In my propulsion notes I always see mention of the importance of minimizing stagnation pressure loss across the inlet shock of a ramjet/turbojet/turbofan. Could anyone explain further why this is the case, possibly with equations to explain certain points in more detail? Thanks very much.

  • $\begingroup$ I have a feeling you want the burning gases inside to head out the back end, instead of popping out the front. $\endgroup$ – user_1818839 Jan 27 '17 at 20:29
  • $\begingroup$ I'l like to answer, having studied compressible flow, but I'm afraid that this answer is out of my experience as of present. I am still studying. One of my readings was this article. I think it may have the answer you are looking for. maecourses.ucsd.edu/~kseshadr/mae113-s109/reference/… $\endgroup$ – MountainClimberi Mar 3 '17 at 1:31

The amount of useful work you can extract out of gas in an engine is proportional to its stagnation pressure. Energy and be stored as thermal, kinetic, or as pressure. The stagnation pressure is the pressure the gas would be at if all the kinetic energy was converted to thermal and pressure storage with 100% efficiency ie. isentropic process.

To generate thrust you want the highest exit velocity, which means you want to convert as much energy as possible from pressure and temperature. This processes is limited by atmospheric pressure (or the size of your nozzle). The pressure of your exit gas can't drop below atmospheric or it will stop exhausting/ there will be a shock to build pressure again so you can keep exhausting.

This means that you want to keep your static pressure as high as possible to get the largest transition to atmospheric, giving you the largest velocity.

A shockwave, while it increases static pressure, decreases stagnation pressure. The discrepancy comes from an irreversible transfer of kinetic energy to thermal energy. While a shock wave increases pressure, it doesn't increase it as much as a smooth isentropic transition to the same velocity.

Some relevant equations:

The relationship between static pressure, pressure, and Mach number during an isentropic process: $$\frac{p}{p_0}=\left(1+\frac{\gamma-1}2M^2\right)^{\frac{-\gamma}{\gamma-1}}$$

Increase in pressure vs. strength of shock:

$$\frac{p_2}{p_1} = \frac{2\gamma M_1^2}{\gamma + 1} - \frac{\gamma - 1}{\gamma + 1}$$

Decrease in static pressure vs. strength of shock:

$$\frac{p_{02}}{p_{01}} = \left(\frac{\frac{\gamma + 1}{2}M_1^2}{1 + \frac{\gamma - 1}{2}M_1^2}\right)^\frac{\gamma}{\gamma - 1}\left(\frac{1}{\frac{2\gamma}{\gamma + 1}M_1^2 - \frac{\gamma - 1}{\gamma + 1}}\right)^\frac{1}{\gamma - 1}$$

Isentropic equations are from Nasa

Shock equations are from Wikipedia

  • 1
    $\begingroup$ Note, the equations are for normal shocks only. For oblique shocks, replace $M$ with $M \sin \mathit{B}$ $\endgroup$ – Phil Sweet Jan 4 '19 at 1:05

The idea is you don't want the shocks to dissipate usable energy in the freestream.

$$\displaystyle \frac{{\mit\Delta}{\cal S}}{{\cal R}} \displaystyle \simeq \frac{(\gamma+1)}{12\,\gamma^{\,2}}\left( \frac{{\mit\Delta}p}{p_1}\right)^{\,3}$$ This the formula for the change in entropy, where $\Delta p$ is the change in pressure across the shock. Since the change in entropy is proportional to the shock strength cubed, Several weak shocks fritter away less energy than a single strong one.

See here for a detailed explanation. You may need to back up a few pages to unpack some of the equations.Supersonic Compression by Turning


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