Workshop 9: Diffusion with surface reaction

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Attempt these problems before opening the worked solutions.

1 Oxidation of a graphite rod

Cylindrical diffusion to a shrinking rod

A cylindrical graphite (pure carbon) rod (length = 25 cm, initial diameter = 2 cm) is placed in a combustion chamber with flowing air stream at 1100 K and 2 atm total pressure. A stagnant gas boundary layer 5 mm thick is formed around the surface of the rod. At 1100 K, solid carbon oxidises to carbon dioxide gas,

C (s)+O2 (g)→CO2 (g)$$\mathrm{C\ (s) + O_2\ (g) \rightarrow CO_2\ (g)}$$

Assuming that the oxidation is limited by molecular diffusion of O2$\mathrm{O_2}$ through the stagnant gas film, and outside of the gas film, the bulk composition of air stream prevails,

a) Estimate the initial rate of CO2$\mathrm{CO_2}$ production from the rod;

b) Write an equation for the rate of change in the radius of the rod as a function of time.

Data: Density of solid graphite = 2.25 g/cm3$\mathrm{g/cm^3}$; 𝐷𝐴𝐵$D_{AB}$ (CO2$\mathrm{CO_2}$-air) @ 273 K and 1 atm = 0.136 cm2/s$\mathrm{cm^2/s}$

Worked solution

Figure 1: Radial diffusion through a gas film around the graphite rod. The radius a decreases during oxidation, while the film thickness remains fixed. The drawing is a cylindrical cross-section; oxygen diffuses inward and carbon dioxide outward.

Data

Quantity	Symbol	Value
Initial radius and rod length	𝑎0,ℓ$a_0,\ell$	0.010,0.25 m$0.010,0.25\ \mathrm m$
Gas-film thickness	𝛿$\delta$	0.0050 m$0.0050\ \mathrm m$
Gas temperature and pressure	𝑇,𝑃$T,P$	1100 K,2 atm$1100\ \mathrm K,2\ \mathrm{atm}$

Property	Value
Graphite density 𝜌𝐶$\rho_C$	2250 kg m−3$2250\,\mathrm{kg\,m^{-3}}$
Carbon molar mass 𝑀𝐶$M_C$	0.012 kg mol−1$0.012\,\mathrm{kg\,mol^{-1}}$
Carbon-dioxide molar mass 𝑀𝐶𝑂2$M_{CO_2}$	0.044 kg mol−1$0.044\,\mathrm{kg\,mol^{-1}}$
Bulk oxygen mole fraction 𝑦𝑏$y_b$	0.21
Reference effective gas diffusivity 𝐷0$D_0$	0.136 cm2 s−1$0.136\,\mathrm{cm^2\,s^{-1}}$ at 273 K and 1 atm

Assumptions

Treat the source’s CO2-air diffusivity as a common effective diffusivity for the oxygen and carbon-dioxide counterdiffusion. This is an approximation to multicomponent air. Nitrogen has zero net flux, and the reaction exchanges one gas mole for one gas mole, so total gas molar flux is zero. Estimate the diffusivity by

𝐷=𝐷01 atm𝑃(𝑇273 K)3/2.$$D=D_0\frac{1\,\mathrm{atm}}{P}\left(\frac{T}{273\,\mathrm K}\right)^{3/2}.$$

Assume isothermal quasi-steady radial diffusion, no gas-phase reaction, constant density and length, and no transport through the ends. The maintained bulk gas has 𝑦𝑂2(𝑎 +𝛿) =𝑦𝑏$y_{O_2}(a+\delta)=y_b$, while the fast reaction gives 𝑦𝑂2(𝑎) =0$y_{O_2}(a)=0$. Keep 𝛿$\delta$ fixed as 𝑎$a$ decreases. Derive the constant 𝑟𝑁𝑂2$rN_{O_2}$ from a cylindrical shell balance, integrate Eq. 1, and relate oxygen consumption to the changing carbon mass. Do not use the initial area as a constant area for the shrinking rod.

