$\def\rmin{{\mathrm{in}}}$ $\def\l{\left}\def\r{\right}$ $\def\tagl#1{\tag{#1}\label{#1}}$ I am using the one-dimensional finite volume method to calculate the air flow in some tube. For subsonic flow one can predefine two quantities at the inlet.

The inlet is a wide opening to the ambient. The overall system deeply inside the tube is a bit more complicated. But, this does not really matter here. The system is described in some PhD and I am coding along the lines of this PhD. The author assumes isentropic flow at the inlet.

For the inlet pressure $p_\rmin$ he gave the formula \begin{align} p_\rmin + \frac{\gamma-1}\gamma\rho_0 \l(\frac{p_\rmin}{p_0}\r)^{\frac1\gamma}\frac{u_\rmin^2}2&= p_0 \tagl{A20amb} \end{align} with the ambient pressure $p_0=10^5 \rm Pa$, the ambient air density $\rho_0=1.23\rm \frac{kg}{m^3}$ and the given inlet velocity $u_\rmin$ . After solving this equation for $p_\rmin$ one can also determine the inlet density $\rho_\rmin$.

Pittyingly, I do not get this equation.

He motivated \eqref{A20amb} through the ideal gas law \begin{align} p=\rho(c_p-c_v)T, \tagl{A1amb} \end{align} the assumption of isentropic state transition \begin{align} \frac{p_\rmin}{p_0} = \l(\frac{\rho_\rmin}{\rho_0}\r)^\gamma \tagl{A3amb} \end{align} where \begin{align} \gamma := \frac{c_p}{c_v}, \tagl{A2} \end{align} and the energy balance \begin{align} c_p T_0 = c_pT_\rmin + \frac12 u_\rmin^2 \tagl{A5amb} \end{align} with the ambient temperature $T_0$.

Substitution of $T=\frac{p}{(c_p-c_v)\rho}$ from \eqref{A1amb} into \eqref{A5amb} delivers with $\rho_\rmin = \rho_0\l(\frac{p_\rmin}{p_0}\r)^{1/\gamma}$ from \eqref{A3amb} the equation \begin{align} c_p\frac{p_0}{(c_p-c_v)\rho_0} = c_p\frac{p_\rmin}{(c_p-c_v)\rho_0\l(\frac{p_\rmin}{p_0}\r)^{1/\gamma}} +\frac12 u_\rmin^2 \end{align} which can be transformed into \begin{align} 1 &= \l(\frac{p_\rmin}{p_0}\r)^{\frac{\gamma-1}\gamma} + (\gamma-1) \frac{\rho_0}{\gamma p_0}\cdot\frac{u_\rmin^2}2\\ 1 &= \l(\frac{p_\rmin}{p_0}\r)^{\frac{\gamma-1}\gamma} + \frac{\gamma-1}2\l(\frac{u_\rmin}{c_0}\r)^2 \tagl{p0} \end{align} where \begin{align} c_0=\sqrt{\frac{\gamma p_0}{\rho_0}} \tagl{speedOfSound} \end{align} is the speed of sound in the ambient.

This formula can be resolved for the inlet pressure: \begin{align} p_\rmin &= p_0 \l(1-\frac{\gamma-1}2\l(\frac{u_\rmin}{c_0}\r)^2\r)^{\frac{\gamma}{\gamma-1}} \tagl{solution} \end{align} I did not check whether \eqref{solution} is equivalent to \eqref{A20amb}. But, I do not think so because then the author would have used the explicite form \eqref{solution} to calculate $p_\rmin$. EDIT: I did check for differences. See below.

What am I missing?? Did I miss-interprete the assumptions?

Note that also \eqref{A20amb} can be re-formulated with the help of $c_0$: \begin{align} \l(\frac{p_\rmin}{p_0}\r) + \frac{\gamma-1}2 \l(\frac{p_\rmin}{p_0}\r)^{\frac1\gamma}\l(\frac{u_\rmin}{c_0}\r)^2 = 1 \tagl{A20mod} \end{align} If one divides by $\l(\frac{p_\rmin}{p_0}\r)^{\frac1\gamma}$ one nicely recognizes the difference between \eqref{A20amb} and \eqref{p0}: \begin{align} \l(\frac{p_\rmin}{p_0}\r)^{\frac{\gamma-1}\gamma} + \frac{\gamma-1}2\l(\frac{u_\rmin}{c_0}\r)^2 = \l(\frac{p_\rmin}{p_0}\r)^{-\frac1\gamma} \tagl{A20mod2} \end{align} Instead of the 1 on the left-hand side of \eqref{p0} the formula from the PhD-thesis has a $\l(\frac{p_\rmin}{p_0}\r)^{-\frac1\gamma}$.

  • $\begingroup$ Sorry, in the equation-references worked in the preview but they do not work in the post. I try to improve them... $\endgroup$
    – Tobias
    Mar 19, 2014 at 18:02
  • 1
    $\begingroup$ Possibly just a typo in your work, but $R=c_p-c_v$, not $c_v-c_p$ as you have above. Looks like you (or the author) carried that mistake through. $\endgroup$
    – Aurelius
    Mar 31, 2014 at 17:04
  • $\begingroup$ @Aurelius Thank you very much. I corrected this. It was my mistake. $\endgroup$
    – Tobias
    Mar 31, 2014 at 18:02

1 Answer 1


I lack the time to dig into it at the moment (sorry), but any chance sections 2.9, 2.10, and 2.11 from Liepmann and Roshko (available in the linked Amazon preview) can help resolve your question? The exposition and derivations in that book are quite good.

  • $\begingroup$ Thanks for the hint. I had a look into the book. It looks good. Nevertheless, I try to attract attention with a bounty. Maybe, someone can point out the origin of the discrepancy. $\endgroup$
    – Tobias
    Mar 31, 2014 at 9:29

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