$\frac{p}{{}^{}}=\frac{p_0}{{}_0^{}}$

$\frac{T}{p^{1-\frac{1}{}}}=\frac{T_0}{p_0^{1-\frac{1}{}}}$

tions apply both to frictionless subsonic and superso ot be used across a shock front.

$10$zzle. In the absence of friction, the mechanical$-$energy aes simply

$\frac{dp}{}=-d(\frac{u^2}{2g_c})$

$$ from Eq$.(6\mathrm{.}25)$ by substitution from Eq$.(6\mathrm{.}23)$ ar limit based on the reservoir, where $p=p_0,={}_0,$ and $t$

${\displaystyle \underset{0}{\overset{u}{}}}d(\frac{u^2}{2g_c})=-\frac{p_0^{\frac{1}{}}}{{}_0}{\displaystyle \underset{p_0}{\overset{p}{}}}\frac{dp}{p^{\frac{1}{}}}$

and substituting the limits yields, after rearrangement,

$u^2=\frac{2g_c{p}_0}{(-1){}_0}[1-(\frac{p}{p_0}{)}^{1-\frac{1}{}}]$