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Jun 5, 2020 at 23:25 comment added Chet Miller And, in the ideal gas limit, $\mu$ approaches a constant. See Transport Phenomena by Bird, et al, Chapter 1.
Jun 5, 2020 at 23:25 vote accept Armadillo
Jun 5, 2020 at 23:22 comment added Armadillo I saw that too (not dimensionally consistent). I realize that the literature to which I refer has made an error. Indeed, using the relation for ideal gas law $\rho=pM_w/(RT)$ will result in (after integrating, and neglecting the pressure dependence of $\mu$), $\bar p \Delta p \rho = \frac{\mu L w}{\beta k A}+\frac{c_f L w^2}{\beta \sqrt{k} A^2}$, where $\beta = 1/p$
Jun 5, 2020 at 22:21 comment added Chet Miller Even with the approximation, I still get the same final result as you (i.e., without the $\beta$s). The presence of the betas isn't even dimensionally correct. Otherwise, their procedure will be OK.
Jun 5, 2020 at 21:59 comment added Chet Miller The problem statement didn't say it was a gas. I naturally assumed it was a liquid. Otherwise, one would use the ideal gas law.
Jun 5, 2020 at 19:14 comment added Armadillo The relation $\rho=\rho(\bar{p})[1+\beta (p-\bar{p})]$ is for "slightly" compressible fluids. Perhaps in this problem the authors assumed gases at low pressures, and thus the isothermal compressibility was approximated by $\beta=1/p$? If so, I'm still not seeing how I approximate $\rho$ as a function of $p$ using this different relation for $\beta$ prior to integration. Hints?
Jun 5, 2020 at 17:47 comment added Armadillo would this be correct then? $-\left(\int_{p_1}^{p_2} \rho(\bar p)[1+\beta(p-\bar p)]\ \text{d}p \right)=\rho(\bar p)[p_1+\beta (\frac{p_1^2}{2} -p_1 \bar p)]-\rho(\bar p)[p_2+\beta (\frac{p_2^2}{2} - p_2\bar p)]$ This is messy. Still not sure how this can be simplified to (1) and used to show why we use the product $\bar p \times \Delta p$ for the plotting/solution method. Thoughts?
Jun 5, 2020 at 12:04 history answered Chet Miller CC BY-SA 4.0