Hydrogen gas (H2) is small enough in size that it can diffuse through metals, ev
ID: 502301 • Letter: H
Question
Hydrogen gas (H2) is small enough in size that it can diffuse through metals, even closed packed FCC structures such as Palladium (Pd). Palladium metal is often used to separate/purify H2. If the diffusion coefficient for H_2 moving through Pd is 1.7 times 10^-8 m^2/s and 600 degree C, how much H_2 in [kg] will pass through 0.5 square meter of a 5 mm thick sheet/membrane of this metal in 1 hour? Assume the flux of hydrogen through the membrane is at steady state, and the upstream surface of the membrane is exposed to a H_2 concentration of 1.0 kg/m^3 and the downstream concentration is 0.4 kg/m^3. 3 pts for Give and Find - avoid rewriting the question, instead symbolically identify the finds. Identify both the solute and solvent species in the above problem What is the likely mechanism for diffusion of H_2 in Pd? Interstitial or Vacancy What is the steady state flux of hydrogen through the Pd sheet in [kg/m^2/s] How much hydrogen in [kg] will pass through the above sheet in 1 hour ? How much time in [hours, h] would have to pass to separate 1 metric ton of H_2 (1000 kg) across that Pd membrane/sheet ? If the flux of H_2 is 1 times 10^-5 kg/m^2/s at 750 degree C with the same surface concentrations, what is the activation energy (Qd) and pre-exponential factor (Do) of H_2 diffusing through the Pd crystal lattice ? Reflect on what attributes of the Pd membrane material can be changed in order to get faster rates of mass transfer?Explanation / Answer
Solution:
IIa) Here solute is palladium and solvent is hydrogen gas.
IIb) The mechanism of diffusion of hydrogen in FCC lattice of Pd is through interstitial voids (octahedral voids).
IIc) Steady state diffusion equation can be given as follows:
J = - D (C/x) [where, J = steady state flux]
= - (1.7x10-8 m2/s) [(0.4 - 1.0 kg/m3)/(5x10-3 m)]
= 2x10-6 kg/m2s
IId) This problem calls for the mass of hydrogen, per hour, that diffuses through a Pd sheet of said thickness. It can be solved by employing the equations of diffusions to calculate the mass (M) as follows.
M = JAt = - DAt (C/x)
= - (1.7x10-8 m2/s)(0.5 m2)(3600 s) [(0.4 - 1.0 kg/m3)/(5x10-3 m)]
= 3.67x10-3 kg/h
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