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I have been flying rocketbelts for 21yrs. I have a new rocketbelt(some call jetpacks); a copy of the Bell rocketbelt.

I believe that the angles are incorrect and the throat opening is too large. My thrust tubes have an inner diameter of 1.75".

What is the the math to determine the dimensions for 95% efficient nozzles? Alternatively, if the specific dimensions may be provided that shall suffice too.

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    $\begingroup$ Wait, what? You are a (recreational?) jet pack pilot? $\endgroup$
    – SAnderka
    Jul 30, 2014 at 12:29
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    $\begingroup$ This is the first time I've read about the Bell Rocket Belt. The Wikipedia article about it is quite fascinating -- not least because of the remarkable kidnapping story in there! $\endgroup$
    – matz
    Feb 4, 2019 at 6:50

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Ideally, a rocket-nozzle behaves isentropic. That's the case if a fluid (=exhaust) experiences changes so rapidly, it can't equilibrate with it's environment. (I.e. it doesn't lose a lot of heat to nozzle walls.)

Supersonic flow has the advantage of having "simple" equations. "Simple" in the sense that once you know the mach-number, you can calculate most properties independently.

That is the case for pressure and area as well. In your specific case, I'd suggest determining the burn chamber pressure and from the ratio chamber pressure to ambient pressure calculating the mach number. That furthermore allows to calculate the ideal area ratio and thereby the ratio of outer to inner diameter.

If you combine the expression for pressure ratio and area ratio to one, it comes out as:

$$ \frac{A_e}{A_t} = \frac{ (\frac{p_t}{p_e})^{\frac{k+1}{2 k}}}{\sqrt{\frac{2}{k-1}((\frac{p_t}{p_e})^{\frac{k-1}{k}}-1)}} * (\frac{k+1}{2})^{-\frac{k+1}{2(k-1)}} $$ with index e for exit and index t for throat. k is the ratio of specific heats, NASA denotes that as $\gamma$. The square root of the area-ratio will deliver the diameter-ratio.

Big thanks to @Christoph for mentioning this handy tool to calculate exhaust conditions based on chemical reactions. This tool to calculate rocket exhaust based on chemical reactions (Big thanks to @Christoph for bringing it to my attention) will show a Cp/Cv ratio as well as a gamma. I do no know what that gamma is, you want to use the Cp/Cv value.

In order to calculate the correct exit area, you need to know this ratio of the exhaust and assume it's constant while traveling through the nozzle or it gets even messier.

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  • $\begingroup$ Normally gamma is the ratio of specific heats. I don't know why that tool uses a different notation. Looking the pdf linked from that page, I guess their "gamma" is "gamma sub s", something I'm not familiar with, but it's equation 2.71 in the pdf. $\endgroup$
    – Organic Marble
    Feb 10, 2019 at 0:57
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    $\begingroup$ Judging from the paper, it seems the $gamma_s$ is $\gamma*M$. (since $a=\sqrt{R*T*n*\gamma_s}$ in the paper, and equation 2.3a) But that doesn't seem to be the gamma mentioned in the results. I'd expect values higher than Cp/Cv, but they are consistently lower. $\endgroup$
    – GammaSQ
    Feb 10, 2019 at 8:50

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