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Provided you know the capacity of a fan (flow rate) at constant speed and at sea level, is there an analytical way to predict what the flow rate would be at altitude? Or is this specific to the fan's design?

docscience
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    What kinds of speeds is the fan rotating at? Compressibility effects could change the approach to this. For instance, a household window fan vs the first stage of a turbofan engine would behave very differently as the altitude changed due to the difference in rotation speed and the resulting compressibility. – tpg2114 Sep 16 '16 at 19:04
  • @tpg2114 subsonic – docscience Sep 16 '16 at 19:06
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    Okay -- but like, $M < 0.3$ subsonic, or $M > 0.3$ but not enough for shocks to form? Compressibility starts to matter around $M > 0.3$ generally, which at sea-level is a fan spinning around 100m/s. And that's peak speed, the tips travel faster than the root, so the size of the fan could matter also. – tpg2114 Sep 16 '16 at 19:07
  • For my specific application, tip speed doesn't exceed 100 m/s – docscience Sep 16 '16 at 19:30
  • @tpg2114 for my specific application, tip speed doesn't exceed 100 m/s – docscience Sep 16 '16 at 21:36
  • If you can ignore compressibility, can't you just use the continuity equation in the appropriate form and the adjustable parameters would be the variation of the density and speed of sound? I am assuming that the fan speed does not change due to the lower density at higher altitudes. Am I missing something? – honeste_vivere Sep 16 '16 at 22:05
  • @honeste_vivere continuity says mass flow rate in equals mass flow rate out. But at lower ambient pressure there is less mass per cubic cm to be 'scooped' up and pushed forward by the fan. So would the reduction in flow rate from sea level to altitude at the same fan speed just be reduced by a factor, the ratio of gas densities at each elevation? – docscience Sep 16 '16 at 22:22
  • @docscience - I have an uncomfortable feeling that the change in the speed of sound matters as well... somehow through an effective sampling volume (e.g., take the limit as the speed of sound goes to zero then after the first fan blade passes, the second would see effectively no particles [but this would be a compressible system]). I will have to think about this more... – honeste_vivere Sep 16 '16 at 22:27
  • @honeste_vivere speed of sound should only be affected by temperature, right? – docscience Sep 16 '16 at 22:31
  • @honeste_vivere ... given the gas is ideal – docscience Sep 16 '16 at 22:32
  • Well it's formally the change in pressure with respect to density so if one assumes an ideal gas law, then one can get a temperature proportionality... Oh I see you already said that... Then yes and temperature decreases with altitude up to ~10 km or so. I have a useful graphic and discussion at http://physics.stackexchange.com/a/266046/59023 – honeste_vivere Sep 16 '16 at 22:34
  • I forgot an answer I wrote that is related, as it was for the surface of Mars, at: http://physics.stackexchange.com/a/217839/59023. – honeste_vivere Sep 16 '16 at 22:38

2 Answers2

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Following answer is speculative.

Flow rate of air ($Q$) is determined once fan's geometry, its angular speed ($\omega$), and thermodynamic state of air (in particular its $\rho,\mu$) is specified. Since geometry of fan is not being changed, let us take any linear dimension associated with it (say, length of fan blade) as a length scale, $d$. Flow rate $Q$ is determined by these variables means that there exists a functional relationship:

$f(Q,\rho,\mu,d,\omega)=$constant

which results in dimensionless groups:

$g(\frac{Q}{\omega d^3},\frac{\omega d^2}{\nu})=$constant

or

$\frac{Q}{\omega d^3}=h(\frac{\omega d^2}{\nu})$

where $f,g,h$ are functions. Now since flow Reynolds number is high (and therefore flow is turbulent), viscosity plays little role in determining the flow (hypothesis). In that case,

$\frac{Q}{\omega d^3}\approx $constant.

Since angular speed of the fan isn't varying either I would guess that flow rate of the fan shall remain constant (to a good approximation).

One may object that density of air hasn't appeared in the final conclusion, so if one were to take the fan to such a height where there is practically no air, the equation still predicts the same flow rate, which is wrong. However in this case continuum approximation breaks down, and the entire analysis would become inapplicable.

Deep
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  • I don't doubt the dimensionless number is correct given the asusmptions; I seem to recall the cube relation with dimension. But perhaps a parameter is missing in the starting assumptions. Because it's a fact according to my data that flow rate does get smaller with increasing altitude – docscience Sep 17 '16 at 15:35
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For a fixed speed, a fan, blower or any turbo-machine in general will deliver the same volumetric flow regardless of the ambient pressure since the machine essentially scoops out a volume of air as each blade of the machine passes the machine's inlet. $$Q_{SL}=Q_{alt}$$ where $SL$ designates 'Sea Level' as reference and $alt$ as some higher altitude

But at higher altitudes there are fewer molecules per unit volume (lower gas density) and so the mass flow rate is lower with increasing altitude and barometric pressure. $$\dot{m}_{alt} < \dot{m}_{SL}$$ So since the volumetric flow rates are the same then $${\dot{m}_{alt}\over{\rho}_{alt}} = {\dot{m}_{SL}\over{\rho}_{SL}}$$ and $$\dot{m}_{alt}={{\rho}_{alt}\over {\rho}_{SL}}\dot{m}_{SL}$$ But if we were to measure these mass flows as volumetric flows relative to sea level then $${{\dot{m}_{alt}}\over {{\rho}_{SL}}}={{{\rho}_{alt}}\over {{\rho}_{SL}}}{{\dot{m}_{SL}}\over {{\rho}_{SL}}}$$ which becomes $$Q_{Malt}={{{\rho}_{alt}}\over {{\rho}_{SL}}}Q_{MSL}$$ And the $Q_M$'s are the measured volumetric flows at altitude and sea level respectively. This result shows the the measured volumetric flow is reduced as altitude increases by the ratio of air density as it decreases. And this is consistent with zero volumetric flow as one moves out of the atmosphere and no more scooping is possible.

docscience
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