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23-Chem-A3 Heat and Mass Transfer · December 2017

Question 5 of 6: Part B: Mass-Transfer Coefficient in an Ammonia Scrubber, and by the Chilton–Colburn Analogy

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

Notes on this paper

National Exams — December 2017 — 16-Chem-A3 Heat and Mass Transfer. Three-hour, open-book exam (one textbook of the candidate’s choice; any non-communicating calculator). Format: two parts — Part A — Heat Transfer (Q1–Q3) and Part B — Mass Transfer (Q4–Q6). All problems are worth 25 points; the rubric requires at least two problems attempted from each part and marks only the first two per part, so a complete paper is four questions (100 points). All six problems are solved below for completeness. Property data are quoted as stated in each problem; where a value applies to a film condition it is used at the stated film temperature.

Reference texts: F. P. Incropera & D. P. DeWitt, Fundamentals of Heat and Mass Transfer (Wiley) — free/forced convection correlations, conduction shape factors, the heat–mass-transfer analogy; J. P. Holman, Heat Transfer (McGraw-Hill) — natural-convection constants $C,m$ and cylinder conduction; R. E. Treybal, Mass-Transfer Operations (McGraw-Hill) — wetted-wall $k_G$, gas absorption and the Chilton–Colburn analogy; R. B. Bird, W. E. Stewart & E. N. Lightfoot, Transport Phenomena (Wiley); supporting property data from Perry’s Chemical Engineers’ Handbook (9th ed.).

Question 5 — Part B: Mass-Transfer Coefficient in an Ammonia Scrubber, and by the Chilton–Colburn Analogy (25 points)

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. Gas enters at 20 mol/hr with $y_\text{in}=0.05$ NH₃ at $P=101.3$ kPa, 20 °C; 90% of the ammonia is absorbed into 2N H₂SO₄ that reacts so fast the interface partial pressure of NH₃ is essentially zero. The column is a wetted-wall tube, ID $=0.015$ m, $L=0.70$ m, so its transfer area is $A=\pi D L$.

DatumValue
Gas rate / NH₃ fraction20 mol/hr / 5%
Column ID / length15 mm / 0.70 m
Diffusivity $D_{\text{NH}_3\text{-air}}$$2.2\times10^{-5}$ m²/s
$Sc$ / $Pr$0.67 / 0.70
Heat-transfer $h$ (part b)75.2 kcal/hr·m²·°C

Find. (a) $k_G$ from the measured absorption; (b) $k_G$ predicted from the Chilton–Colburn analogy.

Wetted-wall column (ID 15 mm, L 0.70 m)gasacid filmgas in: 20 mol/hr, 5% NH₃gas out (10% NH₃ left)2N H₂SO₄ downthe walls90% NH₃absorbedinterfacep(NH₃) ≈ 0
Figure 5 — Wetted-wall column: gas rises through the tube while 2N sulfuric acid films the wall. Because the acid neutralises ammonia instantly, the NH₃ partial pressure at the wall is ≈ 0, so the whole gas-film partial-pressure difference drives the transfer.

Approach. Convert the 90% absorption into a molar transfer rate, evaluate the log-mean NH₃ partial-pressure driving force (interface value zero), and divide by area to get $k_G$ for part (a). For part (b), set the Chilton–Colburn $j$-factors equal, $j_D=j_H$, to convert the heat-transfer coefficient into a mass-transfer coefficient.

  1. Molar transfer rate. NH₃ entering is $0.05\times20=1.0$ mol/hr; 90% absorbed gives $$N_A = 0.9\ \text{mol/hr} = 2.50\times10^{-4}\ \text{mol/s}.$$ The residual gas is 19 mol/hr air + 0.1 mol/hr NH₃, so the outlet fraction is $y_\text{out}=0.1/19.1=0.00524$.
  2. Log-mean partial-pressure driving force. The bulk NH₃ partial pressures are $p_\text{in}=0.05(101.3)=5.07$ kPa and $p_\text{out}=0.00524(101.3)=0.530$ kPa; the wall value is zero at both ends, so $$\Delta p_\text{lm}=\frac{p_\text{in}-p_\text{out}}{\ln(p_\text{in}/p_\text{out})}=\frac{5.07-0.530}{\ln(5.07/0.530)}=2.01\ \text{kPa}.$$
  3. Mass-transfer coefficient (a). With column area $A=\pi(0.015)(0.70)=0.0330\ \text{m}^2$ and $N_A=k_G A\,\Delta p_\text{lm}$, $$k_G = \frac{N_A}{A\,\Delta p_\text{lm}} = \frac{2.50\times10^{-4}}{(0.0330)(2.01)} = \boxed{3.8\times10^{-3}\ \text{mol}\,\text{m}^{-2}\text{s}^{-1}\text{kPa}^{-1}}.$$ As a sanity check this is a Sherwood number $Sh=k_c D/D_{AB}=6.3$ (with $k_c=k_G RT$), which is the right order for laminar wetted-wall flow. That is part (a).
  4. Chilton–Colburn analogy (b). Equating the transfer $j$-factors, $j_D=j_H$, gives $k_c\,Sc^{2/3}=(h/\rho c_p)\,Pr^{2/3}$, so $$k_c = \frac{h}{\rho c_p}\left(\frac{Pr}{Sc}\right)^{2/3}.$$ Converting $h=75.2$ kcal/hr·m²·°C $=87.4$ W/m²K, with air $\rho=1.204$ kg/m³ and $c_p=238\times4.184=996$ J/kg·K: $$k_c=\frac{87.4}{(1.204)(996)}\left(\frac{0.70}{0.67}\right)^{2/3}=0.0751\ \text{m/s}.$$
  5. Convert to $k_G$ (b). Since $N_A=k_c\,\Delta c=k_c\,\Delta p/(RT)$, we have $k_G=k_c/(RT)$: $$k_G=\frac{0.0751}{(8.314)(293.15)}=3.08\times10^{-5}\ \text{mol}\,\text{m}^{-2}\text{s}^{-1}\text{Pa}^{-1}=\boxed{3.1\times10^{-2}\ \text{mol}\,\text{m}^{-2}\text{s}^{-1}\text{kPa}^{-1}}.$$ That is part (b).
QuantityResult
NH₃ transferred0.9 mol/hr ($2.5\times10^{-4}$ mol/s)
Log-mean driving force $\Delta p_\text{lm}$2.01 kPa
(a) $k_G$ from absorption≈ $3.8\times10^{-3}$ mol m$^{-2}$s$^{-1}$kPa$^{-1}$ ($Sh\approx6$)
(b) $k_G$ from Chilton–Colburn≈ $3.1\times10^{-2}$ mol m$^{-2}$s$^{-1}$kPa$^{-1}$ ($Sh\approx51$)
Check
The two estimates differ by roughly an order of magnitude. The absorption value (a) corresponds to $Sh\approx6$, entirely consistent with developing laminar flow in a short wetted-wall tube ($Re\approx760$); the analogy value (b) inherits the supplied $h$, which implies $Sh\approx51$ and therefore describes a much more vigorous (turbulent/entrance-enhanced) condition than the actual absorption run. Both are reported as computed; the discrepancy is the physical lesson — the Chilton–Colburn analogy is only as representative as the heat-transfer coefficient fed into it.