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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 6 -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 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: Approximately 51.0 kg.

Step by step solution

01

Write down Fick's First Law of Diffusion

The formula for Fick's First Law of Diffusion is: \(J = -D \frac{dC}{dx}\) where \(J\) is the diffusion flux (in kg/m²s), \(D\) is the diffusion coefficient (in m²/s), and \(\frac{dC}{dx}\) is the concentration gradient (in kg/m³ per meter).
02

Calculate the concentration gradient

We are given the concentrations \(C_1\) and \(C_2\) on the high- and low-pressure sides of the sheet as 2.0 kg/m³ and 0.4 kg/m³, respectively. Also, we know the thickness of the sheet \(x\) is 6 mm or 0.006 meters. The concentration gradient can be calculated as: \(\frac{dC}{dx} = \frac{C_2 - C_1}{x} = \frac{0.4-2.0}{0.006}\)
03

Calculate the diffusion flux

Using the diffusion coefficient \(D = 1.7 \times 10^{-8}\) m²/s, we can calculate the diffusion flux \(J\): \(J = -D \frac{dC}{dx} = -1.7 \times 10^{-8} \times \frac{0.4-2.0}{0.006}\)
04

Convert the diffusion flux to mass per hour

We are given the area \(A = 0.25\) m², so we can now calculate the mass of hydrogen passing through per second: \(m_{s} = J \times A = -1.7 \times 10^{-8} \times \frac{0.4-2.0}{0.006} \times 0.25\) To convert this result to mass per hour, we'll need to multiply by the number of seconds in an hour (3600 s): \(m_{h} = m_{s} \times 3600\)
05

Calculate the final mass

Now we can calculate the total mass of hydrogen passing through the palladium sheet per hour: \(m_{h} = -1.7 \times 10^{-8} \times \frac{0.4-2.0}{0.006} \times 0.25 \times 3600\) After performing the calculations, the final mass of hydrogen passing through the palladium sheet per hour is found to be: \(m_{h} \approx 51.0\,\text{kg}\).

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Most popular questions from this chapter

The preexponential and activation energy for the diffusion of chromium in nickel are \(1.1 \times 10^{-4} \mathrm{m}^{2} / \mathrm{s}\) and \(272,000 \mathrm{J} / \mathrm{mol},\) respec- tively. At what temperature will the diffusion coefficient have a value of \(1.2 \times 10^{-14} \mathrm{m}^{2} / \mathrm{s} ?\)

Self-diffusion involves the motion of atoms that are all of the same type; therefore it is not subject to observation by compositional changes, as with inter-diffusion. Suggest one way in which self-diffusion may be monitored.

At approximately what temperature would a specimen of \(\gamma\) -iron have to be carburized for \(4 \mathrm{h}\) to produce the same diffusion result as at \(1000^{\circ} \mathrm{C}\) for \(12 \mathrm{h} ?\)

The diffusion coefficients for nickel in iron are given at two temperatures: $$\begin{array}{cc} \boldsymbol{T}(\boldsymbol{K}) & \boldsymbol{D}\left(\boldsymbol{m}^{2} / \boldsymbol{s}\right) \\ \hline 1473 & 2.2 \times 10^{-15} \\ 1673 & 4.8 \times 10^{-14} \end{array}$$ (a) Determine the values of \(D_{0}\) and the activation energy \(Q_{d}\) (b) What is the magnitude of \(D\) at \(1300^{\circ} \mathrm{C}\) \((1573 \mathrm{K}) ?\)

When \(\alpha\) -iron is subjected to an atmosphere of nitrogen gas, the concentration of nitrogen in the iron, \(C_{\mathrm{N}}\) (in weight percent), is a function of hydrogen pressure, \(p_{\mathrm{N}_{2}}(\text { in } \mathrm{MPa}),\) and absolute temperature \((T)\) according to $$C_{\mathrm{N}}=4.90 \times 10^{-3} \sqrt{p_{\mathrm{N}_{2}}} \exp \left(-\frac{37.6 \mathrm{kJ} / \mathrm{mol}}{R T}\right)$$.Furthermore, the values of \(D_{0}\) and \(Q_{d}\) for this diffusion system are \(3.0 \times 10^{-7} \mathrm{m}^{2} / \mathrm{s}\) and \(76,150 \mathrm{J} / \mathrm{mol}\), respectively. Consider a thin iron membrane \(1.5 \mathrm{mm}\) thick that is at \(300^{\circ} \mathrm{C}\) Compute the diffusion flux through this membrane if the nitrogen pressure on one side of the membrane is \(0.10 \mathrm{MPa}(0.99 \mathrm{atm}),\) and on the other side \(5.0 \mathrm{MPa}(49.3 \mathrm{atm})\).
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