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I'm asking if I'm doing this correctly. I've estimated the thrust of a ducted-fan model aircraft neglecting pressure changes across the duct as follows:

$$F = Q * \rho * (V_e - V_0)$$

$$= Q * rho * (Q/A_e - Q/A_0)$$

$$ = 131 [cu. ft. / sec] * 0.00237 [slug/cu. ft] * (131/0.74 -131/0.92)$$ $$ = 0.31 * (177.03 - 142.4)$$ $$= 0.31 * 34.6$$ $$= 10.7 lbf$$

where

  • F = Thrust,
  • Q = volume flow rate of air = 131 [cu. ft. / sec],
  • $\rho$ = density of air = 0.00237 [slug / cu. ft],
  • $A_0$ = an circular inlet area = 0.92 [sq. ft],
  • $A_e$ = an exit area = 0.0.74 [sq. ft].

That seems an awfully high value for thrust to me. Am I missing something?

David Walden
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  • Welcome! The site *can* support Latex formatting, I started it for you, I suggest to make it ready (it is very useful in the sense of the upvotes). – peterh Dec 19 '18 at 01:12
  • Hi, why do you think it is to high? Are you sure about the flowrate? Other than that the equation and assumptions seem to be alright. However please be aware, that you are assuming incompressibility, so there is already some error here. – rul30 Dec 19 '18 at 07:29
  • @rul30 Just my intuition, I guess. There's a lot at $take in having it right before I build a model. Yes, the flowrate is based on rotational speed of the rotor, blade pitch and inlet diameter. I'm not assuming incompressibility, I'm insuring it by limiting velocities to at or below 0.3 mach (standard atmosphere, sea level). At a Q of 131 [cu. ft. / sec] and an exit area of 0.65 [sq. ft] that works out to an exit velocity of about 202 [ft. / sec]. and the tip speeds are likewise limited to at or below 0.3 mach to insure the avoidance of compressibility. Thanks for your question. – David Walden Dec 22 '18 at 01:25
  • I guess that compressibility will not be a major problem (Mach 0.3 is just an arbitrary number which already has an error of 5%). If I understand you correctly you did not measure the flowrate, but calculate it. Could you maybe edit your question and show this calculation as well? Otherwise the answer to your question will be very short: Yes. – rul30 Dec 22 '18 at 12:11
  • You are correct. I have not measured the flowrate nor will I. When I've built a model, then I'll be measuring thrust, not flowrate. Aside from editing the exit area and thrust estimate to correct an error in the original post, I prefer not to show the calculations as I would then be giving away trade secrets. I don't mind short answers. Thanks – David Walden Dec 24 '18 at 16:36

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