Windage calculation for axial motor

AI Thread Summary
The discussion focuses on estimating windage losses in an axial motor design, with the user struggling to apply the correct formulas. Key parameters include the outer and inner disk radii, air gap, and rotational speed, which are used to calculate the friction torque and windage power. Initial calculations yielded a power loss of 3752 watts, significantly higher than the expected 345 watts from a referenced thesis. An error was identified in the friction torque equation, leading to a revised power loss of 1190 watts, but this still does not align with the reference figure. The user remains uncertain about the accuracy of their calculations and the reliability of the reference material.
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.
 

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