2
$\begingroup$

I am trying to understand why normal shocks are caused in Converging Diverging Nozzles. My textbook explains that since the exit pressure must match the back pressure, the normal shock is the only mechanism that can make this happen, which I understand.

However, why does the exit pressure have to match the back pressure? What if it did not? Then simply the jet of fluid exiting the converging diverging nozzle would expand or contract under the pressure driving force induced with the back pressure.

Any clarifications welcome.

$\endgroup$
1
  • $\begingroup$ Well, which textbook for a start? $\endgroup$ – Solar Mike Feb 17 '18 at 19:08
0
$\begingroup$

In order to answer the question, one more question needs to be answered. The exit pressure is present at the downstream of the shock, how does this information is transferred to upstream of the shock?

Normal shocks occur for a supersonic flow. The Euler equations are hyperbolic for supersonic flow but there will be boundary layer at the wall edges. The flow inside the boundary layer is subsonic, making the equations elliptic. Elliptic equations require two boundary conditions in two dimensions one of which is that the pressure at the exit matches with the back pressure. But as dealing with viscous flows is hard, we use inviscid flow theory and say that the exit pressure matches with the back pressure.

So it is the boundary layer, that's helping to communicate this information to the upstream.

$\endgroup$
0
$\begingroup$

It's kind of a matter of semantics: if the fluid is still in a "jet" that's undergoing significant broadening or narrowing, and therefore significant acceleration that can match a pressure difference from the far-field "back pressure", then the fluid hasn't really "exited" the (influence of the) nozzle yet, and the "exit" is somewhere further downstream where the pressure has relaxed to the back pressure. Compare with the fully-subsonic case of an orifice plate: you don't measure the exit pressure right up against the plate, you measure it at a location further downstream where the vena contracta has already done its thing.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.