# What is wrong with my understanding about total pressure for a rocket nozzle?

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?

• $\frac{1}{2} \rho V^2$ is valid for dynamic pressure at low subsonic speed. At sonic speeds and above, compressibility effects are significant and the equation is $\frac{1}{2}\gamma \cdot p \cdot M^2$. Feb 26 '20 at 22:23
• 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 Feb 26 '20 at 22:48
• 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! 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.

• 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. Mar 1 '20 at 11:51
• Yeah that wasn't the full equation. Mar 1 '20 at 13:42