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Is there a way to convert CFM as in a rate of volume of air moved to kilograms of lift generated?

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  • $\begingroup$ Hmm, maybe in particular cases as air flow doesn't always indicate lift (e.g., air flowing in vents). Could you provide a bit more context here (i.e., the specific case you have)? $\endgroup$ – Kyle Kanos Dec 30 '14 at 3:11
  • $\begingroup$ No. You also need the velocity at which the air is flowing. Lift is a force and measured in Newton. $\endgroup$ – CuriousOne Dec 30 '14 at 6:19
  • $\begingroup$ Force is momentum per unit time, so it's not just the mass of air moved, it is mass times the change in velocity. $\endgroup$ – Mike Dunlavey Dec 30 '14 at 15:57
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I'm going to assume that you're referring to the reaction force required to move such a volume. No, there's no direct relationship. A single CFM figure can be met by moving that volume through a small area at high speed or through a large area at low speed. Assuming the air is initially at rest, the first case requires greater acceleration and therefore greater force.

Let's change it from 1CFM to $1m^3/s$. Further I'll assume that your system is accelerating air from rest to the speed necessary to move through a particular size vent. We can calculate the change in momentum and therefore the average force. I'll assume $1.2kg/m^3$ as a density for air.

For case 1, assume the vent is $1m^2$ in cross section. $$v = \frac{\text{flow}}{\text{area}}$$ $$v = \frac{1m^3/s}{1m^2}$$ $$v = 1m/s$$ $$F = \frac{dp}{t} = \frac{m \text{ } dv}{t}$$ $$F = \frac{1.2kg \text{ } m/s}{1s}$$ $$F = 1.2N$$

Do the same calculation for a $1cm^2$ cross section and you get $F = 12000N$

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  • $\begingroup$ 12,000N on 1cm^2 equals a pressure of roughly 1200bar. Using W=pdV we get 1e5Pa*1200*1m^3/s=1.2e8W. Hmmm... world's most powerful compressor running at 120MW? :-) $\endgroup$ – CuriousOne Dec 30 '14 at 6:36
  • $\begingroup$ Heh, yes. I did not emphasize that the calculation is quite simplistic. It is only meant to show that very different forces can come from the same flow rates. $\endgroup$ – BowlOfRed Dec 30 '14 at 7:26
  • $\begingroup$ I didn't say you did anything wrong... just wanted to point out how hard it is to produce significant thrust with small cross sections... that's why practical propellers, jet engines, helicopter blades etc. are large devices. $\endgroup$ – CuriousOne Dec 30 '14 at 7:28
  • $\begingroup$ You seem to be actually using $F=\Delta p/\Delta t$, not $F=dp/dt$ (though you ignore the differential aspect of $dt$ and incorrectly write $t$ in the denominator). $\endgroup$ – Kyle Kanos Dec 30 '14 at 14:11

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