Material 9
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This document discusses diffusion through a membrane and solving related problems. It contains: 1) Information on Fick's first law of diffusion and the Arrhenius equation for calculating diffusion rates. 2) A sample problem calculating the number of nickel ions diffusing through MgO per second given parameters like thickness, temperature, and diffusion coefficient. 3) A multi-part problem designing an iron membrane thickness to allow less than 1% nitrogen loss per hour and 90% hydrogen passage per hour given gas concentrations and diffusion rates at 700C.
Material 9
- 1. DIFFUSION-2 ENG. KAREEM H. MOKHTAR
- 2. ARRHENIUS EQUATION • Rate = c e-E/RT • Rate: rate of diffusion (m2/s or jump per second) • E: activation energy • R: gas constant • T: temperature • ln(rate) = ln(c) – E/RT • Straight line equation Y= m x + cLn (rate) 1/T Slope= -E/R
- 3. FICK’S FIRST LAW • The rate at which atoms, ions, particles or other species diffuse in a material can be measured by the flux J. Here we are mainly concerned with diffusion of ions or atoms. The flux J is defined as the number of atoms passing through a plane of unit area per unit time
- 4. SHEET 4 • Q6) 0.05 cm layer of magnesium oxide (MgO) is deposited between layers of nickel (Ni) and tantalum (Ta) to provide a diffusion barrier that prevents reactions between the two metals. At 1400°C, nickel ions diffuse through the MgO ceramic to the tantalum. Determine the number of nickel ions that pass through the MgO per second. At 1400°C, the diffusion coefficient of nickel ions in MgO is 9 * 10-12 cm2/s and the concentration of Ni on Ni/MgO surface equals 8.573 x 1022 atoms/cm3
- 5. ANSWER • J =-D 𝑑𝑒𝑙𝑡𝑎 𝐶 𝐷𝑒𝑙𝑡𝑎 𝑥 • 𝑑𝑒𝑙𝑡𝑎 𝐶 𝐷𝑒𝑙𝑡𝑎 𝑥 = 0−8.573 𝑥 1022 0.05 = -1.715 * 1024 atoms/ cm3 .cm • J= - (9 * 10-12 ) x( -1.715 * 1024 ) = 1.543*1013 𝑁𝑖 𝑎𝑡𝑜𝑚𝑠 𝑐𝑚2 𝑠 • Total Ni passing in one second = J 𝑁𝑖 𝑎𝑡𝑜𝑚𝑠 cm2 𝑠 x area (cm2) = 1.543*1013 x (2cm x 2cm) • = 6.17 x 1013 Ni atoms/ s
- 6. SHEET 4 • Q7) An impermeable cylinder 3 cm in diameter and 10 cm long contains a gas that includes 0.5 * 1020 N atoms per cm3 and 0.5 * 1020 H atoms per cm3 on one side of an iron membrane . Gas is continuously introduced to the pipe to ensure a constant concentration of nitrogen and hydrogen. The gas on the other side of the membrane includes a constant 1 * 1018 N atoms per cm3 and 1 * 1018 H atoms per cm3. The entire system is to operate at 700°C. Design an iron membrane (BCC) that will
- 7. SHEET 4 – Q7 • allow no more than 1% of the nitrogen to be lost through the membrane each hour • allow 90% of the hydrogen to pass through the membrane per hour. • Satisfy both later conditions • given that for N D0 = 0.0047 cm2/s,, and E= 18300 cal/mol • and for H D0 = 0.0012 cm2/s,, and E= 3600 cal/mol)
- 8. ANSWER
- 9. TO ALLOW NO MORE THAN 1% OF THE NITROGEN TO BE LOST THROUGH THE MEMBRANE EACH HOUR • The total number of nitrogen atoms in the container is • 0.5 * 1020 x (pi/4) x 32 cm2 x 10 cm = 35.34 x 1020 N atom • The maximum number of atoms to be lost = 1% from N per hour • N atoms lost= 35.34 x 1020 N x 0.01 = 35.34 x 1018 N atoms per hour J = 35.34 x 1018 3600 ∗32 ∗ pi 4 = 0.00139x 1018 N atoms per cm2 per second
- 10. TO ALLOW NO MORE THAN 1% OF THE NITROGEN TO BE LOST THROUGH THE MEMBRANE EACH HOUR • Rate of diffusion (D) = c e-E/RT = 0.0047 x e-18300/1.98x973 =3.64x10-7 cm2/s • dx= −𝐷 ∗ 𝑑𝑐 𝐽 = −3.64 ∗ 10−7 ∗ 1∗1018−0.5∗1020 0.00139 ∗1018 = 0.013 𝑐𝑚
- 11. ALLOW 90% OF THE HYDROGEN TO PASS THROUGH THE MEMBRANE PER HOUR. • H atom loss per hour = (0.90)(35.343 * 1020)= 31.80 * 1020 • J = 31.8 x 1020 3600 ∗32 ∗ pi 4 = 1.2 ∗ 1017 H atoms/ cm2 s • D = 0.0012 e-3600/(1.98*973) = 1.86 x 10-4 • dx= ( -1.86* 10-4 * -49 * 1018 )/ 1.25 * 1017 = 0.073 cm
- 12. SATISFY BOTH LATER CONDITIONS • An iron membrane with a thickness between 0.013 and 0.073 cm will be satisfactory.