Calculations

Use molar flux 𝑁𝐴$N_A$ in mol m−2 s−1$\mathrm{mol\,m^{-2}\,s^{-1}}$ and ideal-gas concentration 𝑐 =𝑃/(𝑅𝑇)$c=P/(RT)$, with 𝑅 =8.314 Pa m3 mol−1 K−1$R=8.314\,\mathrm{Pa\,m^3\,mol^{-1}\,K^{-1}}$. With negligible total molar flux, diffusion obeys

𝑁𝐴=−𝑐𝐷𝑑𝑦𝐴𝑑𝑠.$$N_A=-cD\frac{dy_A}{ds}.$$(1)

The coordinate 𝑠$s$ increases away from the reacting surface. Reactant flux is therefore negative. Report removal and production rates as positive magnitudes.

The oxygen consumption balances carbon-dioxide production mole for mole at the rod surface.

A steady cylindrical shell balance gives

𝑑(𝑟𝑁𝑂2)𝑑𝑟=0.$$\frac{d(rN_{O_2})}{dr}=0.$$

The positive oxygen consumption rate 𝑞$q$ satisfies 𝑞 = −2𝜋𝑟ℓ𝑁𝑂2$q=-2\pi r\ell N_{O_2}$.

Integrating Eq. 1 between 𝑎$a$ and 𝑎 +𝛿$a+\delta$ gives

𝑞(𝑎)=2𝜋ℓ𝑐𝐷𝑦𝑏ln⁡(𝑎+𝛿𝑎).$$q(a)=\frac{2\pi\ell cD y_b}{\ln\left(\frac{a+\delta}{a}\right)}.$$(2)

Convert 𝐷0 =1.36 ×10−5 m2 s−1$D_0=1.36\times10^{-5}\,\mathrm{m^2\,s^{-1}}$.

The scaled 𝐷 =5.49989 ×10−5 m2 s−1$D=5.49989\times10^{-5}\,\mathrm{m^2\,s^{-1}}$ and 𝑐 =22.1587 mol m−3$c=22.1587\,\mathrm{mol\,m^{-3}}$.

At 𝑎0$a_0$,

ln⁡(𝑎0+𝛿𝑎0)=ln⁡(1.5)=0.405465.$$\ln\left(\frac{a_0+\delta}{a_0}\right)=\ln(1.5)=0.405465.$$

Using Eq. 2, substitute the length and cylindrical logarithmic factor:

𝑞(𝑎0)=2𝜋(0.25)(22.1587)(5.49989×10−5)(0.21)0.405465$$q(a_0)=\frac{2\pi(0.25)(22.1587)(5.49989\times10^{-5})(0.21)}{0.405465}$$

Evaluating this expression gives

𝑞(𝑎0)=9.91479×10−4 mol s−1,$$q(a_0)=9.91479\times10^{-4}\,\mathrm{mol\,s^{-1}},$$˙𝑚𝐶𝑂2(0)=𝑀𝐶𝑂2𝑞(𝑎0)=4.36251×10−5 kg s−1.$$\dot m_{CO_2}(0)=M_{CO_2}q(a_0) =4.36251\times10^{-5}\,\mathrm{kg\,s^{-1}}.$$

The carbon mass is 𝑚𝐶 =𝜌𝐶𝜋𝑎2ℓ$m_C=\rho_C\pi a^2\ell$.

Its balance is

2𝜋𝜌𝐶𝑎ℓ𝑑𝑎𝑑𝑡=−𝑀𝐶𝑞(𝑎).$$2\pi\rho_C a\ell\frac{da}{dt}=-M_Cq(a).$$

Combining this balance with Eq. 2 yields

𝑑𝑎𝑑𝑡=−𝑀𝐶𝑐𝐷𝑦𝑏𝜌𝐶𝑎ln⁡(𝑎+𝛿𝑎),𝑎(0)=𝑎0.$$\frac{da}{dt}=-\frac{M_C cD y_b}{\rho_C a\ln\left(\frac{a+\delta}{a}\right)}, \qquad a(0)=a_0.$$(3)

Initially

𝑑𝑎𝑑𝑡=−3.36637×10−7 m s−1.$$\frac{da}{dt}=-3.36637\times10^{-7}\,\mathrm{m\,s^{-1}}.$$

The negative sign represents loss of radius.

