Question 6 of 6: NO₂ Adsorption on Silica Gel — Counter- vs Co-current
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — May 2017 — 16-Chem-A3 Heat and Mass Transfer. Three-hour, open-book exam; one textbook and any non-communicating calculator permitted. Format: two parts of three problems each (Part A — Heat Transfer, Part B — Mass Transfer), all problems worth 25 points; candidates attempt at least two per part and only the first two per part are marked. The six problems are renumbered 1–6 and all are solved below for completeness. Property data are taken from the problem statements and standard open-book tables (the paper supplies the air-property table A-9); values used are stated in each Given block.
Reference texts: Incropera, DeWitt, Bergman & Lavine, Fundamentals of Heat and Mass Transfer (Wiley) — conduction resistances, critical radius, fins, convection correlations; Coulson & Richardson, Chemical Engineering Vol. 1 (Butterworth-Heinemann) — the packed-bed, distillation and sublimation problems of this style; Treybal, Mass-Transfer Operations (3rd ed., McGraw-Hill) — differential distillation and gas adsorption; Geankoplis, Transport Processes and Separation Process Principles (4th ed., Prentice Hall) — packed-bed and convective mass-transfer coefficients; supporting data from Perry's Chemical Engineers' Handbook (9th ed.) and the paper's Table A-9 (properties of air).
Part A — Heat Transfer
Question 6: NO₂ Adsorption on Silica Gel — Counter- vs Co-current (25 points)
Given. Gas $0.50$ kg/s, $1.5\%$ NO₂ by volume, silica gel enters NO₂-free; $85\%$ of the NO₂ removed at twice the minimum gel rate; $298$ K, 1 atm. Equilibrium $p$ (mmHg) vs $W$ (kg NO₂/kg gel) as tabulated.
Quantity
Value
Gas rate / NO₂ content
0.50 kg/s / 1.5 vol%
Removal / gel rate
85% / 2 × minimum
Mean gas MW
29.23 g/mol
Total molar flow
17.11 mol/s
Find. Gel leaving rate and loading for (c) countercurrent and (d) cocurrent contact.
Figure 6 — Countercurrent (gas and gel move oppositely; the minimum-gel pinch is where the operating line touches the isotherm) vs. cocurrent (gas and gel enter together; pinch at the lean gas outlet). Countercurrent loads the gel far more heavily, so it needs about eleven times less gel.
Check
The isotherm is used exactly as printed (kg NO₂/kg gel). Those loadings, up to 4.85 kg NO₂ per kg of gel, are far more than silica gel can physically hold (a few per cent by mass), so the table was probably meant as kg NO₂ per 100 kg gel. Every result below is linear in the loading, so that reading simply multiplies the gel rates by 100 and divides the loadings by 100 (countercurrent: 0.533 kg/s gel at 0.0188 kg/kg; cocurrent: 6.03 kg/s at 0.00167 kg/kg). The method and the counter- versus co-current comparison are unchanged.
Approach. Fix the NO₂ mass to be adsorbed from the gas balance and work on an inert basis (kg NO₂ per kg air, kg NO₂ per kg gel), where the operating line is straight. The minimum gel rate is the steepest operating line that still touches the equilibrium curve. For countercurrent flow this isotherm bends upward (loading rises faster than pressure), so the touch point is inside the column rather than at the rich end. For cocurrent flow it is at the lean gas outlet. Double the minimum, then close the solid balance.
Gas flows and NO₂ load. Mean MW $= 0.015(46.0)+0.985(28.97) = 29.23$, so total molar flow $= 500/29.23 = 17.11$ mol/s. NO₂ in $= 0.015(17.11)(46.0)/1000 = 0.0118$ kg/s; with $85\%$ removed,
$$\dot m_{NO_2,\text{ads}} = 0.85(0.0118) = \boxed{0.0100\ \text{kg/s}}.$$
The air (inert) flow is $G_s = 0.985(17.11)(28.97)/1000 = 0.488$ kg/s.
