Windage calculation for axial motor

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SUMMARY

The discussion centers on estimating windage losses in an axial motor design, specifically using equations from the theses by Daily and Nece. The user initially calculated a windage power loss of 3752 watts, which was later corrected to 1190 watts after identifying a misprint in the friction torque equation (Eqn 6.26). Despite the correction, the calculated value still does not align with the expected 345 watts from the reference material, indicating potential discrepancies in the source equations or parameters used.

PREREQUISITES
  • Understanding of axial motor design principles
  • Familiarity with fluid dynamics, specifically air viscosity and density
  • Knowledge of Reynolds number calculations
  • Proficiency in applying torque and power equations in mechanical systems
NEXT STEPS
  • Review the thesis papers by Daily and Nece for detailed equations and assumptions
  • Learn about the implications of misprints in engineering calculations
  • Investigate the effects of air viscosity on windage losses in rotating systems
  • Explore computational fluid dynamics (CFD) tools for simulating axial motor performance
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Mechanical engineers, researchers in motor design, and anyone involved in optimizing axial motor efficiency will benefit from this discussion.

Allen3
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Hi, I'm attempting to estimate the windage losses in an axial motor design, and I'm having trouble with the formulas. It might be a math error, or a misunderstanding of the formulas, but I seem to be stuck.

An axial motor can be modeled as an annular disk spinning in an enclosure. The disk has outer radius R, inner radius r, the distance from each face of the disk to the wall of the enclosure is g, and it is spinning at rotational speed w.

This system was studied by Daily and Nece who came up with equations to describe the air friction on the disk. See the thesis papers at http://alexandria.tue.nl/extra2/200111643.pdf (page 133) and http://lib.tkk.fi/Diss/199X/isbn9512255766/isbn9512255766.pdf (page 17).

I'm trying to duplicate the computation in the first thesis. The basic parameters are:

Outer disk radius R = 190 mm
Inner disk radius r = 110 mm
Air gap g = 1.5 mm
Rotational speed w = 12000 rpm = 200 rev/sec = 1257 radians/sec

I'm also using the following values, which I looked up at http://www.mhtl.uwaterloo.ca/old/onlinetools/airprop/airprop.html :

Air density rho = 1.2 kg/m^3
Air dynamic viscosity u = 1.83E-5 kg-s/m
Air kinematic viscosity v = u/rho = 1.53E-5 m^2/s

From Eqn 6.23: Rotational Reynolds Number Rer = (w*r)*r/v = 1257*0.190^2/1.53E-5 = 2965000

From Figure 6.8: g/R = 0.008 which means the flow is in Regime III

From Eqn 6.28 for Regime III: Cf = 0.08/(g/R)^0.167/Rer^0.25 = 0.08/(1.5/190)^0.167/2965000^0.25 = 0.00433

From Eqn 6.26: Friction Torque T = 0.5*Cf*rho*pi*w^2*(R^5-r^5)
= 0.5*0.00433*1.2*pi*1257^2*(0.190^5-0.110^5)
= 3.0 N-m

From basic physics: Windage power P = T*w = 3*1257 = 3752 watts

The number I'm getting, 3752 watts, seems very high. The power loss shown in Figure 6.9 of the thesis is about 345 watts, which means my calculation differs by about a factor of 10.

Would anyone be able to tell me where I'm going wrong in this calculation? Any assistance you could provide would be greatly appreciated.
 
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Found one error

I found one error. The first reference above contains a misprint. Eqn 6.26 should not contain a factor of pi. Instead, it should be:

From Eqn 6.26: Friction Torque T = 0.5*Cf*rho*w^2*(R^5-r^5)
= 0.5*0.00433*1.2*1257^2*(0.190^5-0.110^5)
= 0.95 N-m

Windage power P = T*w = 0.95*1257 = 1190 watts

That figure of 1190 watts still does not match the value in Figure 6.9 of the first reference. I'm still not completely confident in my calculations, but I'm going to assume for now that reference is incorrect.