Carbon consumption uses 𝑀𝐶$M_C$; carbon-dioxide mass production uses 𝑀𝐶𝑂2$M_{CO_2}$.

The gas-film diffusion time 𝛿2/𝐷 =0.455 s$\delta^2/D=0.455\,\mathrm s$ is much shorter than the initial shrinkage scale

𝑎0∣𝑑𝑎𝑑𝑡∣=2.97×104 s,$$\frac{a_0}{\left|\frac{da}{dt}\right|}=2.97\times10^4\,\mathrm s,$$

supporting quasi-steady transport initially.

The model excludes combustion heating, carbon-monoxide formation, end effects and changes in the flow as the rod shrinks.

Answer

The initial carbon-dioxide production rate is 9.91 ×10−4 mol s−1$9.91\times10^{-4}\ \mathrm{mol\,s^{-1}}$, or 4.36 ×10−5 kg s−1$4.36\times10^{-5}\ \mathrm{kg\,s^{-1}}$. Radius evolves according to Eq. 3, starting from 𝑎0 =0.010 m$a_0=0.010\ \mathrm m$; initially

𝑑𝑎𝑑𝑡=−3.37×10−7 m s−1.$$\frac{da}{dt}=-3.37\times10^{-7}\ \mathrm{m\,s^{-1}}.$$

2 Nitric-oxide removal at a catalyst

Film resistance and finite surface reaction

Nitric oxide (NO) emission from automobile exhaust can be reduced by using a catalytic converter. The following reaction occurs at the catalytic surface:

NO+CO→0.5 N2+CO2$$\mathrm{NO + CO \rightarrow 0.5\ N_2 + CO_2}$$

The concentration of NO is reduced by passing the exhaust gases over the surface of the catalyst, and the rate of reduction at the catalyst surface is as follows:

𝑁𝑁𝑂=−𝑘′′𝐶𝑁𝑂(0),$$N_{NO}=-k'' C_{NO(0)},$$

where 𝐶𝑁𝑂(0)$C_{NO(0)}$ is the concentration of NO at the catalyst surface.

As a first approximation, it may be assumed that NO reaches the surface by one-dimensional diffusion through a thin gas film of thickness L that adjoins the surface, and that the effects of bulk motion is negligible. Determine the mole fraction of NO at the catalytic surface and the NO removal rate for a catalytic surface area of 200 cm2$\mathrm{cm^2}$.

Data: The exhaust gas is at 500 ∘C$500\,{}^\circ\mathrm{C}$ and 1.2 bar, the mole fraction of NO at L is 0.15, 𝐷𝐴𝐵$D_{AB}$ = 10−4$10^{-4}$ m2/s$\mathrm{m^2/s}$, 𝑘′′$k''$ = 0.05 m/s and the film thickness is L = 1 mm.

Worked solution

Figure 2: Nitric oxide diffuses through a nonreacting gas film to the catalyst. Its surface concentration is not zero when surface reaction is finite. The solution measures x from the catalyst toward the bulk.

Data

Property	Value
Bulk NO mole fraction 𝑦𝑏$y_b$	0.15
Film thickness 𝐿$L$	0.0010 m$0.0010\,\mathrm m$
Effective NO diffusivity 𝐷$D$	1.0 ×10−4 m2 s−1$1.0\times10^{-4}\,\mathrm{m^2\,s^{-1}}$
Surface rate constant 𝑘𝑠$k_s$	0.050 m s−1$0.050\,\mathrm{m\,s^{-1}}$
Catalytic area 𝐴𝑠$A_s$	200 cm2$200\,\mathrm{cm^2}$
NO molar mass 𝑀𝑁𝑂$M_{NO}$	0.030 kg mol−1$0.030\,\mathrm{kg\,mol^{-1}}$

Assumptions

Assume steady isothermal one-dimensional diffusion with constant pressure and 𝐷$D$, no gas-phase reaction, sufficient CO, and the source’s first approximation of negligible bulk-motion contribution to NO transport. The reaction changes total gas mole number, so this approximation is not exact equimolar diffusion. Do not interpret the 0.15 mole fraction as realistic dilute exhaust data. It is the supplied idealised problem value. At the surface,

𝑁𝑁𝑂(0)=−𝑘𝑠𝑐𝑦𝑠,𝑦𝑠=𝑦𝑁𝑂(0),$$N_{NO}(0)=-k_s c y_s, \qquad y_s=y_{NO}(0),$$

and 𝑦𝑁𝑂(𝐿) =𝑦𝑏$y_{NO}(L)=y_b$. Integrate Eq. 1 in 𝑥$x$ and equate the film flux to the surface reaction flux.

