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I have been working on modeling liquid rocket engines and have read the book How to Design, Build and Test Small Liquid Fuel Rocket Engines, I had a question about the nozzle area and the units involved and I found this post and it helped me get as far as I am, but the units don't make sense. Need help on rocket nozzle equation

unit Manipulation

The unit I end up with is Sqrt(ft) which makes no sense. If it was a unit squared, you could take the square root of it. I am sure I am missing something. Any help would be appreciated. Thank you.

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I assume you're looking at the "original post" referenced in the linked post, and trying to work out the units.

enter image description here

First of all, I'd strongly recommend you learn how to do this in SI units. The ambiguity of the pound and the lack of a true mass unit in the English system is just going to give you heartburn, and as far as I know, all actual engineers use SI anyway today.

Anyway, based on the equation as laid out in the snippet above, they're treating the foot-pounds and the mass in the gas equation as the same kind of pound. If you don't try to convert between lbm and lbf, it comes out correctly.

(ft-lb/lb/degR * degR) / (ft/s^2)

enter image description here

Do all your cancellation, and you're left with sec^2, so once you do your square root, you'll end up with seconds.

And that makes sense, because the rest of the equation is a flow rate divided by a pressure. Which is to say:

A = lbs/sec / psi * s

Convert psi into the more proper unit lbs/in^2, do your cancellation, and you end up with in^2, which is what we expect for a nozzle throat measurement.

I can't speak to whether that original poster's equation is valid, but going by what he wrote, that's how it's going to work.

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    $\begingroup$ The slug is the mass unit you're missing. en.wikipedia.org/wiki/Slug_(unit) $\endgroup$ Dec 20, 2023 at 0:27
  • $\begingroup$ I don't think replacing the force-derived pound-mass with a force-derived slug actually changes anything. $\endgroup$ Dec 20, 2023 at 14:42
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Going back to the original question of how the units work out, this is how. The equation is

$$A_t=\frac{w_t}{ P_t} \sqrt\frac{RT_t}{\gamma g_c}$$

Now $g_c$ is 32.2 ft-lbf/s$^2$-lbm (rounded numerical value). The units on R are ft-lbf/lbm $deg R$. With no units on $\gamma$, under the square root the units are lbf$^2$s$^2$/lbm$^2$. After taking the square root, the s/lbm cancels the units on $w_t$ and the lbf cancels the pound in pressure leaving square inches from pressure for the unit on area.

Here is a quick derivation of the formula, since the book does not give it and the equation is not really the standard one.

Given flow in the nozzle $$w = \rho V A$$ where V is the flow speed. Use the ideal gas law to eliminate $\rho$, $\rho = P/RT$. The speed of sound is $a=\sqrt{\gamma g_c RT}$. At the throat, $V=a$ so rewrite with subscript t for throat: $$w_t=\frac{P_t A_t}{RT_t}\sqrt{\gamma g_c RT_t} $$

A little simplification of $\sqrt{ RT_t}$ and solve for $A_t$ gives the first equation in this reply.

A note on the slug: A slug is the mass that accelerates at 1 ft/s$^2$ by a force of 1 pound force (lbf). Therefore, F = ma. When pound force and pound mass are used, force, mass and acceleration are related by $$F=\frac1{g_c} ma$$.

If you want to use the slug in the first formula, convert $w_t$ in lbm/s to slugs per second, that is, divide it by 32.2. Convert R, given in the text as 65 ft lb/lb degR by multiplying by 32.2 to get 2093 ft lbf/(slug deg R). The $g_c$ is just 1 (no units) and the lbf in R can be replaced by slug ft/s$^2$. You get the same answer, just more work since given quantities are in lbf and lbm system.

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