Oxygen Diffusion with Metabolic Consumption in Lung Tissue: 23-Chem-B1 Transport Phenomena, May 2013, Q5

Question 5: Oxygen Diffusion with Metabolic Consumption in Lung Tissue (15 marks)

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. Lung tissue modelled as a plane slab of thickness $L$; oxygen concentration held at $C_{Ai}$ on the inner ($z=0$) face and $C_{A0}$ on the outer ($z=L$) face; oxygen consumed by metabolism at a constant (zero-order) volumetric rate $R_A=-k_0$; steady, one-dimensional diffusion with diffusivity $D_{AB}$.

Find. The steady oxygen concentration profile $C_A(z)$ across the tissue.

Oxygen diffusing through a slab of lung tissue of thickness $L$: concentration is fixed at $C_{Ai}$ on the lung-cavity face ($z=0$) and $C_{A0}$ on the blood-side face ($z=L$). Uniform metabolic consumption ($k_0$) makes the profile $C_A(z)$ sag below the straight diffusion line (red, dashed).

Approach. Reduce the species-A continuity equation to the one-dimensional reaction–diffusion form $D_{AB}\,C_A''=k_0$ (a constant zero-order sink), then integrate twice and fix the two constants with the prescribed surface concentrations.

Reduce the species-A equation. For steady, one-dimensional diffusion with a homogeneous zero-order consumption, the species continuity equation becomes $0=D_{AB}\dfrac{d^2C_A}{dz^2}+R_A$ with $R_A=-k_0$, i.e. $D_{AB}\dfrac{d^2C_A}{dz^2}=k_0$.
Integrate twice. $\dfrac{dC_A}{dz}=\dfrac{k_0}{D_{AB}}z+B_1$; $C_A=\dfrac{k_0}{2D_{AB}}z^2+B_1 z+B_2$.
Apply the surface concentrations. $C_A(0)=C_{Ai}\Rightarrow B_2=C_{Ai}$; $C_A(L)=C_{A0}\Rightarrow B_1=\dfrac{C_{A0}-C_{Ai}}{L}-\dfrac{k_0 L}{2D_{AB}}$.
Concentration profile. Substituting the constants and grouping the reaction terms, $$C_A=C_{Ai}+\frac{k_0}{2D_{AB}}\left(z^2-zL\right)+(C_{A0}-C_{Ai})\frac{z}{L},$$ the required result. The last two terms are the straight diffusion field between the two surface concentrations; the quadratic term ($z^2-zL\le0$) is the sag caused by uniform consumption. $\boxed{\,C_A=C_{Ai}+\dfrac{k_0}{2D_{AB}}\left(z^2-zL\right)+(C_{A0}-C_{Ai})\dfrac{z}{L}\,}$

Quantity	Result
Concentration profile	$C_A=C_{Ai}+\dfrac{k_0}{2D_{AB}}(z^2-zL)+(C_{A0}-C_{Ai})\dfrac{z}{L}$
Check at $z=0$ / $z=L$	$C_{Ai}$ / $C_{A0}$

Concept & Theory

Topic. Steady one-dimensional diffusion with a homogeneous zero-order reaction (reaction–diffusion in a slab).

Key relations. Reaction–diffusion equation $D_{AB}\,d^2C_A/dz^2=k_0$; Dirichlet conditions $C_A(0)=C_{Ai}$, $C_A(L)=C_{A0}$; solution = linear diffusion field $+$ parabolic reaction term.

Why this works. A constant volumetric sink adds a constant to the second derivative of concentration, so the profile is necessarily a parabola. The homogeneous part (a straight line) carries the boundary concentrations, while the particular part (the parabola) carries the consumption; superposition gives the full field. Because A is consumed along the path, the flux $-D_{AB}\,dC_A/dz$ varies with $z$ — it is not uniform as in pure diffusion.

Common pitfall. Sign of the source: consumption is $-k_0$, which makes $C_A''>0$ and the profile sag downward; a generation term would bow it upward. Also, do not assume a constant flux — the reaction makes $dC_A/dz$ position-dependent.

Source. Bird, Stewart & Lightfoot, Ch. 18 (diffusion with a homogeneous reaction, Table 5); Welty, Wicks, Wilson & Rorrer, Ch. 26.

Question 5 of 5: Oxygen Diffusion with Metabolic Consumption in Lung Tissue

Question 5: Oxygen Diffusion with Metabolic Consumption in Lung Tissue (15 marks)