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I've performed a CFD of a flow inside a nozzle and I'm evaluating the differences between various settings. I cannot find a reason why after the shock wave formed in the divergence section, the Mach number rises (as can be seen in the blue dotted line) and the pressure decreases.

I don't understand the physical reason behind that. Shouldn't the shock just be a bit downstream with pressure increasing and M decreasing monotonically (after the shock)?

Mach number and pressure field along the centerline

enter image description here

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1 Answer 1

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Much of this answer comes from the following: https://physics.stackexchange.com/a/378783/59023

I cannot find a reason why after the shock wave formed in the divergence section, the Mach number rises (as can be seen in the blue dotted line) and the pressure decreases... Shouldn't the shock just be a bit downstream with pressure increasing and M decreasing monotonically (after the shock)?

If we assume isentropic nozzle flow, we can show that: $$ \frac{du}{u} \ \left( M^{2} - 1 \right) = \frac{dA}{A} \tag{0} $$ where $M$ = Mach number, $u$ is the fluid speed, and $A$ is the nozzle area.

From Equation 0, one can see the following list:

  • for $M$ < 1 and $dA$ > 0 $\rightarrow$ $du$ < 0 (i.e., deceleration);
  • for $M$ < 1 and $dA$ < 0 $\rightarrow$ $du$ > 0 (i.e., acceleration);
  • for $M$ > 1 and $dA$ > 0 $\rightarrow$ $du$ > 0; and
  • for $M$ > 1 and $dA$ < 0 $\rightarrow$ $du$ < 0.

The situation in your case is the third bullet. You want the flow to become supersonic just after the constriction point so that both $M$ > 1 and $dA$ > 0 are satisfied, which leads to $du$ > 0 (i.e., acceleration).

As for the pressure, since we assumed an isentropic flow, we can use the ideal gas approximation for an adiabatic fluid given by: $$ P \ \rho^{-\gamma} = \text{constant} \tag{1} $$ where $P$ is the thermal pressure, $\gamma$ is the polytropic index and $\rho$ is the mass density. We can also use the continuity equation to show that: $$ \rho \ A \ u = \text{constant} \tag{2} $$ where $A$ is again cross-sectional area and $u$ is fluid flow speed. Finally the conservation of energy for an isentropic process is given by: $$ \frac{ 1 }{ 2 } u^{2} + \frac{ \gamma }{ \gamma - 1 } \frac{ P }{ \rho } = \text{constant} \tag{3} $$ Differentiating Equation 1 gives us: $$ dP = \frac{ \gamma \ P }{ \rho } d\rho = C_{s}^{2} d\rho \tag{4} $$ where $C_{s}$ is the speed of sound. We can do some arithmetic on Equations 1, 2, and 3 to show that: $$ \frac{ du }{ u } = -\frac{ 1 }{ M^{2} } \frac{ d\rho }{ \rho } \tag{5} $$

From our discussion above, we know that $du$ > 0 therefore to make sure Equation 5 is consistent this would require $d\rho$ < 0. We can then see from Equation 4 if $d\rho$ < 0 then $dP$ must also be less than zero, i.e., pressure decreases.

More details about this can be found in the answer at: https://physics.stackexchange.com/a/524215/59023.

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  • $\begingroup$ Thank you for your answer. Probably I misunderstood some step but you start saying that my situation is the one in the third bulled. But after the shock I have M<1. Shouldn't this be the first bullet? Hence I should have du<0 and dP>0 according to your answer. Instead, just after the shock, dP<0. Am I wrong? $\endgroup$
    – AMasc
    Feb 5, 2021 at 18:59
  • $\begingroup$ @wilove - No, after the initial place where the flow exceeds M = 1 is what I meant. The flow doesn't necessarily create a shock inside the nozzle (in fact you wouldn't want that) but the flow is supersonic. It won't create a shock until it hits some obstacle (e.g., the slow, background air outside the nozzle). $\endgroup$ Feb 5, 2021 at 19:05
  • $\begingroup$ Ok, thank you. My question was more about the behaviour just after the shock. Because downstream, far from the shock, p increases and M decreases (as expected). Yet, near the shock, for the dotted blue line which represent RANS simulation results, there is a small region where p decreases a M increases (despite the flow being subsonic and dA>0). What is going on there? $\endgroup$
    – AMasc
    Feb 6, 2021 at 14:52
  • $\begingroup$ @wilove - I don't think we are on the same page here. Just because the flow rate satisfies M > 1 does not mean there is a shock... Regardless, I think in the 2nd case where your external pressures are nearly 3 times higher you are running into stiff solutions, i.e., if not handled properly you will get "ringing" which looks like fluctuations in the solution. If this is a collisional fluid, unless you are hypersonic (which is unlikely in a regular nozzle), you will not have a foreshock so something is wrong. $\endgroup$ Feb 12, 2021 at 16:11
  • $\begingroup$ Ok thank you. Then I will look into my simulation settings! $\endgroup$
    – AMasc
    Feb 17, 2021 at 14:16

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