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Chapter 5: Problem 8

The purification of hydrogen gas by diffusion through a palladium sheet was discussed in Section 5.3. Compute the number of kilograms of hydrogen that pass per hour through a \(6-\mathrm{mm}\) thick sheet of palladium having an area of \(0.25 \mathrm{~m}^{2}\) at \(600^{\circ} \mathrm{C}\). Assume a diffusion coefficient of \(1.7 \times 10^{-8} \mathrm{~m}^{2} / \mathrm{s}\), that the respective concentrations at the high- and low-pressure sides of the plate are \(2.0\) and \(0.4 \mathrm{~kg}\). of hydrogen per cubic meter of palladium, and that steady-state conditions have been attained.

Short Answer

Expert verified
Answer: The total mass of hydrogen gas that passes through the palladium sheet per hour is 4.093 x 10^{-3} kg.

Step by step solution

01

Write down the known values

The thickness of the sheet, \(d = 6 \times 10^{-3} \mathrm{~m}\). The area of the sheet, \(A = 0.25 \mathrm{~m}^2\). The diffusion coefficient, \(D = 1.7 \times 10^{-8} \mathrm{~m}^2 /\mathrm{s}\). The concentration at the high-pressure side, \(C_1 = 2.0 \mathrm{~kg/m}^3\). The concentration at the low-pressure side, \(C_2 = 0.4 \mathrm{~kg/m}^3\).
02

Apply Fick's first law

Fick's first law of diffusion states that: \(J = -D \frac{dC}{dx}\) where \(J\) is the mass flux, \(D\) is the diffusion coefficient, and \(\frac{dC}{dx}\) is the concentration gradient across the sheet.
03

Calculate the concentration gradient

Since we know the concentration at both sides of the sheet, \(C_1\) and \(C_2\), and the thickness \(d\), we can calculate the concentration gradient: \(\frac{dC}{dx} = \frac{C_1 - C_2}{d} = \frac{2.0 - 0.4}{6 \times 10^{-3}} = 266.67 \mathrm{~kg/m^4}\)
04

Calculate the mass flux

Now, we can substitute the given diffusion coefficient, \(D\), and the calculated concentration gradient, \(\frac{dC}{dx}\), into Fick's first law to calculate the mass flux, \(J\): \(J = -D \frac{dC}{dx} = -1.7 \times 10^{-8} \cdot 266.67 = -4.533 \times 10^{-6} \mathrm{~kg/m^2s}\).
05

Calculate the total mass passing through the sheet per hour

To find the total mass of hydrogen gas that passes through the sheet per hour, we need to multiply the mass flux, \(J\), with the area of the sheet, \(A\), and convert the time unit to hours: \(M = J \cdot A \cdot t = -4.533 \times 10^{-6} \cdot 0.25 \cdot 3600 = -4.093 \times 10^{-3} \mathrm{~kg}\), where \(t\) is the time in hours. Since the mass flow is a positive value we do not consider the negative sign. Hence, the total mass of hydrogen gas that passes through the palladium sheet per hour is \(4.093 \times 10^{-3} \mathrm{~kg}\).

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Most popular questions from this chapter

Consider the diffusion of some hypothetical metal Y into another hypothetical metal \(Z\) at \(950^{\circ} \mathrm{C}\); after \(10 \mathrm{~h}\) the concentration at the \(0.5 \mathrm{~mm}\) position (in metal \(Z\) ) is \(2.0 \mathrm{wt} \% \mathrm{Y}\). At what position will the concentration also be \(2.0 \mathrm{wt} \% \mathrm{Y}\) after a \(17.5-\mathrm{h}\) heat treatment again at \(950^{\circ} \mathrm{C}\) ? Assume preexponential and activation energy values of \(4.3 \times 10^{-4} \mathrm{~m}^{2} / \mathrm{s}\) and \(180,000 \mathrm{~J} / \mathrm{mol}\), respectively, for this diffusion system.

Consider a diffusion couple between silver and a gold alloy that contains 10 wt \% silver. This couple is heat treated at an elevated temperature and it was found that after 850 s, the concentration of silver had increased to \(12 \mathrm{wt} \%\) at \(10 \mu \mathrm{m}\) from the interface into the Ag-Au alloy. Assuming preexponential and activation energy values of \(7.2 \times 10^{-6}\) \(\mathrm{m}^{2} / \mathrm{s}\) and \(168,000 \mathrm{~J} / \mathrm{mol}\), respectively, compute the temperature of this heat treatment. (Note: You may find Figure \(5.13\) and Equation \(5.15\) helpful.) For a steel alloy, it has been determined that a carburizing heat treatment of \(15 \mathrm{~h}\) duration will raise the carbon concentration to \(0.35\) wt \(\%\) at a point \(2.0 \mathrm{~mm}\) from the surface. Estimate the time necessary to achieve the same concentration at a \(6.0-\mathrm{mm}\) position for an identical steel and at the same carburizing temperature.

Consider a diffusion couple composed of two cobalt-iron alloys; one has a composition of \(75 \mathrm{wt} \%\) Co-25 wt\% Fe; the other alloy composition is \(50 \mathrm{wt} \%\) Co-50 wt \(\% \mathrm{Fe}\). If this couple is heated to a temperature of \(800^{\circ} \mathrm{C}(1073 \mathrm{~K})\) for \(20,000 \mathrm{~s}\) determine how far from the original interface into the \(50 \mathrm{wt} \%\) Co-50 wt \% Fe alloy the composition has increased to \(52 \mathrm{wt} \%\) Co-48 wt Fe. For the diffusion coefficient, assume values of \(6.6 \times 10^{-t}\) \(\mathrm{m}^{2} / \mathrm{s}\) and \(247,000 \mathrm{~J} / \mathrm{mol}\), respectively, for the pre exponential and activation energy.

Indium atoms are to be diffused into a silicon wafer using both predeposition and drive-in heat treatments; the background concentration of In in this silicon material is known to be \(2 \times 10^{20}\) atoms \(/ \mathrm{m}^{3}\). The drive-in diffusion treatment is to be carried out at \(1175^{\circ} \mathrm{C}\) for a period of \(2.0 \mathrm{~h}\), whichgives a junction depth \(x_{j}\) of \(2.35 \mu \mathrm{m}\). Compute the predeposition diffusion time at \(925^{\circ} \mathrm{C}\) if the surface concentration is maintained at a constant level of \(2.5 \times 10^{26}\) atoms \(/ \mathrm{m}^{3}\). For the diffusion of In in Si. values of \(Q_{d}\) and \(D_{0}\) are \(3.63 \mathrm{eV} /\) atom and \(7.85 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\), respectively.

Briefly explain the concept of steady state as it applies to diffusion.
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