Exam task mass transfer 2023-01-14 (KETF40)

[@gloomyend1232]

Wait plzz explain the ending 💀 I can't sleep now 😭😭 i need answers

[@PLE_LU]

😀 A cliff hanger in an exam, is that possible? 😀 Jokes aside, if you really want to know and are not a student of mine (those receive a more detailed explanation in the suggested solutions to the exam), you can enter the LaTeX code below into arachnoid.com/latex/. That page does not understand the LaTeX command \intertext, or indeed ordinary text, so all inline explanation have been removed. The crucial part in the beginning is to realize that although n_A can be expressed through Stefan's equation, it can also be expressed using a time derivative of the film thickness \delta. From there on, it is "just" algebraic manipulations Variable explanations: A: area (m2) n_A: molar transport (mol/m2,s) D_{AB}: Diffusivity of A in B (m2/s) \delta: film thickness (m) P: Pressure (Pa=J/m3) R: Ideal gas constant (J/mol, K) T: Temperature (K) y_A: Molar fraction of A (mol/mol) ho : density (kg/m3) M: molar weight (kg/mol) C: Just a constant with the same unit as \delta^2/t, i.e. m2/s LaTeX code: \begin{align} A n_{\! A}& = A\dfrac{D_{\! AB}}{\delta} \dfrac{P}{RT} \ln \dfrac{1-y_{\! A,2}}{1-y_{\! A,1}} onumber\\ n_A &= \dfrac{\mathrm{d}}{\mathrm{d} t}\left( \delta \dfrac{ ho}{M} ight) onumber\\ A \dfrac{\mathrm{d}}{\mathrm{d} t}\left( \delta \dfrac{ ho}{M} ight)& = A\dfrac{D_{\! AB}}{\delta} \dfrac{P}{RT} \ln \dfrac{1-y_{\! A,2}}{1-y_{\! A,1}} onumber\\ \int_{\delta_0}^\delta \delta \mathrm{d} \delta &= \dfrac{D_{\! AB}M}{ ho} \dfrac{P}{RT} \ln \dfrac{1-y_{\! A,2}}{1-y_{\! A,1}}\int_{t_0}^t \mathrm{d} t onumber\\ \delta^2 - \delta_0^2 &= C\cdot (t-t_0) onumber\\ C &= \dfrac{2 D_{\! AB} M}{ ho} \dfrac{P}{RT} \ln \dfrac{1-y_{\! A,2}}{1-y_{\! A,1}} onumber\\ \delta &= \sqrt{\delta_0^2 + C(t-t_0)} onumber\\ An_{\! A}& = \dfrac{D_{\! AB}}{ \sqrt{\delta_0^2 + C(t-t_0)} } \dfrac{P}{RT} \ln \dfrac{1-y_{\! A,2}}{1-y_{\! A,1}} onumber \end{align}

[@gloomyend1232]

@@PLE_LU thank you ❤️