Chapter 5: Problem 5
(a) Briefly explain the concept of a driving force. (b) What is the driving force for steady-state diffusion?
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Chapter 5: Problem 5
(a) Briefly explain the concept of a driving force. (b) What is the driving force for steady-state diffusion?
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The diffusion coefficients for silver in copper are given at two temperatures: $$ \begin{array}{cc} \hline T\left({ }^{\circ} \mathrm{C}\right) & D\left(\mathrm{~m}^{2} / \mathrm{s}\right) \\ \hline 650 & 5.5 \times 10^{-16} \\ 900 & 1.3 \times 10^{-13} \\ \hline \end{array} $$ (a) Determine the values of \(D_{0}\) and \(Q_{d}\). (b) What is the magnitude of \(D\) at \(875^{\circ} \mathrm{C}\) ?
A sheet of steel \(1.5 \mathrm{~mm}\) thick has nitrogen atmospheres on both sides at \(1200^{\circ} \mathrm{C}\) and is permitted to achieve a steady-state diffusion condition. The diffusion coefficient for nitrogen in steel at this temperature is \(6 \times 10^{-11} \mathrm{~m}^{2} / \mathrm{s}\), and the diffusion flux is found to be \(1.2 \times\) \(10^{-7} \mathrm{~kg} / \mathrm{m}^{2} \cdot \mathrm{s}\). Also, it is known that the concentration of nitrogen in the steel at the highpressure surface is \(4 \mathrm{~kg} / \mathrm{m}^{3} .\) How far into the sheet from this high-pressure side will the concentration be \(2.0 \mathrm{~kg} / \mathrm{m}^{3}\) ? Assume a linear concentration profile.
Briefly explain the difference between selfdiffusion and interdiffusion.
An FCC iron-carbon alloy initially containing \(0.35 \mathrm{wt} \% \mathrm{C}\) is exposed to an oxygen-rich and virtually carbon-free atmosphere at \(1400 \mathrm{~K}\) (1127 \(\left.^{\circ} \mathrm{C}\right)\). Under these circumstances the carbon diffuses from the alloy and reacts at the surface, with the oxygen in the atmosphere; that is, the carbon concentration at the surface position is maintained essentially at \(0 \mathrm{wt} \%\) C. (This process of carbon depletion is termed decarburization.) At what position will the carbon concentration be \(0.15 \mathrm{wt} \%\) after a 10 -h treatment? The value of \(D\) at \(1400 \mathrm{~K}\) is \(6.9 \times 10^{-11} \mathrm{~m}^{2} / \mathrm{s}\).
Briefly explain the concept of steady state as it applies to diffusion.
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