Ammonia Diffusing Through Stagnant Nitrogen: 23-Chem-A3 Heat and Mass Transfer, May 2013, Q1

Question 1: Ammonia Diffusing Through Stagnant Nitrogen (Part A — 20%)

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. Ammonia (A) diffuses through stagnant, insoluble nitrogen (B): $P=1.013\times10^{5}$ Pa, $T=298$ K, $D_{AB}=2.30\times10^{-5}\ \text{m}^2/\text{s}$, path length $z=20\ \text{mm}=0.020$ m, with $p_{A1}=1.333\times10^{4}$ Pa and $p_{A2}=6.666\times10^{3}$ Pa.

Find. The steady molar flux of ammonia, $N_A$, in $\text{mol}/(\text{m}^2\cdot\text{s})$.

Figure 1 — Stagnant-film picture: ammonia (A) diffuses across the 20 mm gap while nitrogen (B) is insoluble at the far boundary and carries no net flux. This is diffusion of A through stagnant B.

Approach. Because nitrogen has zero net flux, apply the steady one-dimensional result for diffusion of A through stagnant B, using the log-mean of the nitrogen partial pressure as the drift correction.

Set up diffusion of A through stagnant B. With $N_B=0$, the flux is $$N_A=\frac{D_{AB}\,P}{R\,T\,z\,p_{B,\text{lm}}}\,(p_{A1}-p_{A2}),\qquad p_{B}=P-p_{A}.$$
Evaluate the nitrogen partial pressures and their log-mean. $p_{B1}=P-p_{A1}=8.797\times10^{4}$ Pa and $p_{B2}=P-p_{A2}=9.463\times10^{4}$ Pa, so $$p_{B,\text{lm}}=\frac{p_{B2}-p_{B1}}{\ln(p_{B2}/p_{B1})}=\frac{9.463\times10^{4}-8.797\times10^{4}}{\ln(9.463/8.797)}=9.126\times10^{4}\ \text{Pa}.$$
Substitute (with $R=8.314\ \text{J/mol}\cdot\text{K}$). $$N_A=\frac{(2.30\times10^{-5})(1.013\times10^{5})}{(8.314)(298)(0.020)(9.126\times10^{4})}\,(1.333\times10^{4}-6.666\times10^{3}).$$ The numerator group $D_{AB}P=2.330$ and the driving pressure difference is $6.664\times10^{3}$ Pa, giving $$N_A=\boxed{3.43\times10^{-3}\ \text{mol}/(\text{m}^2\cdot\text{s})}\;\;(=3.43\times10^{-6}\ \text{kmol}/(\text{m}^2\cdot\text{s})).$$

Quantity	Result
Log-mean N₂ pressure, $p_{B,\text{lm}}$	$9.13\times10^{4}$ Pa
Ammonia flux, $N_A$	$3.43\times10^{-3}\ \text{mol}/(\text{m}^2\cdot\text{s})$

Concept & Theory

Topic. Steady one-dimensional molecular diffusion of one species through a second, stagnant (non-diffusing) species.

Key relations. Fick's law with the convective (drift) term gives $N_A=\dfrac{D_{AB}P}{RTz\,p_{B,\text{lm}}}(p_{A1}-p_{A2})$; the factor $P/p_{B,\text{lm}}$ is the "bulk-flow" enhancement over equimolar counterdiffusion.

Why this works. A stagnant B implies a bulk velocity toward the interface that convects A along with its diffusion; the log-mean $p_{B,\text{lm}}$ is the exact integral of $1/p_B$ across the film for constant $N_A$. When $p_A\ll P$ the enhancement factor $\to 1$ and the result collapses to the simple Fick expression.

Common pitfall. Using the arithmetic mean of $p_B$ (or omitting it entirely) — for a 13% ammonia mole fraction the drift factor is a real ~11% effect. Equivalently, note $N_A=\dfrac{D_{AB}P}{RTz}\ln\!\dfrac{p_{B2}}{p_{B1}}$ gives the identical answer.

Source. Geankoplis, Transport Processes and Separation Process Principles §6.2 (diffusion of A through stagnant B), Example 6.2-2.

Question 1 of 7: Ammonia Diffusing Through Stagnant Nitrogen

Question 1: Ammonia Diffusing Through Stagnant Nitrogen (Part A — 20%)