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?