Calculations

The finite catalytic reaction leaves a nonzero nitric-oxide concentration at the surface.

The steady flux is constant and the profile is linear:

𝑦𝑁𝑂(𝑥)=𝑦𝑠+(𝑦𝑏−𝑦𝑠)𝑥𝐿,$$y_{NO}(x)=y_s+(y_b-y_s)\frac{x}{L},$$𝑁𝑁𝑂=−𝑐𝐷𝐿(𝑦𝑏−𝑦𝑠).$$N_{NO}=-\frac{cD}{L}(y_b-y_s).$$

Matching the reaction flux gives

𝐷𝐿(𝑦𝑏−𝑦𝑠)=𝑘𝑠𝑦𝑠,𝑦𝑠=𝑦𝑏1+𝑘𝑠𝐿𝐷=0.151+0.50=0.100.$$\begin{aligned} \frac{D}{L}(y_b-y_s)&=k_s y_s,\\ y_s&=\frac{y_b}{1+\frac{k_sL}{D}}\\ &=\frac{0.15}{1+0.50}=0.100. \end{aligned}$$(4)

Convert 𝑇 =773.15 K$T=773.15\,\mathrm K$, 𝑃 =120000 Pa$P=120000\,\mathrm{Pa}$ and 𝐴𝑠 =0.0200 m2$A_s=0.0200\,\mathrm{m^2}$.

Then 𝑐 =18.6684 mol m−3$c=18.6684\,\mathrm{mol\,m^{-3}}$ and 𝑁𝑁𝑂 = −0.0933421 mol m−2 s−1$N_{NO}=-0.0933421\,\mathrm{mol\,m^{-2}\,s^{-1}}$.

The positive removal rate is

˙𝑛𝑁𝑂=−𝐴𝑠𝑁𝑁𝑂=1.86684×10−3 mol s−1,$$\dot n_{NO}=-A_sN_{NO} =1.86684\times10^{-3}\,\mathrm{mol\,s^{-1}},$$˙𝑚𝑁𝑂=𝑀𝑁𝑂˙𝑛𝑁𝑂=5.60052×10−5 kg s−1.$$\dot m_{NO}=M_{NO}\dot n_{NO} =5.60052\times10^{-5}\,\mathrm{kg\,s^{-1}}.$$

The equivalent concentration-based coefficient is

𝑘𝑜𝑣𝑒𝑟𝑎𝑙𝑙=(𝐿𝐷+1𝑘𝑠)−1.$$k_{overall}=\left(\frac{L}{D}+\frac{1}{k_s}\right)^{-1}.$$

Its value is 0.0333333 m s−1$0.0333333\,\mathrm{m\,s^{-1}}$.

The film resistance 𝐿/𝐷 =10 s m−1$L/D=10\,\mathrm{s\,m^{-1}}$ and reaction resistance 1/𝑘𝑠 =20 s m−1$1/k_s=20\,\mathrm{s\,m^{-1}}$ add.

Reaction supplies twice the film resistance, but neither can be ignored.

The calculated surface fraction in Eq. 4 confirms finite reaction, rather than a perfectly absorbing surface.

These results apply only to the stated film approximation.

Answer

The surface NO mole fraction is 0.100$0.100$. The positive removal rate is 1.87 ×10−3 mol s−1$1.87\times10^{-3}\ \mathrm{mol\,s^{-1}}$, equivalent to 5.60 ×10−5 kg s−1$5.60\times10^{-5}\ \mathrm{kg\,s^{-1}}$, within the stipulated film approximation.