The equation is 13(EfficiencyxProp diameterxoutlet diameter)^2/3
It seems really high when I use inches and really low when I use feet.
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1$\begingroup$ Please edit your question so it is clear what equation you're actually using - as is, it's not clear. Also, do you have a source for this equation? A properly phrased equation will note what units are used for each term, both input and output... after all, an equation that yields thrust in ounces will have at least one different constant than an equation that gives thrust in pounds. $\endgroup$– Ralph J ♦Jan 1 at 20:02
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2$\begingroup$ This is a quick tutorial on how to write equations on Stackexchange $\endgroup$– sophitJan 1 at 20:14
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$\begingroup$ Ralph J- No I don’t have a source. Thanks for the tutorial sophit $\endgroup$– LelotzJan 2 at 23:06
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2 Answers
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Is my equation for calculating ducted propeller thrust correct?
Almost
It should be something like:
$T=1.3 \eta P^⅔$
but this is valid only on one specific case.
The easiest equation relating thrust $T$ and power $P$ is the one derived from the simple momentum theory:
$T=\sqrt[3]{2\rho A P²}$
where $\rho$ is air density and $A$ is disk area i.e. $πr²$. In this equation you have:
- the "$^⅔$" applied to $P$, just like in your equation;
- the propeller's diameter $d$ (actually its radius $r$), just like in your equation;
- to take into account the duct, the thrust is normally increased of some 30%, the power being the same; and here you get also the 1.3 (not 13) just like in your equation;
- and to take into account all the effects which are simply disregarded in the momentum theory, the right part can be multiplied by an efficiency factor $\eta$ (some 0.6 to 0.8), just like in your equation.
Rearranging the terms and introducing the factors just discussed, you get:
$T=1.3 \eta (Pd)^⅔\sqrt[3]{½\rho π}$
which resembles what you were actually looking for. And if you consider the density at 3km (10'000ft) height and a propeller with 1m diameter, then the equation becomes exactly what you remembered:
$T=1.3 \eta P^⅔$
Note that all these equations are valid only at rest (zero speed).
It seems really high when I use inches and really low when I use feet.
Please use SI units.
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$\begingroup$ This is really helpful and makes a lot of sense. My memory may have been faulty, which explains the 13/1.3. S.I units are metric right? Like meters for length? $\endgroup$– LelotzJan 3 at 6:21
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The short answer is "No" it is not correct. Even though the equation formatting is lost we can still know that it cannot be correct because the quantities shown are efficiency, and two diameters.
Efficiency is dimensionless and the diameters have units of length whereas thrust as units of force. length cannot be combined with itself in any way to get force. Of course there is another number in the equation- a 13. And that could be presumed to have the units required to get force. However, if that were the case it would mean that the resulting thrust would be independent of anything outside the equation. For example engine horsepower. And we intuitively know that for a given efficiency, prop diameter and outlet diameter the thrust can be changed dramatically by using different amounts of engine power.
