Question 5.5: Design of an Iron Membrane An impermeable cylinder 3 cm in d......
Design of an Iron Membrane
An impermeable cylinder 3 cm in diameter and 10 cm long contains a gas that includes 5 × 10^{19} N atoms per cm³ and 5 × 10^{19} H atoms per cm³ on one side of an iron membrane (Figure 5-16). Gas is continuously introduced to the pipe to ensure concentration of nitrogen and hydrogen. The gas on the other side of the membrane contains a constant 1× 10^{18} N atoms per cm³ and 1 × 10^{18} H atoms per cm³. The entire system is to operate at 700°C, at which iron has the BCC structure. Design an iron membrane that will allow no more than 1% of the nitrogen to be lost through the membrane each hour, while allowing 90% of the hydrogen to pass through the membrane per hour.
The total number of nitrogen atoms in the container is
(5 × 10^{19} N atoms/cm³)(π/4)(3 cm)²(10 cm) = 3.534 × 10^{21} N atoms
The maximum number of atoms to be lost is 1% of this total, or
N atom loss per h = (0.01)(3.534 × 10^{21}) = 3.534 × 10^{19} N atoms/h
N atom loss per s = (3.534 × 10^{19} N atoms/h)/(3600 s/h)
= 9.817 × 10^{15} N atoms/s
The flux is then
J =\frac{ 9.817 × 10^{15} \ N \ atoms/s}{(\frac{\pi}{4})(3 \ cm)^2}
= 1.389 × 10^{15} \ \frac{N \ atoms}{cm^2 ⋅ s}
Using Equation 5-4 and values from Table 5-1, the diffusion coefficient of nitrogen in BCC iron at 700°C = 973 K is
D = D_{0} \exp (\frac{-Q}{RT}) (5-4)
D_{N} = 0.0047 \ \frac{cm^2}{s} \exp[\frac{-18,300 \ \frac{cal}{mol}}{(1.987 \ \frac{cal}{mol ⋅ K})(973 \ K)}]
=(0.0047 \ \frac{cm^2}{s})(7.748 × 10^{-5}) = 3.642 × 10^{-7} \frac{cm^2}{s}
From Equation 5-3:
J = -D \frac{dc}{dx} (5-3)
J = -D (\frac{Δc}{Δx})= 1.389 × 10^{15} \ \frac{N \ atoms}{cm^2 ⋅ s}
Δx = – \frac{DΔc}{J}= \frac{(-3.642 × 10^{-7} cm ^2 /s)(1 × 10^{18} \ – \ 5 × 10^{19} \ \frac{N \ atoms}{cm^3})}{1.389 × 10^{15} \ \frac{N \ atoms}{cm^2 ⋅ s}}
Δx = 0.013 cm = minimum thickness of the membrane
In a similar manner, the maximum thickness of the membrane that will permit 90% of the hydrogen to pass can be calculated as
H atom loss per h = (0.90)(3.534 × 10^{21}) = 3.180 × 10^{21} H atoms/h
H atom loss per s = 8.836 × 10^{17} H atoms/s
J = 1.250 × 10^{17} \ \frac{H \ atoms}{cm^2 ⋅ s}
From Equation 5-4 and Table 5-1,
D_{H} = 0.0012 \ \frac{cm^2}{s} \exp [\frac{-3,600 \ \frac{cal}{mol }}{(1.987 \ \frac{cal}{mol ⋅ K})(973 \ K)}]= 1.864 × 10^{-4} \ cm^2 /s
Since
Δx = – \frac{DΔc}{J}
Δx = \frac{(-1.864 × 10^{-4} \ \frac{cm^2}{s})(1 × 10^{18} \ – \ 5 × 10^{19} \ \frac{H \ atoms}{cm^3} )}{1.250 × 10^{17} \ \frac{H \ atoms}{cm^2 ⋅ s}}
= 0.073 cm = maximum thickness
An iron membrane with a thickness between 0.013 and 0.073 cm will be satisfactory.
| Table 5-1 Diffusion data for selected materials | ||
| Diffusion Couple | Q (cal/mol) | D_{0} (cm^2/s) |
| Interstitial diffusion: | ||
| C in FCC iron | 32,900 | 0.23 |
| C in BCC iron | 20,900 | 0.011 |
| N in FCC iron | 34,600 | 0.0034 |
| N in BCC iron | 18,300 | 0.0047 |
| H in FCC iron | 10,300 | 0.0063 |
| H in BCC iron | 3,600 | 0.0012 |
| Self-diffusion (vacancy diffusion): | ||
| Pb in FCC Pb | 25,900 | 1.27 |
| Al in FCC Al | 32,200 | 0.10 |
| Cu in FCC Cu | 49,300 | 0.36 |
| Fe in FCC Fe | 66,700 | 0.65 |
| Zn in HCP Zn | 21,800 | 0.1 |
| Mg in HCP Mg | 32,200 | 1.0 |
| Fe in BCC Fe | 58,900 | 4.1 |
| W in BCC W | 143,300 | 1.88 |
| Si in Si (covalent) | 110,000 | 1800.0 |
| C in C (covalent) | 163,000 | 5.0 |
| Heterogeneous diffusion (vacancy diffusion): | ||
| Ni in Cu | 57,900 | 2.3 |
| Cu in Ni | 61,500 | 0.65 |
| Zn in Cu | 43,900 | 0.78 |
| Ni in FCC iron | 64,000 | 4.1 |
| Au in Ag | 45,500 | 0.26 |
| Ag in Au | 40,200 | 0.072 |
| Al in Cu | 39,500 | 0.045 |
| Al in Al_{2}O_{3} | 114,000 | 28.0 |
| O in Al_{2}O_{3} | 152,000 | 1900.0 |
| Mg in MgO | 79,000 | 0.249 |
| O in MgO | 82,100 | 0.000043 |
Based on several sources, including Adda, Y. and Philibert, J., La Diffusion dans les Solides, Vol. 2, 1966.