- #36
SlowThinker
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- 65
The article is somewhat misleading, they use a formula for zero intake speed and then use is as correct, but obviously, if there is no air going in, the engine has no air to burn fuel. Same with helicopter, you can't have air at zero speed above the propeller, because then there couldn't be any air going down below it. (This doesn't apply to a rocket engine because air/gas is "created" inside).lavoisier said:So I found this article: https://en.wikipedia.org/wiki/Thrust but so far it raised more questions than it answered (sorry, I keep doing that).
I'll try to repeat my derivation, others are welcome to check it... I'm not an expert on this, just using common sense.
v...speed of air going into rotor
v+##\Delta##v...speed of air going out
S...area of the rotor
##\rho##...density of air going in
F...force (lift) generated
P...power used by the rotor
m...mass of air passing through the rotor during an arbitrary time T
a...acceleration of air
s...length that air spends inside propeller
$$m=\rho S v T$$
$$F=m a=m \Delta v/T=\rho SvT\Delta v/T=\rho Sv\Delta v$$
$$s=\frac{1}{2}aT^2+vT=\frac{1}{2}\frac{\Delta v}{T}T^2+vT=\frac{1}{2}\Delta v T+v T$$
$$P=F s/T=F(\frac{1}{2}\Delta v T+v T)/T=F(v+\frac{1}{2}\Delta v)$$
Speed ##v## that minimizes power:
$$P=F(v+\frac{1}{2}\Delta v)=F(v+\frac{1}{2}F/(\rho S v))$$
$$0=P'=1-\frac{1}{2}F/(\rho Sv^2)$$
$$2\rho Sv^2=F$$
$$v=\sqrt{\frac{F}{2\rho S}}$$
$$P=F(\sqrt{\frac{F}{2\rho S}}+\frac{1}{2}\frac{F}{\rho S \sqrt{\frac{F}{2\rho S}}})=F^{3/2}(\sqrt{\frac{1}{2\rho S}}+\frac{1}{2}\sqrt{\frac{2}{\rho S}})=\sqrt{2}\frac{F^{3/2}}{\sqrt{\rho S}}$$
My constant factor is ##\sqrt{2}## instead of Wiki's ##1/2##, not sure if I made a mistake or what. But then, they are doing something different.