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I'm quite sure this is either a small or just plain stupid mistake but my anxiety is getting the better of me. when calculating exhaust velocity this is the equation used:

enter image description here

however I keep getting higher values than that listed, specifically on Astronautix. I should mention I am solving for monopropellant HTP-85%, where $M = 20.7 kg$, $k = 1.34$, $P_c = 10\ bar$, $P_e = 1\ bar$, $T_c = 910\ K$ and $R$ is the universal gas constant. I suspect that it is just the difference between theoretical and actual performance however $1128\ m \cdot s^{-1}$ is very different from $~900\ m \cdot s^{-1}$. Is there something I am missing or am I just using the formula in the wrong context?

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    $\begingroup$ The formula assumes that the exhaust is an ideal gas, which it is not. $\endgroup$
    – WarpPrime
    Commented May 30, 2022 at 11:37
  • $\begingroup$ Related (deleted) question: space.stackexchange.com/q/59232/30164 $\endgroup$
    – WarpPrime
    Commented May 30, 2022 at 11:38
  • $\begingroup$ @fasterthanlight thank you very much is there another formula/means of correction? $\endgroup$
    – R. Hall
    Commented May 30, 2022 at 12:26
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    $\begingroup$ Please remember that Pe is not the ambient pressure, it is the exit plane pressure 1 bar Pe would be a terrible underexpanded booster engine design. $\endgroup$
    – Organic Marble
    Commented May 30, 2022 at 12:33
  • $\begingroup$ @OrganicMarble I understand this is the case with actual boosters which have to optimise for a large envelope but this nozzle is merely designed for sea level use $\endgroup$
    – R. Hall
    Commented May 30, 2022 at 12:58

1 Answer 1

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You are simply calculating the wrong parameter.

You are calculating exhaust velocity. You can't calculate this without knowledge of the nozzle geometry because exit plane pressure is used in the calculation. Hence my comment about you picking exit plane pressure = 1 bar - that means you have assumed a nozzle geometry, and a very poor one for a booster engine at that.

The Astronautix page is giving characteristic velocity. That is primarily a function of the gas properties in the combustion chamber and therefore does not depend on the nozzle geometry.

formula for characteristic velocity

Reference: Mechanics and Thermodynamics of Propulsion, Hill and Peterson, 1970, p. 357

formula for characteristic velocity

Reference: Rocket Propulsion Elements, Sutton, 1976, p. 72

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    $\begingroup$ Thank you very much! I can't tell you how useful this has been; I hadn't realised there was a distinction! Thank you for the quick response $\endgroup$
    – R. Hall
    Commented May 30, 2022 at 13:06

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