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I am seeking an explanation for this graph where the subscript "1" refers to the supersonic region and the subscript "2" refers to the subsonic region present beyond a normal shock.

Flow Properties across a normal shock

Depiction of properties

The static pressure curve shows an increasing trend. Shouldn't the same be applicable to the stagnation pressure Po?

Is the entropy generation associated with the stagnating of the kinetic energy term so high?

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  • $\begingroup$ Can you describe the figure and terms a little more? What are the vertical axes, for instance? What are $(P_{o})_{1}$ and $P_{1}$? Is this for a hydrodynamic, collision-mediated shock? $\endgroup$ – honeste_vivere May 20 '16 at 17:41
  • $\begingroup$ P1 stands for the static pressure in region 1 (before the shock) and (P0)1 stands for the stagnation pressure in the same region. The y-axis show the value of the ratios depicted in the graph. This is not a hydrodynamic case, its for the compressible flow of an ideal gas. $\endgroup$ – DBTKNL May 23 '16 at 7:35
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This can be concluded by reviewing Gibbs equation for upstream and downstream stagnation conditions. $$T_0ds_0=dh_0-\frac 1{\rho_0}dP_0$$

Because across the shock wave is an adiabatic process, $dh_0=0$

Then Gibbs equation becomes $$ds_0=-\frac 1{\rho_0T_0}dP_0=-\frac {R}{P_0}dP_0$$

We know entropy increases. This leads to conclusion that the stagnant pressure decreases.

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  • $\begingroup$ How do you know entropy increases? There is a very critical reason why I ask this and it would help improve your answer. $\endgroup$ – honeste_vivere May 20 '16 at 17:45
  • $\begingroup$ Can we say through a shock wave is an irreversible process? $\endgroup$ – user115350 May 20 '16 at 18:23
  • $\begingroup$ I was going for something like: Irreversibility (e.g., increase in entropy) is required for a nonlinearly steepening wave to halt steepening and form the stable discontinuity we call a shock wave. In other words, shock initiation requires entropy production, which is why people say entropy increases through a shock wave. $\endgroup$ – honeste_vivere May 20 '16 at 21:50

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