kmol/h)liquor out x₁≈0.029293 K, 1 atm, D = 0.747 m
Figure 5 — Counter-current packed absorber: dilute NH₃ gas enters the bottom and is scrubbed by pure water entering the top. The overall gas-phase HTU–NTU method sizes the packed height.

Approach. Close the material balance for the liquid-outlet composition, combine the two film coefficients into an overall gas coefficient, then compute the height as $Z=H_{OG}\,N_{OG}$ using the dilute (log-mean driving force) design method.

  1. Material balance for $x_1$. Inert gas $G_s=V_1(1-y_1)=55.49$ kmol/h; NH₃ absorbed $=G_s\big(\tfrac{y_1}{1-y_1}-\tfrac{y_2}{1-y_2}\big)=2.03$ kmol/h. On solvent $L_s=68.0$, $$x_1=\frac{2.03/68.0}{1+2.03/68.0}=\boxed{0.0290}.$$
  2. Overall gas-phase coefficient. With the equilibrium slope over the operating range $m\approx0.765$, $$\frac{1}{K_y'a}=\frac{1}{k_y'a}+\frac{m}{k_x'a}=\frac{1}{0.0739}+\frac{0.765}{0.169}=18.06\ \Rightarrow\ K_y'a=0.0554\ \tfrac{\text{kmol}}{\text{m}^3\cdot\text{s}}.$$
  3. Height of a transfer unit. Cross-section $S=\tfrac{\pi}{4}(0.747)^2=0.438\ \text{m}^2$; average gas rate $V_\text{av}=56.8\ \text{kmol/h}=0.01577\ \text{kmol/s}$, $$H_{OG}=\frac{V_\text{av}}{K_y'a\,S}=\frac{0.01577}{(0.0554)(0.438)}=\boxed{0.650\ \text{m}}.$$
  4. Number of transfer units (log-mean driving force). With $y_1^{*}=0.0223$ (at $x_1$) and $y_2^{*}=0$, $$(\Delta y)_\text{lm}=\frac{(0.040-0.0223)-(0.005-0)}{\ln[(0.040-0.0223)/(0.005)]}=0.01004,\qquad N_{OG}=\frac{y_1-y_2}{(\Delta y)_\text{lm}}=\boxed{3.49}.$$
  5. Packed height. $$Z=H_{OG}\,N_{OG}=(0.650)(3.49)=\boxed{2.27\ \text{m}}.$$
QuantityResult
Liquid outlet composition, $x_1$0.0290
Overall coefficient, $K_y'a$$0.0554\ \text{kmol}/(\text{m}^3\cdot\text{s})$
$H_{OG}$ / $N_{OG}$0.650 m / 3.49
Packed height, $Z$$\approx2.3$ m

Check: the printed diameter "747 cm" (7.47 m) is taken as $0.747$ m — a 7.47 m tower handling only 57.8 kmol/h of gas is grossly oversized and yields an unphysical ~2 cm packed height, whereas $D=0.747$ m gives a sensible ~2.3 m. Because $T=293$ K, the 20 °C equilibrium column is used ($m\approx0.76$); the 30 °C column would nearly double the required height.