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I’m using the ideal rocket equations and isentropic flow but cannot determine what is wrong in my understanding about total pressure. Can anyone explain where I am wrong using the following example?

Example:

$\gamma = 1.4 $

Converging-diverging nozzle Area ratio $A/A^\star = 12.0$

Inside the rocket chamber $P_0 = P_t = 12 MPa$

Exit area $A_e = 0.002 m^2$

Mass flow rate $\dot m = 2.5kg/s$

$Thrust=5000N$

Supersonic flow

Atmospheric pressure $P_{atm} = 101.3 kPa$


Solution:

Based on isentropic equations - (e.g. https://www.grc.nasa.gov/www/k-12/airplane/isentrop.html)

Mach No. $M = 4.127$

$P/P_t = 0.0056$

$T/T_t = 0.227$

Substituting in $P_t$ gives: $P = P/P_t * P_t = 0.0056*12 MPa = 67200 Pa $

Then based on the ideal rocket equation (i.e. https://www.grc.nasa.gov/WWW/K-12/rocket/thrsteq.html)

$Thrust = v_e * \dot m + A_e (P-P_{atm}) $

Rearranging for exit velocity:

$v_e = \frac {Thrust - A_e (P-P_{atm})}{\dot m} $

$v_e = \frac {5000 - 0.002 (67200 - 101300)}{2.5} = 2027.28 m/s$

Then the exit density can be found by rearranging this (also from the isentropic equations website):

$a = \sqrt{\gamma \frac{P}{\rho}}$

Rearranges into:

$\rho = \frac{\gamma P}{a^2}$

Speed of sound $a = v_e / M = 2027.28 / 4.127 = 491.22 m /s $

Substituting into the density equation:

$\rho = \frac{1.4 * 67200}{491.22 ^ 2} = 0.390 kg/m^3$


And here is the problem/confusion:

From the question the total pressure is $P_0 = P_t = 12 MPa$ This is also supposed to be the total pressure at the nozzle exit because it is reversible/isentropic flow.

But $Total Pressure = Static Pressure + Dynamic Pressure$ $Total Pressure = P + 0.5 \rho v_e^2 $ $Total Pressure = 67200 + 0.5 * 0.390 * 2027.28^2 = 868624 Pa$

And this is nowhere near 12 MPa... so what part am I not understanding correctly?

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    $\begingroup$ $\frac{1}{2} \rho V^2$ is valid for dynamic pressure at low subsonic speed. At sonic speeds and above, compressibility effects are significant and the equation is $\frac{1}{2}\gamma \cdot p \cdot M^2$. $\endgroup$ – Koyovis Feb 26 at 22:23
  • $\begingroup$ Oh, right I didn't realise it had a different equation for compressibility... and I just looked at en.wikipedia.org/wiki/Impact_pressure to see why the $\frac{1}{2}\gamma p M^2$ still isn't the full value $\endgroup$ – waterdragon Feb 26 at 22:48
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    $\begingroup$ I think that this is definitely on topic here but if you don’t get a suitable answer, you might also try Space.SE. Great first question by the way. Keep ‘em coming! $\endgroup$ – dalearn Feb 27 at 2:33
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I'll just answer my own question thanks to Koyovis' comment :)

Koyovis was right - I didn't realise the difference between how dynamic pressure is calculated for compressible vs incompressible flows. There are some useful notes on it on pages 4-8 of this link (http://mae-nas.eng.usu.edu/MAE_5420_Web/section5/section.5.5.pdf)

In summary: Bernoilli’s equation is

$Total Pressure = P_{static} + P_{dynamic}$

For an incompressible flow, $P_{dynamic} = 0.5\rho v^2$

For a compressible flow, use the isentropic equation $P_t / P_{static} = (1 + \frac{\gamma – 1}{2} M^2) ^ {\frac{\gamma}{\gamma-1}}$

To get this in a way you can compare to Bernoilli's equation, make total pressure the subject of the equation, add & subtract $P_{static}$ from the right hand side, and rearrange the equation slightly (shown in the link above) gives:

$P_t = P_{static} +P_{static} ( (1 + \frac{\gamma – 1}{2} M^2) ^ {\frac{\gamma}{\gamma-1}}-1)$

This is the same form as Bernoilli’s equation where the dynamic pressure is: $P_{dynamic}=P_{static} ( (1 + \frac{\gamma – 1}{2} M^2) ^ {\frac{\gamma}{\gamma-1}}-1)$

Putting in the values from the example gives the correct result.

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I don't think Koyovis was right...

$$ q = \frac{\rho V^2}{2} = \frac{1}{2} \frac{P}{R T} \gamma R T M^2 = \frac{\gamma P M^2}{2} $$

and you run the numbers and they are identical... (AC engine Design, Mattingly, pg13)

The link from Utah State uni on slide 7 does give different numbers as you have said in your answer Waterdragon.

$$ q_c = p \left( \left[ 1 + \frac{\gamma - 1}{2} M^2 \right]^{\gamma / (\gamma - 1)} - 1 \right) $$

Happy to be proven wrong and I have to say this was new to me.

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    $\begingroup$ Nick, welcome to Aviation.SE. We use LaTeX formatting for equations. I edited your answer accordingly (if I messed something up, please edit again). You can look at how I did it in the revision history by clicking on side-by-side markdown. $\endgroup$ – Bianfable Mar 1 at 11:51
  • $\begingroup$ Yeah that wasn't the full equation. $\endgroup$ – Koyovis Mar 1 at 13:42

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