Calculating windage for rotating disks in fluid. 3

[dbecker]

Hello,

I am trying to calculate the power loss associated with windage for flat disks about 6 feet in diameter.

If I have disk speed, diameter and fluid properties; is there a generalized formula for calculating power loss?

I have found a document that calculates this value for a cylinder inside a cylinder (rotor inside stator) but not for the back and front of a short flat disk.

This is the NASA document

NASA TN D-4849

I tried looking for something online and was not able to find any information regarding this.

I am assuming this can be derived via integration across the disk radius, but I would rather not go through this process if there exists a general solution backed by experimental data.

Thank you,

- Dan

 

[Jmoore1]

would not be that hard to integrate..... Fdrag=1/2C A p V^2
derive with a changing velocity over the length of the plate (set V=diskspeed*radiusatpoint) to get dFdrag then.... W=F*D and dW=dFdrag*radius +Fdrag*dd(velocity integrated over the distance) ...integrate that and boom you have your amount of work lost per second (c=13/Re for a circular plate parallel to flow)
Could be done in less then 20 minutes.

 

[btrueblood]

It is not a simple subject, though you'd think it would be.

google: "NASA rotating disc drag"

http://ntrs.nasa.gov/search.jsp?Ntt...ll&Ntx=mode+matchall&N=0&Ns=PublicationYear|0

Of these papers, the Theodorsen & Regier, has a lot of good experimental data:

http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19930091870_1993091870.pdf

Later papers explore more complex subjects regarding variations in the flow field of the disc shroud, which can have pretty big impacts on the drag forces. There's a gentleman over at the U. of Bath in England who has a lot of more recent papers on the subject, a google search should turn up his name.

By any chance, would your disc be turning at a tip speed over Mach 1?

 

[dbecker]

Thank you btrueblood.

Disk tip is not passing mach 1. The integration is not that simple, Reynolds number for depends on radius as does tangential velocity so it becomes a very convoluted integration. I will try to make assumptions to simplify it.

 

[Jmoore1]

Your C A V would all be changing...however everything is changing with respect to R..you should be able to simplify it down. If I get some time I can break it out.

 

[Jmoore1]

Pretty easy actually since everything breaks down to a changing R

dW=30.63*dynamicViscosity*DiskSpeed*Integral(From r1-r2)radius^2

integrate over the areas of the disk that are part of the flat disk, add the work from the shaft, and you have your total losses. (And the units work themselves out perfectly to a N*m!!)

 

[Jmoore1]

Note this is for one side of the disk for 1 revolution..

 

[btrueblood]

Jmoore, your analysis assumes laminar flow in the bl, given the form of your equation for the local drag coefficient. For a 6 foot disk, the bl may transition to turbulent (depends on rotation speed, surface roughness, and some other details).

"(c=13/Re for a circular plate parallel to flow)." check against the T&R paper posted, their term for the Cx coefficient is different. Note that the assumption of parallel (flat plate) flow is incorrect; the boundary layer is a true 3d layer, as the flow near the disc surface must include a radial outflow term. Even so, there is a closed form solution for the local Cx term, and allows development of the laminar theory.

Radial flow becomes significant as the far field flow gets restricted (i.e. if a shroud encloses the disc), and this "pumped" flow is inhibited. The T&R paper gives data and coefficients for a disc effectively without a boundary...though this is never clearly shown in the paper.

 

[dbecker]

I just did a calculation for a 60 inch diameter disk rotating at 3600 RPM in air that is at 55 psi and 600F. I am getting about a 100 HP windage loss per side of disk. This is an estimation based off of flat plate flow calculating negative torque by discretizing the disk into 'bands' of radius Ro+R (half inch increments in excel) and calculating force for each band.

Tell me if that sounds reasonable. It seems reasonable to me, tip speed for this disk is about 942 ft/s so nearing Mach 1, drag will be high. The data I am seeing in the document btrueblood gave me is extensive and I have yet to use it until I have a better understanding of the theory.


Thanks for the help.

 

[btrueblood]

Dbecker, sounds like your number is a little on the low side, again because the disk outer rim is clearly going to have a turbulent bl. The T+R paper has moment coefficients, Cm, plotted vs. R (Reynold's number at outer radius). Find R based on your rho, rpm, etc., find Cm from the plot. Then calculate M from their equations. I believe M is for a disc wetted on both sides, but you may want to double check, different authors use different conventions.

Ok, just did the calc. after getting back here to my PC - for your conditions I get a Reynolds no. of 1.65x10^7, which would give a free-air moment coefficient of ~.0053, which grinds down to about 109 HP of windage.

Where it gets tricky: to use pressurized air assumes some type of enclosure or shroud surrounding the disk, which affects the windage (usually increasing it). Bottom line, the Cm data from the T+R paper should probably be taken as a lower bound for the windage drag.

 

[dbecker]

Thanks for the help btrueblood!!