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Chapter 5: Problem 6

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 5-mm-thick sheet of palladium having an area of \(0.20 \mathrm{~m}^{2}\) at \(500^{\circ} \mathrm{C}\). Assume a diffusion coefficient of \(1.0 \times 10^{-8} \mathrm{~m}^{2} / \mathrm{s}\), that the concentrations at the high- and low-pressure sides of the plate are \(2.4\) and \(0.6 \mathrm{~kg}\) of hydrogen per cubic meter of palladium, and that steady-state conditions have been attained.

Short Answer

Expert verified
In one hour, 2.592 kg of hydrogen passes through the specified palladium sheet.

Step by step solution

01

List known parameters and formula

In this problem, we are given: - Diffusion coefficient (D): \(1.0 × 10^{-8} \ \mathrm{m}^2\ \mathrm{s}^{-1}\) - Thickness of the sheet (L): \(5\ \mathrm{mm}\) or \(0.005\ \mathrm{m}\) - Area of the sheet (A): \(0.20\ \mathrm{m}^2\) - Temperature: \(T = 500^{\circ} \mathrm{C} = 773\ \mathrm{K}\) (converted to Kelvin) - Concentrations on the high-pressure side (C1): \(2.4\ \mathrm{kg/m^3}\) - Concentrations on the low-pressure side (C2): \(0.6\ \mathrm{kg/m^3}\) The formula we will use is Fick's first law of diffusion: \(J = -D \frac{dC}{dx}\) where J is the diffusion flux, D is the diffusion coefficient, and \(\frac{dC}{dx}\) is the concentration gradient.
02

Calculate the concentration gradient

The concentration gradient, \(\frac{dC}{dx}\), can be found by taking the difference in concentrations and dividing by the thickness of the sheet, as follows: \(\frac{dC}{dx} = \frac{C1 - C2}{L} = \frac{2.4 - 0.6}{0.005} = \frac{1.8}{0.005} = 360\ \mathrm{kg/m^4}\)
03

Calculate the diffusion flux

Plug the concentration gradient and the diffusion coefficient into Fick's first law of diffusion: \(J = -D \frac{dC}{dx} = -1.0 × 10^{-8} \cdot 360 = -3.6 × 10^{-6}\ \mathrm{kg/m^2s}\) The negative sign indicates that the diffusion flux is in the direction of decreasing concentration.
04

Calculate the total amount of hydrogen diffused

Now we need to find the total amount of hydrogen diffused through the sheet in one hour. To do this, we'll multiply the diffusion flux by the area of the sheet and the time in seconds: Amount of hydrogen = \(J \cdot A \cdot t = -3.6 × 10^{-6}\ \mathrm{kg/m^2s} \cdot 0.20\ \mathrm{m^2} \cdot 3600\ \mathrm{s} = -2.592\ \mathrm{kg}\) Since we cannot have a negative amount of hydrogen, we take the absolute value: Amount of hydrogen = \(| -2.592\ \mathrm{kg} | = 2.592\ \mathrm{kg}\)
05

Final answer

The amount of hydrogen that passes per hour through the specified palladium sheet is: 2.592 kg of hydrogen.

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Most popular questions from this chapter

When \(\alpha\)-iron is subjected to an atmosphere of hydrogen gas, the concentration of hydrogen in the iron, \(C_{\mathrm{H}}\) (in weight percent), is a function of hydrogen pressure, \(p_{\mathrm{H}_{2}}\) (in MPa), and absolute temperature \((T)\) according to $$ C_{\mathrm{H}}=1.34 \times 10^{-2} \sqrt{p_{\mathrm{H}}_{2}} \exp \left(-\frac{27.2 \mathrm{~kJ} / \mathrm{mol}}{R T}\right) $$ Furthermore, the values of \(D_{0}\) and \(Q_{d}\) for this diffusion system are \(1.4 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\) and \(13,400 \mathrm{~J} / \mathrm{mol}\), respectively. Consider a thin iron membrane \(1 \mathrm{~mm}\) thick that is at \(250^{\circ} \mathrm{C}\). Compute the diffusion flux through this membrane if the hydrogen pressure on one side of the membrane is \(0.15 \mathrm{MPa}\) (1.48 atm), and on the other side \(7.5 \mathrm{MPa}\) (74 atm).

The activation energy for the diffusion of carbon in chromium is \(111,000 \mathrm{~J} / \mathrm{mol}\). Calculate the diffusion coefficient at \(1100 \mathrm{~K}\left(827^{\circ} \mathrm{C}\right)\), given that \(D\) at \(1400 \mathrm{~K}\left(1127^{\circ} \mathrm{C}\right)\) is \(6.25 \times\) \(10^{-11} \mathrm{~m}^{2} / \mathrm{s}\)

Carbon is allowed to diffuse through a steel plate \(15 \mathrm{~mm}\) thick. The concentrations of carbon at the two faces are \(0.65\) and \(0.30 \mathrm{~kg} \mathrm{C} / \mathrm{m}^{3}\) Fe, which are maintained constant. If the preexponential and activation energy are \(6.2 \times\) \(10^{-7} \mathrm{~m}^{2} / \mathrm{s}\) and \(80,000 \mathrm{~J} / \mathrm{mol}\), respectively, compute the temperature at which the diffusion flux is \(1.43 \times 10^{-9} \mathrm{~kg} / \mathrm{m}^{2} \cdot \mathrm{s}\).

The outer surface of a steel gear is to be hardened by increasing its carbon content. The carbon is to be supplied from an external carbon-rich atmosphere, which is maintained at an elevated temperature. A diffusion heat treatment at \(850^{\circ} \mathrm{C}(1123 \mathrm{~K})\) for \(10 \mathrm{~min}\) increases the carbon concentration to \(0.90 \mathrm{wt} \%\) at a position \(1.0 \mathrm{~mm}\) below the surface. Estimate the diffusion time required at \(650^{\circ} \mathrm{C}(923 \mathrm{~K})\) to achieve this same concentration also at a) 1.0-mm position. Assume that the surface carbon content is the same for both heat treatments, which is maintained constant. Use the diffusion data in Table \(5.2\) for C diffusion in \(\alpha\)-Fe.

(a) Compare interstitial and vacancy atomic mechanisms for diffusion. (b) Cite two reasons why interstitial diffusion is normally more rapid than vacancy diffusion.
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