Chapter 5: Problem 5
(a) Briefly explain the concept of a driving force. (b) What is the driving force for steady-state diffusion?
Chapter 5: Problem 5
(a) Briefly explain the concept of a driving force. (b) What is the driving force for steady-state diffusion?
All the tools & learning materials you need for study success - in one app.
Get started for freeThe steady-state diffusion flux through a metal plate is \(5.4 \times 10^{-10} \mathrm{~kg} / \mathrm{m}^{2} \cdot \mathrm{s}\) at a temperature of \(727^{\circ} \mathrm{C}(1000 \mathrm{~K})\) and when the concentration gradient is \(-350 \mathrm{~kg} / \mathrm{m}^{4} .\) Calculate the diffusion flux at \(1027^{\circ} \mathrm{C}(1300 \mathrm{~K})\) for the same concentration gradient and assuming an activation energy for diffusion of \(125,000 \mathrm{~J} / \mathrm{mol}\).
Consider a diffusion couple composed of two semi-infinite solids of the same metal, and that each side of the diffusion couple has a different concentration of the same elemental impurity; furthermore, assume each impurity level is constant throughout its side of the diffusion couple. For this situation, the solution to Fick's second law (assuming that the diffusion coefficient for the impurity is independent of concentration) is as follows: $$ C_{x}=\left(\frac{C_{1}+C_{2}}{2}\right)-\left(\frac{C_{1}-C_{2}}{2}\right) \operatorname{erf}\left(\frac{x}{2 \sqrt{D t}}\right) $$ In this expression, when the \(x=0\) position is taken as the initial diffusion couple interface, then \(C_{1}\) is the impurity concentration for \(x<0\); likewise, \(C_{2}\) is the impurity content for \(x>0\). A diffusion couple composed of two silver-gold alloys is formed; these alloys have compositions of \(98 \mathrm{wt} \% \mathrm{Ag}-2 \mathrm{wt} \% \mathrm{Au}\) and \(95 \mathrm{wt} \% \mathrm{Ag}-5 \mathrm{wt} \%\) Au. Determine the time this diffusion couple must be heated at \(750^{\circ} \mathrm{C}\) (1023 K) in order for the composition to be \(2.5 \mathrm{wt} \% \mathrm{Au}\) at the \(50 \mu \mathrm{m}\) position into the 2 wt \(\%\) Au side of the diffusion couple. Preexponential and activation energy values for Au diffusion in \(\mathrm{Ag}\) are \(8.5 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\) and \(202,100 \mathrm{~J} / \mathrm{mol}\), respectively.
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.
For a steel alloy it has been determined that a carburizing heat treatment of 10 -h duration will raise the carbon concentration to \(0.45 \mathrm{wt} \%\) at a point \(2.5 \mathrm{~mm}\) from the surface. Estimate the time necessary to achieve the same concentration at a 5.0-mm position for an identical steel and at the same carburizing temperature.
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).
What do you think about this solution?
We value your feedback to improve our textbook solutions.