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?


$\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 $\dot m = 2.5kg/s$


Supersonic flow

Atmospheric pressure $P_{atm} = 101.3 kPa$


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?

  • 1
    $\begingroup$ $\frac{1}{2} \rho V^2$ 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 '20 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 $\endgroup$ Feb 26 '20 at 22:48
  • 1
    $\begingroup$ 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 '20 at 2:33

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.


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.

  • 2
    $\begingroup$ 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 '20 at 11:51
  • $\begingroup$ Yeah that wasn't the full equation. $\endgroup$
    – Koyovis
    Mar 1 '20 at 13:42

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.