End compositions. Inlet $p_1 = 0.015(760) = 11.4$ mmHg; after removal the outlet mole fraction gives $p_2 = 1.73$ mmHg. On the inert basis $Y = \dfrac{p}{760-p}\cdot\dfrac{46.0}{28.97}$, so $Y_1 = 0.02418$ and $Y_2 = 0.00363$ kg NO₂/kg air (check: $G_s(Y_1-Y_2) = 0.488(0.02055) = 0.0100$ kg/s).
Why the rich end is not the pinch (part c). A line from the lean end $(W=0,\ Y_2)$ to the rich-end equilibrium point $W^\*(p_1) = 4.49$ has slope $(0.02418-0.00363)/4.49 = 0.00458$. At $W = 1.65$ it gives $Y = 0.00363 + 0.00458(1.65) = 0.0112$, but the isotherm point $(6\ \text{mmHg},\ 1.65)$ has $Y^\* = 0.0126$. The line passes below the equilibrium curve, which is impossible for adsorption, so that gel rate is too small.
Tangent pinch and minimum gel rate. The minimum slope is the largest value of $(Y^\*-Y_2)/W$ along the isotherm, which occurs at the tabulated point $(6\ \text{mmHg},\ 1.65\ \text{kg/kg})$:
$$\left(\frac{\dot S_s}{G_s}\right)_{\min} = \frac{0.01264-0.00363}{1.65} = 0.00546,\qquad \dot S_{s,\min} = 0.488(0.00546) = 2.67\times10^{-3}\ \text{kg/s}.$$
That is 19% more than the rich-end estimate. At this minimum rate the gel would leave carrying $0.0100/2.67\times10^{-3} = 3.76$ kg/kg, short of the 4.49 kg/kg in equilibrium with the inlet gas. A smooth curve drawn through the table points instead of straight segments gives $\dot S_{s,\min} \approx 2.53\times10^{-3}$ kg/s, about 5% lower.
Countercurrent gel leaving (part c). Using twice the minimum, $\dot S_s = 5.33\times10^{-3}$ kg/s of NO₂-free gel, and
$$W_{out} = \frac{\dot m_{NO_2}}{\dot S_s} = \frac{0.0100}{5.33\times10^{-3}} = \boxed{1.88\ \text{kg NO}_2/\text{kg gel}},$$
i.e. a leaving-gel NO₂ mass fraction of $1.88/(1+1.88) = 0.653$, in a total leaving stream of $5.33+10.04 = \boxed{15.4\ \text{g/s}}$.
Cocurrent minimum and actual gel rate (part d). Gas and gel travel together, so the operating line runs from $(0,\ Y_1)$ downward and can only meet the isotherm at the outlet: the leaving gel is at most in equilibrium with the lean outlet gas. $W^\*(p_2)$ at $1.73$ mmHg $= 0.333\ \text{kg/kg}$, extending the lowest table segment. Then
$$\dot S_{s,\min} = \frac{0.0100}{0.333} = 3.01\times10^{-2}\ \text{kg/s},\qquad \dot S_s = 6.03\times10^{-2}\ \text{kg/s}.$$
Cocurrent gel leaving.
$$W_{out} = \frac{0.0100}{6.03\times10^{-2}} = \boxed{0.167\ \text{kg NO}_2/\text{kg gel}},$$
a mass fraction of $0.167/1.167 = 0.143$, in a total leaving stream of $60.3+10.04 = \boxed{70.3\ \text{g/s}}$. Drawing the isotherm through the origin instead ($W^\* = 0.2p$) gives 0.346 kg/kg at the pinch, about 4% higher, so the extrapolation hardly matters.
Cocurrent contact needs about 11 times more gel (60.3 against 5.33 g/s) and sends it out far more dilute. In the countercurrent column the loaded gel leaves past the rich incoming gas, so it can pick up a lot of NO₂. In the cocurrent unit the gel leaves beside the already-cleaned gas and can never load beyond equilibrium with that lean stream.
